Download 32 From Galileo to Lorentz transformations

32
From Galileo to Lorentz transformations
[] Reference: The Feynman’s lectures on physics (Addison-Wesley, 1964) : {} —
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} —
[] Reference: R. Resnick, Introduction to special Relativity, (1968, J. Wiley and Sons) : {} — Clear and detailed
introduction
[] Reference: J. D. Jackson, Elettrodinamica Classica, seconda edizione, (1984, Zanichelli) : {} —
[] Reference: H. Goldstein, Meccanica Classica (19xx, Zanichelli) : {} —
[] Reference: W. K. H. Panoffsky - M. Phillips, Classical Electricity and Magnetism, (2◦ ed., 1962, Addison-Wesley)
: {} —
[] Reference: S. Weinberg, Gravitation and Cosmology, (1972, J. Wiley and Sons) : {} —
[] Reference: C. W. Misner - K. S. Thorne - J. A. Wheeler, Gravitation (W. H. Freeman and Co., San Francisco)
1973 : {} —
32.1
Introduction
Classical ElectroMagnetism turned out to be fully consistent with the Theory of Relativity, while classical Mechanics
did not.
Therefore Classical Mechanics had to be modified in order to make it consistent with the Theory of Relativity while
ElectroMagnetism did not require any modification. ElectroMagnetism just required to be re-written via the relativistic
formalism as well as understood and interpreted in the proper context of the Theory of Relativity.
The basic element of the Theory of Relativity are Lorentz transformations for time and space. These affect the timespace coordinates of the world. They are therefore the basic ingredient required for physics, before any other development
and deserve a very detailed discussion.
After defining the Lorentz transformations for time and space, the Lorentz group will be used to define the general
scalars, vectors and tensors (for the Lorentz group) for any other physical quantity.
32.2
32.2.1
Phenomenology
ElectroMagnetism and Galileo invariance
Galileo transformations implies that the accelerations in the two inertial Reference Frame are the same but velocities
are different.
Therefore when one has a velocity dependent force, such as the Lorentz force, one should be careful.
In Galileo invariance the accelerations are the same, so that the Lorentz forces on one point charge are also the same in
both Reference Frame.
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32: From Galileo to Lorentz transformations
v =v+u
E + v ×B = E + (v + u) ×B
v=0
=⇒
32.2.2
=⇒
v× B − B
(32.2.1)
for any u and any v
(32.2.2)
E = E + u ×B
for any v
=⇒
(32.2.3)
B=B
(32.2.4)
Asymmetries in the description of ElectroMagnetic induction phenomena
The example of the bar moving in a constant and uniform magnetic field (section 16.13.12 and 16.13.12) shows that the
observer moving with the bar must see an electric field in order to explain the induced currents. However, even if this is
perfectly correct, the observer is not able to attribute the electric field to any electric charge nor to a time-varying magnetic
field, and therefore the appearance of the electric field is a sort of a mystery to him. Apparently he does not know the orign
of the electric field: one will discover later on that there are indeed electric charges to produce the observed electric field.
Even if the two observers make the same predictions they explain the situation in two different ways: what is seen as a
magnetic field by one observer is seen as an electric plus magnetic field by another observer in relative motion. Therefore:
• either the two observers are not equivalent, as they describe in a different way the same phenomena;
• or electric and magnetic fields are strictly connected.
However the asymmetry is only in the description of ElectroMagnetic induction phenomena, not in the phenomena
themselves, nor in the predictions that the two observers carry on.
Actually the two observers give the same predictions while describing the phenomena in a different way.
Is this a violation of the Principle of Relativity? Is this just an accident or is it the result of the fact that the two
observers are in fact equivalent but this equivalence is a more subtle one?
Beams of charged particles in particle accelerators
The reduction of the repulsive force among identical charged particles forming a straight beam is well-known in the
physics of particle accelerators.
The explanation for this issue has to be found by employing the correct Lorentz transformations for the EM fields and
dynamical quantities, see section 35.4.4.
32.2.3
Relative motion between a point charge and a small magnet
Consider the following situation in the light of sections 35.4.8 and 32.2.3.
A point charge moves along the positive direction of the x axis. A small magnet lies in the xy plane with its magnetic
dipole moment along the positive direction of the y axis and it is located at some y > 0. The charge produces a magnetic
field at the location of the magnet and the magnet therefore experiences a torque whose direction is along the positive
direction of the x axis.
Look at the situations as it is seen by an observer in the Rest Frame of the point charge. In this case the magnet is seen
moving along the negative direction of the x axis. It is not immersed in any magnetic field but only in the electric field of
the point charge. Therefore the magnet does not experience any torque.
Actually the two observers give different predictions and describe the phenomena in a different way; this is an apparent
violation of the Principle of Relativity.
The explanation for this issue has to be found in the fact that, according to Relativity, a moving magnetic dipole also
possesses an electric dipole moment.
32.3
Galileo transformations of space-time coordinates and
Newtonian mechanics
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
32.3.1
Inertial Reference Frame
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
Nowadays we interpret the first Newton’s law as defining an inertial Reference Frame: Inertial Reference Frames are
all and only the Reference Frames such that the principle of inertia applies, that is an object not subject to any force moves
with constant velocity. As a corollary: Inertial Reference Frames do exist.
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32.3: Galileo transformations of space-time coordinates and Newtonian mechanics
When is an object subject to no force? Experience shows that all known real forces are due to the existence of something
which is the source of the force and that the intensity of all real forces decreases when increasing the distance from the
source. In fact, if the forces did not fall off rapidly we could never isolate the interactions of two bodies among those of all
other bodies in the universe.
Therefore: An object is subject to no force when it is far enough from any other body source of forces.
How can one decide whether its own Reference Frame is a inertial one? Operatively: throw force-free massive objects
in various directions and observe their motion; if they move at constant speed: then the Reference Frame is an inertial one;
if they do not: use their motion to correct.
[] Reference: P. W. Bridgman: {Am. J. Phys. 29 (1961) 32} — Excellent; several excerpts below.
A system of three rigid orthogonal axes fixes a Galilean frame if three force-free massive particles projected
along the three axes with arbitrary velocities continue to move along the axes with uniform velocities. Our
terrestrial laboratories do not constitute such a frame, but we may construct such a frame in our laboratories
by measuring how three arbitrarily projected masses deviate from the requirement ... and incorporating these
deviations as negative corrections into our specifications for the Galilean frame. There need be no reference to
the stars, but the behaviour of bodies can be relevantly described in terms of such immediately observable things
as the rotation of the plane of the Foucault pendulum with respect to the earth or the deviation of a falling body
from the perpendicular. Even if the rocket operator who is trying to put a satellite into orbit finds it convenient
to make some of his specifications in terms of observations on the pole star, it is obvious that his apparatus
must eventually be described in terrestrial terms. ... In a Galilean frame a rotating body, after it has been set
into rotation and the forces disconnected, preserves the orientation of its plane of rotation in the frame and,
consequently, preserves the direction of its axis of rotation.
32.3.2
Galileo transformations of space-time coordinates
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
Galileo relativity principle: the law of physics must be the same for all inertial observers linked by Galileo transformations; that is all inertial observers are equivalent.
Consider two Inertial Reference Frames, I and I with origins at O and O. Let R the position of the origin O with
respect to the inertial Reference Frame I. Assume that I is moving, with respect to I, with a speed u. Assume that a
Cartesian Coordinate System and a time coordinate are defined in both Inertial Reference Frames. For simplicity always
use a Cartesian Coordinate System. Assume that the rotation matrix between the two Cartesian Coordinate System is
R. The rotation between the two Coordinate System, if any, is time independent and therefore the rotation matrix R, a
special orthogonal matrix in three dimensions, if any, is time independent. The spatial translation of the origins of the two
Coordinate System at t = t = 0 is: x0 . The time translation of the origins of the two time scales is: t0 .
Galileo transformations in their most general form, read:
(
x = Rx − R = Rx − ut − x0
(32.3.1)
t = t − t0
(
v = Rv − u ,
if no rotation:
v =v−u
=⇒
.
(32.3.2)
a = Ra ,
if no rotation:
a=a
Galileo transformations form a group with ten parameters.
32.3.3
Electromagnetism and Galileo transformations
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} —
Only consider ElectroMagnetism in vacuum, in order to avoid a privileged Reference Frame given by matter.
Electromagnetism is not invariant with respect to Galileo transformations, because Maxwell equations for EM waves
√
implies that EM waves have a fixed speed, c ≡ 1/ ε0 µ0 , univocally determined. So they cannot be valid in any Reference
Frame if Galileo transformations (in particular the law of addiction of velocities) are valid.
Therefore the following possibilities arise:
• c is measured with respect to some privileged Reference Frame: the so-called ether frame;
• ElectroMagnetism is wrong or, at least, is only valid in the ether Reference Frame;
• Classical Mechanics is wrong.
To the physicists of the end of XIX century: the latter could be hardly accepted after centuries of successes for Classical
Mechanics: remember, for instance, the prediction of the existence of Neptune and its discovery as predicted. The second
option seemed to be, anyway false, due to the many correct predictions of ElectroMagnetism.
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32: From Galileo to Lorentz transformations
It should be noted that at the end of the XIX century all known waves had a medium to propagate on as travelling
oscillations in the material medium. It was therefore natural to think of electric and magnetic fields as strains in an invisible
jelly-like unknown medium: the ether.
Therefore people started to look for the ether frame, trying to detect the ether wind.
32.3.4
The speed of light
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
Speed of light should always be understood to mean the speed of light in free vacuum space unless it is explicitly stated
otherwise. The speed of light in a material medium is, in general, different from c.
It is an experimental fact that no faster method of sending signals has ever been discovered.
The speed of light, in the theory of Relativity, was raised to a fundamental constant of nature, in the sense that there
exist a limiting speed, c and that ElectroMagnetic waves in vacuum travel at the limiting speed c.
... See any textbook for the description of different methods for measuring the speed of light. ...
It was understood since a long time that light had some finite, even if very large, speed. One should in fact remember
the experiments carried on by Galileo with a lamp and an distant friend.
The following basic methods had been employed, historically, to measure the speed of light 1 .
• Prediction of the eclipses times of the main satellite of Jupiter by Romër.
• Measurements of the stellar aberration by Bradley.
• Rotating mirrors and toothed wheels.
• Cavity resonator.
• Electro-optic switch via the Kerr cell.
It will be shown, when discussing relativistic dynamics, that an infinite energy is required to accelerate a particle with
non-zero mass to the speed of light. Therefore no massive particle can reach the speed of light. On the other hand relativistic
dynamics shows that massless particles always travel at the speed of light.
Relativistic effects are negligible in real life because the speed of light is so large with respect to every-day speeds. In
fact it will be shown that all relativistic effects disappear when the speed of light tends to infinity.
If there is something mysterious in special relativity this is the invariance of the speed of light. Once the the invariance
of the speed of light is granted everything else follows directly and fairly simply even if many results may appear paradoxical
and it is often difficult to feel confortable with them. Every new situation must be analyzed carefully, without prejudices.
Super-luminal motion
See section 32.8.1 for the explanation in the context of Relativity.
The value of c is the limiting speed for real material bodies and for processes that can be used for the transmission of a
signal, that is the transmission of a certain amount of energy carrying some information.
Super-luminal motion refers to apparent speeds larger than the speed of light. It is only a visual effect, nothing is
running faster than light, nothing carrying information.
Super-luminal motion also refers to motion which is faster than light but not for real objects nor for transfer of
information.
As an example, imagine a machine-gun sending projectiles, at a rate of n Hz while rotating in the horizontal plane
with a fixed angular speed ω. Consider a very far circular wall, centered on the machine-gun, at distance D, and consider
the motion of the track of the impinging projectiles. Determine the speed of the track left by the projectiles on the wall.
Projectiles arrive at the rate they leave the machine-gun, separated by an angular distance ∆θ = ω ∆t = ω/n so that the
distance on the wall is Dω/n and the speed of the track is Dω, which can be as large as one desires provided D is large
enough.
32.3.5
The Michelson-Morley experiment
[] Reference: The Feynman’s lectures on physics (Addison-Wesley, 1964) : {15} —
The Michelson-Morley experiment was designed to try to detect the ether wind. In fact it was realized that any effect
of the ether wind was indeed very small and therefore only very sophisticated experiments might be able to detect it.
The idea was to measure the speed of light in different directions in the ether frame to try to detect the changes in the
speed of light due to the Earth moving with respect to the ether frame, by also exploiting the annual motion of the Earth.
In fact one would expect different speeds in different directions.
1 See
Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill).
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32: From Galileo to Lorentz transformations
32.4: The Postulates of Special Relativity
The experiment clearly showed that the ether does not exist as it did not detected any ether wind because the measured
speed was the same in all directions, while the experiment was in principle sensitive to measure the predicted ether wind.
The null result of Michelson-Morley experiment can only be understood if the speed of light is infinite, but it was known
that this is not the case.
On the basis of experimental evidence Einstein then postulated that Galileo transformations are not correct and must
be replaced with new linear transformation law, Lorentz transformations. As a consequence ElectroMagnetism was all right
while Newtonian mechanics had to be changed.
The new postulate changed drastically the way space and time are conceived.
The invariance of the laws of Electromagnetism (Lorentz invariance) was taken to apply to all physical phenomena.
The new transformation law must be such that the speed of light is invariant when passing from one inertial Reference
Frame to another. Moreover they must transform a motion at uniform velocity into another with uniform velocity and
therefore they have to be linear.
32.3.5.1
Analysis of the Michelson-Morley experiment
... See any textbook for the description of the Michelson-Morley experiment. ...
See also section ??.
Let the speed of light be have module c in the unknown ether frame.
Let the Earth travel at a speed v with respect to the ether frame.
The Michelson-Morley experiment, using the Michelson interferometer, basically compares, with high-precision, the time
for light to travel forth and back a segment, one of the arms of the Michelson interferometer, comparing the travel times
along two orthogonal arms.
Assume, for simplicity, that the two orthogonal arms are one parallel and the other one perpendicular to the speed of
the Earth, v, with respect to the ether frame.
v ≈ 30 km/s
(32.3.3)
L
L
2L
+
=
c−v
c+v
c (1 − v 2 /c2 )
2L
∆t⊥ = p
c 1 − v 2 /c2

!
!
!4 
2
2
v
v
v
Lv 2
 =
≈ 0.3·10−16 s
1+ 2 − 1+ 2 +O
3
2c
c
c
c
∆tk =
∆t⊥ − ∆tk ' 2L c a very small time interval but measurable with a good interferometric technique .
32.4
(32.3.4)
(32.3.5)
(32.3.6)
(32.3.7)
The Postulates of Special Relativity
All the consequences of the special theory of relativity follow from the following two postulates.
32.4.1
The Principle of Relativity
The relativity principle is assumed to be a principle valid for all the laws of physics, not only for mechanics.
The principle of relativity: space is isotropic and uniform; the fundamental laws of physics are identical for any two
inertial observers in uniform relative motion.
Corollary: there exist at least one inertial Reference Frame (and therefore infinite).
32.4.2
The Principle of invariance of the speed of light
The null result of the Michelson-Morley experiment to detect the drift of the Earth through an ether (as well as other
results) can only be understood by assuming the following new principle.
The principle of invariance of the speed of light: the speed of light in empty space is independent of both the motion
of the light source and the receiver; light propagates in free empty space with a definite velocity c, a universal constant of
nature.
Electromagnetic waves or photons are not unique in having a velocity independent of the motion of the source and
receiver. It is believed, with strong evidence, that all other massless particles have velocities equal to c in empty space.
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32.4: The Postulates of Special Relativity
32.4.3
32: From Galileo to Lorentz transformations
The clock hypothesis/postulate
The clock hypothesis states that the tick rate of a clock when measured in an inertial frame depends only upon its
velocity relative to that frame and it is independent of its acceleration or higher derivatives of position.
32.4.4
Some consequences of the postulates and thought experiments
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {} —
Whenever a inertial Reference Frame moves with respect to another Reference Frame with relative velocity u it is useful
to introduce the shorthand notation
1
γu ≡ γ[u] ≡ p
.
(32.4.1)
1 − u2 /c2
Whenever there is no possibility of confusion the notation is usually simplified to: γu ≡ γ.
In this section the three basic and striking results deduced from the new postulates are discussed.
32.4.4.1
Relativity of simultaneity
Consider a light source at the center of a train moving with constant velocity u.
The light source emits a lamp at a certain time and the arrival of the light signals at the two extremes of the train (of
rest length `) are recorded (events E+ and E− ).
The observer inside the train says that the two events are simultaneous, and that the light travel lasts
∆t = `/ (2c) .
(32.4.2)
The observer on the ground measures two different time durations, ∆t+ and ∆t− , as the train is moving during the
light travel:
c ∆t− = `/2 − u ∆t−
c ∆t+ = `/2 + u ∆t+
∆t+ = `/ (2 (c − u))
(32.4.3)
∆t− = `/ (2 (c + u)) .
(32.4.4)
Therefore simultaneity is this a relative concept: two events which happen at the same time at different places for one
observer may be not simultaneous for another observer.
Note that in the case of infinite speed of propagation of light, c u, the simultaneity is recovered as ∆t+ = ∆t− . In
fact the train has to be going very fast in order the discrepancy becomes easily detectable.
Note that the two lengths, ` and ` are not the same, a priori, but this does not affect the conclusion that the two events
E+ and E− are simultaneous in one Reference Frame and are not in the other.
Two events that are simultaneous at different places in one inertial system may be not simultaneous, in general, in
another inertial system.
Note that the relativity of simultaneity makes problematic the application of the action-reaction principle. Consider in
fact the case of time-dependent forces: action-reaction must be equal and opposite at the same time but, as simultaneity
is a relative concept, another observer may not agree on the fact that action and reaction are equal and opposite. Only in
the case of point-contact interactions, with the two forces applied at the same physical point (as well as in the trivial case
where the forces are constant), the action-reaction priciple can be retained.
Note also that the effect is a real one, depending on how time floes, not to be confused with spurious effects such as the
time delay between lighting and thunder.
32.4.4.2
Time dilatation (time interval between the same two events)
Consider a light source at the center of a train moving with constant velocity u.
Consider a light ray which strikes the floor just below the light source for the observer inside the train. Let H be the
rest height of the lamp from the floor.
For the observer inside the train the time between light emission and light reaching the floor is:
∆t = H/c .
(32.4.5)
For the observer on the ground the time between light emission and light reaching the floor is:
p
c2 ∆t2 = H 2 + v 2 ∆t2
∆t = H/ c2 − u2 ,
(32.4.6)
with the same same speed c, but for this observer the light makes a longer travel as the train is moving.
Note that the height of the train is the same for both observers, as demonstrated in section 32.4.4.3: H = H.
Therefore one finds the time dilatation formula:
∆t = γ ∆t
560
.
(32.4.7)
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32: From Galileo to Lorentz transformations
32.4: The Postulates of Special Relativity
the time elapsed between the same two events (light leaves light source and light strikes the floor just below the light source)
is different for the two observers.
Running clocks thus run slow.
The proper time is the time in the Reference Frame where the two events happen at the same place (x1 = x2 ) or, as a
weaker condition applying in this problem, at the same place only in the coordinate along the velocity u.
Note that the result concerns time itself, not the way clocks work, as it is experimentally demonstrated by experiments
on many systems.
Note that in the case of infinite speed of propagation of light, c u, the time dilatation disappears. In fact the train
has to be going very fast in order the discrepancy becomes easily detectable.
It might appear that the time dilatation violates the principle of relativity, but it does not. In fact the two observers
measure two different things:
• the ground observer watches the single moving clock on the train and compares its readings with the readings of two
different synchronised stationary clocks and he finds that the time interval measured by his two different clocks is
longer that the time interval measured by the single moving clock;
• conversely the observer on the train watches the single clock stationary on ground and compares its readings with the
readings of two different synchronised clocks fixed on the train and he finds that the the time interval measured by
his two different clocks is longer that the time interval measured by the single clock stationary on the ground.
Muon lifetime
The lifetime of short-lived particles is increased with their speed, see section 32.8.2.1.
32.4.4.3
Length contraction (distance between the same two events at the same time)
Note that every measure of length must be carried on by taking the position of all the points at the same time: as
simultaneity is a concept relative to the observer this depends on the observer.
32.4.4.3.1
Length contraction along the direction of motion
As the speed of light in free space is, by hypotesis, a universal cosntant, it can be used to mesure lengths. Measure the
length of the train by measuring the travel time of light on the closed path. Note that measuring time on a closed path
allows to avoid any problem with synchronisation of clocks.
Consider a light source at one end of a train moving with constant velocity u. There is reflecting mirror on the other
side.
For the observer inside the train the travel time is related to the length, ∆`, of the train by:
∆t =
2 ∆`
.
c
(32.4.8)
For the observer on the ground the travel time is related to the length, ∆`, of the train by:
∆t = ∆t+ + ∆t− =
∆`
∆`
+
(c − u)
(c + u)
=⇒
∆t = 2 ∆` c/ c2 − u2
(32.4.9)
as given by the sum of two travel times ∆t+ and ∆t+ with mutual velocity c + u and c − u between the light and the train.
These two intervals are related by the time dilatation formula 32.4.7, so that we obtain the length contraction formula:
∆` = ∆` /γ
.
(32.4.10)
Moving objects are thus shortened.
Note that in the case of infinite speed of propagation of light, c u, the length contraction disappear. In fact the train
has to be going very fast in order the discrepancy becomes easily detectable.
Note that it does not matter that the path is asymmetrical for the moving observer one just uses the correct formula
relating time and space.
32.4.4.3.2
Length contraction perpendicularly to the direction of motion
It should be emphasised that moving objects are shortened only along the direction of motion (that is velocity u). There
is no shortening in the direction perpendicular to motion.
The following argument applies 2 . Suppose the train is running near a wall where a horizontal blue line is painted at
one meter height from the ground. A woman inside the train leans out the window and paints a horizontal red line one
meter above the ground. Which of the two lines is higher? If, for instance, length in direction perpendicular to the velocity
contracts then the man on ground would predict the red line is lower than the blue line. On the other hand the woman on
the train would predict the red line is higher than the blue line (as the wall is moving for the woman). The principle of
relativity allows us to conclude that the two lines must coincide.
Note that no subtlety of synchronisation enters this reasoning.
2 See
D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall), and references therein.
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32.5
32: From Galileo to Lorentz transformations
Lorentz transformations of space-time coordinates
Consider the special case when the Cartesian axes of both Coordinate System are parallel and with the same orientation,
that the two origins coincide at times t = t = 0 and that the velocity of Reference Frame I with respect to Reference Frame
I is parallel to both z and z axes and let its component along z be u ≡ ue3 .
Assume that at t = t = 0 a short light lamp is emitted at the coincident origins of the two Reference Frames: r = r = 0.
The equation of motion of the light front is a spherical one, in both Reference Frames, thanks to the isotropy of space
and the speed of light is the same, for both Reference Frames:
x2 + y 2 + z 2 = c2 t2
x2 + y 2 + z 2 = c2 t
2
.
(32.5.1)
In fact every inertial observer must see a spherical light front, independently on the velocity of the light source.
The correct transformations must transform the first expression into the second one.
This special case contains all the essential ingredinates of Relativity without too much non essential mathematical
complexity.
Note that the transformation must be a linear one, as the inertia principle requires that a uniform rectiliner motion is
transformed into a uniform rectilinear motion.
32.5.1
Galileo transformations are not compatible with the new postulates
When using Galileo transformations 32.3.1 in the second equation one finds:
x2 + y 2 + z 2 + u2 t2 − 2zut = c2 t2 .
(32.5.2)
It is therefore clear that Galileo transformations cannot manage the equality of the speed of light in both Reference
Frames, due to the Galilean law of addition of the velocities.
Moreover in order to cancel the undesired time dependence u2 t2 − 2zut one needs to invoke a dependence on time of
the space components. However since it has been found that the distances perpendicular to the speed u do not change.
It is therefore assumed that t only depends on z.
As a result of the previus reasonings we are led to section 32.5.2.
32.5.2
One-dimensional special Lorentz transformations
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {11} — A more general Ansatz.
Let’s continue as in section 32.5.1.
The relativity of time intervals, derived in section 32.4.4, unavoidably compels us to assume a transformation which
includes and mixes space and time coordinates.
In order to be consistent with the principle of inertia the space-time transformations between any two inertial Reference
Frames must be a linear one, as any uniform motion must be mapped into a uniform motion.
In order to find the one-dimensional Lorentz transformation (and its inverse) one can try the following educated guess,
in terms of the so far unknown parameter γ:
u ≡ u ·e3 R 0
z = γ (z − ut)
⇔
(32.5.3)
z = γ z + ut
(32.5.4)
by inverting the second one above to find t and using the first one above to replace z one finds:
!
!
z 1 − γ2
z 1 − γ2
t = γt +
⇔
t = γt −
.
u
γ
u
γ
(32.5.5)
The, so far unknown, factor γ is necessarily a function of the module of the veolcity u. It is a function of the module
only thanks to the isotropy of space.
In order to preserve the sense of time we must impose: γ ≥ 0.
Note that the equivalence of all inertial observers impose the reciprocity so that the
⇔
applies in equations 32.5.4
and 32.5.5 as the observer I sees the inertial Reference Frame I moving with velocity −u.
The constancy of the speed of light implies (using the above equation 32.5.5):
z = ct
⇔
z = ct
(32.5.6)
using equations 32.5.4 and 32.5.5 in the first one above and comparing the result with the second one above
=⇒
γ2 =
1
≥1
1 − u2 /c2
562
(32.5.7)
(32.5.8)
.
(32.5.9)
December 23, 2011
32: From Galileo to Lorentz transformations
32.5: Lorentz transformations of space-time coordinates
Note that the above relation 32.5.6 cannot be satisfied by any Galileo transformation (except for c → ∞).
It follows that:
!
!
uz
uz
t=γ t− 2
⇔
t=γ t+ 2
.
c
c
(32.5.10)
Moreover we already know (see section 32.4.4.3) that:
x=x
y=y .
(32.5.11)
Let
β≡
u
c
All in all we have the special Lorentz transformations:
t=γ
γ≡ p
1
1 − β2
uz
t− 2
c
.
(32.5.12)
!
(32.5.13)
x=x
(32.5.14)
y=y
(32.5.15)
z = γ (z − ut)
.
(32.5.16)
And the inverse Lorentz transformations:
t=γ
32.5.2.1
uz
t+ 2
c
!
(32.5.17)
x=x
(32.5.18)
y=y
(32.5.19)
z = γ z + ut
.
(32.5.20)
Special Lorentz transformations in matrix form
The special Lorentz transformations read, in matrix form:
r ≡ {ct; x, y, z} ≡ {ct; r}
32.5.2.2
r = Lr
with

γ

 0
L=
 0

−βγ
0
0
1
0
0
1
0
0
−βγ


0 

0 

γ
.
(32.5.21)
Inverse Special Lorentz transformation
Inverse special Lorentz transformations can be obtained by the replacement β → −β. It can be shown by direct
calculation that the inverse Lorentz transformation, given replacing β → −β, is described by the matrix inverse of the
Lorentz transformation one (according to 32.5.43):
{β → −β}
⇔
L → L−1
.
(32.5.22)
32.5.2.3
Composition of two Lorentz transformations along the same direction
Imagine to apply two different Lorentz transformations along the same z direction, from the Reference Frame I1 to I2
(with velocity v12 along z) and from the Reference Frame I2 to I3 (with velocity v23 along z).
Let’s determine the Lorentz matrix corresponding to the composite Lorentz transformation from the Reference Frame
I1 to I3 (with velocity along z).
By explicit calculation of the product of the two matrixes, L23 [v23 ] and L12 [v12 ], and remembering that the matrix of a
special Lorentz transformation has the main diagonal {γ, 1, 1, γ} one finds:
L13 = L23 [v23 ]L12 [v12 ] = L13 [
v12 + v23
] .
1 + v12 v23 /c2
(32.5.23)
The following algebraic identity must be used:
1
γ[v] ≡ p
1 − v 2 /c2
γ[v12 ]γ[v23 ] 1 + v12 v23 /c2 = γ[
563
v12 + v23
] .
1 + v12 v23 /c2
(32.5.24)
December 23, 2011
32.5: Lorentz transformations of space-time coordinates
32.5.2.4
32: From Galileo to Lorentz transformations
Special Lorentz transformations and rapidity
The special Lorentz transformations can be also written in the alternative form:
r = Lr
with



L=


+ cosh χ 0
0
− sinh χ
0
1
0
0
0
0
1
0
− sinh χ
0
0
+ cosh χ






,
(32.5.25)
where the χ parameter, defined by relations
γ ≡ cosh χ ≡
exp [+χ] + exp [−χ]
2
βγ ≡ sinh χ ≡
exp [+χ] − exp [−χ]
2
exp [+χ] − exp [−χ]
,
exp [+χ] + exp [−χ]
(32.5.26)
β ≡ tanh χ ≡
is called rapidity.
Special Lorentz transformations expressed in terms of rapidity have the nice property that the special Lorentz transformation corresponding to two special Lorentz transformations in series, all with the velocities parallel to the z = z 0 = z 00
axes, has a rapidity given by the sum of the rapidities of the two transformations in series. This is analogous to the fact
from elementary geometry that when composing two rotations in the same plane in series the overall rotation has a rotation
angle given by the sum of the angles of the two rotations. This is manifest when using the rotation angle as a parameter for
the rotation. It is not at all obviuos if one would use, for instance, the cosinus or the sinus of the angle as the parameter:
in this case the overall rotation would not be described by the sum of the parameters of the two rotations.
Note the similarity between the Lorentz matrix in terms of rapidity and the one describing ordinary rotations in euclidean
space.
32.5.3
Pure Lorentz transformations for the time-space 4-vector
As a second particular case consider the case when when the Cartesian axes of both Coordinate System are parallel
and with the same orientation, that the two origins coincide at times t = t = 0 but the velocity of Reference Frame I with
respect to Reference Frame I is arbitrary, u.
These pure Lorentz transformations thus read:
t = γ (t − β ·r/c)
r =r+β
(32.5.27)
!
γ2
( β ·r) − γct
γ+1
!
γ−1
( β ·r) − γct
β2
=r+β
.
(32.5.28)
The following identity is often useful to deal with the above expressions:
γ2
γ−1
=
γ+1
β2
.
(32.5.29)
The Lorentz transformation matrix has the following form for a pure Lorentz transformation along
a direction indicated
by the unit vector (for three-dimensional vectors the position of the index is not relevant) n ≡ nk = {nk }:
L=
L00 = γ
Lk0 = −βγnk
L0j = −βγnj
Lkj = (γ − 1) nk nj + δ kj
!
.
(32.5.30)
This expression is coherent with equation 32.5.43 as to take the inverse matrix one must change n −→ −n.
One should not forget that the vectors in equations 32.5.27 and 32.5.28 must be considered in a pure algebraic sense,
that is as triples of real numbers as they actually belong to different Euclidean spaces.
32.5.3.1
Transformation of the parallel and transverse components of the time-space 4-vector
From the equations 32.5.27 and 32.5.28 one can derive the following relations.
r k = γ r k − cβt
r⊥ = r⊥ .
(32.5.31)
(32.5.32)
They are consistent with the special case of the special Lorentz transformations.
564
December 23, 2011
32: From Galileo to Lorentz transformations
32.5.3.2
32.5: Lorentz transformations of space-time coordinates
Events and Lorentz intervals
An event is a well defined and univocally identifiable circumstance happening at a well defined time and at a well defined
place, in any Reference Frame. It is thus characterised by a time and space coordinate: E ≡ {t, x}.
Given two events, E1 ≡ {t1 , x1 } and E2 ≡ {t2 , x2 } their Lorentz invariant interval is defined as:
2
2
2
2
2
(∆s) ≡ c2 (∆t) − (∆x) = c2 (t2 − t1 ) − (x2 − x1 )
.
(32.5.33)
The quantity ∆s has the same value for any inertial observer when calculated between the same to events.
The invariance of the Lorentz interval can be checked for the special cases discussed above.
32.5.4
The general Lorentz transformation
Consider two Inertial Reference Frames, I and I. Assume that I is moving, with respect to I, with a speed u. Assume
that a Cartesian Coordinate System and a time coordinate are defined in both Inertial Reference Frames.
Lorentz transformations give the relation between the time-space coordinates of the Inertial Reference Frame I, the
µ
4-tuple of real numbers
Reference Frame I, the 4-tuple
x ≡ {ct, x1, x2 , x3 }, and the time-space coordinates of the Inertial
µ
µ
of real numbers x ≡ ct, x1 , x2 , x3 . Note that the greek 4-dimensional indexes {...} are placed high.
Define the real 4 × 4 matrix of the linear transformation between xµ ≡ {ct, x1 , x2 , x3 } and xµ ≡ ct, x1 , x2 , x3 as L:
x −→ x
x = Lx
.
(32.5.34)
x0 ≡ ct
x0 ≡ ct
.
(32.5.35)
Let:
The Lorentz transformation can be written, with explicit indexes, as:
xµ = Lµν xν
.
(32.5.36)
The general Lorentz transformations between xµ ≡ {ct, x, y, z} and xµ ≡ ct, x, y, z are derived from the basic postulates of the relativity theory as the set of linear transformations leaving invariant the matrix η (the metric tensor):
ηµν ≡ η
µν

+1

0
≡
0

0
0
0
−1
0
−1
0
0
0
0


0

0

−1
.
(32.5.37)
As a matrix η has the following properties (by direct calculation):
η = η−1 = ηT
ηη = I
.
(32.5.38)
The matrix η with high indexes is defined as the inverse of the η with low indexes which therefore equals η with high
indexes as η = η−1 :
η µν ≡ η−1 µν
no index convention .
(32.5.39)
In components one finds:
η αρ ηρβ = δ αβ ≡ η αβ
.
(32.5.40)
In fact the η matrix, the metric tensor, is used to raise and lower the indexes (see section 8).
Lorentz transformations are defined by the set of real 4 × 4 matrixes, L, satisfying the Lorentz condition:
η = LT ηL
(32.5.41)
In terms of explicit indexes condition 32.5.41 becomes:
η = LT ηL
=⇒
α
ηµν = LT µ ηαβ Lβ ν = Lαµ Lβ ν ηαβ
.
(32.5.42)
The inverse Lorentz transformation matrix can be derived in a straightforward way from the condition 32.5.41 by right
multiplying by L−1 and left multiplying by η:
L−1 = ηLT η
.
(32.5.43)
The Lorentz matrix inversion thus amounts to just transposing and changing the sign of the elements on the first row and
first column, except for the L00 element as implied by equation 32.5.43.
565
December 23, 2011
32.6: Some consequences of the Lorentz transformations
32: From Galileo to Lorentz transformations
From equation 32.5.43, after left multiplication by L and right multiplication by η, one also obtains:
η = LηLT
ηµν = Lµα ηαβ LT
=⇒
β
ν
= Lµα Lν β ηαβ
.
(32.5.44)
Finally, from equation 32.5.43, one obtains:
L−1
T
−1
= LT
= ηLη
.
(32.5.45)
When handling Lorentz matrixes and indexes one should carefully remember the rules/conventions stated in section 8.2.3.
The following additional conditions define the orthochronous proper Lorentz transformations:
DetL = +1
L00 ≥ 1
,
(32.5.46)
which corresponds to exclude from Lorentz transformations both parity inversion and time inversion.
32.6
Some consequences of the Lorentz transformations
Consider special Lorentz transformations along the z axis for simplicity.
32.6.1
Time dilatation and proper time (special Lorentz transformations)
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {12} —
Consider an object fixed in I: r 1 = r 2 .
The time interval between any two events happening at the same place is so-called proper time.
More generally: the proper time is the time between two events whenever:
∆r = 0
same position along the direction of u is enough
.
(32.6.1)
Therefore one needs to relate: ∆t, ∆t and ∆r = 0. From the appropriate expressions of the Lorentz transformations
(equation 32.5.17) one finds:
∆τ ≡ ∆t = ∆t /γ
.
(32.6.2)
Any clock in motion runs slower. However the effect is only noticeable when the clock travels at relativistic speeds.
Moreover: ∆z = u ∆t, as it is obvious from the definition of velocity for a point fixed in I.
The observer who sees the clock moving, I, will not agree that the positions have been recorded at the same place but
it will measure a difference of position ∆z = u ∆t, as it is obvious from the definition of velocity for a point fixed in I.
Note that r is held fixed, because we are watching one specific clock moving. If one would keep r fixed then one would
watch a whole series of different clocks in I passing by and this would not tell us whether one of them is running slow or
not.
It can be shown that the time dilatation is a reciprocal effect among the two observers, such that any of the two sees
the clocks of the other observer running slower 3 .
Note that moving clocks are not synchronised among each other as synchronisation is not an invariant concept so that
if the clocks are synchronised in the frame where they are at rest they are not in other systems; therefore it is essential
when checking time dilation to focus attention on a single moving clock.
Proper time is normally denoted as ∆τ or dτ . Proper time of an object, that is time measured by a clock at rest with
the object, is clearly an invariant, by definition. In many cases it is a quantity more useful than time.
Simultaneity synchronization and time dilation
Suppose that at time t = 0 the observer I decides to examine all the clocks in I. He will find that they all read different
times, depending on their location: those at negative z are ahead, and those at positive z are behind, by an amount that
increases in proportion to their distance. Only the master clock at the origin reads t = 0.
How does the observer examines the clocks at rest in I? He needs to compare at any time the moving clock with its
own clock at rest at the position of the moving clock.
3 D.
J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall), cap. 12.1.2
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December 23, 2011
32: From Galileo to Lorentz transformations
32.6.1.1
32.7: Lorentz transformation of physical quantities
Proper time an proper Rest Frame
Consider any Reference Frame, I, moving arbitrariliy without rotations with respect to the observer I. That is assume
that the two Reference Frames are always linked by a pure Lorentz transformation, at any time.
At any instant the relations between the times measured in the two Reference Frame is given by:
dt = γ dt ≡ γ dτ
.
(32.6.3)
The invariant dτ is the proper time that is the time measured in the proper Rest Frame, because that gives the time
interval when dx = 0, that is the time measured by an observer at rest with respect to the system.
In fact it is a basic postulate of Relativity that time only depends on the speed of the clock, not on acceleration nor any
other higher-order derivative, see section 32.4.3.
32.6.2
Length contraction and proper length (special Lorentz transformations)
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {12} —
In order to measure the length of a moving segment one needs to pick-up the position of the two ends of the segment at
the same time. It is necessary to understand precisely what the length represents: it is the distance between the two points
at the same instant as judged in the Reference Frame of the observer. However the criterion of simultaneity is different in
any other Reference Frame, and hence the result of measuring the length will be different, too.
Consider an object at rest in I, whose ends are fixed at r 1 and r 2 .
When observing from the Reference Frame I the observer must take the position of the two ends at the same time in
order to measure its length, so that:
∆t = 0 .
(32.6.4)
Therefore one needs to relate: ∆r, ∆r and ∆t = 0. From the appropriate expressions of the Lorentz transformations
(equation 32.5.16) one finds:
∆z = γ ∆z .
(32.6.5)
On the other hand we don’t need to care about times in I because the segment is at rest in I. In fact for the I
observer the positions of the two ends are a function of its time, while for the I observer the positions of the two ends are
time-independent and therefore the time in I does not enter.
The observer at rest with the object, I, will not agree that the positions have been recorded at the same time but it
will measure a difference of time ∆t = −u ∆z /c2 .
Any object in motion appears to be shorter along the direction of motion. However the effect is only noticeable when
the obiects travelss at relativistic speeds.
Of course no physical effect has happened to the rod: it is only the process of measurement in the moving Reference
Frame which has given a different result.
It can be shown that the length contraction is a reciprocal effect among the two observers, such that any of the two sees
length of the objects at rest with respect the other observer being shorter 4 .
32.6.3
Transformation law for the volume
Consider a parallelepiped, fixed with respect to the moving Reference Frame I, with its sides either parallel or perpendicular to the speed u. The sides perpendicular to the speed u do not contract. The side along the speed u contracts by a
factor γ. Therefore volumes change as:
∆V = ∆V /γ
32.7
32.7.1
.
(32.6.6)
Lorentz transformation of physical quantities
Notations and conventions
Note that the system of units such that c = 1 will be often used except for some of the basic formulas.
A 4-vector has one index (always a greek letter) which can have four values: 0, 1, 2, 3. The index zero denotes the
time-component of a 4-vector. Note that the position of the vector/tensor indexes (high/low) are important and high/low
indexes are not equivalent. Any equality must carry indexes in the same position. Whenever a pair of indexes are summed
one of them must be high while the other one must be low.
Three dimensional space position vectors will be denoted either as r ≡ rk ≡ rk = {x, y, z} or x ≡ xk ≡ xk = {x1 , x2 , x3 }.
The position of the vector/tensor indexes (high/low) is not relevant.
Four dimensional time-space 4-vectors will be denoted either as r ≡ rµ = {ct, r} = {ct, x, y, z} or x ≡ xµ = {ct, x} =
{ct, x1 , x2 , x3 }. The position of the vector/tensor indexes (high/low) are important and high/low indexes are not equivalent.
4 D.
J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall), cap. 12.1.2
567
December 23, 2011
32.7: Lorentz transformation of physical quantities
32.7.2
32: From Galileo to Lorentz transformations
Scalars or invariants under Lorentz transformations
By definition a Lorentz scalar is any physical quantity which has the same value in any inertial Reference Frame, that
is any physical quantity which is invariant under Lorentz transformations:
S=S
.
(32.7.1)
Examples of Lorentz scalar quantities are: the Lorentz interval between any two events, the mass of any system and the
electric charge.
32.7.3
4-vectors under Lorentz transformations
A 4-vector is any set of four physical quantities transforming, under Lorentz transformation:
• either as the differentials of the time-space coordinates (a controvariant vector);
• or as the partial derivatives with respect to the time-space coordinates (a covariant vector);
See also section 8.
In fact two different possible transformations laws exists for 4-vectors under Lorentz transformations, as it is the case
for general coordinate transformations, see section 8. The two kinds of 4-vectors are distinguished by upper (high) indexes,
controvariant vectors, and lower (low) indexes, covariant vectors.
As Lorentz transformations mixes the different componets it is necessary that the four components have the same
physical dimensions; in fact the Lorentz matrix is adimensional. When grouping together two different physical quantities
to form a 4-vector we will use the convention of retaining the physical dimensions of the tri-vector and modifying with
suitable constants the physical dimensions of the zero-component.
32.7.3.1
Controvariant and covariant form and transformations of 4-vectors
The following heuristic considerations can be carried on.
It can be shown that the η matrix can be used to raise and lower the indexes (see section 8), so that controvariant and
covarinat form of a 4-vectors are just different ways to represent the 4-vector. The relation between controvariant, V α , and
covariant, Vα , form of a 4-vector is:
Vα = ηαβ V β
V α = η αβ Vβ .
(32.7.2)
By definition the time-space 4-vector is (controvariant/covariant forms):
xα ≡ {ct, +x}
⇔
xα ≡ {ct, −x}
.
(32.7.3)
By definition the time-space 4-gradient is (controvariant/covariant forms)::
(
) (
)
(
) (
)
1 ∂
∂
1 ∂
∂
1 ∂
∂
1 ∂
∂
α
∂α ≡
≡
,+
≡
, +∇
⇔
∂ ≡
≡
,−
≡
, −∇
.
∂xα
c ∂t
∂x
c ∂t
∂xα
c ∂t
∂x
c ∂t
(32.7.4)
The direct and inverse Lorentz transformations in matrix form read:
x = Lx
x = L−1 x
⇔
.
(32.7.5)
The direct and inverse Lorentz transformations with explicit indexes read:
xα = Lαβ xβ
xα = L−1
⇔
The Jacobian matrixes of the two transformations are:
∂xα
= Lαβ
∂xβ
∂xα
⇔
∂x
β
α
β
= L−1
xβ
α
β
.
(32.7.6)
.
(32.7.7)
The differentials of the space-time coordinates transform, in matrix form, as:
dx = L dx
.
(32.7.8)
Let’s introduce the useful notation:
∂
≡ ∂µ
∂xµ
∂
≡ ∂µ
∂xµ
.
(32.7.9)
The partial derivative of the space-time coordinates transform as:
∂
∂x
β
=
α
∂ ∂xα
∂
=
L−1 β
β
α
α
∂x ∂x
∂x
568
=⇒
∂ β = ∂α L−1
α
β
.
(32.7.10)
December 23, 2011
32: From Galileo to Lorentz transformations
32.7: Lorentz transformation of physical quantities
It is therefore clear that for a given Lorentz transformation the partial derivatives transform in a different way from the
T
differentials: in matrix form the row 4-gradient transformed vector, ∂ , is found from the product between the row 4-gradient
vector, ∂T , and the inverse Lorentz matrix L−1 .
The partial derivative of the space-time coordinates transform, in matrix form, as:
T
∂ = ∂T L−1
.
(32.7.11)
In order to distinguish between the two different way of transformations of 4-vectors under the same Lorentz transformation:
1. the first type of 4-vectors, those transforming as the differentials of the coordinates, have their greek indexes in the
high position and called controvariant 4-vectors;
2. the second type of 4-vectors, those transforming as the gradient of the coordinates, have their greek indexes in the
low position and called co-variant 4-vectors.
32.7.3.2
Propeties of 4-vectors
It can be shonw that the relations between the controvariant and covariant components of a four vector are:
Vµ = ηµν V ν
V µ = η µν Vν
.
(32.7.12)
Controvariant vectors, by definition, transforms as the differentials of the coordinates, that is:
V
µ
= Lµν V ν
V = LV
in matrix form
.
(32.7.13)
Covariant vectors, by definition, transforms as the gradient of the coordinates, that is:
V µ = Lµν Vν
T
V = V T L−1
in matrix form
.
(32.7.14)
Note that the above result just comes from lowering-rising the indexes. It is a particular case of the more general
treatment is section 8 which would be valid for any change between Coordinate System.
See section Problem - 32.16.
32.7.3.3
Scalar product of 4-vectors
It can be shown that in order to produce a scalar as product of two 4-vectors one needs to use the controvarinat form
of one 4-vector and the covariant form of the other one:
A ·B ≡ Aµ Bµ = Aµ B µ
.
(32.7.15)
The invariance of the scalar product is clear from the above relations 32.7.13 and 32.7.14. In fact by considering the
controvariant form as a column vector and the covariant form as a row vector the scalar product can be written in matrix
form as:
Aµ B µ = AT ·B = BT ·A = Bµ Aµ ,
(32.7.16)
as the transpose of a 1 × 1 matrix is the same number.
It is then clear from equations 32.7.13 and 32.7.14 that applying a Lorentz transformation the result does not change
as the Lorentx matrix and its inverse cancel out.
It immediately follows that in order to have a scalar one needs to sum one high and one low index.
Direct check of invariance
Check by direct calculation for a special Lorentz transformation that the scalar product of any two 4-vectors is invariant.
32.7.4
Invariant quantities versus conserved quantities
One has to be careful to avoid confusion between the concept of a scalar (invariant) quantity and the concept of a
conserved quantity.
• A Lorentz scalar (Lorentz invariant) is a quantity which has the same numerical value in any inertial Reference Frame.
• A conserved quantity is a quantity which does not change before and after some process. Conserved quantities are
assumed to be additive.
The rest mass is a scalar. However the rest mass of a complex system (also called the invariant mass) is not, in general,
conserved. The rest mass (just called mass) of an elementary entity (particle) is an intrinsic properties of the entity: if it
changes we assume that the identity of the elementary entity has changed.
Energy is a conserved quantity, but it is not a scalar; it is the time-component of a 4-vector.
Electric charge is both a scalar and a conserved quantity.
Velocity is neither conserved nor invariant.
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December 23, 2011
32.8: Some Examples and Physical Applications
32.7.5
32: From Galileo to Lorentz transformations
Scalars, vectors and tensors for Lorentz versus Galileo
Rotation transformations (leave I invariant):
I = RT R = RRT
=⇒
R−1 = IRT I
δpq = Rpa Rq b δab
(32.7.17)
.
(32.7.18)
Note that equation 32.7.18 actually means that δ is a second order tensor for the rotations group, actually an invariant
one, that has the same components in any Coordinate System.
Lorentz transformations (leave η invariant):
η = LT ηL = LηLT
=⇒
ηµν = Lµα Lν β ηαβ
L−1 = ηLT η
(32.7.19)
.
(32.7.20)
Note that equation 32.7.20 actually means that η is a second order tensor for the Lorentz group, actually an invariant
one, that has the same components in any Reference Frame.
32.8
32.8.1
Some Examples and Physical Applications
The speed of light as a limiting speed
[] Reference: WRindler: {Relativity} — Excellent
The γ factor of Lorentz transformations becomes infinty, for u = c, and imaginary, for u > c. Therefore the relative
velocity between two different Reference Frame must be less than the speed of light, if finite real-valued coordinates in
one Reference Frame must correspond to finite real-valued coordinates in any other Reference Frame. It follows that no
particle can move superluminally relative to any Reference Frame, because any set of such particles moving parallelly would
constitute a Reference Frame moving superluminally relative to the first Reference Frame.
Other indications show that the speed of particles and of all physical signals, is upper-limited by c, such as violation
of causality in presence of superluminal signal tramsmission. Consider, in fact, any signal or process whereby a first event
causes a second event (or whereby information is sent from the first event to the second event) at superluminal speed,
U , relative to some inertial Reference Frame. By applying Lorentz transformations it is then possible to find an inertial
Reference Frame, moving at speed u, such that c2 /U < u < c, where the time-ordering of the two events is inverted.
Therefore there would exist some inertial Reference Frame in which the second event precedes the first one, so that cause
and effect are reversed and the signal is considered to travel in the opposite spatial direction.
The whole concept of superluminal signaling is seen to produce paradoxes. We are thusled to accept the axiom that no
superluminal signals can exist, so that c is an upper-limit to the speed of macroscopic information-conveying signals. In
particular, this speed limit must apply to particles, since they can convey messages.
Relativistic mechanics provides a speed limit by having the mass of particles increase beyond all bounds as their speed
approaches the speed of light. Any massive particle that constantly accelerates, no matter how strongly, approaches but
never attains the speed of light.
One consequence of the upper relativistic speed limit is that such things as rigid bodies and incompressible uids have
become impossible objects, even as idealizations or limits: in fact, by denition, they would transmit signals instantaneously.
Note, however that arbitrarily large velocities are possible for moving points that carry no information. The sweep of a
light spot from the Earth on the Moon surface, where a movable laser beam from Earth impinges, or the intersection point
of two rulers that cross each other at an arbitrarily small angle are two such examples.
32.8.2
Some experimental tests of special relativity
Experimental tests of special relativity are particularly importan as special relativity has many counter-intuitive predictions and in order to appreciate relativistic results it is necessary to have objects moving close to the speed of light, which
is difficult to attain.
Experiments on the contraction of lengths are clearly impossible, as this would require to run physical objects close to
the speed of light.
Experiments on the dilatation of time are easier.
32.8.2.1
Muon lifetime
The lifetime increase of a high-energy particle is often used in High-Energy Physics to observe the decay point of a
short-lived particle away from the production point.
Muons have a lifetime τ0 = 2.2·10−6 s and mass m = 0.105 GeV/c2 .
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December 23, 2011
32: From Galileo to Lorentz transformations
32.8: Some Examples and Physical Applications
Which energy is necessary to a muon to reach the Earth surface once it was produced in the high part of the atmosphere
at H = 10 km?
How is the kinematics described by the observer at the Earth and by muon?
Worked Solution
τ0 p
τ0 βε
=
−→ ε = 1.6 GeV
assuming β ' 1 .
(32.8.1)
m
mc
Description of the observer at the Earth: the muon lives τ0 γ and travels τ0 γβc (time dilatation).
Description of the muon: the muon lives τ0 and travels τ0 βc (length contraction). The muon feels a shorter life, by a
factor γ, but it sees distances contracted, by a factor γ, so that the descrition of the muon is consistent with the description
of the ground observer.
H = τ0 γβc =
32.8.2.2
Time dilatation and the clock hypotesis and the twin paradox
The so-called twin paradox occurs when two clocks are synchronized, separated, one of them takes a long journey,
rejoined and theri readings are finally compared. If one clock remains in an inertial frame then the other must be accelerated
sometime during its journey and it will display less elapsed proper time than the inertial clock at the end of the journey.
Therefore the situation is not symmetrical between the two clocks.
This is a paradox only in that it appears to be inconsistent, but is not.
This has been tested using muons stored in a storage ring and measured their lifetime. When combined with measurements of the lifetime of muons at rest (the twin at rest) this becomes a twin paradox experiment.
Atomic clock on commercial airplanes
...Argomento trattato a lezione...
Proposal: Hafele,J.andKeating,R.(July14,1972);http://www.sciencemag.org/cgi/content/abstract/177/4044/
166.
Experiment: Hafele,J.andKeating,R.(July14,1972);http://www.sciencemag.org/cgi/content/abstract/177/
4044/168.
They flew caesium atomic clocks on commercial airliners around the world in both directions, and compared the time
elapsed on the airborne clocks with the time elapsed on an earthbound clock.
The effect includes both an effect of special relativity and an effect of general relativity.
Lifetime of muons in storage rings
...Argomento trattato a lezione...
Bailey et al., Measurements of relativistic time dilation for positive and negative muons in a circular orbit, Nature 268
(July 28, 1977) pag. 301.
They stored muons in a storage ring and measured their lifetime. When compared with measurements of the muon
lifetime at rest this is a highly relativistic twin scenario as v ' 0.9994c, for which the stored muons are the traveling twin
and return to a given point in the lab every few µs.
The experiment confirms the clock hypothesis stating that the tick rate of a clock when measured in an inertial frame
depends only upon its velocity relative to that frame, and it is independent of its acceleration or higher derivatives with
respect to time. In fact muons in the storage ring are accelerated and their lifetime could be measured to only depend on
the velocity.
32.8.3
Experimental test of the second postulate
[] Reference: J. D. Jackson, Elettrodinamica Classica, seconda edizione, (1984, Zanichelli) : {} —
...Argomento trattato a lezione...
Experiment performed at CERN, Geneva, Switzerland, in 1964: the speed of photons with ε = 6 GeV (speed 0.99975c),
produced in the decay of very energetic neutral pions, was measured by time of flight over paths up to 80 meters.
Within the experimental errors it was found that the speed of the photons emitted by the extremely rapidly moving
source was equal to c.
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December 23, 2011
32.9: Exercises, problems and physical applications
32.9
32: From Galileo to Lorentz transformations
Exercises, problems and physical applications
Problem - 32.1
The invariant quantity of Lorentz transformations
Show, by direct calculation, that, for a special Lorentz transformation:
2
2
c2 (∆t) − (∆z) = c2 ∆t
2
− (∆z)
2
− (∆x)
2
.
(32.9.1)
2
.
(32.9.2)
Any general Lorentz transformation leaves invariant the quantity:
2
2
c2 (∆t) − (∆x) = c2 ∆t
Problem - 32.2
An algebraic identity
Show that:
γ2 − 1
u2
=
γ2
c2
.
(32.9.3)
Problem - 32.3
The non-relativistic limit
Show that Lorentz transformations reduce to Galileo transformations in the limit: c → ∞.
Problem - 32.4
Definition of rapidity
Show that the above definitions 32.5.26 are consistent, using the definitions of hyperbolic functions and of the beta and
gamma functions.
Problem - 32.5
Special Lorentz transformations and rapidity
Demonstrate, using the hyperbolic trigonometric relations, that the transformation corresponding to two special Lorentz
transformations in series has a rapidity given by the sum of the rapidities of the two transformations.
Problem - 32.6
Composition of Lorentz transformations
Consider three Inertial Reference Frames, I1 , I2 and I3 . Let u21 be the velocity of I2 with respect to I1 , let u32 be
the velocity of I3 with respect to I2 and let u31 be the velocity of I3 with respect to I1 . Choose the Coordinate Systems
in such a way that both transformations (1) → (2) and (2) → (3) are pure transformations.
Show that:
!
u21 ·u32
γu31 = γu21 γu32 1 +
.
(32.9.4)
c2
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December 23, 2011
32: From Galileo to Lorentz transformations
32.9: Exercises, problems and physical applications
Problem - 32.7
Group properties of the Lorentz transformations
Demonstrate that the Lorentz transformation form a group.
Problem - 32.8
Inverse Lorentz transformation
Prove that the inverse Lorentz transformation in 32.5.30 is obtained by replacing n with −n.
Problem - 32.9
Infinitesimal Lorentz transformation
Show that for an infinitesimal Lorentz transformation,
h
i
2
L = I + B dθ + O (dθ)
,
(32.9.5)
the matrix B satisfies:
ηB = −ηBT ,
(32.9.6)
that is the matrix ηB is anti-symmetrical.
Problem - 32.10
Composition of pure Lorentz transformation
Show that the composition of two pure Lorentz transformations is a pure Lorentz transformation if and only if the
velocities of the transformations are parallel.
Problem - 32.11
Change of angles
The Reference Frame I is related to the Reference Frame to I by the special Lorentz transformation 32.5.13.
Calculate in I the length and the angle with the z axis of a rod of length L at an angle θ with the z axis in I.
Problem - 32.12
♠
Change of direction in a Lorentz transformation
Show that, because of the length contraction in the direction of motion, only oriented segments either parallel or
perpendicular to the velocity fo the Lorentz transformation keep the same direction after a Lorentz transformation.
Worked Solution
The formula of transformation of directions is:
1
−1
γ
!
u ·λ
λ+u
u2
q
λ=
2
1 − u ·λ/c
573
!
.
(32.9.7)
December 23, 2011
32.9: Exercises, problems and physical applications
32: From Galileo to Lorentz transformations
Problem - 32.13
♠
Change of angles in a Lorentz transformation
Show that the concept of parallelism is invariant under Lorentz transformations, but the concept of perpendicularity it
is not.
Show that the angle does not change when both directions are either perpendicular or parallel to the velocity.
Worked Solution
The formula of transformation of angles is:
cos [α] = cos [α] −
u ·λ1
u ·λ2
c2
2 2 1 − u ·λ1 /c
1 − u ·λ2 /c
.
(32.9.8)
Problem - 32.14
Check of the reciprocity relation
Use equation 33.1.3 and consider its inverse equation. Replace the inverse equation into 33.1.3 and check that the result
is an identity.
Problem - 32.15
Raising/lowering indexes and lowering/raising indexes
Show that raising a previously lowered indexes as well as lowering a previously raised index recovers the original 4-vector.
Problem - 32.16
♠
Relation between controvariant and co-ariant transformations
Find the transformation law for covariant vectors from the one for controvariant vectors.
Worked Solution
Vµ ≡ ηµβ V β
=⇒
V µ = ηµβ V
β
= ηµβ Lβ α V α = ηµβ Lβ α η αν Vν ≡ Lµν Vν
.
(32.9.9)
Note that the transformation matrix is exactly obtained by Lµν by lowering the first index and raising the second one.
574
December 23, 2011
33
Relativistic Kinematics
33.1
Transformation laws of velocity
The transformation law of velocities is a rather complex one, and it is a non-linear one. In fact the transformation law
of the velocities requires to transform both the numerator and the denominator of the derivative and this makes the law
complex.
33.1.1
Transformation laws for the velocity (special Lorentz transformation)
Consider two Inertial Reference Frames, I and I. Assume the two Reference Frames are connected by a special Lorentz
transformation and that I is moving, with respect to I, with a speed u ≡ βc along the z axis (u = ue3 ).
∆z
∆t→0 ∆t
∆(x|y)
≡ lim
.
∆t
∆t→0
v z ≡ lim
v (x|y)
(33.1.1)
(33.1.2)
When ∆t → 0 then also ∆x → 0, ∆y → 0 and ∆z → 0.
Therefore Lorentz transformations imply that ∆t → 0 and also ∆x → 0, ∆y → 0 and ∆z → 0.
γ (∆z − u ∆t)
∆z
vz − u
= lim
=
2
∆t→0 γ (∆t − u ∆z /c )
1 − uvz /c2
∆t→0 ∆t
v(x|y)
∆(x|y)
∆(x|y)
= lim
≡
= lim
.
2
∆t→0 γ (1 − uvz /c2 )
∆t
∆t→0 γ (∆t − u ∆z /c )
v z ≡ lim
v (x|y)
(33.1.3)
(33.1.4)
The non-relativistic limit applies when all velocities (on the right-hand side of the above equations) are non relativistic:
vz c, v(x|y) c and |u| c. In this limit the above relations reduce to the one predicted by Galileo transformations
and the resulting velocities on the left-hand side turns out to be non-relativistic.
These equations implement, among other things, the fact that the maximum possible velocity is the speed of light, so
that no composition of velocities can lead to a speed greater than the speed of light. Consider, for instance, the special case
vz = c, which implies vx = vy = 0, one has v z = c, which implies v x = v y = 0, whatever u. The general case, for arbitrary
velocity, is discussed in section Problem - 33.2.
In summary, in the case of a special Lorentz transformation along the z axis (u = ue3 ):
vz − u
1 − uvz /c2
(33.1.5)
vx
γ (1 − uvz /c2 )
(33.1.6)
vz =
vx =
vy =
vy
γ (1 − uvz /c2 )
.
(33.1.7)
Note that the transverse components of the velocity transform, because even if the transverse components of the position
vector do not transform, one has to take into account the transformation law for the time at the denominator.
575
33.1: Transformation laws of velocity
33.1.2
33: Relativistic Kinematics
Transformation law for the velocity (pure Lorentz transformation)
Consider two Inertial Reference Frames, I and I. Assume the two Reference Frames are connected by a pure Lorentz
transformation and that I is moving, with respect to I, with a speed u ≡ βc.
It can be straightforwardly derived, for a pure Lorentz transformation, from the transformation laws 32.5.27 and 32.5.28:
v≡
dx
dx dt
=
=
dt dt
dt
v
+β
γ
1
1−
γ
!
β ·v
−c
β2
!
!
γ2
v+β
( β ·v) − γc
γ+1
!
=
,
β ·v
γ 1−
c
(33.1.8)
β ·v
+c
β2
!
!
γ2
v+β
( β ·v) + γc
γ+1
!
.
=
β ·v
γ 1+
c
(33.1.9)
β ·v
1−
c
and the inverse:
v≡
dx
dx dt
=
=
dt
dt dt
v
+β
γ
1
1−
γ
!
β ·v
1+
c
The expressions in the case the velocity v is either parallel or orthogonal to the relative velocity β can be derived by
direct computation from equation 33.1.8:
vku
=⇒
v⊥u
=⇒
v−u
=⇒
1 − uv/c2
v
v=
−u
=⇒
γ
vku
v=
(33.1.10)
v is not perpendicular to u, as for Galileo transformations
.
(33.1.11)
These equations are consistent with equations coming from a special Lorentz transformation 33.1.1.
Equation 33.1.10, in fact, is the same as equation 33.1.5.
Equation 33.1.11, when vz = 0, in fact, implies that:
• the component of v parallel to u is −u, which is the same result coming from equation 33.1.5 when vz = 0;
• the component of v perpendicular to u is v/γ, which is the same result coming from equations 33.1.6 and 33.1.7 when
vz = 0.
33.1.3
Lorentz transformation of the module of speed
See section Problem - 33.2 to demonstrate that the Lorentz transform of any speed less than c is less than c:
2
2
c −v =
33.1.4
c2 − v 2
c2 − u2
(c2 − u ·v)
2
.
(33.1.12)
Mutual and relative position, velocity and acceleration
The concept of mutual vector involves one observer and two different observed points; the concept of relative vector,
based on the composition of vectors involves two different observers and one observed point. Consequently, the concept of
mutual vector cannot involve the Galileo/Lorentz transformations.
The most intersting case is the case of velocity.
There are, therefore, two different concepts:
• mutual velocity between two Reference Frames, as seen by an external observer: u2 − u1 (one observer but two
different observed points);
• the relative velocity of the Reference Frame (2) with respect to the Reference Frame (1), that is the velocity of the
Reference Frame (2) as measured by Reference Frame (1) (two different observers but one observed point).
The two concepts turn out to coincide in classical mechanics due to Galileo invariance.
While the expression v 1 −v 2 is invariant under Galileo transformations it is not invariant under Lorentz transformations.
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December 23, 2011
33: Relativistic Kinematics
33.1.5
33.2: Transformation laws for the acceleration
Ordinary velocity and proper velocity
dx
Ordinary velocity is defined as: v ≡
; from now on it will be simply called velocity. Both quantities, dx and dt are
dt
measured with respect to the inertial Reference Frame of the observer, for a moving object.
Sometimes it might be useful to consider the distance per unit proper time, which defines a hybrid quantity, the so-called
dx
. In this case dx is measured with respect to the inertial Reference Frame of the observer but the
proper-velocity: V ≡
dτ
proper time is measured in the Reference Frame at rest with the moving object.
In fact, while remembering the relation between time and proper time
dt = γ dτ
.
(33.1.13)
one can define, alongside with the velocity, the proper velocity, V :
dx
dt
|u| c
v≡
V = γv v
V ≡
=⇒
dx
dτ
(33.1.14)
V =v .
The definition can be extended to the 4-dimensional quantity:

dx

V ≡
= γv v
dx
dτ 0
V ≡

dτ
V 0 ≡ dx = γ c
v
dτ
(33.1.15)
,
(33.1.16)
which is called the proper velocity 4-vector.
The proper velocity, V , will transform according to Lorentz transformations, that is a much easier transformation law
than the transformation law for velocity.
It immediately follows from its definition that:
V ·V = c2 .
(33.1.17)
Note that the proper velocity is a hybrid quantity: it is the ratio of a displacemetn as measured by the Reference Frame
of the observer, I, and the time, as measured by an observer at rest with the moving object, that is moving with respect
to Reference Frame of the observer, I.
33.2
Transformation laws for the acceleration
The transformation law for ordinary acceleration is rather complex. Therefore proper acceleration is most often used,
which transforms as a 4-vector:
dV
d2 x
A≡
.
(33.2.1)
≡
dτ
dτ 2
It turns out that:
V ·A = 0 .
33.3
33.3.1
(33.2.2)
MPWs: transformation of frequency and wavevector
Lorentz invariance of the phase of MPWs
Consider a MPW (a scalar one, for simplicity), with angular frequency ω and wavevector k:
ψ[t, x] = ψ 0 cos [ k ·x − ωt − φ]
Φ[x, t] ≡ k ·x − ωt − φ ,
(33.3.1)
where ψ 0 is a scalar, giving the scalar character to the MPW.
Thanks to the linearity of Lorentz transformations the phase-factor, cos [ k ·x − ωt − φ], will transform into a similar
phase factor in any other Reference Frame, and therefore the monochormatic plane wave will appear as a MPW.
As the phase of the MPW must be a Lorentz invariant, as demonstrated below, the transformed monochormatic plane
wave will have the same scalar/vector/tensor character in any Reference Frame, the one given by the multiplicative factor
to the phase-factor (the scalar ψ 0 in our case).
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December 23, 2011
33.3: MPWs: transformation of frequency and wavevector
33: Relativistic Kinematics
The phase of the wave must be a scalar (that is Lorentz invariant) because the measurement of the difference of the
phase at two different space-time events can be reduced to the number of counts of the wave-crests in between the two
space-time events, which must give the same result in any Reference Frame.
As a first example consider the difference of the phase at two different space points at the same time:
∆Φ = k ·(x2 − x1 ) = k |x2 − x1 | cos [θ] ,
(33.3.2)
where θ is the angle between the vectors k and x2 − x1 . The difference of the phase is just the fractional number of
wavelengths (space periods) between the two space points (times 2π).
As a second example consider the difference of the phase at the same space point between two different times:
∆Φ = −ω (t2 − t1 ) .
(33.3.3)
The difference of the phase is just the fractional number of time periods passing away at the fixed space point between the
times t1 and t2 (times 2π).
33.3.2
Lorentz transformation of frequency and wavevector
If one defines, for a wave with phase-velocity v, the object with 4-components k µ :
k µ ≡ {ω/c, k} ≡ k 0 , k
v |k| = ω ,
(33.3.4)
The fact that the phase is invariant implies that k µ is a 4-vector (thanks to the quotient law), the frequency-wavevector
4-vector. Note that the time-like component is always ω/c and not ω/v, as it follows from the expression of the phase in
equations 33.3.1. Its transformation laws give the Doppler effect and the aberration of wave-fronts.
33.3.3
EM radiation
In the case of ElectroMagnetic radiation:
c |k| = ω
=⇒
k µ kµ = 0 .
(33.3.5)
In the EM case the phase-velocity is always c in any Reference Frame.
Consider the Reference Frame I, where a MPW has 4-vector
k}, and an observer moving with a velocity u
k ≡ {ω/c,
with its Reference Frame I, seing the MPW with 4-vector k ≡ ω/c, k . A light ray is described by the unit vector n, in
Reference Frame I, and by the unit vector n, in I.
Relativistic optics is rather simple as there is no privileged Reference Frame. Light rays in vacuum follow straight paths
with speed c in any Inertial Reference Frame.
Let’s consider the pure Lorentz transformation between the two inertial Reference Frames, I and I, the latter moving
with velocity u with respect to the former, and denote the velocities of the same light ray in the two Reference Frames
with v ≡ cn and v ≡ cn.
33.3.3.1
Aberration of light
The law of addition of the velocities gives:
v
≡n=
c
n
+u
γu
1
1−
γu
u ·n
1
−
2
u
c
!
u ·n
1
+
u2
c
!
u ·n
1−
c2
with inverse law:
v
≡n=
c
!
n
+u
γu
1
1−
γu
!
,
(33.3.6)
.
(33.3.7)
u ·n
c2
Alternatively the aberration can be described via the law of transformation of the wave vector, k, such that for a pure
transformation one has:
!
!
γ−1
ω
γ2
ω
k =k+β
( β ·k) − γ
=k+β
( β ·k) − γ
.
(33.3.8)
β2
c
γ+1
c
1+
The relation between n and n can be deduced from equations 33.3.6 and 33.3.7, describing light aberration.
Let’s define:
n ·u ≡ −u cos [α]
n ·u ≡ −u cos [α] .
578
(33.3.9)
(33.3.10)
December 23, 2011
33: Relativistic Kinematics
33.4: Some Examples and Physical Applications
Using equations 33.3.6 and 33.3.7 one finds:
cos [α] +
cos [α] =
33.3.3.2
u
1 + cos [α]
c
.
(33.3.11)
The Doppler effect
ω = γ (ω − u ·k)
33.3.3.2.1
u
c
.
(33.3.12)
The particular case of source at rest in I
If the source is at rest in I, so that the k ≡ ω/c, k is the proper frequency-wavevector of the source, the observer will
see the source moving with velocity u with respect to I. Assume that the incoming light wave-vector, k, is perpendicular
with respect to the velocity u.
Therefore:
ω = γω ≥ ω ,
(33.3.13)
k ·u = 0
=⇒
so that the observer sees a smaller frequency than the proper frequency of the source, that is a red-shift with respect to the
proper frequency of the source.
Therefore the transformation law for the frequency gives the so-called transverse Doppler effect when the source is
moving perpendicularly the line of sight in the opinion of the observer, which is a consequence of time dilatation.
This is a purely relativistic effect, for classically one would not expect any frequency shift from a source that moves by
right angles with respect to the line of sight.
33.3.3.2.2
The transverse Doppler effect
The transverse Doppler effect comes in when a light ray is reaching the observer Reference Frame, I, perpendicularly
to the velocity, u, of I: k ·u = 0.
Consider in fact the pure Lorentz transformation, equation 32.5.27, with k ·u = 0, one can write:
(
ω = γω
k ·u = 0
=⇒
.
(33.3.14)
k = k − βγω/c
As a consistency check one can use the previous result to evaluate:
ω = γ ω + c β ·k = γ ω + β 2 γω
=⇒
ω = ω/γ .
(33.3.15)
Note that the presence of the source is, so-far, inessential.
33.4
33.4.1
Some Examples and Physical Applications
Relativity and absolute quantities
Note that there actually exist absolute quantities in relativity, indeed: the rest length of a rod is an absolute quantity,
the same for all inertial observers; the proper time of a clock, which might better have been called the local time, is an
invariant quantity.
Also, in non-relativistic physics, it is a common notion that measured physical quantities depend on the inertial Reference
Frame: velocity, kinetic energy, frequency....
33.4.2
Relative velocity between particles in a symmetric particle collider
Two particles collide head-on with velocities ±v in the laboratory Reference Frame. Determine the velocity of one
particle as seen from the other particle.
Worked Solution
Let’s consider the one-dimensional problem: ±v is set along the z axis.
Assume to transform from the laboratory Reference Frame to the Rest Frame of particle 1, the one moving with speed
+v.
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December 23, 2011
33.4: Some Examples and Physical Applications
33: Relativistic Kinematics
u = +v
v1 = +v
v2 = −v
(33.4.1)
v1 = 0
(33.4.2)
−v−v
− 2v
=
1 + v 2 /c2
1 + v 2 /c2
lim v 2 = −c .
v2 =
(33.4.3)
(33.4.4)
v→c
Note that in the laboratory Reference Frame:
REL
v2 − v1 ≡ vLAB
= −2v .
(33.4.5)
Note that relativity does not forbid that v2 − v1 is larger than c as this is not, in relativity, the relative velocity of
particle 2 as seen by particle 1 (which is given by the above equation).
33.4.3
c
A light ray seen from another Reference Frame
Assume: v z = c. It follows that: vz = c.
This is clearly a consistency check as the starting postulates would predict that the speed of light is always c: v z =
=⇒
vz = c.
33.4.4
Laser cooling
...Argomento trattato a lezione...
33.4.5
Experimental test of special relativity via the stellar redshift
33.4.6
Experimental test of special relativity via the Ives-Stilwell experiment
In 1938 Ives and Stilwell observed the spectral lines of a high-velocity beam of hydrogen atoms. The spectral lines are
blue shifted when looking in front of the incoming beam while they are red-shifted when looking at the beam from behind.
Show that the average forward-backward shift is blue shifted by a factor γ.
33.4.7
Astronomical apparently super-luminal objects
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {Problem 12.6} —
Gli astronomi hanno scoperto oggetti che emettono getti di materia ad altissima velocità. Se la distanza dell’oggetto
dalla Terra è nota (sia d) la misura della velocità angolare alla quale si vede il getto muoversi sulla sfera celeste permette di
determinare la velocità con cui il getto di materia è emesso dal nucleo centrale. In alcuni casi sono stati osservati getti in
moto con velocità apparente superiore alla velocità della luce, fatto in apparente contraddizione con la teoria della relatività
speciale. Mostrare come sia possibile ottenere velocità apparenti superiori a quella della luce e come ciò non contrasti con
la relatività speciale considerando l’emissione del getto ad un angolo θ rispetto alla linea di vista dell’osservatore terrestre.
1. Determinare la relazione tra velocità apparente e quella reale del getto di materia.
2. Determinare la velocità apparente di un sistema in cui la direzione di emissione del getto forma un angolo θ = 9.2◦
con la direzione di vista supponendo la velocità di emissione pari a v = 0.987c (questa si può determinare dallo studio
dello spostamento Doppler delle linee spettrali).
3. Determinare il valore dell’angolo θ per cui la velocità apparente del getto sulla sfera celeste è massima e il valore di
tale velocità massima.
Worked Solution
Sia t0 l’istante di emissione del getto di materia in direzione θ rispetto alla linea di vista. L’emissione del getto viene
osservata da Terra al tempo t1 :
t1 = t0 + d/c
(33.4.6)
Il getto viene osservato da Terra, dopo che ha percorso la distanza v∆t, all’istante
t2 = t0 + ∆t +
d − v∆t cos θ
c
580
(33.4.7)
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33: Relativistic Kinematics
33.5: Exercises, problems and physical applications
L’intervallo di tempo ∆t tra le due osservazioni da Terra è
v cos θ
) < ∆t
c
t2 − t1 = ∆t(1 −
(se cos θ > 0)
(33.4.8)
L’astronomo conosce la distanza d dell’oggetto e misura uno spostamento angolare ∆θ sulla sfera celeste da cui deduce uno
spostamento apparente ∆x del getto. Lo spostamento apparente ∆x sulla sfera celeste vale
∆x ≡ ∆θd =
v∆t sin θ
d = v∆t sin θ
d
(33.4.9)
È da notare il fatto che la relazione precedente è esatta essendo d v∆t per cui non è necessario distinguere fra la distanza
dell’oggetto che emette il getto, d, e la distanza del getto al tempo t2 che vale ' d − v∆t cos θ. La velocità apparente
osservata risulta dunque:
∆x
d∆θ
v sin θ
vapp (θ) =
=
=
(33.4.10)
t2 − t1
t2 − t1
1 − (v/c) cos θ
che può essere maggiore di c.
Il massimo della velocità al variare dell’angolo di emissione θ si trova derivando rispetto a θ e ponendo la derivata uguale
a zero:
dvapp (θ)
v cos θ − v 2 /c
=0
(33.4.11)
=
dθ
(1 − v/c cos θ)2
da cui segue che l’angolo a cui si ottiene la velocità apparente massima e la velocità apparente massima valgono
vapp |max = p
cos θmax = v/c
Con i dati del problema si ottiene vapp = 6.14c.
v
(33.4.12)
1 − v 2 /c2
[] Answer:
1. vapp (θ) = v sin θ/ (1 − (v/c) cos θ);
2. vapp = 6.14c;
3. cos θmax = v/c.
33.4.8
GPS
...Argomento trattato a lezione...
33.5
Exercises, problems and physical applications
Problem - 33.1
Composition of two velocities close to the speed of light
Assume |u| ≤ c and u ' c as well as |v z | ≤ c and v z ' c. Show that vz ' c.
Problem - 33.2
Relation between the modules of the velocities
Show that the modules of the velocities are related by equation 33.1.12:
2
2
2
c −v =c
c2 − v 2
c2 − u2
2
(c2 − u ·v)
.
(33.5.1)
Note that the expression is symmetrical in u and v.
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33.5: Exercises, problems and physical applications
33: Relativistic Kinematics
Use this equation to show that the module of the velocity v 2 will always be smaller than c as long as both modules of
v and u are smaller than c.
Show that the above equation can be also written as:
2
v 2 = c2
c2 (v − u) − ( v ×u)
2
.
2
(c2 − u ·v)
(33.5.2)
Problem - 33.3
Relation between the modules of the velocities
Show that:
γv =
v ·u
1− 2
c
!
.
γv γu
(33.5.3)
Problem - 33.4
Relative velocity
See section 33.1.4.
If one observer measures two objects to be travelling at velocities v 1 and v 2 respectively what is the relative speed
between the two objects, that is what speed observer (1) would measure observer (2) to be travelling at, or vice versa.
In Galilean mechanics the relative velocity is just:
v 21 = v 2 − v 1
.
(33.5.4)
Show that in relativistic mechanics the relative velocity is:
v 21 ≡
q
2
2
(v 2 − v 1 ) − ( v 1 ×v 2 ) /c2
(1 − v 1 ·v 2 /c2 )
.
(33.5.5)
Problem - 33.5
Properties of the 4-acceleration
Show that:
2
A ·A = Aα Aα = γv4 a ·a + γv2 ( v ·a) /c2
.
(33.5.6)
Show that in the particle rest-frame the module of the 4-acceleration is the square of the proper acceleration, a0 .
Show that:
2
A ·A = Aα Aα = |a0 | = γv4 a2⊥ + γv2 a2k
.
(33.5.7)
Problem - 33.6
The relationship between the 3-velocity and the 4-velocity
For events along the world-line of a particle traveling with 3-velocity u we have the following result for the link between
the 3-velocity and the 4-velocity:
Vα ≡
dxα
dt dxα
dxα
=
= γv
≡ γv v α = {γv c, γv v}
dτ
dτ dt
dt
.
(33.5.8)
Show that:
V ·V = V α Vα = Vα V α = c2 .
582
(33.5.9)
December 23, 2011
33: Relativistic Kinematics
33.5: Exercises, problems and physical applications
Problem - 33.7
The relationship between the 3-acceleration and the 4-acceleration
The relationship between the 3-acceleration and the 4-acceleration is less straightforward the corresponding one among
velocities.
Introduce the 4-acceleration:
Aα ≡
d2 xα
dτ 2
=
dV α
dV α
d
d
= γv
= γv
(γv v α ) = γv
{γv c, γv v}
dτ
dt
dt
dt
Show that:
α
A = γv
dγv
dγv
c
, γv a + v
dt
dt
!
a≡
where
dv
dt
.
.
(33.5.10)
(33.5.11)
Show that in the Rest Frame of the particle one has:
V α = {c, 0}
Aα = {0, a} .
(33.5.12)
Moreover show that:
V ·A = 0 .
(33.5.13)
Problem - 33.8
Transformation of the acceleration from/to the Rest Frame
Show that the transformation of v̇ 2 from the Rest Frame to a general frame is:
2
u̇2 −→ γ 6 u̇2 − ( u̇ ×β)
583
.
(33.5.14)
December 23, 2011
33.5: Exercises, problems and physical applications
33: Relativistic Kinematics
584
December 23, 2011
34
Relativistic Dynamics
[] Reference: The Feynman’s lectures on physics (Addison-Wesley, 1964) : {} —
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} —
[] Reference: J. D. Jackson, Elettrodinamica Classica, seconda edizione, (1984, Zanichelli) : {} —
[] Reference: H. Goldstein, Meccanica Classica (19xx, Zanichelli) : {} —
[] Reference: W. K. H. Panoffsky - M. Phillips, Classical Electricity and Magnetism, (2◦ ed., 1962, Addison-Wesley)
: {} —
[] Reference: S. Weinberg, Gravitation and Cosmology, (1972, J. Wiley and Sons) : {} —
[] Reference: C. W. Misner - K. S. Thorne - J. A. Wheeler, Gravitation (W. H. Freeman and Co., San Francisco)
1973 : {} —
[] Reference: E. Massa: {Appunti di Fisica Matematica} —
34.1
Relativistic energy/momentum and power/force
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {12} —
Third Edition
The laws of classical mechanics are not compatible with the postulates of relativity, as a constant force can make the
speed of any object arbitarility large, if acting long enough. This is incompatible with the speed of light being a limiting
speed, see section 32.8.1.
Morevoer, for speeds approaching the speed of light, total energy and total momentum of an insulated system, as Newton
defined momentum and energy, are not conserved; this is an experiental fact.
34.1.1
Relativistic Energy and Momentum
... Alternative introduction to relativistic dynamics can be found in the standard textbooks. ...
Momentum and energy different concepts but in relativity they are combined and put on equal footing as for space and
time.
In classical mechanics momentum is a vector, defined as the product of mass times the velocity, p = mv. Therefore it
transforms as the velocity, not according to a Lorentz transformations: it is not, in fact, the space part of a 4-vector. The
transformation law of Newtonian momentum under Lorentz transformations is the same as the transformation of velocity.
Therefore when trying to extend the definition of mementum into relativistic physics one wonders wheter ordinary or
proper velocity needs to be used. Any extension of this concept must reduce to the Newtonian expression at low velocities.
Any extension of the concept of momentum must ensure momentum conservation in insulated systems, regardless of the
velocities of the particles relative to Reference Frame; this must be an invariant concept. Any extension of the concept of
momentum must be linked to the resultant of the external forces.
585
34.1: Relativistic energy/momentum and power/force
34: Relativistic Dynamics
If one attempts to extend the concept of momentum using the 4-velocity, instead of the ordinary velocity, one finds an
expression whose space part transforms as the space part of a 4-vector and reduces to the Newtonian expression of the
ordinary velocity in the low velocity limit. Moreover the time component of the 4-velocity is proportional to γ: it is not
clear, so far what it is.
One can then try the following definitions, waiting for confirmation that the resulting quantity is useful to describe the
dynamics, that is it is conserved for an insulated system and linked to the resultant of external forces for a non-insulated
system.
Let us define the tentative 4-momentum as the controvariant 4-vector:
pµ ≡ {ε/c, p} ≡ mV µ
whatever pµ is
p ·p = m2 c2
=⇒
(34.1.1)
p0 ≡ ε/c = mγc
(34.1.2)
p ≡ mV = mγv
γ ≡ γv ≡ 1 − v 2 /c2
(34.1.3)
.
(34.1.4)
All in all we have the special Lorentz transformations for the 4-momentum is:
ε = γu (ε − upz )
(34.1.5)
px = px
(34.1.6)
py = py
(34.1.7)
p z = γu
uε
pz − 2
c
!
.
(34.1.8)
And the inverse Lorentz transformations:
ε = γu (ε + upz )
(34.1.9)
px = px
(34.1.10)
py = py
(34.1.11)
p z = γu
uε
pz + 2
c
!
.
(34.1.12)
One needs to understand what is the meaningn of p0 , the time component of the 4-vector defined above. It is what
Einsten called relativistc mass but today it is usually called just the (relativistic) energy.
Squaring equation 34.1.2 and 34.1.3 and subtracting side by side one finds:
ε=
p
p2 c2 + m2 c4
.
(34.1.13)
Dividing side by side equation 34.1.2 and 34.1.3 one finds:
c2 p = εv
⇔
v = c2
p
ε
|v| ≤ c
.
(34.1.14)
When the object is at rest we obtain the rest energy, which is non-zero, for m > 0:
ε0 = mc2
⇔
p=0
⇔
v=0
.
(34.1.15)
The remainder of the total energy when the rest energy is subtracted is defined as the kinetic energy, as it reduces to
the kinetci energy of classical mechanics at low velocities:
k ≡ ε − mc2
k'
1
3mv 4
mv 2 +
+ ...
2
8c2
for low velocities |v| c
.
(34.1.16)
Note that in classical mechanics there is no such thing as a massless particle. In fact for any massless system classical
mechanics predicts that the resultant of the external forcse must be zero and therefore the force applied by the massless
system onto any other system is zero as well. Moreove it has zero momentum and zero kinetic energy. Therefore a massless
system in classical mechanics has no interactions with nothing else.
Conversely in relativistic mechanics one has a meaningful:
m=0
=⇒
{ε = pc
586
|v| = c}
.
(34.1.17)
December 23, 2011
34: Relativistic Dynamics
34.1: Relativistic energy/momentum and power/force
The question arises what distinguishes two photons with different energies. Relativity provides no answer. Quantum
physics shows that they differ in the frequency of the associated wave, which, in the case of photons, relates to the wavelength
of the EM radiation.
The non-relativistic and extreme relativistic limits are:

p2


p mc
=⇒
ε ' mc2 +
2m
(34.1.18)
m 2 c3 .


=⇒
ε ' pc +
p mc
2p
34.1.1.1
Controvariant and covariant form of 4-momentum
By definition the energy-momentum 4-vector of any system is defined in terms of the energy and momentum as:
pα ≡ {ε/c, +p}
⇔
pα ≡ {ε/c, −p} ,
(34.1.19)
with
2
p ·p = pα pα = pα pα = ε2 /c2 − p ·p = m2 c2 .
(34.1.20)
Note that the rest mass, ε0 = mc , is a Lorentz invariant but not conseervd.
34.1.2
Relativistic 4-momentum conservation for insulated systems
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {Problem 12.02} —
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {Problem 12.28} —
So far what has been described in section 34.1 may be just a useful notation, and nothing more with any physics content.
If all the velocities are so small that the non-relativistic approximation is valid, then the classical laws of conservation
of mass and momentum are equivalent to the conservation of the time and space parts of 4-momentum.
This suggests the use of 4-momentum as an extension of the classical momentum.
Physics lies in the experimental fact that 4-momentum is conserved for every insulated system and its changes are
related to the resultant external force.
34.1.2.1
Galileo transformations and classical momentum conservation
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {Prob. 12.2} —
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
Consider, for simplicity, a one-dimensional collision problem with two particles in the initial state and two particles in
the final state, a priori different particles from the ones in the initial state: A + B → C + D.
Show, using Galileo transformations, that if momentum is conserved in one inertial Reference Frame it is conserved in
another inertial Reference Frame if and only if the sum of the masses (the total mass) is conserved. Note that there is no
assumption that the number of incoming particles is equal to the number of outgoing particles, only that the total mass is
conserved.
Even if it is known that Galileo transformations are non correct in general this result shows that classical mechanics is
self-consistent when it is valid, that is for velocites small with respect to the speed of light.
Worked Solution
34.1.2.2
(pA + pB ) − (pC + pD ) = 0
(34.1.21)
(mA (v A + u) + mB (v B + u)) − (mC (v C + u) + mD (v D + u)) = 0
(34.1.23)
(mA vA + mB vB ) − (mC vC + mD vD ) = 0
(34.1.22)
((pA + pB ) − (pC + pD )) + u ((mA + mB ) − (mC + mD )) = 0 .
(34.1.24)
Lorentz transformations and classical momentum non-conservation
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {Prob. 12.28} —
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
It can be shown that the classical expression for momentum, p = mv is incompatible with momentum conservation and
Lorentz transformations.
As it is known that Lorentz transformations are the correct trasnformation rules among different Reference Frame, not
Galileo transformations, this implies that the notion of classical momentum must be abandoned.
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December 23, 2011
34.1: Relativistic energy/momentum and power/force
34.1.2.3
34: Relativistic Dynamics
Lorentz transformations and 4-momentum conservation
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {Prob. 12.28} —
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
Consider, for simplicity, a one-dimensional collision problem with two particles in the initial state and two particles in
the final state, a priori different particles from the ones in the initial state: A + B → C + D.
Show, using Lorentz transformations, that if momentum is conserved in one inertial Reference Frame it is conserved in
another inertial Reference Frame if and only if the sum of the energies is conserved.
As we know that Lorentz transformations are the correct transformations in general, this result shows that in order to
generalize the definition of momentum to something meaningful one needs to use the 4-momentum defined as the prduct
of mass times the proper velocity.
This is necessary in order to have an invariant conservation fo 4-momentum.
Worked Solution
Direct calculation via the velocity transformations
The direct calculation is long and tedious, because the transformation law for the velocity is complex. The following
identity can be used:
γv v = γu γv (u + v) .
(34.1.25)
Therefore:
vX + u
1 + uv X /c2
(pA + pB ) − (pC + pD ) = 0
(34.1.26)
vX =
m A γ vA
vA + u
1 + uv A /c2
!
(34.1.27)
(mA γvA vA + mB γvB vB ) − (mC γvC vC + mD γvD vD ) = 0
!!
!
vB + u
vC + u
+ mB γvB
− m C γ vC
+ mD γvD
1 + uv B /c2
1 + uv C /c2
vD + u
1 + uv D /c2
!!
(34.1.28)
=0
(34.1.29)
...
by developing the calculations one is led to use:
γv v = γu γv (u + v)
(mA γvA vA + mB γvB vB ) − (mC γvC vC + mD γvD vD ) = 0
(mA γu γvA (u + v A ) + mB γu γvB (u + v B )) − (mC γu γvC (u + v C ) + mD γu γvD (u + v D )) = 0
(34.1.30)
γu ((mA γvA v A + mB γvB v B ) − (mC γvC v C + mD γvD v D )) + uγu ((mA γvA + mB γvB ) − (mC γvC + mD γvD )) = 0 .
(34.1.31)
Using the 4-momentum transformation law
By using the transformation law for the 4-momentum the calculation is simpler:
p = γu p + uε/c2
The immediate result is:
.
(pA + pB ) − (pC + pD ) = 0
((pA + pB ) − (pC + pD )) +
34.1.2.4
u
((εA + εB ) − (εC + εD )) = 0 .
c2
(34.1.32)
(34.1.33)
(34.1.34)
Galileo transformations and 4-momentum non-conservation
One might also show, by direct calculation, that the conservation of 4-momentum is not compatible with Galileo
transformations.
34.1.2.5
Conservation of energy
[] Reference: TaylorWheeler: {} —
All this talk of reconciliation at low speeds obscures an immensely powerful feature of the relativistic expression for
The total energy of an insulated system is conserved. Total energy is conserved in all interactions: elastic and inelastic
collisions as well as creations, transformations, decays and annihilations of particles. In contrast the total kinetic energy
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December 23, 2011
34: Relativistic Dynamics
34.1: Relativistic energy/momentum and power/force
of a system calculated using Newtonian mechanics for low-speed is conserved only for elastic collisions, which are in fact
defined as collisions in which kinetic energy is conserved. In inelastic collisions kinetic energy transforms into other forms
of energy, such as internal energy, chemical energy, potential energy or other forms of energy. In Newtonian mechanics each
of these forms of energy must be treated separately and conservation of energy must be invoked as a separate principle, as
something beyond Newtonian analysis of mechanical energy.
34.1.3
Relativistic power and force for non insulated systems
Newton’s law are still valid provided the relativistic expressions of momentum and energy are used.
π ≡ v ·f =
f=
34.1.3.1
dp
dt
dε
dt
(34.1.35)
.
(34.1.36)
Relation between force and acceleration
dp
dv
dγ
f=
=m γ
+v
dt
dt
dt
!
dv
γ 3 v dv
=m γ
+v 2
dt
c dt
!
γ 3 v dv
= m γa + v 2
c dt
!
.
(34.1.37)
The component perpendicular to the velocity, which has contributions from the first term only in 34.1.37, is:
!
dv
f ⊥ = mγ
= mγa⊥ .
(34.1.38)
dt
⊥
The component parallel to the velocity, which has contributions from both terms in 34.1.37, is (see section 34.2.9):
!
dv
3
= mγ 3 ak .
(34.1.39)
f k = mγ
dt
k
Therefore:
fk =
f⊥ =
dp
dt
dp
dt
!
!
= mγ 3 ak
k
⊥
= mγa⊥
(34.1.40)
.
Note that the inertia to forces parallel and perpendicular to the velocity are different.
The relation can also be written as:
!
2 v ( v ·a)
f = mγ a + γ
.
c2
(34.1.41)
(34.1.42)
The relation between force and acceleration is not that simple as it is in non-relativistic mechanics. Note that cceleration,
even though it is caused by a force, does not generally point in the direction of the force: at high speeds the force is not,
in general, parallel to the acceleration.
For slow motions the second term in the parentetsis is negligible with respect to the first one and the expression reduices
to the non-relativistic one.
34.1.3.2
Transformation laws for force and power
Let v be the velocity of the particle with respect to the Reference Frame I and v the velocity of the particle with
respect to the Reference Frame I. Let and u the velocity of the Lorentz transformation from the Reference Frame I to the
Reference Frame I.
Let:
u = cβ
γ ≡ γu .
(34.1.43)
The transformation law for force and power can be derived in the same way as the transformation law for the velocities.
In case β k e3 one finds:
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December 23, 2011
34.1: Relativistic energy/momentum and power/force
34: Relativistic Dynamics
π − uz fz
1 − uz vz /c2
fz − βπ/c
fz =
1 − βvz /c
fx|y
f x|y =
.
γ (1 − βvz /c)
π=
(34.1.44)
(34.1.45)
(34.1.46)
In the general case:
π=
π − c β ·f
1 − β ·v/c
f +β
f=
34.1.3.3
(34.1.47)
γ2
π
( β ·f ) − γ
γ+1
c
!
.
γ (1 − β ·v/c)
(34.1.48)
The Minkowsky 4-force
Similarly to what is done when passing from ordinary velocity to 4-velocity one can formalu introduce the quantities:
f M ≡ γv f
πM ≡ γv π = v ·f M
.
(34.1.49)
The so-called Minkowsky power and force, πM and f M ,
fMµ ≡ {πM /c, f M } ,
(34.1.50)
transforms then like a 4-vector.
34.1.3.4
On the practical use
The Minkowsky 4-force is easier to handle than ordinary power-force. This hybrid quantity relating 4-momentum,
measured in the Reference Frame of the observer, to the proper time, measured in the Rest Frame of the system is often
less useful than ordinary power-force, in practice, because the latter relates quantities measured in the Reference Frame of
the observer.
34.1.3.5
Dynamical invariants
Relativistic invariants often lead to surprisingly simple calculations, avoiding long, tedious and error prone algebra.
If a question is of such a nature that its answer will be the same in all Lorentz systems it must be possible to formulate
its answer entirely using the invariants which can be formed with the available Lorentz tensors. One can find the answer in
a particular Lorentz system, freely chosen, in such a way that the answer is simple. One then looks at how the invariants
appear in that system and expresses the answer in term of the invariants. This gives the unique general answer. In fact if
a Lorentz Invariant expression assumes the correct value in a specific frame it assumes that same value in any other frame,
because it is invariant.
Refer to a single particle with 4-momentum p = (ε, p). The following are invariants quantities.
p2 = ε2 − p ·p = m2
dp /ε
the mass of a particle
the phase-space factor
.
(34.1.51)
(34.1.52)
For any given system of particles, k = 1, 2, . . . n, having 4-momenta pk , the invariant mass of the system is defined as:
?? M0 ≡
X
k
pk
!2
.
(34.1.53)
The invariant mass of a system of particles corresponds to the total energy in the CM-frame. It is also the total energy
available to produce new particles.
34.1.4
Energy and Momentum and relativistic mometum-energy 4-momentum
In classical mechanics momentum is defined as the product of mass times ordinary velocity. When trying to extend
momentum in relativistic physics should one use ordinary velocity or proper velocity? In classical physics the two quantities
are identical, so that there is no reason a priori to favor one over the other. In relativistic physics one is led to conjecture
to use proper momentum in oder to deal with a 4-vector. However this might just be a conjecture, not physics. But the
590
December 23, 2011
34: Relativistic Dynamics
34.1: Relativistic energy/momentum and power/force
previous discussion has confirmed that, in the context of relativity, it is essential that one uses the proper velocity to define
momentum or otherwise the law of conservation of momentum would be inconsistent with the principle of relativity.
In classical mechanics energy is a scalar (invariant under changs of the Coordinate System) and not invariant under
changes of Reference Frame; momentum is a vector. They are entirely different quantities. Energy remains energy and
momentum remains momentum under Galileo transformations.
In relativistic physics energy and momentum are mixed by Lorentz transfomations, the same way as time and space are
mixed.
In classical mechanics the total mass is always conserved and therefore there is no such thing as energy conversion
into/from rest mass. However in relativistic mechanics there is no such thing as rest mass conservation, but only energy
conservation. This opens the way to enery conversion into/from rest mass.
34.1.4.1
Energy versus relativistic mass
In the old times, in an attempt to keep the classical defintion of mometum as the product of mass times velocity, it was
said that mass changes with velocity (the relativistic mass).
Nowadays it is preferred to consider mass as a (relativistic invariant) intrinsic property of any specific system and assume
that the expression of momentum as a function of velocity is different from the classical one. This approach perfeclty fits
the concept of momentum as a four-vector.
34.1.5
The Center-Of-Momentum (or Zero-Momentum) frame
L’estensione del concetto classico di sistema di riferimento del centro di massa è il concetto di CM-frame: Center of
Momentum frame or Zero-Momentum frame.
34.1.5.1
The CM-frame: the general case
Let’s consider a pair of particles having, in a certain frame of reference, the 4-momenta (ε1 , p1 ) and (ε2 , p2 ). The
derivation will show that the extension to the general case of an arbitrary number of particles is straightforward.
The CM-frame is defined as that frame where the total 3-momentum of the system of particles is zero. Let’s call u
the velocity of CM-frame with respect to the current frame and let’s denote with ? the quantities in the CM-frame. Let:
E ≡ ε1 + ε2 and P ≡ p1 + p2 . The equation to solve in order to find u is thus:
p?1 + p?2 = 0 .
(34.1.54)
From equation ?? one has:
!
γ
( u ·(p1 + p2 )) − (ε1 + ε2 ) = 0
(p1 + p2 ) + γu
γ+1
!
γ
P + γu
( u ·P − E) = 0
γ+1
(34.1.55)
(34.1.56)
The equation can be solved for u in the following way. First take the scalar product of the equation with u and use the
identity
u2 γu2
γu = 1 +
(34.1.57)
γu + 1
to obtain
u ·P = u2 E .
(34.1.58)
Secondly use the latter equation in 34.1.55 to obtain the final result:
u=
P
E
.
(34.1.59)
One then immediately finds that:
2 −1/2
γu ≡ 1 − u

= 1 −
P
E
!2 −1/2

=
E
E2 − P 2
E
= √
s
.
Note that equations 34.1.59 and 34.1.60 are consistent with the single particle case because
mass of the particle system.
591
(34.1.60)
√
s is actually the invariant
December 23, 2011
34.2: Some Examples and Physical Applications
34.2
34.2.1
34: Relativistic Dynamics
Some Examples and Physical Applications
Kinematics of ultra-relativistic particles
The Large Electron-Positron (LEP) Collider operated at CERN from year 1989 to 2000. In the first phase of operation
of the Collider electrons and positrons were colliding head-on, with opposite momenta, at a CM-energy of ECM ' 92 GeV.
Determine the velocity of the electron/positron. If the energy is known with an uncertainty of ∆ε determine the uncertainty
in the velocity.
34.2.2
LHC parameters
The Large Hadron Collider (LHC) at CERN will collide head-on two proton beams at ε = 7 TeV ⊕ 7 TeV. Every beam
is made of nb = 2808 bunches of protons (m = 938 GeV/c2 ) and every bunch has np = 1.15·1011 protons. The total length
of the collider, which is approximately a circular one, is L = 26.659 km.
1. Calculate the gamma factor and velocity of the protons.
2. Determine the current.
3. Determine the average magnetic field required to keep the protons in the orbit.
4. Determine the total energy stored in the proton beams.
√
5. Compare the available energy in the CM ( s = 2ε = 14 TeV) with the one that would be available by sending a
proton beam at ε0 = 14 TeV against a stationary target.
34.2.3
The hydrogen atom
34.2.4
A relativistic totally anaelastic collision
Two objects, each one with mass m, collide head-on with equal and opposite speeds and stick together.
1. If the speed is a typical speed of a gun projectile, v = 300 m/s = 1·10−6 c, what is the relative increase of mass of the
composite lump with respect to the sum of the masses of the two projectiles?
Use:
1
γ−1'
2
v
c
!2
for v c .
(34.2.1)
2. If the speed is v = 0.6c, what is the mass of the composite lump?
Worked Solution
Conservation of momentum is trivial.
Conservation of energy gives:
2mγc2 = M c2 .
(34.2.2)
The final mass is larger than the sum of the initial masses: mass is not conserved; energy is conserved. Kinetic energy
was converted into rest energy, so mass increased.
In a classical analysis one says that kinetic energy has been converted into internal energy: the composite object is
hotter than the two original colliding objects. This is true also in the relativistic analyses. Internal energy is the sum of
the random kinetic and potential energies as well as rest masses of all the particles, atoms and molecules of the objects.
Relativity tells us that these microscopic energies are represented in the mass of the object: a hot object is heavier than
a cold object as well as a compressed spring is heavier than a relaxed spring. This is normally not a large macroscopic
quantity as internal energy, U, contributes an amount U/c2 to the mass, and c2 is a very large number when compared
to our typical units of measure. Therefore such kind of relativistic effects are hardly observable in the macroscopic world.
However in the realm of elementary particles the effects can be very striking. Nuclear energy production, either by fission
or fusion, rely on these kind of phenomena.
1.
M − 2m
= 0.5·10−12 .
2m
(34.2.3)
2.
2mγc2 = M c2
=⇒
592
M = 2.5m > (m + m) .
(34.2.4)
December 23, 2011
34: Relativistic Dynamics
34.2.5
34.2: Some Examples and Physical Applications
Stato composto
It is an example of conversion of energy into mass.
Si abbia una particella di massa m0 = 1 GeV/c2 ed energia cinetica k = 2m0 c2 . La particella interagisce con una
particella fissa nel Reference Frame del laboratorio di massa 2m0 .
Le due particelle formano una stato legato: derminarne la massa e l’impulso.
34.2.6
Neutral pion at rest decaying into two photons
It is an example of conversion of mass into energy.
...Argomento trattato a lezione...
A neutral pion at rest (mass Mπ0 = 135 GeV/c2 ) decays into two massless photons. Determine the energy and momentum of the outgoing photons.
In this process pure mass is transformed into pure energy.
34.2.7
Einstein box and the inertia of energy
...Argomento trattato a lezione...
The equivalence of energy and mass is such an important and striking result of Relativity that Einstein himself, after
his derivation of the result, sought and found an alternative elementary physical line of reasoning that leads to the same
conclusion. It is called Einstein box. The purpose of this thought experiment is to suggest that energy must have associated
with it a certain inertial mass equivalent.
Consider the following thought experiment. One photon leaves one side of a freight-car, which is free to move on a
rail, directed towards the other side. Due to momentum conservation, after the photon has started the train recoils. The
train stops again when the photon is absorbed at the other side. However, the Center of Mass seems to have moved for a
massless photon. Assume that the train moves at low velocity and therefore its dynamics can be described by the newtonian
dynamics.
A simplified solution
Assume the veocity of the freight-car is |V | c and the photon mass-equivalent, if any, is m M , much less that the
freight-car mass M .
Assume the photon has zero mass-equivalent. It follows:
ε
= −M V = M |V | > 0
c
VL
pγ L
∆Z = V ∆t =
=−
<0 .
c
Mc
pγ =
(34.2.5)
(34.2.6)
The CM has moved.
Now imagine that the photon carries some kind of mass-equivalent, m. It follows that in order to keep the CM fixed
one needs to have:
M ∆Z + mL = 0
m=−
M ∆Z
pγ
ε
=
= 2
L
c
c
The mass-equivalent, m, associated to the photon is therefore,
593
(34.2.7)
.
(34.2.8)
ε
.
c2
December 23, 2011
34.2: Some Examples and Physical Applications
34: Relativistic Dynamics
A more precise solution
Assume the photon has zero mass-equivalent. It follows:
ε
= −M V = M |V | > 0
c
L
zCM =
2
L
L
∆t =
=
c−V
c + |V |
L
∆Z = V
<0
c + |V |
L
zCM 0 =
+ ∆Z .
2
pγ =
(34.2.9)
(34.2.10)
(34.2.11)
(34.2.12)
(34.2.13)
Therefore the Center of Mass seems to have moved.
Now imagine that the photon carries some kind of mass-equivalent, m. It follows:
ε
= −M V = M |V | > 0
c
M L/2
zCM =
m+M
L
L
=
∆t =
c−V
c + |V |
L
∆Z = V
<0
c + |V |
!
pγ =
M
0
zCM =
L
+ ∆Z
2
+ m (L + ∆Z)
= zCM + ∆Z + L
m+M
zCM 0 = zCM
=⇒
=⇒
∆Z = −L
ε = mc2
(34.2.14)
(34.2.15)
(34.2.16)
(34.2.17)
m
m+M
(34.2.18)
m
<0
m+M
.
(34.2.19)
(34.2.20)
Therefore consistency is ensured by assuming the the photon carries a mass equivalent to its energy (divided by c2 ). In
other words: even if the rest mass of the photon is zero its energy has the role of a mass for the calculation of the Center
of Mass position.
A simplified derivation assuming m M and |V | c gives the same result.
The transport of mass by the photon can be understood only as a new feature of the raaiation itself.
34.2.8
Charged pion at rest decaying into a muon and a neutrino
...Argomento trattato a lezione...
A charged pion at rest (mass M ) decays into a muon (mass m) and a (massless) neutrino. Determine the energy of the
outgoing muon in terms of the two masses.
Find the velocity of the muon.
Worked Solution
ε = c2
M 2 + m2
2M
594
.
(34.2.21)
December 23, 2011
34: Relativistic Dynamics
34.2.9
34.2: Some Examples and Physical Applications
Component parallel to the velocity of the force
vak ≡ v ·a = v ·
ak =
dv
dt
dv
1 d
1 dv 2
dv
=
( v ·v) =
=v
dt
2 dt
2 dt
dt
(34.2.22)
as it is well-known from the representation in intrinsic coordinates of the acceleration
!
!
!
v
v2 2
γ 3 v dv
v2 γ 3
· = m γak + 2 ak = mγak 1 + 2 γ
= mγ 3 ak
f k ≡ m γa + v 2
c dt
v
c
c
!
dv
mγ 3
= mγ 3 ak .
dt
(34.2.23)
(34.2.24)
(34.2.25)
k
34.2.10
Relativistic motion from rest subject to a constant force
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {12.10} —
An object of mass m is subject to a constant force F = F e3 , starting from rest at time t = t0 . Determine the law of
motion.
Worked Solution
It can be solved by using the relations in section 34.1.3.1.
Use:
w
γ 3 dv = γv .
(34.2.26)
The law of motion is:
mc
F
α≡
p
z[t] = c t2 + α2 − αc
=⇒
vz [t] = √
ct
t2
+ α2
≤c .
(34.2.27)
While in classical mechanics a constant force gives a parabolic dependence of the position on time, in relativistic
mechanics the position as a function of time is hyperbolic.
The asymptote of the hyperbola z[t] ≈ c (t − α) gives the limiting speed:
z[t] ≈ ct
for very large times t α ≡
mc
.
F
(34.2.28)
The short-time behaviour reproduces the classical result, as it is a low-speed result:
z[t] ≈
1 c 2
1 F 2
t =
t
2 α
2 m
for very short times t α ≡
mc
.
F
(34.2.29)
Note that both energy and momentum increse indefinitely, but the speed never reaches the speed of light. In fact the
momentum increase at a constant rate, as from the equation of motion, while the energy is always greater or equal to the
momentum times the speed of light.
34.2.11
Motion subject to a constant force perpendicular to the initial velocity
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
34.2.12
Motion subject to a force always perpendicular to the velocity
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
See section Problem - 34.8.
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December 23, 2011
34.3: Exercises, problems and physical applications
34.2.13
34: Relativistic Dynamics
The Head-Light effect (relativistic beaming)
Consider a small particle moving at constant velocity V with respect to the Laboratory Frame (Reference Frame I).
Assume that the particle is emitting photons with an angular distribution isotropic in its rest frame, (Reference Frame I 0 ).
Determine the angular distribution of the photons as it is seen by the Laboratory Frame.
Neglect the Doppler shift of the emitted radiation on the basis of the fact that the velocity of the grain of dust is assumed
to be non relativistic.
Worked Solution
Consider the inverse special Lorentz transformation from the Reference Frame I 0 to the Reference Frame I, choosing
the axes z = z 0 parallel to V . Use the notation: β ≡ βV and γ ≡ γV . Assume c = 1.
ε = γ (ε0 + βp0z )
(34.2.30)
pz = γ (p0z + βε0 )
(34.2.31)
px =
py =
p0x
p0y
(34.2.32)
.
(34.2.33)
Let θ be the angle between the 3-vector p and the z axis (an I Reference Frame quantity). Let p⊥ and p0⊥ be the
projections of p on the plane perpendicular to V . For the special Lorentz transformation considered:
p⊥ = p0⊥ .
(34.2.34)
Consider any photon emitted with p0z = 0. These photons divide the set of all emitted photons into two hemispheres.
Let’s consider how these photons appear in the I frame. We have:
p0z = 0
(34.2.35)
ε = γε0
(34.2.36)
pz = γβε0 = βε 6= 0
0
ε =
One then has:
tan θ ≡
It follows:
sin2 [θ] =
|p0⊥ |
(34.2.37)
= |p⊥ | .
(34.2.38)
|p0⊥ |
1
|p⊥ |
=
=
0
pz
γβε
γβ
tan2 θ
1 + tan2 θ
=⇒
.
sin [θ] =
(34.2.39)
1
γ
.
(34.2.40)
Therefore the photons with p0z = 0 appear, to the I Reference Frame observer, to move in the forward direction, making
an angle sin [θ] = 1/γ with the velocity vector V . On average, therefore, the I Reference Frame observer, will see the
photons moving in the forward direction with a average angle sin [θ] = 1/γ with the velocity vector V . Note that the
average angle, or average momentum seen by the I Reference Frame observer, are just estimated by considering the fate of
a photon with p0z = 0. The exact calculation would require to average the full velocity distribution seen by the I Reference
Frame observer.
34.3
Exercises, problems and physical applications
Problem - 34.1
Kinetic energy and Galileo transformation in classical mechanics
Show that in classical mechanics the transformation law for the kinetic energy of a system of point particles, assuming
that the mass of a point particle is Galileo invariant, is:
K=K+
1
M u2 + u ·P
2
.
(34.3.1)
Show, using the above relation, that a collision is elastic if and only if the total mass and total momentum are conserved.
596
December 23, 2011
34: Relativistic Dynamics
34.3: Exercises, problems and physical applications
Problem - 34.2
Non-relativistic and Ultra-relativistic approximation
What is the maximum value of v and γ if the non-relativistic approximation for energy, ε ≈ mc2 + p2 / (2m), is to be
used with an error less than α, that is with an absolute error on the energy equal to αp2 / (2m). Hence show that for an
error of α = 0.01 the momentum satisfies p ≈ 0.2mc.
What is the minimum value of v and γ if the ultra-relativistic approximation for energy, ε ≈ pc, is to be used with an
error less than α, that is with an absolute error on the energy equal to αpc. Hence show that for an error of α = 0.01 the
momentum satisfies p ≈ 7mc.
Problem - 34.3
Relation between kinetic energy and momentum
Determine the relativistic relation between kinetic energy and momentum.
L’approssimazione non relativistica vale per k mc2 , equivalente a p mc:
k = ε − mc2 = mc2 (γ − 1) =
p
p2 c2 + m2 c4 − mc2 '
p2
2m
(34.3.2)
L’approssimazione non relativistica corrisponde a v c, o equivalentemente, ε mc2 .
Problem - 34.4
♠
The Head-Light effect; the exact velocity distribution
Determine the exact velocity distribution of the photons as it is seen by the I Reference Frame observer in problem 34.2.13.
Problem - 34.5
Mass of a insulated system
Does conservation of the 4-momentum of an isolated system implies that collisions and interactions within an isolated
system cannot change the systern mass?
Does conservation of the 4-momentum of an isolated system implies that the constituents that enter a collision are
necessarily the same in individual mass and in number as the constituents that leave that collision?
Is the mass of an isolated system composed of freely-moving objects obtained by simply adding the masses of the
individual objects?
Problem - 34.6
Relation between force and power
Show that equations 34.1.35 and 34.1.36 imply:
π = c2
d
(mγ) .
dt
(34.3.3)
Show that equations 34.1.35 and 34.1.36 imply:
mγ
dv
π
=f − 2v .
dt
c
(34.3.4)
The latter equation shows that in the relativistic case, in general, force and acceleration are not parallel, at variance
with respect to the classical case. Force and acceleration are parallel if and only if either f k v or f ⊥ v.
Problem - 34.7
The Minkowsky force
Show that the Minkowsky force is a 4-vector.
597
December 23, 2011
34.3: Exercises, problems and physical applications
34: Relativistic Dynamics
Problem - 34.8
A charged particle in a uniform and constant magnetic field
A charged particle is moving in a constant and uniform magnetic field. At a certain instant its velocity is perpendicular
to the magnetic field.
The Lorentz force does no work and therefore the energy, velocity as well as the gamma factor, are constant:
ak =
dv
=0 .
dt
(34.3.5)
The motion lies in plane as at the initial instant its velocity is perpendicular to the magnetic field.
The Lorentz force is perpendicular to the velocity, therefore equation 34.1.41 can be used at once as
f⊥ =
dp
dt
!
⊥
= mγa⊥ = mγ
v2
eN = q v ×B
ρ
=⇒
ρ=
mγv
p
=
= constant
qB
qB
.
(34.3.6)
Note that, for a uniform circular motion:
dp
dθ
v
=p
eθ = p eθ .
dt
dt
R
(34.3.7)
Problem - 34.9
Dynamic properties of a particle as seen from the rest system of another particle
Two particles have 4-momenta p1 and p2 . Determine the energy, module of momentum and velocity of particle 2 as
seen from the rest system of particle 1: E21 , |p21 | and |v 21 |.
Problem - 34.10
Invariant mass of a pair of particles
Show that for any pair of particles having 4-momenta p1 and p2 and masses m1 and m2 , one has the identity:
p1 ·p2 =
M02 − m21 + m22
2
in terms of the invariant mass of the pair, M0 =
√
!
=
s − m21 + m22
2
!
,
(34.3.8)
s.
Problem - 34.11
Outrun a light ray
Suppose you are subject to a constant force and start a race with a light ray.
How much head start do you need in order that the light ray will never reach you?
Problem - 34.12
Electron-positron annihilation
The positron is the anti-particle of the electron, in all identical to the electron except for having its charges of the
opposite sign.
A positron having kinetic energy equal to its mass strikes an electron at rest. They annihilate, creating two high-energy
photons. One photon enters a detector placed at right angle with respet to the direction of the initial positron with respect
to the interaction point.
Determine the energies of both photons and the direction of motion of the second photon.
598
December 23, 2011
34: Relativistic Dynamics
34.3: Exercises, problems and physical applications
Problem - 34.13
Energy production in the sun
Energy from the Sun reaches the outer atmosphere of the Earth at a rate of about 1.4 kW/m2 (this is called the solar
constant).
1. How much of the mass of the Sun is converted into energy every second to supply the energy that reaches the Earth.
2. What total mass is converted to energy every second in Sun?
3. Most of the energy comes from burning hydrogen nuclei (mostly protons) into helium nuclei (mostly a two-protontwo-neutron combination). How much hydrogen must be converted into helium per second?
4. Estimate how long Sun will continue to warm Earth, neglecting all other processes in Sun and emissions from Sun.
Problem - 34.14
Isolated photon
Demonstrate that an isolated photon cannot split into two photons going into two different directions other than going
both into the original direction.
Problem - 34.15
Pair production
A high-energy photon may carry an energy larger than the rest energy of an electron-positron pair. Nevertheless it
cannot give rise to an electron-positron pair in the absence of other particles
1. Prove that this process is incompatible with the laws of conservation of momentum and energy as employed in the
laboratory frame of reference.
2. Repeat the demonstration in the CM frame of the electron-positron pair.
Problem - 34.16
Muon in the atmosphere
Refer to section 32.8.2.1.
Assume, as a gross approximation, that the Earth magnetic field is 0.5 Gauss radially directed.
Determine how much the horizontal component of the velocity of the muon changes in direction if its energy is α = 1/100
more than the one determined in section 32.8.2.1, just enough to reach the Earth surface and the αε additional energy is
in carried by the horizontal component of the velocity.
599
December 23, 2011
34.3: Exercises, problems and physical applications
34: Relativistic Dynamics
600
December 23, 2011
35
ElectroMagnetism and relativity
[] Reference: The Feynman’s lectures on physics (Addison-Wesley, 1964) : {} —
[] Reference: Berkeley Physics Course, Vol. 1, Mechanics, (1965, McGraw-Hill) : {} —
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} —
[] Reference: J. D. Jackson, Elettrodinamica Classica, seconda edizione, (1984, Zanichelli) : {} —
[] Reference: H. Goldstein, Meccanica Classica (19xx, Zanichelli) : {} —
[] Reference: W. K. H. Panoffsky - M. Phillips, Classical Electricity and Magnetism, (2◦ ed., 1962, Addison-Wesley)
: {} —
[] Reference: S. Weinberg, Gravitation and Cosmology, (1972, J. Wiley and Sons) : {} —
[] Reference: C. W. Misner - K. S. Thorne - J. A. Wheeler, Gravitation (W. H. Freeman and Co., San Francisco)
1973 : {} —
[] Reference: E. Massa: {Appunti di Fisica Matematica} —
[] Reference: http://www.batmath.it/eng/a_relativity/campo_em.htm: {} —
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} —
35.1
Introduction
See section 32 for introductory remarks.
Note that c = 1 will be often used except for some of the basic formulas.
35.1.1
EM and relativity
ElectroMagnetism is consistent with the pricimple of relativity from the beginning: all inertial observers describe physical
phenomena, including ElectroMagnetic ones, in the same way.
However the interpretations may be different and what appears to be an electric/magnetic phenomenon to one observer
may appear as a magnetic/electric phenomenon to another observer.
Relativistic invariance of Electromagnetism is not demonstrated but, on the other hand as for any physical law, it is an
assumption arrived at via a process of induction from the experiments.
35.1.2
Some comments on the relativistic formulation of EM
It is easy to be carried away by the mathematical elegance of the relativistic formulation of ElectroMagnetism. However
one should not loose the physical meaning of the equations and the complex physical relations underlying the equations
and their relativistic invariance.
The importance of the ordinary three-dimensional representation of physical quantities must not be neglected even if
complex and compact relativistic invariant relations might embrace different physical concepts. For instance the Maxwell
equations without charges/currents will be compacted into a single 4-dimensional equation, but this single equation embraces, at the same time, the lack of magnetic Monopoles and the Faraday-Neumann-Lenz law of EM induction.
601
35.2: Charge and Current: the current 4-vector
35.2
35.2.1
35: ElectroMagnetism and relativity
Charge and Current: the current 4-vector
Invariance and conservation of the electric charge
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} —
The electric charge, in addition to be a Lorentz invariant and a conserved quantity, comes in multiples of the elementary
charge of the electron and the proton. The absolute values of the electric charge of proton and electron are the same as far
as we know today.
The electric charge is an intrinsic property of elementary particles, together with the mass, the spin and any other
charges (such as baryon charge and lepton charge) which one discovers in fundamental physics.
35.2.1.1
Some experimental evidence
The equality of the absolute values of the electric charge of the electron and the proton is an experimental fact. These
experimental facts 1 can be also used to support the fact that the electrical charge is a Lorentz invariant quantity, or, in
other words, the invariance of the charge of the particle from its velocity. In fact neutral atoms at rest with different atomic
number have electrons moving with different velocities, depending on the energy levels, which in turn strongly depend on
the atomic number. The fact the atoms are neutral provide support for the independence of the electric charge of the
electrons from their velocity.
Invariance of the electric charge means that every observer will measure the same value for the flux of the electric field:
{
{
[t] E ·dS .
(35.2.1)
[t] E ·dS =
Σ
Σ
One must be very careful that the electric fields, surfaces and times must be taken for the two different obsrvers. In
particular the distinction between t and t must not be overlooked because one knows that events that are simultaneous in
I need not be simultaneous in I. Each of the surface integrals in must be evaluated at one precise instant in its Reference
Frame. If charges lie on the boundary one has to be rather careful about ascertaining that the charges within I at t are the
same as those within I at t. If the charges are well away from the boundary there is no problem in this respect.
35.2.2
The 4-current
35.2.2.1
Transformation law of the charge density
As the electric charge is a relativistic invariant the transformation of charge volume density is dictated by the law of
transformation of volumes. In particular the relation between the charge volume density in the proper Rest Frame, ρ0 , and
in any other Reference Frame, ρ, is given by:
dV =
35.2.2.2
dV 0
γ
=⇒
dq
≡ ρ = γρ0
dV
.
(35.2.2)
Transformation law of the current density
As the electric charge is a relativistic invariant the transformation of current volume density is dictated by the law of
transformation of volumes. In particular the relation between the current volume density in the proper Rest Frame, ρ0 ,
and in any other Reference Frame, ρ, is given by:
dV =
35.2.2.3
dV 0
γ
=⇒
j ≡ ρv =
dq dx
dq dτ dx
dq 1 dx
dq dx
=
=
=
= ρ0 V
dV dt
dV dt dτ
dV γ dτ
dV0 dτ
.
(35.2.3)
The current 4-vector
The two relations 35.2.2 and 35.2.3 allows one to define the current 4-vector.
Consider, to start with, a set of all identical particles moving with the same velocity.
The current is a 4-vector defined as
(
α
α
α
α
dx
dt
dx
dx
dx
j 0 ≡ ρc = ρ0 γc
j α ≡ ρ0
= ρ0
= ρ0 γ
=ρ
=
dτ
dτ dt
dt
dt
j ≡ ρv = ρ0 γv = ρ0 V
,
(35.2.4)
showing explicitly the 4-vector character of the current, in terms of the Lorentz-invariant volume charge density in the Rest
Frame of the charge, ρ0 , and the Lorentz-invariant proper time, τ .
1 J.
D. Jackson, Elettrodinamica Classica, seconda edizione, (1984, Zanichelli)
602
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35.3: Elementary considerations on the transformation of EM fields
Note that:
dxα
dτ
α
dx
jα ≡ ρ
dt
j α ≡ ρ0
expresses j α in terms of quantities in the proper Rest Frame only
(35.2.5)
expresses j α in terms of quantities in the Laboratory only .
(35.2.6)
In the proper Rest Frame the definition reduces to:
j α = {cρ0 , 0}
in the proper Rest Frame .
(35.2.7)
The current 4-vector so defined is time-like:
j µ jµ = c2 ρ20 = invariant ≥ 0 .
(35.2.8)
In the general case where the current is due to either the motion of different species of charges and/or a different sets
of charges with different kinematics the 4-current must be defined as the algebraic sum of the different components, all of
them being convective currents:
X
X
dxα [k]
jα ≡
,
(35.2.9)
j α [k] ≡
ρ[k]
dt
k
k
possibly generating non convective currents, that is conduction currents.
35.2.2.4
Conduction and convective currents
The motion of charges produces a current.
• The convection current is the one due to the motion of the Center of Mass of the element of matter. In the simplest
case that is a number of all identical particles moving with the same velocity.
• The conduction current is the one due to the motion relative to the Center of Mass. As an example one can quote
a plane circular loop of current whose Center of Mass is at rest (no convection current) while it has a conduction
current due to the motion of the free charges with respect to the Center of Mass.
35.2.2.5
Some comments on the current 4-vector
Note that the definition of j µ is totally analogous to the definition of the 4-momentum, pµ , of a particle, with the volume
charge density, ρ, replacing the energy and the volume charge density in the Rest Frame of the particle, ρ0 , replacing the
mass of the particle.
One can also write explicitly the definition of current as:
j µ ≡ {cρ, j} ≡ {cρ, ρv}
with
ρ = γu ρ0
,
(35.2.10)
where ρ0 is the volume charge density in the rest frame of the particle.
A stationary charge in the I Reference Frame is described by a time-like 4-vector. As it is expected when the charge is
seen from the I 0 Reference Frame, moving with velocity V with respect to I, the 4-vector has both a time component and
a space component. This is not a surprise as we know that a moving charge gives both a charge and a current. What is less
obvious is the behaviour of a space-like current 4-vector and the fact that a globally neutral system of currents will show a
total net charge when it is seen from a Reference Frame moving with respect to the other system. In general one can say
that, as volume charge density and current density are components of a 4-vector, they can mix in a Lorentz transformation.
One consequence, as discussed in section 35.4, is that a moving magnetic dipole appears as having an electric dipole and
vice-versa.
35.2.2.6
Charge conservation and the current 4-vector
The charge continuity equation itself becomes, in terms of the 4-current:
∂µ j µ = 0
35.3
.
(35.2.11)
Elementary considerations on the transformation of EM
fields
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} —
603
December 23, 2011
35.3: Elementary considerations on the transformation of EM fields
35: ElectroMagnetism and relativity
[] Reference: D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed., (1999, Prentice Hall) : {} —
It might be thought that the electric and magnetic fields are the space part of a 4-vector but the situation is actually
more complex than that.
It is not unexpected that while a stationary charge distribution only produces an electric field a moving charge distribution produces an electric field plus a magnetic field.
It is less obvious that while an electrically neutral stationary current distribution only produces a magnetic field, a
moving observer will see a magnetic field plus a electric field. This applies, for instance, to an observer moving with respect
to a real magnet.
Several arguments in this section are adapted from the references 2 .
35.3.1
Transformations of electric fields
Consider two infinite parallel sheets of charge, perpendicular to the z axis at z = ±d/2, and charged with surface charge
density ±ρS (ρS > 0) at rest for the inertial observer I.
The electric field is zero outside the plates and uniform along the z axis in between the two plates and the magnetic
field is zero:
ρS
B=0 .
(35.3.1)
E = − e3
ε0
An inertial observer I moving with velocity u perpendicularly to the z axis will see the surface charge density increased
by a factor γ, as charge is invariant and the distances are contracted by γ along the direction of motion. Any inertial
observer moving perpendicularly to the z axis with velocity u will reach the same conclusion.
Therefore, if the electric field is still perpendicular to the xy plane:
{e3 = e3
ρS = γρS }
=⇒
E ⊥ = γE ⊥ .
(35.3.2)
Moreover the moving observer will also see a surface current density ∝ −u. In fact he will see two infinite parallel sheets
of currents with surface current:
±
j S = ∓ρS u = ∓γρS u .
(35.3.3)
He will then see a magnetic field in between the two sheets, given by the sum of the magnetic fields produced by the two
sheets:
−
+
j S ×(−e3 )
j S ×(+e3 )
+ µ0
= µ0 ρS u ×e3 .
B = µ0
(35.3.4)
2
2
Note the appropriate signs for e3 in order to pick up the region internal to the two sheets and the factor two in the
formula for the magnetic field of one sheet of current removed by summing up the magnetic fields produced by the two
sheets. All in all:
jS
±
= ∓γρS u
e3 = e3
B = µ0 ρS γ ( u ×e3 ) = −ε0 µ0 γ ( u ×E ) = −
γ
u ×E .
c2
(35.3.5)
Therefore:
B⊥ = −
γ
γ
u ×E = − 2 u ×E ⊥
c2
c
.
(35.3.6)
It is not strange that charges in motion produce a current and therefore a magnetic field.
An inertial observer I moving along the z axis will see the surface charge density unchanged. She would see the distance
between the two plates contracted, but this has no effect on the electric field.
Therefore, if the field is still seen as a field perpendicular to the plane:
Ek = Ek .
35.3.2
Transformations of magnetic fields
35.3.2.1
A solenoid
(35.3.7)
Another simple example can be taken from a ideal infinite solenoid, with axis the z axis, whose nl (helical windings
per unit length) carry a constant current I. The constant and uniform magnetic field at the interior of the solenoid is:
Bz = µ0 nl I.
An observer in motion along the z axis, who sees the solonoid moving along the negative z axis, will see the number of
windings per unit length increase by a factor γ, as the number of windings does not change but the length of the solenoid
decreases by a factor γ. Therefore: nl = γnl .
The current I, assumed to be running exactly perpendicularly to the z axis, is given by the product between the volume
current density, which does not change as it runs perpendicular to the relative velocity of the two Reference Frame, and
2 Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill),D. J. Griffiths, Introduction to Electrodynamics, 3rd Ed.,
(1999, Prentice Hall)
604
December 23, 2011
35: ElectroMagnetism and relativity
35.4: Some Examples and Physical Applications
the cross-section of the wire, which shrinks by a factor γ, when the solenoid is moving. Therefore the current is reduced,
in the judgement of the observer who sees the solenoid moving: I = I/γ.
Alternatively, one can reach the same conclusion on the current by observing that the moving observer will see the same
charge across any section of the wire in a time longer than the time measured by the stationary observer (who measures the
proepr time). Therefore the current measured by the moving observer is reduced by a factor γ. Therefore, again: I = I/γ.
Therefore, all in all, the magnetic field does not change, as the two factors cancel.
35.3.3
Transformations mixing-up Electric/Magnetic fields
See section 35.4.2 for one example of magnetic field transofrming into an electric field.
35.3.4
The relativistic transformation laws of ElectroMagnetic fields
In relativity electric and magnetic fields have no separate meaning: the single concept of an ElectroMagnetic field exists.
However the electric and magnetic field vectors are not the space component of any 4-vector. In fact they are members
of a 4-tensors with two indexes.
The full transformation laws can be derived from the study of other simple situations, as the one described above, as
well as from a more general treatment (see section ??). The results are:
Ek = Ek
(35.3.8)
Bk = Bk
E⊥ = γ E + c
+1
−1
B⊥ = γ B − c
35.4
35.4.1
β ×B
β ×E
(35.3.9)
(35.3.10)
⊥
.
⊥
(35.3.11)
Some Examples and Physical Applications
The 4-current of a ohmic conductor wire
Consider the 4-current in a typical globally neutral Ohmic metallic rectilinear conductor with conduction current 4-vector
equal to:
j = {cρ+ , 0} + {cρ− , ρ− v − } = {0, ρ− v − } ,
(35.4.1)
in terms of the charge density, ρ− , and drift velocity, v − , of the electrons and of the charge density, ρ+ , of the positive ions.
All quantities are measured in the Laboratory Rest Frame.
Show that the Current 4-vector, j, is space like with
2
j µ jµ = − (ρ− ) v − 2 = − ρ0+
2
v− 2
(35.4.2)
in terms of the charge density of the positive ions at rest in the conductor, ρ0+ .
35.4.2
EM fields for an observer moving with respect to straight infinite neutral
current wire (1)
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} — Alternative
detailed demonstration with velocities of the charges
See also section 35.4.3, for a different solution.
Consider a very long rectilinear electrically neutral current-carrying wire and consider it as an infinite rectilinear electrically neutral current-carrying wire (aligned with the z axis). Suppose the current I > 0 is generated by electrons moving
with velocity u < 0 inside a reticle of fixed positive ions. Let γ ≡ γu .
Let us also consider an observer, I, moving with the negative electrons.
The wire is neutral, in the Laboratory Reference Frame, with positive ions at rest and negative electrons moving in the
negative z direction. The positive ions are at rest and the free electrons move at constant speed. The system is electrically
neutral bacause the measured separation beetween positive ions and electrons are identical.
The Laboratory Reference Frame is the Rest Frame for the positive ions, but it not the Rest Frame for the electrons
and therefore:
−
ρ− = γρ−
0 > ρ0
ρ+ = ρ+
0
+
−
ρ +ρ =0
(35.4.3)
as the wire is electrically neutral in the Laboratory Reference Frame
ρ≡
ρ+
0
+
−
= ρ = −ρ > 0
605
just a definition .
(35.4.4)
(35.4.5)
December 23, 2011
35.4: Some Examples and Physical Applications
35: ElectroMagnetism and relativity
The wire will generate a tangential magnetic field and no electric field:
B = B[r]eθ
B[r] =
µ0 I
2πr
and
E=0 .
(35.4.6)
Note that the cross-section of the wire as well as the radial distance do not change, when passing from the Laboratory
Reference Frame to I:
S=S
r=r .
(35.4.7)
We have, in the Laboratory Reference Frame:
jz = ρ− u = −ρu > 0 .
I ≡ −ρuS = −ρcβS > 0
(35.4.8)
Lorentz transformations imply
ρ = −γβjz /c = γρβ 2 > 0
j z = γjz
(35.4.9)
I = γI .
(35.4.10)
The law of transformation of the charge density can be also derived, with more physical insight, directly as follows. An
observer moving with the electrons will see a charge density:
−
+
−
2
ρ = ρ+ + ρ− = γρ+
0 + ρ0 = γρ0 + ρ /γ = ρ (γ − 1/γ) = ρβ γ > 0 .
(35.4.11)
Therefore he will see a radial electric field (to within the sign, which will be determined later on):
E[r] =
ρS
ρβ 2 γS
1
µ0 ρcβS
=
= γβc
= γβcB .
2
2πε0 r
2πε0 r
ε0 µ0 c
2πr
(35.4.12)
It turns out, introducing vectors (all the fields are perpendicular to u):
E ⊥ = γ u ×B = γ u ×B ⊥
=⇒
ργβ 2 S
E = E ⊥ =
2πε0 r
.
(35.4.13)
The moving observer will then see also an electric force on a point charge, not only a magnetic force.
It is important to stress that the electric field seen by the moving observer is due to the appearance of a real charge
density from the point of view of this observer. Compare with the discussion of section 32.2.2.
The magnetic field is:
µ0 I
.
(35.4.14)
B ⊥ = γB ⊥ =
2πr
Note that the apparent non-invariance of electric charge, due to the fact that the wire appears neutral to one observer
and charged to the other one, is an artifact of the idealisation of infinite wire with infinite charge. It might seem to be
paradoxical, in view of the invariance of the total electric charge, that the wire appears to be charged to one observed and
neutral to another one. Actually there is no problem because the rectilinear infinite wire is an idealisation with charges
entering from infinity and exiting to infinity, such that it is not clear what is the total charge (which is, in any case, infinite).
See section 35.4.8 for a more realistic situation.
One should consider a finite loop of current to avoid such kind of problems (see section 35.4.8). In the real case there
will be parts of the wire positively charged and parts of the wire negatively charged, thus retaining the overall neutrality
of the wire.
However, when considering a finite piece of wire inside a closed finite surface, one might still wonder what happens
with the total internal charge, which should be an invariant quantity. In this case one should remember the discussion in
section 35.2.1, about the relativity of simultaneity.
Note that it as been shown not only that an observer in motion with respect to a stationary neutral current distribution
will see an electric field but also that the sources of the electric field are real charges (as any real electric field must be
traced to specific source charges). This fact was not clear from the discussion in section 32.2.2.
35.4.3
EM fields for an observer moving with respect to straight infinite neutral
current wire (2)
[] Reference: The Feynman’s lectures on physics (Addison-Wesley, 1964) : {} —
[] Reference: Berkeley Physics Course, Vol. 2, Electricity and Magnetism, (1965, McGraw-Hill) : {} — Alternative
detailed demonstration with velocities of the charges
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December 23, 2011
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35.4: Some Examples and Physical Applications
See also ?? for a different solution.
Consider the central part of a long rectilinear and uniform wire of radius r, section S, carrying a constant current I and
electrically neutral, in global.
Assume that the conduction current is provided by electrons moving inside a reticle of fixed positive ions.
Calculate the ElectroMagnetic fields at a distance R from the wire according to an observer at rest with respect to the
wire (an observer in the Laboratory Frame).
Calculate the ElectroMagnetic fields at a distance R from the wire according to an observer moving with the same
velocity as the average velocity of the (current producing) electrons inside the wire.
Verify that the ElectroMagnetic fields according to the moving observer are produced by the charge density and current
as seen from that system.
Verify that the two invariants of the ElectroMagnetic fields (section ??) are actually invariant in the transformation.
Introduce a cylindrical Coordinate System with the wire on its axis and choose for the z axis the same orientation as
the current flow.
For the stationary observer the electric field is zero, because the wire is electrically neutral. The magnetic field is
tangential. It is:
µ0 I
B[R] =
eθ
j ·ez > 0 .
(35.4.15)
2πR
Consider an observer moving with the same velocity as the average velocity of the (current producing) electrons inside
the wire. Use the transformation formulas in section ??, use the fact that for this Lorentz transformation we have
B[R] = B ⊥ =
µ0 I
eθ ,
2πR
(35.4.16)
use the fact that the observer is moving like the electrons, that is with a negative component of the velocity, Vz < 0, to
find:
E 0k = E k = 0
B 0k
E 0⊥
B 0⊥
(35.4.17)
= Bk = 0
+1
= γV E + c
β ×B
= γV B − c−1 β ×E
(35.4.18)
⊥
⊥
= γV c β V ×B ⊥
(35.4.19)
= γV B ⊥ .
(35.4.20)
A radial electric field has appeared while the perpendicular magnetic field has been modified.
The relation between the charge volume density in the Rest Frame, ρ0 , and the one in an arbitrary frame, moving with
velocity −V with respect to the Rest Frame, is given by:
ρ = γV ρ0
.
(35.4.21)
In the Rest Frame the volume of an element of charge is maximum and the charge density is thus minimum, due to the
invariance of the charge.
Let ρ+ and ρ− be the charge volume density in the Laboratory Frame. Let ρ0+ and ρ0− be the charge volume density in
the Reference Frame moving with the same velocity as the average velocity of the electrons in the wire. Let ρ0+ and ρ0− be
the charge volume density in the Rest Frame of the corresponding charge.
We have:
ρ+ + ρ− = 0
globally neutral in the Laboratory system
|ρ+ | = |ρ− | ≡ ρ > 0
0
ρ− = γρ−
0
ρ+ = ρ+
0
definition
(35.4.23)
from equation 35.4.21
(35.4.24)
from equation 35.4.21
(35.4.25)
0
ρ− = ρ− = ρ− /γ
(35.4.26)
ρ+ = γρ+ = γρ+
(35.4.27)
0
0
0 ≤ I = ρ− Vz S = −ρ− |Vz | S = ρ |Vz | S ≡ ρVz S .
35.4.3.1
(35.4.22)
(35.4.28)
First method
From equation 35.4.21 the following relations can be derived:
ρ0 ≡ ρ0+ + ρ0− = γρ+ + ρ− /γ = ρ (γ − 1/γ) = ρ
γ2 − 1
γ
!
= ργβ 2 > 0 .
(35.4.29)
Note that ρ− decreases in the transformation, as we are moving onto the Rest Frame, while ρ+ increases, as we are moving
away from the Rest Frame. Therefore we expect to see the wire electrically charged with positive charge.
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The moving observer will see a positive charge density and thus a radial electric field. Its calculation of the (radial)
electric field gives:
Er0 =
ρ0 S
ργβ 2 S
=
2πε0 R
2πε0 R
,
(35.4.30)
µ0 c2 ργβ 2
ργβ 2 S
er =
er ,
2πR
2πε0 R
(35.4.31)
At the same time equation 35.4.19 gives:
E 0⊥
µ0 I
2πR
= γV c β V ×B ⊥ = γβc
!
( −e3 ×eθ ) =
in agreement with 35.4.30. The two methods provide then the same value for the electric field, as it should.
35.4.3.2
Second method
It is obvious that the wire appears charged because the current transforms in the Lorentz transformation.
Write the special Lorentz transform, along the z axis, from the Laboratory Frame to the frame of an observer moving
with the same velocity as the average velocity of the (current producing) electrons inside the wire:
β = −V < 0
(35.4.32)
ρ0+ = γ (ρ+ − βj+ ) = γρ+
(35.4.33)
ρ0− = γ (ρ− − βj− )
(35.4.34)
0
(35.4.35)
0
(35.4.36)
j+ = γ (j+ − βρ+ )
j− = γ (j− − βρ− ) = 0 ,
where the last passages come from the fact that in this problem:
j+ = 0
0
j− = 0
=⇒
(35.4.37)
j− = βρ− .
(35.4.38)
It follows:
ρ0+ = γρ+
0
0
0
2
ρ− = γ ρ− − β ρ− = γρ− 1 − β
0
2
(35.4.39)
= ρ− /γ
2
ρ = ρ+ + ρ− = γρ+ + γρ− = γρ+ + ρ− /γ = ργβ > 0 ,
(35.4.40)
(35.4.41)
where the last passages are the same as in equation 35.4.29.
This second method is actually more general than the first one, as it can be applied to any transformation, not only
to the specific one considered in this problem, moving to a Reference Frame where the electrons are at rest. In this case
the generalisation requires to distinguish between the velocity of the Lorentz transformation and the drift velocity of the
charges. It is than better to stick to the more general relation (ρ = 0):
ρ0 = −γβj .
(35.4.42)
It might seem to be paradoxical, in view of the invariance of the total electric charge, that the wire appears to be
charged to one observed and neutral to another one. Actually there is no problem because the rectilinear infinite wire is
an idealisation with charges entering from infinity and exiting to infinity, such that it is not clear what is the total charge
(which is, in any case, infinite).
See section 35.4.8 for a more realistic situation.
35.4.4
A beam of identical charged particles
Consider a cylindrical beam of identical charged particles moving at constant velocity, u, on straight line with uniform
charge density and a circular section of radius R. This is often a good approximation for a beam of particles in a particle
accelerator.
Show that the repulsive force between particles is damped by a factor γ 2 , as a consequence of the attractive effect of
the magnetic field, with respect to the case when the particles are at rest.
Alternatively show that the same effect can be explained by the relativistic time dilatation.
Worked Solution
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35.4: Some Examples and Physical Applications
In the Laboratory Frame let the beam density be ρ. Let the z axis be on the direction of the beam and let the current
density along z be j z = ρv. Therefore in the Laboratory Frame:
ρr
at an point internal to the beam: r ≤ R ,
2ε0
µ0 ρvr
µ0 jz r
=
,
Bφ [r] =
2
2
f = q (E + v ×B) = q Er [r]er + ve3 ×Bφ [r]eφ
,
Er [r] =
fr [r] = q (E + v ×B) ·er = q
ρr 1
2ε0 γ 2
(35.4.43)
(35.4.44)
(35.4.45)
.
(35.4.46)
The repulsive force in the Laboratory Frame is trying to scatter away the particles in the beam but in the Laboratory
Frame the time evolution is slowed down.
Let’s look at the situation as it is seen by the Reference Frame where the charged particles are at rest, I. In this system
the force is of electrical nature only. The charge density is:
ρ = ρ/γ ≤ ρ .
(35.4.47)
The perpendicular component of the electric field is:
E ⊥ = γ (E + c β ×B)⊥ = γ
µ0 ρv 2 r
ρr
er + e3 ×eφ
2ε0
2
!
γρr
=
2ε0
v2
1− 2
c
!
er =
ρr
ρr
er =
er .
2ε0 γ
2ε0
(35.4.48)
As the electric charge is invariant the force on one particle in Reference Frame I is thus:
qρr
qρr
er =
er = γf .
2ε0
2ε0 γ
f=
(35.4.49)
This can be written as the Minkowsky force as:
fM ≡ f =
qρr
er ,
2ε0
(35.4.50)
where γ = 1 in the definition of the Minkowsky force because particles are at rest in the I Reference Frame.
On the other hand the force on one particle in Reference Frame I is:
f=
qρr
qρr
er .
er =
2
2ε0 γ
2ε0 γ
(35.4.51)
This can be written as the Minkowsky force as:
f M ≡ γf =
qρr
er .
2ε0
(35.4.52)
It is thus evident that:
fM = fM ,
(35.4.53)
as it is required for the transformation of the transverse component of the spatial part of a 4-vector as the Minkowsky force
is.
35.4.5
Force on a charge near a straight infinite neutral current wire
Consider the central part of a long rectilinear and uniform cylindrical wire of radius R, cross-sectional area S, carrying
a constant current I and electrically neutral globally.
Consider a positive point charge moving parallel to the wire with speed v and determine the force it feels.
Compare the description of the observatory in the Laboratory Frame (I) and the observatory moving with the charge
(I).
Worked Solution
Let the wire be along the z axis.
The current, charge and current density in the Laboratory Frame are given by:
ρ=0
j ≡ je3
609
I = jπR2 .
(35.4.54)
December 23, 2011
35.4: Some Examples and Physical Applications
35: ElectroMagnetism and relativity
The charge and current density in the I Reference Frame are given by:
ρ = −γ β ·j /c = −γvj/c2
j = γj .
(35.4.55)
A positive charge moving equiverse with the current is attracted by the wire by the Lorentz force.
In the Laboratory Frame one has:
Ek = 0
(35.4.56)
E⊥ = 0
(35.4.57)
Bk = 0
B⊥ =
(35.4.58)
µ0 I
eφ .
2πr
(35.4.59)
Actually not that E k = 0 is only approximatively valid at large distance from the central part of a long wire. In fact
the conservation of the tangential component of the electric field implies that the electric field just outside the wire is the
same as inside it. Moreover a radial component of the field exists in a real finite wire.
In the I Reference Frame one has the transformed fields:
Ek = Ek = 0
E ⊥ = γ v ×B = γ v ×B ⊥
(35.4.60)
it appears
(35.4.61)
Bk = Bk = 0
B ⊥ = γB ⊥ = γ
µ0 I
eφ
2πr
(35.4.62)
it is modified .
(35.4.63)
The electric field which appears in the I Reference Frame requires that the wire appears as charged. In fact:
E ⊥ = γvB = γv
µ0 I
γvµ0 πR2 j
c2 ρµ0 R2
ρR2
=
=−
=−
2πr
2πr
2r
2ε0 r
,
(35.4.64)
attractive .
(35.4.65)
as it should be for an attractive electric field due to a circular cylindrical wire of charges.
The Lorentz force in the Laboratory Frame is:
B = B⊥
f = q v ×B =
µ0 qvI
µ0 qvI
e3 ×eφ = −
er
2πr
2πr
The Lorentz force in the I Reference Frame (the velocity of the charge is zero) is:
E = E⊥
f = q E = qγ v ×B = γf .
(35.4.66)
As the Lorentz transformation is perpendicular to the force the Minkowsky force should be unchanged by the transformation. In fact:
f M = γf
(35.4.67)
fM = f ,
(35.4.68)
which are actually identical, as demonstrated by the previous equations.
35.4.6
ElectroMagnetic Fields of a point charge moving at constant velocity
We are going to use Lorentz transformations and the transformation laws for the ElectroMagnetic fields (section ??) to
determine the EM fields produced by a point charge q moving at constant velocity V (with β ≡ V /c) in a certain Reference
Frame, I.
Let I be the inertial Reference Frame where the charge is seen to move at constant velocity V . Let I 0 be the inertial
Reference Frame where the charge is at rest.
Let the fixed point P , a fixed point in I, be the point where the ElectroMagnetic fields have to be calculated.
Let r P and r C be the positions, in the Reference Frame I, respectively, of the point P and of the moving point charge
C.
Let r[t] be the vector from the position of the moving point charge to the point P , so that the kinematics of the motion
of the point charge is:
r[t] ≡ r P − r C [t] = r P − V (t − t0 ) − r C [t = t0 ] .
(35.4.69)
Note that the relation relating the position of P to the origin of the Coordinate System in I and to C are the same in
terms of the coordinates of both I and I 0 :
rP = r + rC
r 0P = r 0 + r 0C ,
610
(35.4.70)
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35: ElectroMagnetism and relativity
35.4: Some Examples and Physical Applications
as all the coordinates are related by the linear Lorentz transformation:
x0 = Lx .
(35.4.71)
Let θ[t] be the angle between the velocity vector V and the position vector r:
cos [θ[t]] =
V ·r
.
|r| |V |
(35.4.72)
We can show that the (time dependent) electric field is:
r
q 1 − β2
1
E[r P ] =
3/2
2
4πε0 1 − β 2 sin [θ]
r3
,
(35.4.73)
while the (time dependent) magnetic field is:
q 1 − β2
1
1
µ0
β ×r
B = β ×E ⊥ = β ×E = c
3/2
c
c
4π 1 − β 2 sin2 [θ]
r3
.
(35.4.74)
Note that the fields are expressed in terms of the instantaneous position of the point charge, that is they are radial with
respect to the instantaneous position of the point charge. They are not expressed in terms of the retarded position (see
section 29 for the precise definition of the retarded position of the charge, that is its apparent position).
Choose the z and z 0 axes parallel to V and denote as n ≡ e3 the unit vector associated with this direction.
Let the moving charge be at the origin of the Coordinate System of I 0 and let the two origins of the two Coordinate
System of I and I 0 be coincident at the times: t = t0 = 0.
We are going to calculate the EM fields as a function of the instantaneous position of the point charge, not of the
retarded one (see section 29 for the precise definition of the retared position of the charge, that is its apparent position).
We will therefore calculate, at a generic time t, the ElectroMagnetic fields at a point P , with coordinates r P and fixed in I.
The point P is located at a distance H from the velocity vector V and has a distance d from the origin along the direction
of V , that is:
n ≡ e3 ≡ V /V
r P = nd + (H/ |r|) n ×( r ×n) .
(35.4.75)
The distance H is the so-called impact parameter of the point particle with respect to the point where the field is being
calculated.
Moreover, with our conventions of I and I 0 , we have:
r 0C = 0 ,
rC = V t
(35.4.76)
and
r P = {xP , yP , zP = d}
H≡
q
x2P + yP2 .
(35.4.77)
Let’s then transform to the Rest Frame of the moving point charge, the inertial Reference Frame I 0 whose z 0 axis is
parallel to the z axis (and to V ).
The Lorentz transformation of the time-space coordinates of the moving point charge, C, and of the point P are:
t0 = γ t − V z/c2
(35.4.78)
z 0 = γ (z − V t)
0
r⊥
= r⊥ = H
0
zC
=0
0
zP0 [t] = γ (zP − V t) = γ (d − V t) ≡ zC
+ z0 = z0
.
(35.4.79)
(35.4.80)
We can now calculate, first, and transform to the I Reference Frame, later, the EM fields produced at the point P in
the rest Reference Frame of the point charge, I 0 .
It must be emphasised that this transformation involves both transformation of the EM fields and transformation of
the time-space coordinates where the fields are to be calculated. In particular equations 35.3.8 and 35.3.10, for the electric
field, as well as equations 35.3.9 and 35.3.11, are valid provided the EM fields are those at the same time-space point which,
however, has different coordinates in the two different Reference Frame.
The components of the EM fields parallel to V do not change.
35.4.6.0.1
Electric Field
The components of the Electric field parallel to V do not change.
The transformation of component of the Electric field perpendicular to V can be calculated by using the inverse transform
of equation 35.3.10:
E ⊥ = γ E 0 − c+1 β ×B 0 ⊥ ,
(35.4.81)
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35.4: Some Examples and Physical Applications
35: ElectroMagnetism and relativity
and the fact that B 0 = 0 one finds:

q
1
q
1

0

Ez0 [r 0P , t0 ] =
(γ (d − V t))

3 (z ) = 4πε
0
03

4πε
0 r
0 r



1
q



=⇒
Ez [r P , t] = Ez0 [r 0P , t0 ] =
(γ (d − V t))
1 qr 0
03
0 0
0
4πε
0 r
=
E [r P , t ] =
3
1
1
q
q

4πε0 r0

0
0
E⊥
(H)
[r 0P , t0 ] =

3 (r⊥ ) 4πε
0
03

4πε

0 r
0 r


1
q



=⇒
E ⊥ [r P , t] = γE 0⊥ [r 0P , t0 ] =
(γH)
4πε0 r0 3
.
(35.4.82)
Note that in both components a multiplicative factor γ appears, in one case due to the transformation of abscissa in
the other case due to the transformation of the electric field.
Moreover one needs to transform r0 :
q
q
2
2
2
2
0
0
0
0
r ≡ (x ) + (y ) + (z ) = x2 + y 2 + (γ (d − V t)) .
(35.4.83)
We observe that all the dependence on d and t just happens in the combined variable d − V t. Therefore the EM fields,
for the observer I, translate rigidly with time. We can therefore limit to calculate the fields at t = 0 and generalise the
final result for any time by replacing, in the final result, d → d − V t.
The expression for r0 , at t = 0, can be written as:
!1/2
!!1/2
1
H2
0
2
2 2 1/2
2
2
2
2
=γ d + 2
=γ d +H +H
−1
=
(35.4.84)
r = H +γ d
γ
γ2
!1/2
1/2
1/2
H2
2
2
2 1/2
.
(35.4.85)
1−β 2
1 − β 2 sin2 [θ]
≡ γ d2 + H 2
γ d +H
2
d +H
Finally, letting θ be the angle between r and V in the stationary Rest Frame, one finds, at t = 0:
1
q
qγd
1
qd 1 − β 2
1
Ez [r P , t = 0] =
=
(γd) =
4πε0 r0 3
4πε0 γ 3 (d2 + H 2 )3/2 1 − β 2 sin2 [θ]3/2
4πε0 (d2 + H 2 )3/2 1 − β 2 sin2 [θ]3/2
(35.4.86)
2
q
qγH
1
qH 1 − β
1
1
=
.
E⊥ [r P , t = 0] =
(γH) =
3
3/2
0
3/2
2
4πε0 r
4πε0 γ 3 (d2 + H 2 )
4πε0 (d2 + H 2 )3/2 1 − β 2 sin2 [θ]3/2
1 − β 2 sin [θ]
(35.4.87)
Putting together the above equations and replacing d → d − V t one finds:
q 1 − β2
r
1
r[t] ≡ r P − r C [t] = r P − V t
E[r P , t] =
3/2
4πε0 1 − β 2 sin2 [θ]
r3
θ = θ[t]
.
(35.4.88)
Note the electric field is not conservative, because it is a radial field whose module depends on the angle. However as
the situation is not a static one this is not a problem and the curl of the electric field will be compensated by a changing
magnetic field.
The result shows that the electric field, when one considers the moving charge at the present position, is still radial, but
its module depends on θ. This is not a violation of the finite velocity of propagation of signals because the point charge
has been known to move at constant velocity since an infinite time.
At fixed distance from the charge the θ dependence shows that the field is weaker in the forward and backward directions.
This anisotropy increases at increasing velocities.
In ultra-relativistic limit, β −→ 1, the electric field becomes basically a field transverse to the velocity of the charge. In
fact:
(
θ = 0 −→ E ∼ 1 − β 2
E −→ 0
for β −→ 1
.
(35.4.89)
2 −1/2
E −→ ∞
for β −→ 1
θ = π/2 −→ E ∼ 1 − β
This behaviour is not incompatible with Gauss law, because the field increases in the plane perpendicular to the velocity
but the region in solid angle where the field increases is decreasing at high speeds.
35.4.6.0.2
Magnetic Field
The components of the Magnetic field parallel to V do not change.
It is therefore zero in each Rest Frame. The transformation of component of the Magnetic field perpendicular to V can
be calculated by using the inverse transform of equation 35.3.11:
B ⊥ = γ B 0 + c−1 β ×E 0 ⊥ .
(35.4.90)
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35.4: Some Examples and Physical Applications
By using equation 35.4.82 to express E 0⊥ and observing that E 0⊥ can be replaced with E without changing the result of
the vector product one finds:
B[r P , t] = B ⊥
1
q 1 − β2
β ×r
1
µ0
= β ×E ⊥ = β ×E = c
c
c
4π 1 − β 2 sin2 [θ]3/2 r3
r[t] ≡ r P − r C [t] = r P − V t
θ = θ[t]
.
(35.4.91)
The result shows that the magnetic field, when one considers the moving charge at the present position, is still radial,
but its module depends on θ. This is not a violation of the finite velocity of propagation of signals because the point charge
has been known to move at constant velocity since an infinite time.
At fixed distance from the charge the θ dependence shows that the field is weaker in the forward and backward directions.
This anisotropy increases at increasing velocities.
In ultra-relativistic limit, β −→ 1, the magnetic field has the same behaviour than the electric field, and becomes
basically a field transverse to the velocity of the charge. In fact:
(
θ = 0 −→ B ∼ 1 − β 2
B −→ 0
for β −→ 1
.
(35.4.92)
2 −1/2
θ = π/2 −→ B ∼ 1 − β
B −→ ∞
for β −→ 1
35.4.6.0.3
The ElectroMagnetic fields
The electric and magnetic fields are always perpendicular: E ·B.
In ultra-relativistic limit, β −→ 1, the modules tend to become identical, in a system of units where c = 1.
The ElectroMagnetic fields produced by a ultra-relativistic charge resembles a impulse of a plane wave.
Note that the uniformly moving charge is not radiating EM waves.
The ElectroMagnetic fields move as rigid fields because the charge is supposed to be moving with constant velocity since
an infinite time so that any effect of past accelerations has already reached an infinite distance.
35.4.7
Magnetic field produced by a rectilinear infinite line of charges moving at
constant velocity
Use the results of section 35.4.6 to calculate the magnetic field produced by a rectilinear line of charges moving at
constant velocity, like the moving charges in a infinite rectilinear current wire of section A carrying a current I.
Introduce a cylindrical Coordinate System having as the z axis the current wire, oriented as the current. Use equation 35.4.74 for an infinitesimal element dz of the current wire. Let R be the distance of the point P from the wire.
Then:
1
λ dz β 1 − β 2
sin3 [θ]
dB =
eφ
(35.4.93)
4πε0 1 − β 2 sin2 [θ]3/2 R2
z = −R cot θ
=⇒
B=
wπ
dB =
0
λ = ρA
=⇒
B=
dz =
R
sin2 [θ]
dθ
(35.4.94)
µ0 λβ
eφ
4π R
(35.4.95)
µ0 βρA
µ0 jA
µ0 I
µ0 βρA
eφ =
eφ =
eφ =
eφ .
4π R
4π R
4π R
4π R
(35.4.96)
The indefinite integral:
w
was used.
35.4.8
sin [θ]
1−
2
β 2 sin [θ]
3/2 dθ =
√
2 cos [θ]
p
,
(−1 + β 2 ) 2 − β 2 + β 2 cos [2θ]
(35.4.97)
A simplified model for a moving Magnetic Dipole
According to section 35.4.3 a current carrying long rectilinear stationary wire, globally neutral, will appear as charged
to an observer moving along the wire. This is an apparent violation of the Lorentz invariance of the charge, as the two
different observers will measure a different total charge. The problem lies in the unphysical infinite wire.
In this section a real (possibly super-conducting) closed current loop is modeled. The use of a super-conducting loop
avoids nuisances such as batteries, external wires and geometrical asymmetries.
Use the results of problem 35.4.3 to cross-check the results of problem ?? in the simple case of a real magnetic dipole
made of a plane rectangular current loop, with sides a and b and negligible dimensions of its cross-section (of area S), where
a current I passes in its own Rest Frame.
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35.4: Some Examples and Physical Applications
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Assume that the loop is moving, with respect to an inertial Reference Frame, with constant velocity u parallel to its
sides of length a along the axes z ≡ z. Let the sides with length a be parallel to the axes z ≡ z and the sides with length
b be perpendicular to the axes z ≡ z. Look at the loop so that the current runs in counterclock-wise direction. Therefore
the magnetic dipole moment of the loop is directed toward the observer.
The four sides of the rectangle appear neutral in its own Rest Frame: ρ = 0. On the other hand: j 6= 0.
In the upper side of length a the current goes in the negative z ≡ z direction, so that it appears negatively charged to
the observer who sees the loop moving.
In the lower side of length a the current goes in the positive z ≡ z direction, so that it appears positively charged to the
observer who sees the loop moving.
Whatever the velocity u, the two sides of length a will appear charged to the observer who sees the loop moving. The
Lorentz transformation of the current 4-vector allows one to calculate the charge density seen by the moving observer, as
discussed in problem 35.4.3.
In fact:
in the upper side with j z < 0
(35.4.98)
ρ = γ ρ + βj z /c = γβj z /c < 0
ρ = γ ρ + βj z /c = γβj z /c > 0
in the lower side with j z > 0 .
(35.4.99)
Therefore the observer who sees the loop moving will see on the two sides of length a the charge densities:
ρ = ±γ |β| j /c ,
(35.4.100)
the ± sign, different for the two sides, being decided by the current being parallel or anti-parallel to u, according to the
above equations.
Note that there is no effect on the charge and currents the other two sides, of length b, because for those sides j ·u = 0,
that is the current density is transverse to the velocity of the Lorentz transformation and there is therefore no transformation
at all (it remains neutral). However the cross-section of the other two sides, of length b, is shrinked by a factor γ due to
the contraction of the length in the direction of motion.
The loop appears now charged, in its two sides of length a, which appear contracted by a factor γ, generating an electric
dipole moment with absolute value:
(35.4.101)
|p| = γ |β| j /c (Sa/γ) (b) = |β| I /S /c Sab = |β| I ab/c = |β| |m| /c .
Conduction current
Determine the conduction current in the loop seen by the stationary observer.
Verify that the stationary observer sees the same current crossing each section of each side of the loop, coherently with
the fact that he is seing a stationary situation.
Determine the convection current seen by the stationary observer.
35.4.9
Relative motion between a point charge and a small magnet
See section 32.2.3.
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35: ElectroMagnetism and relativity
35.5
35.5: Exercises, problems and physical applications
Exercises, problems and physical applications
Problem - 35.1
ElectroMagnetic fields of a charge moving at constant velocity
Consider a point charge moving at constant velocity in an inertial Reference Frame. Use the explicit expression of the
electric field (section 35.4.6), with its time dependence, to show that the electric field is divergence-less at any point outside
the charge. Use the explicit expression of the magnetic field (section 35.4.6), with its time dependence, to show that the
magnetic field is divergence-less at any point outside the charge.
Problem - 35.2
Charge invariance
Consider a point charge moving at constant velocity in an inertial Reference Frame. Use the explicit expression of the
electric field (section 35.4.6), with its time dependence, to show that Gauss theorem is satisfied by calculating the flux on
a sphere centered on the charge and with arbitrary radius.
Note that if the sphere is moving along with the charge Gauss law must still apply.
Problem - 35.3
Internal forces of a pair of charges moving at constant velocity
Use the results of this chapter to calculate the force between two identical charges, such that one charge fixed at the
origin of an inertial Reference Frame while the second one is moving with constant velocity V with respect to the same
inertial Reference Frame with impact parameter b > 0 with respect to the fixed charge. Calculate the sum of the internal
forces of the two charges.
Problem - 35.4
Transformation laws for Polarisation and Magnetisation
Use the results of section ?? as well as charge invariance and volume contraction to write down the relativistic transformation laws of the Polarisation and Magnetisation vector fields, analogous to equations 35.3.8, 35.3.9, 35.3.10 and 35.3.11.
Problem - 35.5
A cross-check
Verify that equations 35.4.17, 35.4.18, 35.4.19 and 35.4.20 are consistent with the invariants of the ElectroMagnetic field.
Problem - 35.6
A variation
Solve the problem 35.4.3 or the case of arbitrary velocity of the observer, parallel to the field, interpreting it in terms of
relativistic transformations of charges and currents. Solve the problem using both methods used in section 35.4.3.
Problem - 35.7
Properties of the ElectroMagnetic fields produced by a charge in motion at constant velocity
Refer to section 35.4.6.
• Show that Gauss theorem is satisfied by equation 35.4.88 at fixed time on an arbitrary sphere centered on the charge.
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35: ElectroMagnetism and relativity
• Show that equation 35.4.88 gives zero flux of the magnetic field at fixed time on an arbitrary sphere centered on the
charge.
• Show that for β −→ 1 one finds: β ·E = β ·B = 0 and the ElectroMagnetic fields tend to zero at any point θ 6= π/2.
• Consider a fixed point and the moving charge. Show that an estimate of the length of time during which a significant
field is seen at the fixed point is of the order of:
∆t ≈
H
βγ
.
(35.5.1)
Problem - 35.8
Biot-Savart law
Refer to section 35.4.6.
Show that equation 35.4.74 gives, in the ultra-relativistic limit, the Biot-Savart law.
Problem - 35.9
Poynting vectoer for a charge moving at constant velocity
Find the Poynting vector for a point charge in uniform motion.
Problem - 35.10
Force between the armatures of a plane capacitor
Refer to the plane infinite capacitor described in section 35.3.1.
The observer at rest with respect to the capacitor measures an attractive force between the two armatures. The observer
in motion sees a larger electric field, so that he measures a larger attractive force. However he also measures a repulsive
force due to the magnetic field. Show that the overal forces is consistent with the force measured by the observer at rest
taking into account the law of transformation fo the forces.
Problem - 35.11
Lorentz transformation of the ElectroMagnetic field
Use equations 35.3.8 and 35.3.10, for the electric field, and equations 35.3.9 and 35.3.11, for the magnetic field and
combine each of the two pairs to write in a single vector equation the transformation law for both the electric and the
magnetic field.
Worked Solution
Use c = 1 to simplify the formulae. One finds:
γ 2 β ( β ·E )
1+γ
2
γ
β
( β ·B)
B 0 = γ (B − β ×E ) −
.
1+γ
E 0 = γ (E + β ×B) −
(35.5.2)
(35.5.3)
Problem - 35.12
Invariants of the ElectroMagnetic fields
Using the relations in section 35.3.4 show that the two quanties
E ·B
c2 B 2 − E 2 ,
(35.5.4)
are invariant under Lorentz transformations.
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35: ElectroMagnetism and relativity
35.5: Exercises, problems and physical applications
Problem - 35.13
Transformation laws for a purely electric or purely magnetic field
Use the Lorentz invariants
E ·B
c2 B 2 − E 2 ,
(35.5.5)
to show that a purely electric (magnetic) field in one Reference Frame cannot transform to a purely magnetic (electric)
field in another Reference Frame.
Problem - 35.14
Transformation laws for an ElectroMagnetic field
In a certain inertial Reference Frame the electric field and the magnetic field are neither parallel nor perpendicular at
a particular space-time point.
1. Show that in a different inertial Reference Frame moving velocity u given by
!
E ×B
u2
,
u= 1+ 2
2
c
B + E 2 /c2
(35.5.6)
2. Is there a Reference Frame in which the two fields are perpendicular?
Problem - 35.15
Electric and Magnetic fields around a straight current wire
Start from a point charge at rest in a Reference Frame which moves with velocity u with respect to the laboratory
Reference Frame and where there is a pure electric field.
Transform to fields to the Laboratory Reference Frame to derive the Lorent force law, by also trnasforming the force
from one Reference Frame to the other.
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December 23, 2011