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Electromagnetism and
special relativity
We have nearly finished our program on special relativity… And then we
can go on about how it affects electromagnetism, and how we have to
change our familiar quantities –charges, currents, potentials, fields,
energy (and momentum!) density and flow, etc. Lots of vectors in
electromagnetism….!
We still have to discuss only the concept of force in connection with
special relativity, and its aspects at high velocity.
We shall do that one of the next lessons. For now, what we need to
anticipate of the force and related subjects is that, at low energy, we
can write, in three dimensions, the Newton law under the form
dP
F
dt
Which can easily be rewritten in 4 dimensions since we only need to
substitute τ for t, while P… we already know how to use its 4-vector
form.
Let us summarize the cornerstones of special relativity that we shall
make use of, as we did when sorting out 4-velocity etc.
PRINCIPLES
•Equivalence of Inertial Reference Systems (IRFs)
•The fundamental physics laws have the same form in all IRFs. This means
that the mathematical expression is the same independently of IRF, that
the two terms on the sides of the “equal” sign have the same transformation
properties for a change of IRF. AND… that the basic physics constants
(including the speed of light) have the same value in all IRFs.
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RESULTS of Special Relativity
•Physics phenomena take place in a pseudo-euclidean 4-dimensional
space, called Space-Time.
•Basic physics laws have the same form in all IRFs.
•The coordinate transformation between two IRFs (for a change
between two systems with the same direction of axes but relative
uniform motion) is the Lorentz transformation LT.
We shall use the representation of 4-vectors, 4-tensors etc. based on
the use of contravariant and covariant vectors. The convention is that
the 4 components are expressed with indices 0, 1, 2, 3 where index 0
indicates the “time” component. The LT – for a relative motion along
axis x1, i.e. “x”, is
 x0   
 1 
 x    
x2   0
 3 
 x   0
 0 0  x 0 '
 
 0 0  x1 ' 
 2
0
0
1 0

0 1
 x '
 3
 x '
where we have used for time and space the same units (c=1).
The “norm” of a 4-vector A is:
A 2  A  A  A0  A0  A1  A1  A2  A2  A3  A3
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Covariance of the physical laws
Physical laws are written in the form of Equalities between two terms
which:
• Have well-defined transformation properties for the LT.
• Such properties are the same for the two members of the equality
(scalars, 4-vectors etc.)
• A LT transforms them into equations that have the same form:
p.ex.:
pμ  pμ  m2
  
p'μ  p'μ  m2
If we are talking fundamental laws of physics, not applications to
special systems with a preferred direction, then if we find that a
particular component of a vector is equal to the same component
of another vector then the two vectors are equal!
Electrostatics and special relativity
Starting assumptions:
1. The force on a charged particle at rest is due only to
the electric field at the particle’s position.
2. IF – in the particle’s rest frame, charges and currents
which generate fields are static, the electric field is:
E(r ) 
 r  r '
 r  r'
3
dV '
3. The electric charge is an invariant for Lorentz
Transformation.
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What has been called “Starting assumptions” in the previous slide are
in fact experimentally established facts. The Coulomb law and the
principle of superposition are proved experimentally. And so is the
invariance of the electric charge for Lorentz transformations . This
is demonstrated by the neutrality of matter independent of
temperature.
As we heat some matter, electrons’ velocity increases much faster
than protons’, but the total charge does not change!
Charge density ρ, Current density J, and
Special Relativity
In a given IRF,
J = ρ·v
where
v is the velocity of the charges. Now, the question:
How do ρ and J transform under a LT?
Now, ρ is defined as

charges’ rest system is
and J will be zero.
dQ
dQ

dV dx  dy  dz
0 
which, in the
dQ
dx  dy  dz
If I look at that system from an IRF’ in motion (β,
γ) the charge will
not change. What will change is dx, which will become
and therefore
dx '  dx / 
 '    0
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Moreover, the observer in IRF’ will record a charge density ρ’ moving
at speed – β. Let us use now a system in which c is not equal to one.
A current distribution
J’ = -ρ’·v
will be seen in IRF’. And it will be precisely
J'    'v    0  v
So, our charge density at rest in its own system becomes a different
charge density when seen in IRF’, and on top a current density is
generated. The whole thing with coefficients that seem to be
borrowed by the LT.
Let us then check if an object made with 4 components:
{ , J x / c, J y / c, J z / c}  { , J / c}
does transform like a 4-vector, and then is a 4-vector?
Well, under a LT such object in the charge’s rest frame is
{ 0, 0} and  { 0, 0}  {   0,       0, ,0,0}
where
Λ is the matrix of the LT. The new values the
LT gives for the
new charge and current distributions are exactly what is seen! Then….
The charge density and current distribution are
components of a 4-vector.
This makes a big change wrt electromagnetism in 3-dimensional space.
We used to consider charge density as a scalar, but now it turns out to
be the zero-component of a 4-vector!
This fact will have important consequences.
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First, how is modified the equation of the charge conservation?
J 

0
t
The derivatives are the components of the 4-dimensional nabla,
while J and ρ are the components of the 4-current. And, in
Minkowski space language, the equation of charge conservation
becomes:
μ J μ  0
Well then, now that we have found the 4-current, we can go on
with adjusting the EM formulas to spacetime. The charge and
current densities appear in many EM equations, let us start with
the simplest:
 2  4
Let us examine with Mr. Lorentz’s eye how this equation behaves:
pretty badly. In the first term, we have a laplacian which in 4space is not a physical quantity – but could easily become a
D’Alambertian, since (remember? we are in electrostatics) the
time-derivatives are null. And… the D’Alambertian is a scalar.
The potential Φ, well, in electrostatics it is a scalar but…… What
settles it is the second term. The equation is obviously a scalar
equations, i.e. both members are just one number. But they are
obviously not scalars from the point of view of LT!!! Because the
second term is, apart from a multiplying constant, the
zero_component of a 4-vector.
We have here found a case of an equality between the 2
corresponding components of two 4-vectors: the equality
must therefore hold for the 4-vectors.
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“Two” 4-vectors? That in the second term there must a 4-vector,
can be agreed upon. But the first term?? Φ in 3-D electrostatics is
a scalar. Well, the second term is the zero-component of a 4vector, so there must be three other equations for the other
three components of the 4-current; and on the left side
there will be the three space-components of a “vector
potential” that has still to be defined (which is the least
problem), and especially understood in terms of physics:
what does it represent?
Well, the Poisson equation of electrostatics is now written:
2
  4J 0
The relativity then entails the existence of this vector-potential
A in 3 dimensions, and that it satisfies the three equations
2
A  4J / c
Remark how with this formula we have also found that the components
of vector A are determined as functions of J same way that Φ was
determined by ρ. In the time-independent case, it is:
 (r ) 

 r  r '
r  r'
dV '
Then 
So far for the potential… and now:
The Electric Field.
Well, we now know what to do: examine the
electrostatics equation, recognize the terms whose
relativistic properties we know, i.e.: are they scalars, 4vectors or what? We start from the field equation in
electrostatics:
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E  
The equation above can now be re-written in Minkowski space terms:
Ex  1 A0 Ey  2 A0 Ez  3 A0
These equations tell us that the electric field is not a 4-vector, since
it has 2 indices: it is a part of a two-indices object, i.e. a 4-tensor. And
since those three terms in the right side transform as elements of a
tensor, so must do the components of the electric field.
Ex is really a
F10
Ey is really a
F20
Ez is really a
0
F3
}
components of the first line
of a Tensor
A tensor of rank 2 has 16 components – of which we already know
as many as three!
So, we have only 13 components to find out. Now, rank 2 tensors
come in three types: symmetrical ( Tij = Tji ); antisymmetric
(
Tij = -Tji ); and generic, i.e. neither the one nor the other.
Any such tensor can anyway be written as the sum of a symmetric
plus an antisymmetric tensor.
It is easy to find that a symmetric tensor has 10 independent
components, while an antisymmetric one has only six. If the field
tensor were an antisymmetric one we would need only 3 new
components. Remark, btw, that the LT transforms symmetric
tensors into symmetric ones, and the same for antisymmetric ones.
Of course, it would simpler if our tensor were antisymmetric.
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That the field tensor is an antisymmetric one can be demonstrated.
To start with, let us use the Newton law F=ma. In relativistic
matters the quantity “force” loses meaning, so we will replace it by
dP/dτ. The Newton’s law is now written as
Fi 
dPi
0
 qFi
d
There must then exist other components that satisfy the equations:

p 
 qF 0

Now in the rest system pμ
equation becomes
=[m0c, 0, 0, 0]= pμ,and the previous
p 
q

p F 

m0 c
This equation is now written as an equality between 4-vectors and,
as such, it is valid in all IRFs. We can then make the scalar product
with the covariant momentum pμ.

p 
2q
2 p

p p F  

m0 c

 2
2q
[p ]
p p F   0

m0 c
The derivative wrt τ is the derivative of a constant and therefore
always null.

2q

Then also m c p  p F
is null, which can only happen if
0
μν = - ν μ
F
F
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Since that equality holds for any value of
it is the field which is antisymmetric.
p p
then it must be that
Since, on the other hand, the field is some sort of
  A ,
then a
term that satisfies both this requirement and the requirement of
antisymmetry is :
F     A   A
where A is what we have called “the vector potential”, without
having the slightest idea of what it might be. But we know that this
formulas, when applied to the first row of the field tensor reads:
E   
1 A
c t
Of course adding in the elements that second term  A  , which
for the electric field becomes (1/c)dAi/dt, the complete field
tensor now reads:
0
 1 Ax

x c t
 1 Ay


y c t
 1 Az


z c t
 1 Ax

x c t

0
Ax Ay

y
x
Az Ax

x
z
 1 Ay

y c t
Ax Ay

y
x

0

Ay
z

Az
y
 1 Az

z c t
Ax Az

z
x
Ay Az

z
y
0
We find again in row 1 and again in column 1 the well known formula
for the electric field E, valid now also in the case of charges and
currents varying in time. But we have also 3 new terms, for which
we need an explanation. We remark that they have the structure
of the curl of the 3-vector A. Let us call them “B“.
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We have just defined a new vector, and called it B.
B   A
And we can re-write the field tensor:
F 
μν
0
E
x

Ey

 Ez
 Ex
 Ey
0
Bz
 By
 Bz
0
Bx
 Ez 
By 
 Bx 

0 
What does the “B“ vector is still to be understood. What is known
is that under a LT it transform as the components of a tensor.
F '   F 
μν
μν
1
If the LT represented by Λ has β aligned along the X axis then
F ' 
μν
0
 Ex
  ( E y  Bz )   ( Ez  By )



E
0


(
B


E
)

(
B


E
)
x
z
y
y
z 

 ( E y  Bz )  ( Bz  E y )

0
 Bx



(
E


B
)


(
B


E
)
B
0
 z

y
y
z
x
Which can be re-written as
{
B'x  Bx
E'x  Ex

E 'y   ( E y  Bz )  E y  ( v  B) y
c

E 'z   ( E z  B y )  E z  ( v  B) z
c
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Which can be written in the new, more explicit, form:
E 'L  EL
E 'T  ET 
where the indices “L “ and “T “ stand

c
| vB |
for longitudinal, i.e. parallel to v and
transverse, i.e. perpendicular to it.
Note that the vector product is already
perpendicular to v.
The Lorentz force
We have now seen the fields – we must use the plural now, since we
have two fields to work with. And…. two fields which are the same
thing, since they mix when a LT is applied as a consequence of
change of IRF. Note that we not only have two fields, but also have
the formulas to compute them as functions of the charge and
current distributions. Now what we still miss is what one of these
fields, the one we called B, is doing to the charges. In order to find
what this field does in a generic situation, i.e. with moving
charges, we can do (actually, will do) the following:
•Study the interaction with one charge only. It will be only too easy
to extend this case to the presence of many charges.
•The general case studied is that of a charge moving in an environment
in which there are fields, E and B.
•We want to find the force acting on that charge in that IRF
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How we shall proceed is the following:
We have the velocity of the charge: we shall make a LT to the IRF’ in
which the charge is at rest: in that system, for one of the preliminary
assumptions, the force acting on the particle is only that due to E’. We
shall use as indicator of the force the instantaneous momentum change
dP' / dt .
Then we will transform again the
dP' / dt
original IRF
frame in which the particle is in motion, and see what corresponds to
the values of dP / dt . Since the LT has different values for
longitudinal and transverse coordinates, we shall keep the two separate.
In the charge’s
rest frame:
{
dP ' dP '

 qE '
d
dt '
dP'T
with especially
 qE'T
d
F' 
1.
What we want to calculate is dP/dt as a function of the
fields the charged particle sees in the IRF (the lab).
2.
We express the force in IRF’ (Coulomb law): we will have to
separate the longitudinal and transverse components of
force and fields since they behave differently under a LT.
3.
We establish the relation between dP/dt and dP’/dt’ (LT).
Again, keep transverse and longitudinal components
separate.
4.
We write the LT of longitudinal and transverse electric
fields E’ as a function of E, B
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E 'L  EL

E 'T  ET  ( v  B)
c
dP' dP'

 qE'
d
dt '
Fields in Rest Frame (IRF’)
dP'L dP'L

 qE'L
d
dt '
dP' dP
dP

 
d
d
dt
dPL
dP'L
dP / 
dP

 L
 L
d
d
dt / 
dt
NB : (dPL    dP'L )
Equations of motion in IRF’
Lorentz Transformation of
motion from IRF’ (rest) to
IRF.
Putting everything together:
dPL dP'L

 qEL
dt
d
dP dP'
v

 q (E T   B )
dt
  d
c
This is the (3-dimensional) force applied by the fields on a moving
charge. Observe the force generated by the “vector” B. The
Lorentz force which had been found experimentally is deduced
from electrostatics plus special relativity plus charge conservation.
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