Download Chapter 26 – Relativity

College Physics 150 Chapter 26 – Relativity
•  The Postulates of Special Relativity
•  Simultaneity
•  Time Dilation
•  Length Contraction
•  Relativistic Velocity Addition
•  Relativistic Momentum
•  Relativistic Rest Mass Energy, Kinetic Energy, and Total Energy
Postulates of Relativity
Postulate 1: The laws of physics are the same in all inertial reference frames (the principle of relativity).
An inertial reference frame is one in which no accelerations are observed in the absence of external forces. (Recall Newton’s first law).
Postulate 2: The speed of light in vacuum is the same in all inertial reference frames. The value is independent of the motion of the source of light or of the observer.
What follows does not show that Newtonian physics is wrong; it is just not complete. Newtonian physics breaks down at speeds near c; this is where special relativity is needed. All of the familiar relationships from Newtonian physics are contained within special relativity in the limit of low speeds. This is known as the correspondence principle.
Simultaneity and Ideal Observers
The location of an event can be specified by four coordinates: the three spatial coordinates (x,y,z) and a time coordinate t. Taken together (x,y,z,t) are the four coordinates of space-­‐‑time.
Consider a high speed train with AC = BC & A’C’ = B’C’. The marks ACB are on a stationary train platform and the marks A’C’B’ are on the moving train.
At t = 0: AA’, BB’, and CC’ are lined up and two lightning bolts strike at AA’ and BB’.
A
’
C
’
B
’
A
C
B
A
’
C
’
B
’
A
C
B
The observer at C’ (on the train) will see the strike at AA’ first followed by the strike at BB’.
An observer standing at point C (on the platform) will see both strikes simultaneously.
The observers do not agree on what happened, but both are correct.
This experiment shows that events that are simultaneous in one frame (the platform) are not simultaneous in another frame (the train).
Time Dilation
Consider a light clock. A light pulse reflects back and forth between two mirrors. One complete trip can be considered one “tick” on the clock.
Mirror
L
Light pulse
Mirror
The time interval for a round trip by the light pulse is
d 2L
Δt0 = =
.
v
c
As long as an observer is stationary with respect to the clock, he will measure a time interval of Δt0 between clock ticks. Now put the light clock on a moving train. What does a stationary observer outside the train see?
The motion of the train is to the right with speed v.
The light pulse appears to have the path shown below. How long does it take the light pulse to return to the bo`om mirror?
⎛ 1
⎞
2 L2 + ⎜ vΔt ⎟
d
⎝ 2
⎠
Δt = =
c
c
L
vΔt
Know
d 2L
Δt0 = =
v
c
2
Eliminating L gives
Let,
Δt =
Δt0
⎛ v ⎞
1 − ⎜ ⎟
⎝ c ⎠
v
1
β = and γ =
c
1− β 2
2
= γΔt0 .
(Lorena Gamma factor)
The person outside the train will measure a longer time interval between ticks compared to the observer in the train.
The relative motion between the observers (train and outside the train) changes the rate at which time passes. The effect is known as time dilation.
Moving clocks run slower.
The quantity Δt0 is known as the proper time. This is time interval between two events that occur in the same place. Example (text problem 26.6): An unstable particle called the pion has a mean lifetime of 25 ns in its own frame. A beam of pions travels through the laboratory at a speed of 0.60c. (a) What is the mean lifetime of the pion as measured in the laboratory? Given: Δt0 = 25 ns and v = 0.60c (β=0.60)
γ=
1
1− β
2
=
1
1 − 0.60
2
= 1.250
Δt = γΔt0 = (1.250)(25 ns ) = 31 ns
(b) How far does a pion travel (as measured by laboratory observers) during this time?
d = vΔt = (0.6c)(31 ns) = 5.6 m
Length Contraction
To measure the length of an object, its ends must be located simultaneously. The proper length is measured when an object is at rest relative to you.
A person on a train platform will measure the length of the platform to be L0. A moving object will pass by the platform in a time Δt so L0 = vΔt (note: Δt is not the proper time). A person riding on a train will measure the length of the platform to be L = vΔt0 (this person measures the proper time).
L0 vΔt
Δt
=
=
=γ
L vΔt0 Δt0
or L =
L0
γ
where L0 is the proper length.
The two observers measure different lengths for the platform.
Moving meter sticks are shorter.
This effect is known as length contraction and only applies to lengths parallel to the direction of motion.
Example (text problem 26.15): A cosmic ray particle travels directly over a football field from one goal line to the other, at a speed of 0.50c. (a) If the length of the field between goal lines is 91.5 m, what is the length measured in the rest frame of the particle?
Given: L0 = 91.5 m and v = 0.50c (β = 0.50)
γ=
1
1− β
2
=
1
1 − 0.50
2
= 1.155
L0
91.5 m
L=
=
= 79 m.
γ
1.155
(b) How long does it take the particle to go from one goal line to the other according to earth observers? L0 91.5
Δt =
=
= 6.1×10 −7 s
v 0.50c
(c) How long does it take in the rest frame of the particle?
L 79 m
Δt0 = =
= 5.3 ×10 −7 s
v 0.50c
Example (text problem 26.17): Two spaceships are moving directly toward each other with a relative velocity of 0.90c. If an astronaut measures the length of his own spaceship to be 30.0 m, how long is the spaceship as measured by an astronaut in the other ship?
Given: L0 = 30.0 m and v = 0.90c (β=0.90)
γ=
1
1− β
2
=
1
1 − 0.90
L0
2
= 2.294
30.0m
L=
=
= 13 m.
γ
2.294
The Barn Paradox
Since simultaneity is relative, the two doors did not need to be shut at the same time, and the ladder did not need to fit inside the garage.
The back of the pole enters the barn before the front of the pole leaves, so a 1 ns gate could be closed on both ends, containing the entire pole.
The front gate closes just as the back of the pole enters. The gate closings are not simultaneous, and they permit the pole to pass through without hi`ing either gate.
The Twin Paradox
One twin goes off into space at close to the speed of light while the other twin stays on Earth. When the twin returns, the space twin is younger than the Earth twin due to time dilation.
But, if motion is relative…why can’t the Earth twin be younger.
Solution:
The non-­‐‑moving twin is in an inertial reference frame (Observes time dilation).
The moving twin is in a non-­‐‑inertia reference frame (No time dilation). The effect of the speed of light on space-­‐‑time
We typically think that time and space are not related to each other. However, Einstein realized that time and space are actually woven together and that space-­‐‑
time must constant, but they can change individually when you are moving.
Therefore, when you have a velocity, you bend space-­‐‑
time to keep the speed of light constant.
Velocities in Different Reference Frames
What is the velocity of the probe as measured by Abe? According to Galilean relativity, it is vPA= vPB+vBA. Once the speeds get large enough, the relativistic velocity transformation formula must be used.
vPA
vPB + vBA
=
vPB vBA
1+
c2
Example (text problem 26.26): The rogue starship, Galaxa, is being chased by the ba`lecruiser, Millenia. The Millenia is catching up to the Galaxa at a rate of 0.55c when the captain of the Millenia decides it is time to fire a missile. First the captain shines a laser range finder to determine the distance to the Galaxa, and then he fires a missile that is moving at a speed of 0.45c with respect to Millenia. What speed does the Galaxa measure for (a) the laser beam and (b) the missile as they both approach the starship? The laser beam will be measured to have speed c.
The velocity of Millenia relative to Galaxa is vMG= 0.55c. The velocity of the missile relative to Millenia is vmM = 0.45c. The velocity of the missile relative to Galaxa is
vmG =
vmM + vMG
0.45c + 0.55c
=
= 0.802c.
vmM vMG
0.45c ) ( 0.55c )
(
1+
1+
2
c
c2
Relativistic Momentum
p = γmv
Note: it is true that Δp = ΣFΔt, but ΣF≠ma.
The ordinary definition of momentum with the mass replaced by the relativistic mass. It follows from the Lorena transformation when collisions are described from a fixed and moving reference frame, where it arises as a result of conservation of momentum. Example: Show that the expression for relativistic momentum reduces to its classical form when v << c.
When v << c, β << 1
⎛ 1
p = γmv = ⎜
⎜ 1 − β 2
⎝
⎞
⎟mv
⎟
⎠
⎛ ⎛ 1 ⎞ 2 ⎞
1 2
≈ ⎜⎜1 − ⎜ − ⎟ β ⎟⎟mv ≈ mv + β mv
2
⎝ ⎝ 2 ⎠ ⎠
⎛ v ⎞ 1 ⎛ v ⎞
≈ mc⎜ ⎟ + mc⎜ ⎟
⎝ c ⎠ 2 ⎝ c ⎠
≈ mv
3
To first order in v/c
Example (text problem 26.36): A body has a mass of 12.6 kg and a speed of 0.87c. (a) What is the magnitude of the momentum?
Given: m = 12.6 kg and v = 0.87c (β = 0.87)
γ=
1
1− β
2
=
1
1− 0.87
2
= 2.028
p = γmv
= (2.028)(12.6 kg )(0.87 c ) = 6.7 ×10 9 kg m/s.
(b) If a constant force of 424.6 N acts in the direction opposite to the body’s motion, how long must the force act to bring the body to rest? Δp
6.7 ×10 9 kg m/s
Δt =
=
424.6 N
∑F
= 6.58 ×10 7 s = 0.50 years
Mass and Energy
The rest mass energy of a particle is its energy measured in its rest frame.
E0 = mc 2
A proton has m = 1.67×10-­‐‑27 kg; its rest mass energy is
E0 = mc 2
= 938 MeV
= 0.938 GeV
The mass of the proton can be wri`en as 938 MeV/c2.
1 eV (electron volt) is the change in energy that a charge e (the fundamental unit of charge) experiences when accelerated through a 1 volt potential difference.
W = qΔV = e ⋅1 Volt = 1 eV
Relativistic Kinetic Energy
A quantity is conserved if its value is unchanged in a given reference frame.
A quantity is invariant if its value is the same in all reference frames. K = (γ − 1)mc 2
Kinetic energy
E = E0 + K
= mc 2 + (γ − 1)mc 2
= γmc 2
Total energy
Example: Show that the expression for relativistic kinetic energy reduces to its classical form when v << c.
When v << c, β << 1
⎛ 1
⎞ 2
K = (γ − 1)mc = ⎜
− 1⎟mc
⎜ 1 − β 2
⎟
⎝
⎠
⎛ ⎛ 1 ⎞ 2 ⎞ 2 1 2 2 1 2
≈ ⎜⎜1 − ⎜ − ⎟ β − 1⎟⎟mc ≈ β mc ≈ mv .
2
2
⎝ ⎝ 2 ⎠
⎠
2
The relationship between relativistic momentum and total energy is
2
E = ( pc ) + mc
Other useful relationships are
2 2
( ).
2
2
( pc )
= K 2 + 2KE0
v pc
= .
c E
Example (text problem 26.48): When an electron travels at 0.60c, what is its total energy in MeV?
Given: v = 0.60c (β = 0.60)
γ=
1
1− β
2
=
1
1− 0.60
2
= 1.250
E = γmc 2
(
)
= (1.250) 0.511 MeV/c 2 c 2 = 0.65 MeV.
Example (text problem 26.51): An electron has a total energy of 6.5 MeV. (a) What is its momentum (in MeV/c)?
The energy-­‐‑momentum relationship isE 2
Solving for p: pc
2
2 2
( ).
= ( pc ) + mc
2 2
( )
= E − mc
=
2
(6.5 MeV )2 − (0.511 MeV )2
p = 6.5 MeV/c
(b) What is the value of γ for this electron?
E
6.5 MeV
γ= 2=
= 12.7
2
2
mc
(0.511 MeV/c ) c
(c) What is the kinetic energy of this electron?
K = (γ −1) mc 2
= (11.7) ( 0.511 MeV/c 2 ) c 2 = 6.0 MeV
= 6.5 MeV