SPECIAL RELATIVITY II

1 2 1 2
1. Recall the Lorentz Transformation: v and v . Suppose we take
the space interval between two events , and the time interval .
Then, these intervals will b
x x t t t x
c
x x x t t t
′=γ − ′=γ⎛⎜ − ⎞⎟
⎝ ⎠
Δ = − Δ = −
( ) 2
e seen in S as v and v .
Now consider two particular cases:
a) Suppose the two events occur at the same place (so 0) but at different times
(so 0). Not
t t x x x t
c
x
t
′ Δ′=γ⎛⎜Δ − Δ ⎞⎟ Δ ′=γ Δ − Δ
⎝ ⎠
Δ =
Δ ≠ e that in S they do not occur at the same point: (0 v )!
b) Suppose the two events occur at the same time (so 0) but at different places
(so 0). Note that in S they are not simul
x t
t
x
′ Δ′=γ − Δ
Δ =
Δ ≠ ′ 2
taneous: 0 v .
2. As seen in the frame S, suppose a particle moves a distance in time . Its velocity is
then (in S-frame). As seen in the S -frame, meanwhile, it has move
t x
c
dx dt u
u dx
dt
Δ ′ =γ ⎜⎛ − Δ ⎟⎞
⎝ ⎠
= ′
( )
( )
2
2 2 2
d a distance
where v and the time that has elapsed is v . The velocity
in S -frame is therefore vv 1 v/ /v 1 vv. Thi
dx
dx dx dt dt dt dx
c
u ddtx ddtx ddtx dx ddt x dt u u
c c c
γ γ
γ
γ
′
′= − ′= ⎛⎜ − ⎞⎟
⎝ ⎠
′ ′= ′= − = − = −
′ ⎛⎜ − ⎞⎟ − −
⎝ ⎠
2
s is the
Einstein velocity addition rule. It is an easy exercise to solve this for u in terms of ,
1 vv .
3. Note one very interestin
u
u u u
c
′
′ +
= ′
+
g result of the above: suppose that a car is moving at speed v and
it turns on its headlight. What will the speed of the light be according to the observer on
the ground? If we use the Galilean
2
transformation result, the answer is v+c (wrong!). But
using the relativistic result we have and 1 vv 1 vv vv . In other
words, the speed of the source makes no differenc
u c u cc c ccc c
c c
′= = + = + = + =
+ + +
2
e to the speed of light in your frame.
Note that if either or v is much less than , then reduces to the familiar result:
1 vv v, which is the Galilean velocity addition rule.
u c u
u u u u
c
′
′= − → −
−
x
ct rocket
photon
body at
rest
uu>c
u=c
ct
present t = 0 present t = 0x
your future
your past
your world line
4. The Lorentz transformations have an interesting property that we shall now explore. Take
the time and space intervals between two events as observed in frame S, and the
corresponding quantit
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 2 2 2
2 2 2 2 22 2
ies as observed in S . We will now prove that the quantities defined
respectively as and are equal. Let's start with :
v / 2v
I c t x I c t x I
I c t x γ c t x c t x
′
= Δ − Δ ′ = Δ′ − Δ ′ ′
′= Δ′ − Δ ′ = ( Δ + Δ − Δ Δ ( )2 ( )2 − Δx −vΔt +2vΔtΔx)
( )(( )( ) ( ) ( ))
( ) ( )
2 2 2 2 2 2
2 2
2 2
1 v 1 v /
1 v /
This guarantees that all inertial observers measure the same speed of light !!
5. If t
c t x c
c
c t x I
= − Δ − Δ −
−
= Δ − Δ =
he time separation is large, then 0 and we call the interval
If the space separation is large, then 0 and we call the interval
If 0 and we call the interval
I timelike.
I spacelike.
I lightl
>
<
=
: a) If an interval is timelike in one frame, it is timelike in all other frames as well.
b) If interval between two events is timelike, their time ordering is absolute.
ike.
Note
c) If the interval is spacelike the ordering of two events depends on the frame from
which they are observed.
6. It is sometimes nice to look at things graphically. Here is
a graph of position versus time for objects that move with
different speeds along a fixed direction. First look at an
object at rest. Its position is fixed even though time (plotted
on the vertical axis) keeps increasing. Then look at the
rocket moving at constant speed (which has to be less than
c), and finally a photon (which can only move at c).
7. The trajectory of a body as it moves through spacetime
is called its world-line. Let's take a rocket that
is at 0 at 0. It moves with non-constant speed,
and that is why its world-li
x= t=
ne is wavy. A photon that
moves to the right will have a world line with slope
equal to +1, and that to the right with slope -1. The
upper triangle (with positive) is called the future
t
light cone (don't forget we also have and ). The
lower light cone consists of past events.
y x
8. Earlier on we had discussed the Doppler effect in the context of sound. Now let us do so
for light. Why are they different? Because light always moves with a fixed speed while
the speed of sound is different according to a moving and a fixed observer. Consider the
case of a source of light at rest, and one that is moving to the right as shown below:
0 0
0
Let 1 be the frequency measured in the source's rest frame S, where is the time for
one complete cycle in S. We want to calculate , the frequency as seen by the observer
in S mov
T
T
ν
ν
=
′ ing to the right at speed v. Call the distance between two successive wave
crests (i.e. the wavelength according to the observer). In time the crests ahead of the
source move a distance
T
cT
λ
( ) ( )
0
, even as the source moves a shorter distance v in the same
direction. Hence v and so . Now, as discussed earlier, the
v
time measured by observer will not be because of time
T
c T c c
c T
T
λ ν
λ
= − = =
−
( )
2 2
0 0
2 2 2 2 0 0
0
dilation. Instead,
. Hence, v v .
1 v / v v v v
( If source moves away from the observer just change the sign of v: v .)
v
This is the famous Doppl
T T cT c c c c
c c c T c c c
c
c
ν ν ν
ν ν
= − = − = − =⎛⎜⎝ − ⎞⎟⎠ − = +−
−
=
+
er effect formula. In the lecture I discussed some applications
such as finding the speed at which stars move away from the earth, or finding the speed
of cars or aircraft.
9. We now must decide how to generalize the concept of momentum in Relativity theory.
The Newtonian definition of momentum is . The problem with this definition is
that we are used to having momentum conserved
p􀁇 =mu􀁇
when particles collide with each other,
and this old definition will simply not work when particles move very fast. Consider the
collision of two particles as in the diagrams below:
In the frame fixed to the lab (S-frame) conservation of momentum implies:
. Mass is also conserved: . Now suppose
we wish to observed the collision from a fram
A A m u +mBuB=mCuC+mDuD mA+mB=mC+mD
2 2
e S moving at speed v. Then, from the
relativistic addition of velocities formula, 1vv and, from it, 1 vv. Insert
this into the equation of conservation of momentum (using
u uu u uu
c c
p m
′
′= − = ′+′
− +
=
2 2 2 2
v as the definition):
v v v v
1 v/ 1 v/ 1 v/ 1 v/
This is clearly not the equation So momentum wil
A B C D
A B C D
A B C D
A A B B C C D D
m u m u m u m u
u c u c u c u c
m u m u m u m u
⎛ ′+ ⎞ ⎛ ′+ ⎞ ⎛ ′ + ⎞ ⎛ ′ + ⎞
⎝⎜ + ′ ⎠⎟+ ⎝⎜ + ′ ⎠⎟= ⎝⎜ + ′ ⎠⎟+ ⎝⎜ + ′ ⎠⎟
′ + ′ = ′ + ′ l not be
conserved relativistically if we insist on using the old definition!!
10. Can we save the situation and make the conservation of momentum hold by finding some
suitable new definition of momentum? The new definition must have two properties:
1) At low speeds it
2 2
must reduce to the old one.
2) At all speeds momentum must be conserved.
Let's see if the definition will do the job. Obviously if u<1 /
get , so requireme
p mu mu
u c
p mu
= =γ
−
=
􀁇 􀁇 􀁇
􀁇 􀁇 nt 1 is clearly satisfied. . Let's now see if the conservation of
momentum equation will hold if the new definition of momentum is used:
. A A A B B B C C C D D D mγ u +mγ u =mγ u +mγ u
After doing some algebra you find,
1 v 1 v 1 v 1 v
This gives , which is just wha
A AA A B BB B C CC C D DD D
A A A B B B C C C D D D
m u m u m u m u
m u m u m u m u
γ γ γ γ γ γ γ γ
γ γ γ γ
γ γ γ γ
⎛⎜ ′ ′ + ⎞⎟+ ⎛⎜ ′ ′ + ⎞⎟= ⎛⎜ ′ ′ + ⎞⎟+ ⎛⎜ ′ ′ + ⎞⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
′ ′+ ′ ′= ′ ′+ ′ ′ t we want.
A u
B u
C u
D u
During Collision
After Collision
Before Collision
11. We shall now consider how energy must be redefined relativistically. The usual expression
for the kinetic energy 1 is not consistent with relativistic mechanics (it does not
2
satis
K= mu
fy the law of conservation of energy in relativity). To discover a new definition, let
us start from the basics: the work done by a force when it moves through distance
adds up to an in
F dx
( )
( ) ( )
( )
2 2
2 2 2 2 2 2 3 / 2 2 2 3/ 2
2
2 2 3 / 2
0 0
crease in knetic energy, . Now use:
/
1 / 1 / 1 / 1 /
This give
1 / 1
u u
K Fdx dp dx dx dp udp
dt dt
dp md u mdu m u c du mudu
u c u c u c u c
K udp m udu mc
u c
= = = =
⎛ ⎞
= ⎜ ⎟= + =
⎝ − ⎠ − − −
= = =−
−
∫ ∫ ∫ ∫
∫ ∫
( )
2
2 2
2
0 2 2
2 22 1/ 2 2
0 2 2 2
2
2 2 2
2
/
Note that K is zero if there is no motion, or where and
1 /
. Now expand 1 1 / 1
1 / 2
Hence, 1 1 as
2 2
mc
u c
K E E E mc
u c
E mc u c u
u c c
K mc u mc mu u
c
−
−
= − =
−
= = − = + + ⋅ ⋅ ⋅
−
⎛ ⎞
= ⎜ + + ⋅ ⋅ ⋅⎟− →
⎝ ⎠
2
2
2 2 2 2
2
/ 0.
12. Now that we have done all the real work, let us derive some alternative expressions
using our two main formulae: and .
1 / 1 /
a) or
c
p mu mu E mc mc
u c u c
u p pc pc E
m E
γ γ
γ
→
= = = =
− −
= = =
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇
( ) ( ) 2
2 2 2 2 4 2 2 2 4 2 4 2 2 4
2 22 24
2
b) Clearly 1 1/ . Hence,
.
c) For a massless particle ( 0), .
13. A particle with mass has energy even tho
u
c
pc E u m c m c m c E m c
c
E pc mc
m E pc
m mc
γ γ γ
⎛ ⎞
⎜ ⎟
⎝ ⎠
= ⎛⎜ ⎞⎟ = − = − = −
⎝ ⎠
= +
= =
ugh it is at rest. This is called its rest
energy. Since is a very large quantity, even a small corresponds to a very large
energy. We interpret this as follows: suppose all the mass coul
c m
2
d somehow be converted
into energy. Then an amount of energy equal to mc would be released.
I
Vacuum
chamber
Metal
plate
Collecting
plate
Ammeter
Potentiostat
Light, frequency ν
e−