SPECIAL RELATIVITY I

1. The Special Theory of Relativity was created by Albert
Einstein, when he was a very young man in 1905. It is
one of the most solid pillars of physics and has been tested
thousands of times. Special Relativity was a big revolution
in understanding the nature of space and time, as well as
mass and energy. It is absolutely necessary for a proper
understanding of fast-moving particles (electrons, photons,
neutrinos..). Einstein's General Theory of Relativity -
which we shall not even touch here - goes beyond this
and also deals with gravity.
2. Relativity deals with time and space. So let us first get some understanding of how we
measure these two fundamental quantities:
a) Time : we measure time by looking at some phenomenon that repeats itself. There are
endless examples: your heart beat, a pendulum, a vibrating quartz crystal, rotation of
the earth around its axis, the revolution of the earth about the sun,... These can all be
used as clocks. Of course, an atomic clock is far more accurate than using your heart
beat and is accurate to one part in a trillion. Although different systems of measurement
have different units it is fortunate that time is always measured in seconds.
b) intuitively we know the difference between short and long. But to do a
m
Distance :
easurement, we first have to agree on what should be the unit of length. If you use
a metre as the unit, then you can use a metre rod and measure any length you want.
Of course, sometimes we may use more sophisticated means (such as finding how
high a satellite is) but the basic idea is the same: the distance between point A and
point B is the number of metre rods (or fractions thereof) that can be made to fit in
between the two points.
3. Newton had believed that there was one single time for the entire universe. In other words,
time was absolute and could be measured by one clock held somewhere in the centre of
the universe. Similarly, he believed that space was absolute and that the true laws of
physics could be seen in that particular frame which was fixed to the centre of the universe.
(As we shall see, Einstein shocked the world by showing that time and space are not
absolute quantities, but depend on the speed of your reference frame. Even more shocking
was his proof that time and space are not entirely separate quantities!)
y
z z ′
y ′
O O ′
S x S ′ P
v t x ′
A B
4. An is something that happens at some point in space at some time. With respect to
the frame S below, the event P happened at (x,y,z,t). Now imagine a girl running to the
right at fixed
event
speed v in frame S . According to her, the same event P happened at
(x ,y ,z ,t ). What is the relation between the two sets of coordinates?
′
′ ′ ′ ′
If you were Newton, then you would look at the above figure and say that obviously it is
the following: v , , , . These are called
. Note
x x t y y z z t t Galilean coordinate
transformations
′= − ′= ′ = ′ =
that it is assumed here that the time is the same in both frames
because of the Newtonian belief that there is only one true time in the world.
5. There are certain obvious consequences of using t
( )
he Galilean transformations. So, for
example a rod is at rest in S-frame. The length in S-frame , while the length in
the S -frame v . Since , the length is the
B A
B A B A B A B A
x x
x x x x t t t t
= −
′ = ′ − ′ = − − − = same in
both frames: . (As we shall see later this will not be true in Einstein's
Special Relativity).
B A B A x′ −x′ = x −x
6. Let us now see what the Galilean transformation of coordinates implies for transformations
of velocities. Start with v and differentiate both sides. Then v. Since
, it f
x x t dx dx
dt dt
t t
′
′ = − = −
= ′ ollows that and so v. Similarly, and .
Now, is the x-component of the velocity measured in S -frame, and similarly
is the x-com
x
x
dx dx dx dx dy dy dz dz
dt dt dt dt dt dt dt dt
dx u
dt
dx u
dt
′ ′ ′ ′ ′
= = − = =
′ ′ ′ ′
′
=′ ′
′
= ponent of the velocity measured in S-frame. So we have found that:
v, , (or, in vector form, v)
Taking one further derivative,
x x y y z z
x
u u u u u u u u
du d
dt d
′ = − ′ = ′ = ′= −
′
=
′
􀁇 􀁇 􀁇
( v) (remember that v constant),
we find that the components of acceleration are the same in S and S :
, , .
x
x
x x y y z z
u du
t dt
du du du du du du
dt dt dt dt dt dt
− = =
′
′ ′ ′
= = =
′ ′ ′
d=9×1010m
A c=3×108m/sec B
8
7. We had learned in a previous chapter that light is electromagnetic wave that travels at a
speed measured to be 2.907925 10 / sec. Einstein, when he was 16 years old,
asked himself the qu
c= × m
estion: in which frame does this light travel at such a speed? If I
run holding a torch, will the light coming from the torch also move faster? Yes, if we use
the formula for addition of vel
8
ocities derived in the previous section! So if light actually
travels at 2.907925 10 / sec then it is with respect to the frame in which the "aether"
(a massless fluid which we cannot feel) is
× m
at rest. But there is no evidence for the aether!
8. Einstein made the two following postulates (or assumptions).
1) The laws of physics have the same form in all inertial frames. (In other words, there
is no constant-velocity frame which is preferred, or better, than any other).
2) The speed of light in vacuum has the same value in all inertial systems, independent
of the relative motion of source and observer.
[Note: as stressed in the lecture, no postulate of physics can ever be mathematically
proved. You have to work out the consequences that follow from the postulates to know
whether the postulates are good ones or not.]
9. Let's first get one thing clear: in any one inertial frame, we can imagine that there are
rulers and clocks to measure distances and times. We can synchonize all the clocks to
read one time, which will be called the time in that frame S. But how do we do this? If
are two clocks, then we can set them to read the same time (i.e. synchronize them) by
taking account of the time light takes to travel between them.
Example: The observer with clock A sees the time on clock B as 2:55pm. But he knows
that light took 5 minutes to travel from B to A, and therefore A and B are actually reading
exactly the same time.
10
8
(Time taken by light in going from B to A is equal to 9 10 300sec 5min.)
3 10 /sec
10. Now I shall derive the famous formula which shows that a moving clock runs slow. This
will be some
d m
c m
×
= = =
×
thing completely different from the older Newtonian conception of time.
Einstein derived this formula using a "gedanken" experiment, meaning an experiment
which can imagine but not necessarily do. So imagine the following: a rail carriage has a
h
vΔt
d
v
a bulb that is fixed to the ceiling. The bulb suddenly flashes, and the light reaches the
floor. Time taken according to the observer inside the train is Δt′ =h/c.
Now suppose that the same flash is observed by an observer S standing on the ground.
According to S, the train is moving with speed v. Let's consider the same light ray. Clearly,
the tra
2 ( )2
in has moved forward between the time when the light left the ceiling and when it
v
reached the floor. According to S, the time taken is . Now
t d h t
c c
+ Δ
Δ = =
( )
2
2
2
2
2 2 2 1
v
1
1
v
1
square both sides: ( ) v , which gives . Here is
the relativistic factor, and is a number that is always bigger than one. As v
gets closer and close
c
c
c t h t t h t
c
γ γ
γ
−
−
Δ = + Δ Δ = = Δ ′
=
r to c, the value of gets larger and larger. For v=4c/5, =5/3. So, if
1 sec elapses between the ticks of a clock in S (i.e. 1), the observer in S will see
5/3 seconds between the tick
t
γ γ
′ Δ′=
-18
s. In other words, he will think that the moving clock is slow!
11. The muon is an unstable particle. If at rest, it decays in just 10 seconds. But if traveling
at 3/5 the speed of the light,
( )2
it will last 25% longer because 1 5. If it is
1 3/5 4
traveling at v=0.999999c, it will last 707 times longer. We can observe these shifts due
to time dilation quite easily, and they are
γ = =
−
an important confirmation of Relativity.
12. Another amazing prediction of Relativity is that objects are shortened (or contracted) along
the direction of their motion. Einstein reached this astonishing conclusion on the basis of
yet another gedanken experiment. Again, consider a moving railway carriage with a bulb
h
vΔt1 2 vΔt
Lamp
Mirror
at one end that suddenly flashes. Let be the length of the carriage according to ground
observer S, and be the length according to the observer S inside the carriage. So,
2
x
x
t
Δ
Δ′ ′
Δ ′ = Δ ′ is the time taken for the light to go from one end to the other, and then return
after being reflected by a mirror. Now let's look at this from the point of view of S, who is
fixed to the g
x
c
1
1
1 2
round. Let be the time for the signal to reach the front end. Then,
because the mirror is moving forward, v . Call the return time. Then,
t
t x t t
c
Δ
Δ + Δ
Δ = Δ
( )
2
2 1 2 1 2
1 2 2 2
2 2
v . Solving for and : and . The total time
v v
is therefore 2 / . Now, from the time dilation result derived
1 v /
earlier, 1 v / ,
t x t t t t x t x
c c c
t t t x c
c
t c t
Δ − Δ Δ Δ
Δ = Δ Δ Δ = Δ =
− +
Δ
Δ = Δ + Δ =
−
Δ ′ = − Δ ( )
2 2
2 2 2 2
2 2
and so 1 v / 2 / 2 / . This
1 v / 1 v /
gives, , or / . This an astonishing result! Suppose that there
1 v /
is a metre rod. Then the observer riding with it ha
t c x c x c
c c
x x x x
c
γ
Δ ′ = − ⎛⎜ Δ ⎞⎟= Δ
⎜⎝ − ⎟⎠ −
Δ′= Δ Δ = Δ′
−
s 1. But according to someone
who sees the metre rod moving towards/away from him, the length is 1/ . This is less
than 1 metre!
x
γ
Δ ′ =
13. Although an object shrinks in the direction of motion (both while approaching and
receding), the dimensions perpendicular to the velocity are not contracted. It is easy
to conceive of a gedanken experiment that will demonstrate this. One of the exercises
will guide you in this direction.
x′
y y′
vt d
x
v
14. We shall now derive the "Lorentz Transformation", which is the relativistic version of
the Galilean transformation discussed earlier. Consider an event that occurs at position
x (as measured in ) at a distance (again, as seen in ) from the origin of S . Then,
vt. From the Lorentz contraction formula derived earlier, where is the
distance measured in S .
S d S
x d d x x
γ
′
′
= + = ′
′ This gives x′=γ(x−vt). Now, by the same logic, x′=d′−vt′.
( ) ( ( ) )
2
Here . This gives v v v . We find that the time in S is
related to the time in S by v . Note that if we make very large, then .
c
To summarize:
d x x x t x t t
t t x c t t
γ γ γ
γ
γ
′= = ′+ ′ = − + ′ ′
′= ⎛⎜ − ⎞⎟ ′=
⎝ ⎠
( )
LORENTZ TRANSFORMATION
v
x x t
y y
′=γ −
′ =
2
v .
(Note: in various books you will find slightly different derivations of the above Lorentz
transformation. You should look at on
z z
t t x
c
γ
′ =
′= ⎛⎜ − ⎞⎟
⎝ ⎠
e of your choice and understand that as well.)