Lecture-XIX
Special Theory of Relativity
Lorentz transformation
The question is how (x, y, z; t) and (x0; y0; z0; t0) are related?
If both coordinate systems are inertial (that is, no relative acceleration), then a particle moving
along a straight line in one system must move along a straight line in the other. Otherwise we
would introduce spurious forces into the system with a curved trajectory. Hence we require
linear transformations.
• A reasonable assumption:
The most general, linear transformation between (x, t) and (x’, t’) can be written as
where a1; a2; b1; b2 are constants that can only depend on v, the velocity between the
coordinate systems, and on c.
Step 1: note that the origin of the primed frame (x0 = 0) is a point that moves with speed v as
seen in the unprimed frame. Therefore,
and hence
Step 2:
The only way this is consistent with
is if,
solving we get,
Then substituting a1, we get
Inverse Transformation
So Galilean transformations are a limiting case of the Lorentz transformations
when
Alternate method
x′ = γ
(x
− v t ), y ′ = y , z ′ = z b u t t ′ ≠ t.
 1 − γ ′γ 
′v t ′, ⇒ t ′ = γ t + 
 x
′
γ
v


A t t − t ′ = 0 , th e o rig in s o f th e tw o s y s te m s c o in c id e .
x′ = ct′ and x = ct
x = γ ′ ( x ′ + v t ′ ) = γ ′γ
γ
(x
− vt
)=
(x
)+ γ
− vt

 1 − γ ′γ  
c γ t + 
 x ⇒
′
γ
v

 

 1
 c
v
= 1− 
− 1 , ⇒
c
 γγ ′
 v
H ence, 1 +


v


1+
c


x = ct

 1
 c 
1
−
−
1




′
γ
γ

 v 

γγ ′ =
1
v2
1− 2
c
γ γ ′ d e p e n d s o n v 2 a n d n o t o n v , th e r e la tiv e v e lo c ity o f S ' w ith
re s p e c t to S . O n e c a n n o t d is tin g u is h S a n d S ' e x c e p t fo r th e s ig n
o f v w h i c h d o e s n o t e f f e c t t h e d e p e n d e n c e o f γ γ ′.
H e n c e ,γ = γ ′ =
x′ =
x − vt
1− v
2
c
2
1
1− v2 c2
, t′ =
.
t − vx c 2
1− v
2
c
2
, y ′ = y , z′ = z.
Some consequences of Lorentz Transformations
• Simultaneity:
Consider a railway-man standing at the middle of a freight car of length 2L. He flicks on his
lantern and a light pulse travels out in all directions with the velocity c. Light arrives at the
two ends of the car after a time interval L/c. In this system, the freight car's rest system, the
light arrives simultaneously.
Now let us observe the same situation from a different frame, one moving to the right with
velocity v.
In the rest frame: Take the origin of coordinates at the
center of the car, and take t = 0 at the instant the lantern
flashes. The two events are
In this system, the freight car's rest system, the light arrives simultaneously at A and B.
Same events in the moving frame:
Message: It makes no sense to say that one event happens at the same time as another, unless
you state which frame you’re talking about. Simultaneity depends on the frame in which the
observations are made.
Two events A and B have the following coordinates in the x, y system.
Event A:
Event B:
The distance L and time T separating the
events in the x, y system
The coordinates in the x', y' system,
with L > 0 and T > 0.
The distance L' between the events in primed frame:
• Length Contraction: Imagine that a rod is lying at rest along the x’-axis in the S’ frame. Its
end points are measured to be at xA’ and xB’, therefore its rest or proper length is xB’ – xA’= l0.
Now what is the rod’s length as measured by the S frame observer, for whom the rod moves
with the relative speed v? Suppose the two end points of the rod are at xA and xB as measured
at the same instant of time t from S.
Example: A rod of length lies in the x'y' plane of its rest system and makes an angle with
the x' axis. What is the length and orientation of the rod in the lab system x, y in which the rod
moves to the right with velocity v?
The coordinates of the end points A and B in rest frame:
The coordinates of the end points A and B in Lab frame:
The angle that the rod makes with the x axis is: