The Lorentz transformation

The figure shows inertial reference frame S1 moving with speed v relative to frame S, in the
common positive direction of their horizontal axes (marked x and x 1

Side note v=d/t hence d=vt

An observer in S reports space time co-ordinates x,y,z, t for an event and an observer in S1
reports x 1, y 1, z 1 , t 1for the same event. How are these sets of numbers related.

We claim at once that y and z co-ordinates, which are perpendicular to the motion, are not
affected i.e y = y 1 and z = z 1

Our interest then reduces to the relation x and x 1 and between t and t 1

The Galilean transformation equations

Prior to Einstein publication of his special theory of relativisty, the four co-ordinates of
interest were assumed to be related by the Galilean transformation equations
1
x =x−vt
1
t=t

These are the Galilean transformation equations approx valid at low speeds

These equations are written with assumptions that t=t 1 = 0 when the origins of S and S1
coincide. You can verify the first equation from the figure.

The second equation effectively claims that the time passes at the rate for observers in both
reference frames.

This would have been so obvious true to a scientist prior to Einstein.

When speed v is small compared to c, the equations above work generally well
Lorentz Transformation equations

We state without proof that the correct transformation equations which remain valid for all
speeds up to the speed of light, can be derived from postulates of relativity

The results are called Lorentz transformation equations and they are

1 1 vx
x =γ ( x −vt ) t =γ (t− 2
)
c

1
y =y
1
z =z

1 1
Where γ = =
√¿ ¿ ¿ ¿ ¿

The equations are written with assumptions that t=t 1 = 0 when the origins of S and S1
coincide

Note that the spatial values x and the temporal values t are bound together in the first and last
equations

This entanglement of space and time was prime message of Einstein’s theory. A message that
was long rejected by many of his contemporaries.

It is a formal requirement of relativistic equations that they should reduce to familiar classical
equations if we let c approach infinity.

That is if speed of light were infinitely great, all finite speeds would be low and classical
equations would never fail.

If we let c → ∞ in Lorentz transf equations then γ =1 and the equation reduce to Galilean
equations.

Lorentz transformation equations are written in a form that is useful if we are given x 1 and t 1
and wish to find x and t, then we write Lorentz transformation equations as
1
1 1 1 vx
x=γ ( x +v t ) and t=γ (t + 2
)
c

Problem

An observer S reports that an event occurred on the x axis of his reference frame at x = 3x108
m at time t = 2.5 s. If observer S1 and her frame are moving in the positive direction of x at a
speed of 0.4c. Further x= x 1 = 0. What co-ordinates does observer S1 report for an event.

Solution
 1
x =γ (x−vt) putting x=3x108 m and t= 2.5 s and v= 0.4c

First compute

1 1
γ= = = 1.09
√¿ ¿ ¿ √¿ ¿ ¿
1 8
x =1.09(3 x 10 −0.4 cx 2.5) =

Also compute

1 vx
t =γ (t− 2
)
c

Problem

An observer s assign S the space time co-ordinates x =100km and t=200 μs to an event. What
are the co-ordinates of this event in frame S1, which moves in the positive direction of x with
speed of 0.95c relative to S?

Assume x=x = 0 at t=t 1 = 0

Consequences of Lorentz-transformation equations

1. Time dilation

2. Lorentz-Fitzgerald contraction (also called space contraction)

1. Time dilation (also called time dilatation)

As an object approaches the speed of light an observer at rest observes that its time interval
become longer relative to time interval when the object is at rest.

Consider a vehicle moving to the right hand side with speed v. Observer O^1 is travelling
with the vehicle while O is standing on the ground.

A mirror is fixed on the ceiling of the vehicle and O^1 holds a laser light source.

At some instant the laser emits a pulse of light towards the mirror (event 1) and at some later
time reflecting from the mirror, the pulse arrives back to the laser (event 2)

a)

Observer O^1 carries a clock c^1 and uses it to measure the time interval 〖∆t〗_p between
these 2 events. P stands for proper.
Because the light pulse has speed c, the time it takes the pulse to travel from O^1 to the
mirror and back to O^1 is

〖∆t〗_p= distance/speed = 2d/c

Now consider the same pair of events as viewed by observer O in a second frame. According
to this observer, the mirror and laser are moving to the right with speed v and as a result the
sequence of events appears entirely different.

By the time the light reaches the mirror, the mirror has moved to the right a distance

v∆t/2

Where ∆t is the time it takes the light to travel from O^1 and back to O^1 as measured by O.
In other words, O concludes that, because of the motion of the vehicle, if the light is to hit the
mirror, it must leave at an angle with respect to the vertical direction.

b)

Comparing figure a and b we see that the light must travel further in b) than in a)

Note that neither observer knows that he or she is moving. Each is at rest in his/her own
inertial frame.

O is longer than time interval 〖∆t〗_p measured by O^1

To obtain a relationship between these two time intervals we use Pythagoras thm

(c∆t/2)^2=(v∆t/2)^2+ d^2

∆t= 2d/√(c^2-v^2 ) = 2d/(c√((1-v^2/c^2 ) ))

Because 〖∆t〗_p = 2d/c

∆t= (∆t_p)/√(1-v^2/c^2 )

∆t= γ∆t_p where γ=(1-v^2/c^2 )^(-1/2)

Example (twins paradox)

Speedo and Goslo are twins aged 20 yrs old. Speedo sets out on an epic journey to planet
Silicius located 20 ly from the earth, Furthermore, her spaceship is capable of cruising at
0.95c relative to inertial frame of her twin sister, after reaching planet silicius, she returns
immediately due to homesickness. Upon her return, Speedo is shocked to see Goslo is now
very old. Find how a) long Speedo travelled b) How old Goslo is now?
Note one light year is the distance travelled by light in one year which is 9.4607 x10^12 km

a) 〖∆t〗_p = 2d/v = (2x20x9.46x10^12 x1000)/0.95c=13 years

b) ∆t= (∆t_p)/√(1-v^2/c^2 ) = 13/√(1-〖(0.95c)〗^2/c^2 ) = 41 yrs

Goslo is 41 +20 = 61 years now while speedo is 20+13=33yrs

Example

Marrium and James are twins aged 18 yrs old. Marrium sets out on an epic journey to planet
Titan located 5 Ly from the earth, Furthermore, her spaceship is capable of cruising at 0.95c
relative to inertial frame of her twin brother, after reaching planet Titan, she returns
immediately due to homesickness. Find how long Marrium flew as observed by a) James and
as observed by b) herself?

〖x3.0x10〗^8)-(0.95x3.0x10^8 x3.0x10^8))) =
a) ∆t= 2d/(c√(c^2-v^2 )) = (2x5x9.4607x10^12 x10^3 metres)/(c√(〖(3.0x10〗^8

b) 〖∆t〗_p = 2d/c =(2x5x9.4607x10^12 x10^3 metres)/(〖(3.0x10〗^8))

2. Length Contraction

Lorentz-Fitzgerald contraction (also called space contraction)

It is the shortening of an object along the direction of its motion relative to an observer.
Dimensions in other directions are not contracted.

The concept of contraction was proposed by irish Physicist George Fitzgerald in 1889 and
therafter developed by Hendrik Lorentz of Netherlands

Length contraction is only in the direction parallel to the direction in which the observed
body is travelling.

This effect is negligible in everyday speeds and can be ignored. But at greater speeds it
becomes relevant. At speeds of 13400000 m/s the contracted length is 99 % of the length at
rest.

As magnitude of velocity approaches speed of light, the effect becomes dominant

Consider a spaceship travelling with speed v from one star to another. There are two
observers, one on earth and the other in the spaceship. The observer at rest on the earth (and
also assumed to be at rest with respect to the two stars) measures the distance between the
stars to be the proper length L_p

According to this observer the time it takes the space ship to complete the voyage is

∆t= L_p/v
Because of time dilation, the space traveller measures smaller time to travel by the spaceship
clock

〖∆t〗_p= ∆t/γ

The space traveller claims to be at rest and sees the destination star moving towards the
spaceship with speed v.

Because the space traveller reaches the star in time ∆t_p, she concludes that the distance L
between the stars is shorter than L_p

This distance measured by space traveller is

L=v∆t_p= v∆t/γ

Because L_p=v∆t

L= L_p/γ = L_(p ) (1-v^2/c^2 )^(1/2)

L= L_p (1-v^2/c^2 )^(1/2)

Interpretation; if an object has proper length 〖(L〗_p) i.e rest length, when it moves with
speed v in a direction parallel to its length, it contracts to

L=L_p/γ = L_(p ) (1-v^2/c^2 )^(1/2)

Example

A spaceship is measured to be 120m long and 20 m in diameter while at rest relative to an

Solution

What will observer record? It will appear shorter, so L= L_p/γ

L= L_p √((1-v^2/c^2 ) ) = 120√(1-(0.99c)^2/c^2 ) = 17 m

Diameter remains same because contraction occurs only along the direction of motion.