Let us go over how the Lorentz transformation was derived and

[Pages:24]1

Relativity notes Shankar

Let us go over how the Lorentz transformation was derived and

what it represents.

An event is something that happens at a definite time and place,

like a firecracker going off. Let us say I assign to it coordinates (x, t)

and you, moving to the right at velocity u, assign coordinates (x, t).

It is assumed that when you, sitting at x = 0 passed me (sitting at

x

=

0), we set our clocks

to zero:

t

=

t

=

0.

Thus

our origins in

space-time coincide.

Let us see how the coordinates would have been related in pre-

Einstein days.

First,

once

we

synchronize

clocks

at

t

=

t

=

0,

they

remain

synchronized for all future times t = t. This is the notion of absolute

time we all believe in our daily life.

You will assign to the event the spatial coordinates

x

=

x

-

ut.

(1)

This relationship is easy to understand: your origin is ut meters to the right of mine (Figure (2)) because you have been moving to the right at velocity u for time t so the x-coordinate you assign to an event will be less by ut compared to what I assign. You can invert

S

S'

ut x

x'

Figure 1: The same event (solid dot) is assigned coordinates (x, t) by me (S) and (x, t) by

you

(S').

In

the

pre-Einstein

days

t

=

t

not

only

initially

(when

our

origins

crossed)

but

always.

2

this relation to say

x

=

x

+

ut.

(2)

How are these two equations modified post-Einstein? If the ve-

locity of light is to be same for both you and me, it is clear we do

not agree on lengths or times or both. Thus if I predict you will say the event is at x = x - ut, you will say that my lengths need to be

modified by a factor so that the correct answer is

x

=

(x

-

ut).

(3)

Likewise

when

you

predict

I

will

say

x

=

x

+

ut

I

will

say,

"No,

your lengths are off, so the correct result is

x = (x + ut)."

(4)

Note two things. First, I leave open the option that the time elapsed between when we synchronized clocks and when the firecracker went off is t for you and t for me, with two times being possibly different. Next, the "fudge factor" for converting your lengths to mine and mine to yours are the same . This comes from the postulate that both observers are equivalent.

So let us look at the equations we have:

x = (x + ut)

(5)

x

=

(x - ut).

(6)

We proceed to nail down as follows. The event in question was a fire cracker going off. Suppose when our origins coincided we sent

off a light pulse that this pulse set off the firecracker. Since the light

pulse

took

t

seconds

to

travel

x

meters

according

to

me

and

took

t

seconds to go x meters according to you and we both agree on the

value of c, it must be true for this particular event that

x = ct

and

x

=

ct .

(7)

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Let us multiply the LHS of Eqn 5 by the LHS of 6 and equate the result to the product of the RHS's to get

xx

=

2(xx

+

xut

-

x ut

-

u2tt),

and

upon

setting

x

=

ct,

x = ct we g(e

2

c tt

=

2(c2tt

+

uctt

-

uct t

-

u2tt)

and now upon cancelling tt

(

2

=

1 1 - uc22

(1

1

=

1

-

uc22

.

(1

Note that once we have found it does not matter that we got

it form this specific event involving a light pulse. It can applied to

Eqns. (5,6) valid for a generic event. Putting back into Eqn. (6)

we obtain

x

=

x - ut

1

-

uc22

.

(12)

If

we

now

go

to

Eqn

5

and

isolate

t

we

find

t

=

t1--uc2uxc22

(13)

upon remembering that

1

-

1 2

=

u2 c2 .

(14)

To summarize, if I am frame S and you are S', and you are moving

to the right (increasing x direction) at speed u and my coordinates for an event are (x, t) and yours are (x, t) the Lorentz transformation

tells us that

x

=

x - ut

1 - u2/c2

(15)

t

=

t - cu2 x 1 - u2/c2

(16)

4

If we consider two events labelled 1 and 2, then the coordinate differences obey

x

=

x - ut

1 - u2/c2

(17)

t

=

t

-

cu2 x

(18)

1 - u2/c2

where x = x2 - x1 etc., and differences are not necessarily small. If you want to get my coordinates in terms of yours, you need to

invert Eqns. (15) and 16. The answer is we get the same equations but with u -u. The same goes for the differences. As a result the differences will be related as follows;

x

=

x + ut

(19)

1 - u2/c2

t

=

t

+

cu2 x

1 - u2/c2

(20)

Now for the velocity transformation law .

Let us follow a particle as it moves by an amount x in time t according to me and x in time t according to you. Let us agree

that velocities are defined as follows once and for all:

v

=

x t

according to me

(21)

w

=

x t

according to you

(22)

u = your velocity relative to me

(23)

In the above all s better be infinitesimals going to zero, as we are talking about instantaneous velocities and the derivative needs to be taken.

5

Suppose I see a particle with velocity v. What is the velocity w according to you? To get this we take the ratio of the equations (37

-38) that give the primed coordinate differences in terms of unprimed

ones:

w

=

x t

=

x - ut t - cu2 x

=

v-u 1 - uc2v

(24)

where in the last step we have divided top an bottom by t. It is

good to check that for small velocities (dropping the terms that go

as 1/c2) we get results agreeing with common sense.

Let us get used to going from your description to mine. Suppose

you think a particle has velocity w. What will I think its velocity is? Now we use Eqns.(19-20). Taking the ratios as before and recalling

the definition of w we get

v

=

w+u 1 + vcw2

,

(25)

which just has the sign of u reversed in Eqn. (24) as expected and

w v.

Suppose you see an object moving at w = 3c/4 and you yourself

are moving relative to me at u = 3c/4. In the old days I would

expect that object to be moving at 1.5c. However the correct answer

is

v

=

3c 4

+

3c 4

1

+

9 16

=

24 25c.

(26)

It is interesting to see that if you choose to apply this to a light pulse

seen by you (w = c) the speed I will find (for any relative velocity u

between us)

v

=

c+u 1 + u/c

=

c.

(27)

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Time dilatation. Let me carry a clock with ticks t = 0

seconds apart. If the two events are two successive ticks, the spatial

separation is x = 0 since the clock is at rest for me. You think the

time difference between ticks is

t

=

0

-

0

2

uc2

1 - u2/c2

=

1

0 - u2/c2

(28)

Thus you say my clock is slow. This is what I will hear from anyone moving relative to me.

Let us rederive this using Eqns (19-20). Thus set

t

=

0

=

t

+

cu2 x

1 - u2/c2

(29)

and note that the other equation says

x

=

0

=

x

+

ut

1 - u2/c2

(30)

which means x = -ut. (This just means you think I and my

clock are moving to the left at speed u) Feeding this into Eqn.29 to

get

0

=

t(1 - u2/c2) 1 - u2/c2

(31)

which agrees with Eqn. 28.

Length contraction Suppose you are carrying a rod of length L0. This is its rest length. Suppose you are moving relative to me at velocity u. How long will I say it is? To find its length I have two of my co-moving assistants measure the location of its front and

back ends at the same time. These two measurements, which can

be seen as two events, have x = L, the length according to me, and t = 0 since the measurements of a moving rod's ends have to

7

be simultaneous. (What happens if I measure one end, take a break

and them measure the other?)

The separation in space between the two events is the rest length

L0 of the rod according to you. Thus

L0

=

L-0

or

1 - u2/c2

L = L0 1 - u2/c2.

(32)

that is I will say your meter stick is actually 1 - u2/c2 meters long,

i.e., shortened.

Likewise you will see my meter sticks are contracted. Suppose the contraction factor 1 - u2/c2 = .5. I say your meter sticks are

.5m long. You say yours are fine and mine are .5m long. How do

you understand a guy with a shortened meter stick accusing you of

having a shortened meter stick? Should I not say your are 2m long? The answer is this : You will say that I measured the leading end of

your meter stick first and then after a delay , during which time the

stick slid by, I measured the other end. That is, you will say I did

not measure the ends simultaneously. I will insist I did, and both are

right since simultaneity is not absolute.

Order of events and causality If event 1 causes event 2, no one should see the cause come after the effect for this leaves time

for some one to prevent the cause from itself occurring. On the

other hand if the two events are not causally related, a reversed

order, while strange from our daily experience, will not lead to logical

contradictions.

According to the LT

t

=

t

-

cu2 x

(33)

1 - u2/c2

Let t = t2 - t1 > 0, so that I think 2 occurred after 1. Then we

8

can find an observer for whom t < 0 provided

u c2

x

>

t

(34)

u ct c > x

(35)

If ct > x the order can be reversed only if u/c > 1 which is

impossible. In other words if a light signal can go a distance greater

than the spatial separation between events, their order cannot be

reversed in this theory. This is because if a light signal has enough

time to interpolate between the events, event 1 could have been

the cause of event 2 and we do not want the time- order of such

(potentially) causally related pairs reversed. On the other hand if

ct < x, even a light pulse could not have made it from one event

to the other and the allows us to find frames with u/c < 1 in which

the order is reversed. This is harmless since if a light signal cannot

connect the events, neither can any other means, and the events could

not have been causally connected.

In summary:

If there is time for a light signal to go from the first event to the

second, the theory assumes they could have been causally connected

and does not allow us to find a frame with the order reversed, while

if there is not enough time for a light signal to go from the first event

to the second, the theory assumes they could not have been causally

connected and allows us to find frames with the order reversed.

Thus if I am at (0, 0), space-time is divided into three regions:

absolute future (which has ct > |x| which I can affect from the origin

using light or something slower)) absolute past (ct < |x|, which could

have affected me here and now using light or something slower ) and

the "elsewhere" region where events I say happen later (earlier) could

have happened earlier (later) for another observer moving at a speed

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