# I Derivation of length contraction

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1. Jun 16, 2017

### Pushoam

Problem : To measure length of a scale
Rest frame : The frame w.r.t. which scale is at rest
Moving frame : The frame w.r.t. which the scale is moving with speed v along +ve x-axis

In rest frame ,the positions of the two ends of the scale are (measured simultaneously ) x1 and x2. So length L = x2 -x1
and t1 =t2=t

In moving frame,
the positions of the two ends of the scale are (measured simultaneously ) x'1 and x'2. So length L' = x'2 -x'1
and t'1 =t'2=t

Now,
x'2 -x'1 = γ ( x2 -x1 - v(t2 - t1) )
⇒ L' = γ L as t2 = t1

and

x2 -x1 = γ ( x'2 -x'1 - v(t'2 - t'1) )
⇒ L= γ L' as t'2 = t'1

What is wrong here?
I am asking this as I often do this mistake.

I think,

What I am doing here is I am measuring (x2 ,x1),and (x'2 ,x'1) simultaneously.
So, I am having four events.

What I should do:
I am supposed to have only two events for measuring the length and these two events should be such that these are simultaneous in the moving frame while non - simultaneous in the rest frame. Let's denote the two events as A and B.
Now, the coordinates of A and B are in rest frame [(t2,x2 ),(t1,x1)] respectively and the coordinates of A and B are in moving frame [(t'2,x'2),(t'1,x'1)] respectively and t'1 =t'2
Now,
x'2 -x'1 = γ ( x2 -x1 - v(t2 - t1) )
Now, expressing t2 and t1in prime coordinates using Lorentz transformation and doing further calculation gives L= γ L'
and
x2 -x1 = γ ( x'2 -x'1 - v(t'2 - t'1) )
⇒ L= γ L' as t'2 = t'1

Is this correct?

2. Jun 16, 2017

### Staff: Mentor

I think that the main problem is that you are thinking of the ends of the rod as two events, but they are two worldlines. So you should be thinking of transforming the worldlines rather than transforming events.

3. Jun 16, 2017

### Staff: Mentor

No, you are supposed to have two pairs of events. As Dale has already pointed out, you have two worldlines - one for each end of the rod.

To do your length measurements, you need three events: A, which is somewhere on the worldine of one end of the rod, and B and C which are events on the worldline of the other end. B and C are chosen so that A and C are simultaneous in one frame and A and B are simultaneous in the other. Now A and B give us the position of both ends of the rod at the same time in one frame, and A and C give us the position of both ends at the same time in the other frame.

4. Jun 16, 2017

### Pushoam

The book is considering only two events.
How to apply the concept of three events?

5. Jun 16, 2017

### Vitro

This is the error x'2-x'1 is not L' when t2 = t1. Because then t'2 <> t'1 so the measuring of the end positions in S' is not at the same time and that's not a correct length measurement.

6. Jun 16, 2017

### Mister T

What is your description of those two events?

7. Jun 16, 2017

### Staff: Mentor

Look at what they were doing when they said "Now as measured in $S$ these same two point events have $x$ coordinates equal to $x_1$ and $x_2$, respectively, independent of the time as measured in $S$." They are taking advantage of the fact that in the rest frame, and only the rest frame, you can skip the "at the same time" constraint.

8. Jun 17, 2017

### morrobay

Consider equating two spacetime intervals for S and S' , A and B
A at rest at x1
B moving to right at v from x0 to x1 l = x1 - x0
From A the time between events is l/v and l = 0
From B the time between events is l0/v and distance = l0
So l2/v2 = lo 2 /v2 - lo2
l = lo √1-v2