# Special Relativity- Lorentz contraction

## Homework Statement

Q1 http://www.maths.ox.ac.uk/system/files/private/active/0/b07.2_c7.2.pdf [Broken]

## The Attempt at a Solution

How do I do the final bit that lead to the discrepancy and why does this occur? I have no idea where to begin.

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tiny-tim
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Hi sebb1e! How do I do the final bit that lead to the discrepancy and why does this occur? I have no idea where to begin.

It has to do with the difference in time when the rays start. I can't seem to get it to come out but am I along the right lines?

First ray leaves from (-r1*u-L/g) and second ray leaves from (L/g-r2*u) so apparent length if 2L/g+u(r1-r2). To find r1 and r2 I need to use Pythag theorem on the right angled triangles, (r1c, lamda, -r1u-L/g) and (r2c, lamda, L/g-r2u) I'm getting (r1-r2) to be 4u^2L/gc^2.

tiny-tim
Homework Helper
Hi sebb1e! (have a gamma: γ and a lambda: λ and try using the X2 and X2 tags just above the Reply box )

I'm not sure I follow what you're doing …

wouldn't it be easier to use the simple case where the angles are the same?

Essentially I've tried to work out the different lengths of time that light rays from the rod take to reach the observer. From the left end it leaves r1 before t=0 and from the right end r2 before t=0. r1>r2 as left end of rod is further from origin.

I've drawn two right angled triangles, one with sides (r1c, r1u+l/γ,λ) and the other (r2c, -r2u+l/γ,λ) This then means the apparent length is 2L/γ+u(r1-r2).

r1 and r2 are both positive so (r1-r2) cancels the roots and leaves 2uL/(γ(c2-u2)

This multiplied by u and then added to 2L/γ gives 2Lγ so I must have made an algebraic mistake before.

Does this make sense?

The angles are the same in O' and the question mentions angles. If correct, my method shows that the angle between the two rays implies a length of 2Lγ but I guess there's an easier way?

tiny-tim
Homework Helper
I've drawn two right angled triangles, one with sides (r1c, r1u+l/γ,λ) and the other (r2c, -r2u+l/γ,λ) This then means the apparent length is 2L/γ+u(r1-r2).

r1 and r2 are both positive so (r1-r2) cancels the roots and leaves 2uL/(γ(c2-u2)

Sorry, still not following it. Assume that the centre of the rod crosses the y-axis at t = 0 in both frames.

Take any two points A and B with (x',t') coordinates in O'.

Find their (x,t) coordinates in O.

OK, so use the Lorentz transformation to show that two points that are simultaneous in O' and 2L apart are 2Lγ apart in O but not simultaneous.

Does this mean that the light rays hitting O at t=0 will imply a length of 2lγ then?

What I've done before is draw the actual light rays coming in from the end of the rod.

To reach O at time=0 the light must have left one end of the rod at r, which is before time 0. So at that time in O, the rod was actually at -l/γ-ru, the light ray then travels a distance of rc (the hyp) so it reaches the origin at time=0. Solve for r to get the time.

Similarly the other end of the rod was actualy at l/γ-su when the light ray left it to reach the origin at time=0 after travelling sc. Solve for s to get this time

So the apparent length of the rod will be 2l/γ+u(r-s) as this is the rays O sees at t=0.

Edit: Ignore

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