Why Does Length Contraction Differ When Calculated at an Angle?

saadhusayn
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Hi,

My question concerns the following problem from chapter 4 of Special Relativity by A.P. French. A statement of the problem:

4-15 A flash of light is emitted at point O and is later reabsorbed at point P. In frame S, the line OP has a length l and makes an angle θ with the x axis. In a frame S' moving relative to S with a constant velocity v along the x axis:

(a) How much time τ' elapses between emission and absorption of the light?

(b) What is the spatial separation l' (l primed) between the point of emission and the point of absorption of the light?I solved part (a) correctly using the transformation equation t' = γ(t – vx / c^2), taking t to be l/c (i.e. the time the light takes to make it across OP in the S frame and x to be lcos θ. My answer is γl(1- β cos θ)/c, where β = v/c. I solved part b correctly by multiplying the answer to part a by c. I get γl(1- βcos θ).

The problem is, when I try to solve part (b) using an alternate method, I get a different answer. My reasoning is this:

The horizontal x-component of OP (lcos θ) looks contracted in the S' frame while the vertical component (lsin θ) is unaffected. So (l')^2 = (lcosθ /γ)^2 + (lsinθ)^2. So my final answer is

l' = l(1- β^2cos^2 θ)^1/2

Why is the second result different from the answer to part (b)? Thank you in advance.
 
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saadhusayn said:
The problem is, when I try to solve part (b) using an alternate method, I get a different answer. My reasoning is this:

The horizontal x-component of OP (lcos θ) looks contracted in the S' frame while the vertical component (lsin θ) is unaffected. So (l')^2 = (lcosθ /γ)^2 + (lsinθ)^2.

Did you take into account the fact that the point P moves relative to frame S' while the light is propagating from O to P?
 
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