Question

Time dilation implies that when a clock moves relative to a frame S, careful measurements made by observers in S will find that the clock is running slow. This is not at all the same thing as saying that a single observer in S will see the clock running slow, and this latter statement is not always true. To understand this, remember that what we see is determined by the light as it arrives at our eyes. Consider an observer standing close beside the x axis as a clock approaches her with speed V along the axis. As the clock moves from position A to B, it will register a time

ΔtoΔt_o

, but as measured by the observer's helpers, the time between the two events ("clock at A" and "clock at B") is

Δt=γΔto.Δt = γΔt_o.

However, since B is closer to the observer than A is, the light from the clock at B will reach the observer in a shorter time than will the light from A. Therefore, the time

ΔtsecΔt_{sec}

between the observer's seeing the clock at A and seeing it at B is less that Δt. (a) Prove that

Δtsec=Δt(1β)=Δto1β/1+βΔt_{sec} = Δt(1 - β) = Δt_o √1-β/1+β

(which is less than

ΔtoΔt_o

). Prove both equalities. (b) What time will the observer see once the clock has passed her and is moving away?

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