Leading clocks lag, the twin paradox and the Doppler effect?

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I have an obvious understanding failure here, so hopefully someone can help me clear this up. Thanks for reading this obnoxious drivel.


Leading clocks lag

So, if two clocks are fixed to a the ends of a barn, and they are set off with light pulses from the midpoint, in the frame of the barn the clocks start simultaneously.

Now a rod moves to in the +x direction at speed v relative to the barn. To the observer on rod, the "leading clock" is the clock on the -x side (when the observer at the midpoint of the rod hits the origin of the barn). The pulses, which are shot when the origins coincide, will hit at different times according to the rod observer, notably, the -x side clock will be struck later from the rod observer's perspective, and so will be "lagging" behind the trailing clock (which is on the +x side of the barn).

Is that correct on how that works?



Twin paradox

If so, what about the situation with the twin paradox, where light from an approaching clock results in the appearance of the clock moving faster (is that right?).


Twin paradox with returning twin with two clocks, one in the front of the spaceship and one in the back.

So now consider the standard twin paradox scenario, and we're on the return trip, except the returning twin has two clocks, one in the front of his spaceship (nearest the earth twin) and one in the back of the spaceship. Now we have a situation where, from the earth's frame, light from the nearest clock (the "leading" clock) will actually reach earth first, won't it? So each tick would take less time to be seen, so wouldn't in this case the leading clock appear to run faster?


Clearly I still have some confusion about what clocks actually measure and what distant clocks appear to measure, but can anyone help me clear up this confusion? Thanks.
 

Answers and Replies

  • #2
Nugatory
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now consider the standard twin paradox scenario, and we're on the return trip, except the returning twin has two clocks, one in the front of his spaceship (nearest the earth twin) and one in the back of the spaceship. Now we have a situation where, from the earth's frame, light from the nearest clock (the "leading" clock) will actually reach earth first, won't it? So each tick would take less time to be seen, so wouldn't in this case the leading clock appear to run faster?
Both clocks are approaching the earth at the same speed, so will be equally Doppler-shifted; the time interval between the arrival of successive flashes from either one of the clocks will be the same, and therefore both will be ticking at the same rate in the earth frame (and in the spaceship frame, although the two frames will disagree about what that tick rate is).

The lead/lag effect is something different: at the same time in the earth frame, the two ship clocks will not display the same elapsed time, and the difference between the two readings will be proportional to the distance between them (in this case, the length of the ship). Assume for simplicity that the ship clocks were zeroed at the turnaround. In the ship frame both clocks start their return journey at the same time, but the "front clock zeroed and starts back towards earth" event and the "rear clock zeroed and starts back towards earth" events are not simultaneous in the earth frame, and this explains the constant offset in the earth frame between the two clocks.
 
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Both clocks are approaching the earth at the same speed, so will be equally Doppler-shifted; the time interval between the arrival of successive flashes from either one of the clocks will be the same, and therefore both will be ticking at the same rate in the earth frame (and in the spaceship frame, although the two frames will disagree about what that tick rate is).

The lead/lag effect is something different: at the same time in the earth frame, the two ship clocks will not display the same elapsed time, and the difference between the two readings will be proportional to the distance between them (in this case, the length of the ship). Assume for simplicity that the ship clocks were zeroed at the turnaround. In the ship frame both clocks start their return journey at the same time, but the "front clock zeroed and starts back towards earth" event and the "rear clock zeroed and starts back towards earth" events are not simultaneous in the earth frame, and this explains the constant offset in the earth frame between the two clocks.
So, if I understand this correctly, the calculated time of the clocks will be the same, but from earth, visually there will be an offset dependent upon the distance between them, and presumably the speed of the ship as well: I’m guessing the offset is something like Lv/c2 where L is the distance between the clocks and v is speed? I am pretty sure I’ve seen that formula before for something like this.

So the time dilation is the same but the synchronization is off by an amount dependent upon distance between them and speed?
 
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PeroK
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So the time dilation is the same but the synchronization is off by an amount dependent upon distance between them and speed?
Mathematics can help here. Imagine two clocks a distance ##L## apart in their rest frame. A light source half-way between the clocks sends out a pulse in both directions and illuminates the clocks simultaneously. If the clocks are synchronised in their rest frame, then the clocks read the same time when they are briefly illuminated. Let's call this reading ##T##.

Now, if this setup is moving with speed ##v## to the right in a second reference frame, then the clocks will not be illuminated at the same time. The rear clock will be illuminated first (and show time ##T##), then the leading clock will be illuminated some time later (and show time ##T##). Hence the leading clock lags in this frame. How much is this lag?

In the second frame, from length contraction, the light source is ##\frac{L}{2\gamma}## from both clocks. The pulse reaches the rear clock in a time ##t_1 = \frac{L}{2\gamma (c+v)}##, as (in this frame) the light and the rear clock are moving towards each other. The pulse reaches the front clock in a time ##t_2 = \frac{L}{2\gamma (c-v)}##.

The difference is:

##\Delta t = t_2 - t_1 = \frac{L}{2\gamma}(\frac{1}{c-v} - \frac{1}{c+v}) = \frac{L}{2\gamma}(\frac{2v}{c^2-v^2}) = \frac{L}{\gamma}(\frac{v}{c^2})\gamma^2 = \frac{\gamma Lv}{c^2}##

Hence, in the second frame, the leading clock is illuminated a time ##\frac{\gamma Lv}{c^2}## after the rear clock.

Finally, both clocks are time dilated, so when the leading clock is illuminated, the rear clock has advanced from ##T## to ##T + \frac{\Delta t}{\gamma} = T + \frac{Lv}{c^2}##.

That, then, gives us simultaneous readings for the two clocks in the second frame:

The rear clock reads ##T + \frac{Lv}{c^2}## when the leading clock reads ##T##.
 

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