Stainless' Paradox

[Lincoln]

First post:

Stopped Clock Paradox

Scenario
The stationmaster has set two synchronized stop-clocks along side the track.
The stationmaster placed the clocks exactly 8 µls apart.
Exactly half way between the clocks is a stop-button.
The stop-button functions by simultaneously sending a photon toward each stop-clock.

Einstein is approaching a station in his train traveling at 0.5c.
The stationmaster pressed the stop-button when the train was 6 µls from the first clock.
The stationmaster noted that the clocks read exactly 10 µs when he pressed the button.
At the exact moment the clocks stop, they light up flashing the time they stopped.

Questions
What time will Einstein see flashed from each clock and why?
What time will the stationmaster see flashed from each clock and why?
What time will the clocks later show as the time they actually stopped?

[Lincoln]

Second relevant post. Before we move on, was there anything else important left out?

Okay guys, let's get back to the subject.

My method of analyzing the OP scenario came in 4 stages;

Stage 1;
The station frame should report that both clocks would stop at 18 μs (assuming the stationmaster had to perceive that clocks to be at 10 μs before he pressed the button).

From Einstein’s POV, the clocks are traveling at .5c.
The first clock (the closest to Einstein) is moving away from the button flash and thus it takes longer for the light to get to that clock.

t1 = time concerning the first stop-clock photon as it leaves the button.
x1 = position of the first photon.
x2 = position of the first stop-clock
Length/time contraction factor for .5c = 0.866

x1 = t1 * c
x2 = t1 * .5c + .866 * 4*10^-6

x1 = x2: the moment the photon strikes the first clock, E's POV
t1 * c = t1 * .5c + .866 * 4*10^-6
t1(c - .5c) = .866 * 4*10^-6
t1 = .866 * 4*10^-6/ (.5c)
t1 = .866 * 8*10^-6
= 6.928*10^-6

From Einstein’s POV, that is how long it should have taken for the photon to get to the first clock.

Stage 2;
The second clock is moving toward the flash and thus it takes less time for the light to get to that clock;
t2 * (-c) = t2 * .5c - .866 * 4*10^-6
t2(-c - .5c) = .866 * -4*10^-6
t2 = .866 * -4*10^-6/ (-1.5c)
t2 = .866 * 2.667*10^-6
= 2.3096*10^-6


So from Einstein’s POV for the clocks to show the same 18, they would have to have been out of sync by;
18 – 6.928*10^-6 = 11.072 μs for clock1, and
18 – 2.3096*10^-6 = 15.690 μs for clock2

That leaves 4.6184 μs time difference between them at the button press moment.

Stage 3;
At some point from Einstein's POV;
Clock1 = 11.072
Clock2 = 15.690


So the question becomes;
For what time t1 and other clock time t1+4.6184 separated by 6.928 μls would there be an x location from which to observe the difference of 4.6184 μs?

x' = γ(x-tβ)

x' = 1.1547(x - t1/2) ;for clock1
x’ = 1.1547((x+6.928) - (t1+4.6184)/2) ;for clock2

(x-t1/2) = ((x+6.928) - ( t1+4.6184)/2)
(x) = x + 6.928 - ( t1+4.6184)/2 + t1/2
at this point x drops out meaning that any distance will do.

6.928 = ( t1+4.6184)/2 + t1/2
2*6.928 = ( t1+4.6184) + t1
2*t1 = 2*6.928 - 4.6184
t1 = 6.928 - 4.6184/2
= 4.6188

So at any location x, as long as t1 = 4.6188, the proper difference in time can be observed by Einstein at the button press.

Stage 4;
If Einstein must see clock1 at 4.6184 at button press time. And the light coming from it telling the stationmaster to press the button had to be 4.6184 - 3.464 = 1.1544 by Einstein’s POV.

So in the station’s POV clock1 had to be reading 10 when it sent the message and in Einstein’s POV that same clock1 had to be reading 1.1544.

Lorentz would suggest that Einstein’s POV would require that a clock in the station’s POV that reads 10 in that frame, must be reading 8.66 by Einstein’s POV, not 1.544.
.

[Lincoln]

In addition, I add this image as what I think you want. You insert L as 4 micro light seconds, which I take to be 4E-6 x 3E8 = 1200 meters. You have a velocity of 0.5 c.

I am having trouble seeing why the clocks would have a time of 10 microsecond when the button is pushed. This seems to be an arbitrary number. Wouldn't they start at zero? I mean, you can start with anything, but setting it to zero means there is less baggage to carry around in the equations.

But the answer to all three of your questions is 14 microseconds...they start at 10 and have 4 microseconds of transit time for the photons to get there.

I actually think you aren't asking the question you want to ask. You've not even done any frame swapping. You're asking a moving guy what the stopped guy sees. This doesn't require any relativity at all.

[flannel jesus]

But the answer to all three of your questions is 14 microseconds...

Actually I think the answer is 18, as he said before, but I could be mistaken. I thought it was 14 until he said 18, and then I rethought it, and I think 18 is correct. Takes 4 microseconds from the time the clock actually reads 10 for the guy to perceive it as 10, then he immediately presses the button (at which time the clock is already at 14) and then another 4 microseconds to get to the clock and stop it, at which time the clock reads 18. That's how I understood it anyway.

[Lincoln]

I think it depends on how you read it. The question is how the information gets to the person pressing the button.

In any event, I'm not doing anything until I get a verification that I understand the question. Disagreements in relativity often stem from (a) assuming simultaneity when you can't and (b) forgetting transit times or putting them in unnecessarily. After years of doing this, I long ago learned that step one is tie down the problem...preferably symbolically.