Relativity and The Stopped Clock Paradox

[James S Saint]

Might there be anyone here who knows the math to work this one out?

Stopped Clock Paradox Simplified

"The speed of light is the same for all observers."

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 flashing a photon toward each stop-clock, stopping them.
At the exact moment the clocks stop, they light up flashing the time they stopped.

Einstein is approaching the 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.

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?From Einstein’s POV;
The station and clocks are moving toward him.
After the button is pressed, one clock is moving toward the stop-button’s flash. The other is moving away. Thus it seems that the clocks must stop at different time readings unless they are out of sync. And though they would be out of sync as perceived by Einstein, the math doesn't seem to work out such that both of them could agree with the actual stopped reading as revealed later.

 

[ghwellsjr]

I guess we are to surmise that the stationmaster is colocated with the stop button because he can see the same time on the stop clocks before they are stopped and flash the stopped time, correct?

If so, then it will take 4 μs for the photons to reach the stop clocks at which point they will read 14 μs, and that is the answer to all your questions.

You sure threw in a lot of extraneous junk that has nothing to do with your questions.

EDIT: Opps, I didn't do this right.

When the stationmaster sees the clocks reading 10 μs, they will actually be reading 14 μs. It will then take 4 μs for the photons to get to the stop-clocks at which point they will be reading 18 μs and that is the answer to all your questions.

 

[James S Saint]

"extraneous junk"?
Really?
You did notice it was a relativity question?

You showed the obvious and trivial. Care to show the math involved in Einstein's POV?

 

[jtbell]

the math doesn't seem to work out such that both of them could agree with the actual stopped reading as revealed later.

How so? Show us your calculations.

 

[ghwellsjr]

"extraneous junk"?
Really?
You did notice it was a relativity question?

You showed the obvious and trivial. Care to show the math involved in Einstein's POV?

Your scenario is ill-defined. It doesn't have enough information to fill in the details. I'm sure it makes sense in your mind but I can't read your mind. I even had to surmise the location of the stationmaster based on what I would guess was an off-hand comment of yours. So let me ask you some questions before I can address whatever other concerns you have:

Where is the stationmaster located?

How does the stationmaster know when to press the stop button?

Can the stationmaster read the time on the stop-clocks before they are stopped?

Am I supposed to assume that the distance the train is from the first clock is in the rest frame of the station?

Where is Einstein located in the train?

Why did you ask your three questions as if they could have different answers?

Ditto what jtbell asked: what math are you using that creates a paradox?

The proper way to deal with problems like this is to define in detail all the events in one frame of reference and then transform them into another frame of reference using the Lorentz Transform. It's guaranteed to get the correct answer without any fear of a paradox. So if you can provide the information for the details of the events in the station frame, I will show you the numbers in Einstein's frame.

 

[James S Saint]

Okay;

Where is the stationmaster located?[you guessed right..at the stop-button]

How does the stationmaster know when to press the stop button?[He doesn't "know when". It was stated that he did press it as 6μls, obviously from the station's POV]

Can the stationmaster read the time on the stop-clocks before they are stopped?[Yes]

Am I supposed to assume that the distance the train is from the first clock is in the rest frame of the station?[Yes]

Where is Einstein located in the train?[On the toilet. It doesn't matter]

Why did you ask your three questions as if they could have different answers?[Why would I ask them if they didn't?]

Ditto what jtbell asked: what math are you using that creates a paradox?[I didn't want to disrupt anyone's potential answer with my meanderings since mine don't seem to work]

I will show you the numbers in Einstein's frame.[Go for it]

 

[ghwellsjr]

Since there are enumerable ways in which to select a frame and since you didn't specify one, I will define one here:

It is a frame in which the station is at rest. Along the x-axis we have the following items. (Distance units are in μls and time units are in μs.)

At x=0, we have Einstein at the start of the scenario.
At x=7, we have the location of Einstein when the stop-button is pressed, at t=14. Note that he is traveling at 0.5c.
At x=13, we have the first stop-clock. Note that it is 6 μls from where Einstein was when the stop-button is pressed.
At x=17, we have the location of the station-master and the stop-button.
At x=21, we have the second stop-clock. Note that there is 8 μls between the two stop-clocks and the stop-button is midway between them.

I believe this adheres to your description of the scenario.

So now we can define some significant events. I will use the nomenclature of [t,x].

Since Einstein is traveling at 0.5c, all of his events will have the x-coordinate equal to one-half of the t-coordinate.

[0,0] Einstein at the start of the scenario.
[0,17] Station-master at the start of the scenario.
[14,7] Einstein when the stop-button is pressed.
[14,17] Station-master when the stop-button is pressed.
[14,13] First stop-clock when the stop-button is pressed.
[14,21] Second stop-clock when the stop-button is pressed.
[18,9] Einstein when the photons reach the stop-clocks.
[18,13] First stop-clock when photon hits it.
[20.667,10.333] Einstein first sees image of first stop-clock reading 18.
[18,21] Second stop-clock when photon hits it.
[26,13] Einstein first sees image of second stop-clock reading 18.
[26,13] Einstein colocated with first stop-clock (when it would have read 26).
[42,21] Einstein colocated with second stop-clock (when it would have read 42).

I think a little explanation is in order for how I arrived at the events for which Einstein sees the two stop-clocks reading 18. It based on the fact that light travels twice as fast as Einstein, therefore Einstein will cover 1/3 of the distance while the image covers 2/3. The distance between Einstein and the first stop-clock is 13-9 = 4 so Einstein will have traveled 4/3 or 1.333 beyond 9 where he was when the photon stopped the first clock which equals 10.333. The distance between Einstein and the second stop-clock is 21-9 = 12 so Einstein will have traveled 12/3 or 4 beyond 9 where he was when the photon stopped the second clock which equals 13.

Now we use the Lorentz Transform to convert the coordinates of these events into Einstein's rest frame. First we need to calculate gamma for beta equal to one-half:

γ = 1/√(1-β²) = 1/√(1-0.5²) = 1/√(1-0.25) = 1/√(0.75) = 1/0.866 = 1.1547

Now we use the version of the Lorentz Transform that applies to units where c=1 and distances are in μls and time is in μs:

t' = γ(t-xβ)
x' = γ(x-tβ)

The first event [0,0] transforms into the same coordinates so I will show the calculation for the second event and the rest of the events I show just the final results:

t' = γ(t-xβ) = 1.1547(0-17*0.5) = 1.1547(-8.5) = -9.815
x' = γ(x-tβ) = 1.1547(17-0*0.5) = 1.1547(17) = 19.630

[0,0] Einstein at the start of the scenario.
[-9.815,19.630] Station-master at the start of the scenario.
[12.124,0] Einstein when the stop-button is pressed.
[6.351,11.547] Station-master when the stop-button is pressed.
[8.660,6.928] First stop-clock when the stop-button is pressed.
[4.041,16.166] Second stop-clock when the stop-button is pressed.
[15.588,0] Einstein when the photons reach the stop-clocks.
[13.279,4.619] First stop-clock when photon hits it.
[17.898,0] Einstein first sees image of first stop-clock reading 18.
[8.660,13.856] Second stop-clock when photon hits it.
[22.517,0] Einstein first sees image of second stop-clock reading 18.
[22.517,0] Einstein colocated with first stop-clock (when it would have read 26).
[36.373,0] Einstein colocated with second stop-clock (when it would have read 42).

You will note that all of the events corresponding to Einstein have an x-coordinate of 0 since he is at rest in this frame.

The first thing to notice is the time and distance that each photon has to travel in Einstein's rest frame. For the first stop-clock, the event of the start of the photon is [6.351,11.547] and the arrival of the photon at the first stop-clock is [13.279,4.619]. If we take the difference between the components of these two events, we get [6.928,-6.928] which shows us that the photon traveled a distance of -6.928 μls in 6.928 μs which means it traveled at c in Einstein's frame. Doing the same thing for the second stop-clock we use the same start event [6.351,11.547] but the arrival of the photon at the second stop-clock is [8.660,13.856] for a difference of [2.309,2.309], again showing that this photon traveled at c in Einstein's frame but has a shorter distance to travel and took less time, as you noted it would in your first post.

The second thing to notice is regarding the stopped times on the two stop-clocks. First off, you have to be aware that the times in the events are coordinate times so the events pertaining to the stop-clocks will not show you the times on the stop-clocks. One way to figure out from Einstein's frame what time will be on the stop-clocks, is to see what time would have been on the stop-clocks when Einstein is colocated with them and then determine Einstein's coordinate time deltas from when the photons hit the clocks and then use the time dilation factor to determine the proper time delta on those clocks. It sounds complicated and it is complicated but it will give us the same answer that we got in the station rest frame and that's all that matters.

So for the first stop-clock, we note that in Einstein's frame, he is colocated with it at [22.517,0] when it would have read 26 (the same as the coordinate time in its rest frame). The time at which the the photon hits the stop-clock in Einstein's frame is 13.279. The delta is 9.238. Dividing this by gamma we get 8. Subtracting 8 from 26 we get 18, the flashed time on the first stop-clock.

And for the second stop-clock, we note that in Einstein's frame, he is colocated with it at [36.373,0] when it would have read 42 (the same as the coordinate time in its rest frame). The time at which the the photon hits the stop-clock in Einstein's frame is 8.660. The delta is 27.713. Dividing this by gamma we get 24. Subtracting 24 from 42 we get 18, the flashed time on the second stop-clock.

Did I cover all your concerns? Do you still think there is a paradox?

 

[James S Saint]

I am going to have to take this a little at a time, especially since you seem to be adding unnecessary elements. But we'll see..

It is a frame in which the station is at rest. Along the x-axis we have the following items. (Distance units are in μls and time units are in μs.)

So it is the station's frame, okay..

At x=0, we have Einstein at the start of the scenario.
At x=7, we have the location of Einstein when the stop-button is pressed, at t=14. Note that he is traveling at 0.5c.

Okay, being the station's frame, we know that Einstein traveled 7 from x₀ during his .5c run because we know that the stationmaster saw the clocks reading 10, thus they had to actually be at 14, when he pressed the button. And we know that the train (Einstein) was at 6 from the first clock at 4 from the button when the button was pressed. The train then was at 10 from the button when it was pressed.

So, I have to assume that you have x=0 at t=0 (which wasn't intended, but I think is okay).
At t=0;
xE (x location of Einstein) = 0
At t=14
xE = 7

So at that point, the button gets pressed (t=14), thus
Button Press == (14,7)_S (station frame)

At x=13, we have the first stop-clock. Note that it is 6 μls from where Einstein was when the stop-button is pressed.

Adding 6 to the (14,7)s = (14,13)s as the state for the clock when the button is pressed.

At x=17, we have the location of the station-master and the stop-button.

(14,13)s + 4 = (14,17)s = button state.

At x=21, we have the second stop-clock.

(14,13)s + 8 = (14,21) = second clock state.

I believe this adheres to your description of the scenario.

I think it will do.

[0,0] Einstein at the start of the scenario.
[0,17] Station-master at the start of the scenario.
[14,7] Einstein when the stop-button is pressed.
[14,17] Station-master when the stop-button is pressed.
[14,13] First stop-clock when the stop-button is pressed.
[14,21] Second stop-clock when the stop-button is pressed.

Verified.

[18,9]s Einstein when the photons reach the stop-clocks.
[18,13]s First stop-clock when photon hits it.
[20.667,10.333]s Einstein first sees image of first stop-clock reading 18.
[18,21]s Second stop-clock when photon hits it.
[26,13]s Einstein first sees image of second stop-clock reading 18.
[26,13]s Einstein colocated with first stop-clock (when it would have read 26).
[42,21]s Einstein colocated with second stop-clock (when it would have read 42).

The orange parts, I don't believe are relevant, but otherwise okay.

I think a little explanation is in order for how I arrived at the events for which Einstein sees the two stop-clocks reading 18. It based on the fact that light travels twice as fast as Einstein, therefore Einstein will cover 1/3 of the distance while the image covers 2/3. The distance between Einstein and the first stop-clock is 13-9 = 4 so Einstein will have traveled 4/3 or 1.333 beyond 9 where he was when the photon stopped the first clock which equals 10.333. The distance between Einstein and the second stop-clock is 21-9 = 12 so Einstein will have traveled 12/3 or 4 beyond 9 where he was when the photon stopped the second clock which equals 13.

Where Einstein was when he saw the flashed stop-time is irrelevant. If he was 20 meters back, he would merely see the same image a little later.

I will get to the rest shortly..

 

[James S Saint]

Now we use the Lorentz Transform to convert the coordinates of these events into Einstein's rest frame. First we need to calculate gamma for beta equal to one-half:

γ = 1/√(1-β²) = 1/√(1-0.5²) = 1/√(1-0.25) = 1/√(0.75) = 1/0.866 = 1.1547

Okay, basic Lorenz giving us the dilation factor of .866 or 1.1547 (depending).

The first event [0,0] transforms into the same coordinates so I will show the calculation for the second event and the rest of the events I show just the final results:

t' = γ(t-xβ) = 1.1547(0-17*0.5) = 1.1547(-8.5) = -9.815
x' = γ(x-tβ) = 1.1547(17-0*0.5) = 1.1547(17) = 19.630

[0,0] Einstein at the start of the scenario.
[-9.815,19.630] Station-master at the start of the scenario.

Okay, so you are saying there, that “if Einstein could have seen a clock at the stop-button, that clock would have been reading [-9.815] at Einstein’s t=0 because he would see a distance of [19.630] and was traveling at .5c.”

But wait. The station sees Einstein at a distance of 17 from the button at t=0 as per prior post. But now Einstein sees the button at a different distance of 19.630? A length expansion?

Why would Einstein see the button further away than the button would see Einstein?
I think the notation conversion might need a little explaining.

[12.124,0] Einstein when the stop-button is pressed.

Okay, now we begin. The button, assumed to be pressed at 14 station time, is seen as 12.124 by Einstein due to the dilation factor 0.866 of 14. But you have changed notation now to a different perspective. The location “x” within [t,x] now refers to the object’s location from Einstein and the time, t, represents the perceived time reading at that distance?

[6.351,11.547] Station-master when the stop-button is pressed.
[8.660,6.928] First stop-clock when the stop-button is pressed.
[4.041,16.166] Second stop-clock when the stop-button is pressed.

So now with those, you are saying that the t values are what time Einstein would see on the clocks (if he could see them) due to the distances, x, from him at that time. Hmm..

The stop button is pressed when Einstein sees the stop-clocks reading 12.124.
At that moment, the stationmaster/button is 11.547 away. Where does the 6.351 come from?

Need some help here with your notation changeover.

 

[ghwellsjr]

Okay, basic Lorenz giving us the dilation factor of .866 or 1.1547 (depending).

I was calculating gamma which in this case has the value 1.1547, never .866. Time dilation is always a number greater than one. "Dilation" means getting larger.

Okay, so you are saying there, that “if Einstein could have seen a clock at the stop-button, that clock would have been reading [-9.815] at Einstein’s t=0 because he would see a distance of [19.630] and was traveling at .5c.”

No, that's not what I'm saying. We are not now talking about clocks that are at rest in the station frame but clocks that are at rest in Einstein's frame which are traveling with respect to the stop-button, or more precisely, the stop-button is traveling past a whole bunch of Einstein's coordinate clocks. The event in the station frame that I called "Station-master at the start of the scenario" is [0,17] and corresponds to a clock located at a distance of 19.630 in front of Einstein but at a much earlier time (-9.815) than what Einstein would consider the start of the scenario.

But wait. The station sees Einstein at a distance of 17 from the button at t=0 as per prior post. But now Einstein sees the button at a different distance of 19.630? A length expansion?

Why would Einstein see the button further away than the button would see Einstein?
I think the notation conversion might need a little explaining.

If you want to see how far away the stop-button is when Einstein's clocks read zero, you have to calculate how far the stop-button has moved in 9.815 μs which, at 0.5c, is 4.908 μls. So this would put it at 19.630-4.908 which is 14.722 μls. This shows length contraction if we divide 17 by 1.1547 (which equals 14.722).

Okay, now we begin. The button, assumed to be pressed at 14 station time, is seen as 12.124 by Einstein due to the dilation factor 0.866 of 14. But you have changed notation now to a different perspective. The location “x” within [t,x] now refers to the object’s location from Einstein and the time, t, represents the perceived time reading at that distance?

Yes, they are Einstein's distance coordinate and time coordinate not to be confused with the station's coordinate distance and time.

So now with those, you are saying that the t values are what time Einstein would see on the clocks (if he could see them) due to the distances, x, from him at that time. Hmm..

Just remember, they are Einstein's coordinate clocks not the station's coordinate clocks or the stop-clocks.

The stop button is pressed when Einstein sees the stop-clocks reading 12.124.

No, the stop button is pressed when Einstein sees his own coordinate clock reading 12.124.

At that moment, the stationmaster/button is 11.547 away. Where does the 6.351 come from?

That's a different moment in Einstein's frame. If you want to calculate how Einstein determines what the time was on the stop-clocks when the button was pressed, you need start with the event of Einstein being colocated with the stop-button. In the station frame this event is [34,17] and transforms to [29.445,0] in Einstein's frame. The delta between this time and the event of the button press in Einstein's frame is 29.445-6.351=23.094. Dividing this by gamma gives us 20. Now we subtract 20 from 34 and we get 14. Do you see the significance of the 6.351?

Need some help here with your notation changeover.

I hope I have given you the needed help but it's not my notation-it's Einstein's and Lorentz's.

 

[James S Saint]

Okay, so you are saying there, that “if Einstein could have seen a clock at the stop-button, that clock would have been reading [-9.815] at Einstein’s t=0 because he would see a distance of [19.630] and was traveling at .5c.”

No, that's not what I'm saying. We are not now talking about clocks that are at rest in the station frame but clocks that are at rest in Einstein's frame which are traveling with respect to the stop-button, or more precisely, the stop-button is traveling past a whole bunch of Einstein's coordinate clocks.

How is Einstein's "coordinate clocks" any different than "what Einstein would see of the other clocks"? It seems to be the same thing to me. If there is a difference, I seriously need to know what that is.

If you want to calculate how Einstein determines what the time was on the stop-clocks when the button was pressed, you need start with the event of Einstein being colocated with the stop-button. In the station frame this event is [34,17] and transforms to [29.445,0] in Einstein's frame. The delta between this time and the event of the button press in Einstein's frame is 29.445-6.351=23.094. Dividing this by gamma gives us 20. Now we subtract 20 from 34 and we get 14. Do you see the significance of the 6.351?

I had asked for the derivation of the 6.351. Your explanation of its "significance" makes it seem like merely a number injected so as to justify the chosen conclusion. How did you calculate it?

..and btw, my method for getting to the conundrum is much simpler. I am looking for anything that would indicate what my error might be.

 

[ghwellsjr]

How is Einstein's "coordinate clocks" any different than "what Einstein would see of the other clocks"? It seems to be the same thing to me. If there is a difference, I seriously need to know what that is.

Do you understand Einstein's concept of a Frame of Reference that he described in his 1905 paper introducing Special Relativity?

In the station frame, we assume that there are synchronized clocks at every possible location. These are coordinate clocks and remain fixed at coordinate locations. The time on a coordinate clock plus the x, y, z coordinates for the location define an event. I have ignored the y and z coordinates because they are zero. I described several events and gave their coordinates in the station frame.

Then I used the Lorentz Transform to see what the coordinates in Einstein's frame would be. His Frame of Reference is created in the same way as the station frame, by setting up synchronized clocks at every possible location but these clocks and locations are stationary with respect to Einstein but they are all moving with respect to the station's Frame of Reference.

All the clocks and locations are different in the two frames and that's why we need to use the Lorentz Transform to get from one to the other. We have to keep them straight. We have to make the calculations of the type I showed you to see what Einstein will actually see during the scenario but the main point is that when doing these calculations based on Einstein's frame, we will get the same answers that we get from the station frame, except that it is generally a lot more work, as should be obvious by now.

So to answer your question, Einstein's "coordinate clocks" are stationary with respect to him and all have the same time on them and all run at the same rate, so he can tell what any of them has on them by looking at the one that is next to him at location zero. On the other hand, the other clocks are moving with respect to him so they run at a different rate and don't have the same time on them so he can't know what time is on anyone of them unless he does some calculations. All he knows is the time on the clock that is nearest him as it goes flying by him at one-half the speed of light. All other clocks, he has to wait for an image of it to reach him at which point it will have a different time on it and be in a different location.

I had asked for the derivation of the 6.351. Your explanation of its "significance" makes it seem like merely a number injected so as to justify the chosen conclusion. How did you calculate it?

That number is calculated using the Lorentz Transform for the event in the station frame [14,17] for the station-master when the stop-button is pressed. Here's the calculation for the time coordinate:

t' = γ(t-xβ) = 1.1547(14-17*0.5) = 1.1547(14-8.5) = 1.1547(5.5) = 6.351

Is that clear?

..and btw, my method for getting to the conundrum is much simpler. I am looking for anything that would indicate what my error might be.

I haven't seen your method and I am not aware of a conundrum so I can't point out your error.

 

[jadrian]

pictures. that is all...

 

[James S Saint]

Einstein's "coordinate clocks" are stationary with respect to him and all have the same time on them and all run at the same rate

It would be a bit pointless to have a half dozen clocks spread out that all show the same time and run at the same rate.

We have t as the station time,
and we have t' as Einstein's time.

We have [t,x]s and,
we have [t',x']e

That number is calculated using the Lorentz Transform for the event in the station frame [14,17] for the station-master when the stop-button is pressed. Here's the calculation for the time coordinate:

t' = γ(t-xβ) = 1.1547(14-17*0.5) = 1.1547(14-8.5) = 1.1547(5.5) = 6.351

So with that, you are saying that when t=14 (the stop-time for the station POV) and x=17 (the distance to the train from the button for the station POV),

[14,17]s = [6.351,x']t for the moment and distance of stop-time at distance x'.

x' = γ(x-tβ) = 1.1547(17-14*0.5) = 1.1547(17-7) = 1.1547(10) = 11.547.
thus;
[14,17]s = [6.351,11.547]e

[6.351,11.547] Station-master when the stop-button is pressed.

Now how did the train get up to 11.547 from the button from anyone's POV at press time?

The original stipulation was that when the button was pressed, the train was only 10 from the button from the station POV. Now we are talking about a 17 and an 11.547. I accepted that you took it that the stationmaster saw "10" and thus pressed the button, which wasn't the actual stipulation, but we can work with that. So now we have the stop-clock at 14 when the button is pressed.

Now the stipulation was also that the button was pressed when the train was 6 from the first clock, or 10 from the button. If again, you assume that the stationmaster must perceive the train to be at 6 from the first clock, then it would take 10 more for him to perceive that (complicating it further) and leading to the assessment that the button was pressed when the train was 5 closer to the button, making it 5 away.

The button is pressed at t=10 and Einstein is only x=10 away (from station POV) or x=5 if you assume the perception issue with the stationmaster. Either way, I am not seeing how you are now coming up with him being 11.547 away from his POV. He cannot be perceiving himself any further way than the stationmaster perceives him.

You seem to be saying that the station sees Einstein 10 away from the button when Einstein sees the button 11.547 away from himself. Those distances can't be different.

I haven't seen your method and I am not aware of a conundrum so I can't point out your error.

If we can clear up your notation enough to see what is what, or if it gets too confusing, I'll just show you my method and we can go from there.

 

[PeterDonis]

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?

Two observations which seem to me to be critical:

(1) The readings shown by the stop-clocks must be the same for all observers, since they are stopped by definite events (photons hitting them after traveling from a common starting event) and the time they flash after they are stopped is the time of those definite events in a particular frame (the clocks' rest frame). So the answers to all three of the questions above must be the same. (Note that Einstein, who is moving relative to the stop-clocks and the stop button, will assign different *coordinate times* to the events of photons hitting each stop-clock, but that's not what the above questions ask for--they ask specifically for the time "flashed from each clock", not the coordinate time assigned to the event "photon hits clock".)

(2) Since the stop-clocks and the stop button are all at rest with respect to each other, and each stop-clock is specified to be the same distance (4 uls) from the stop button in the frame in which they are mutually at rest (this isn't explicitly stated in the OP but I can't see any other way of making sense of the OP), then the time shown by both stop-clocks must be the same, and must be 4 us after the time the stop button is pushed (again, "time" being in the frame in which the button and both clocks are mutually at rest).

 

[James S Saint]

Two observations which seem to me to be critical:

(1) The readings shown by the stop-clocks must be the same for all observers, since they are stopped by definite events (photons hitting them after traveling from a common starting event) and the time they flash after they are stopped is the time of those definite events in a particular frame (the clocks' rest frame). So the answers to all three of the questions above must be the same. (Note that Einstein, who is moving relative to the stop-clocks and the stop button, will assign different *coordinate times* to the events of photons hitting each stop-clock, but that's not what the above questions ask for--they ask specifically for the time "flashed from each clock", not the coordinate time assigned to the event "photon hits clock".)

(2) Since the stop-clocks and the stop button are all at rest with respect to each other, and each stop-clock is specified to be the same distance (4 uls) from the stop button in the frame in which they are mutually at rest (this isn't explicitly stated in the OP but I can't see any other way of making sense of the OP), then the time shown by both stop-clocks must be the same, and must be 4 us after the time the stop button is pushed (again, "time" being in the frame in which the button and both clocks are mutually at rest).

This is a relativity question.
Obviously the issues of relativity are the issue.
The question proposes that you calculate the simple case for the station frame.
And it also proposes that you calculate from the train's frame.
Of course it is expected that the times would be the same.

But I have yet to see that both calculations work out to be the same.
I haven't finished with ghwellsjr's argument yet, but it appears that he is simply reverse calculating from the presumed station's POV so as to make the answer come out right. In the long run, I'm pretty sure that fails. But thus far, it feels like a shell game of switching notations around. We are still working on it.

 

[ghwellsjr]

It would be a bit pointless to have a half dozen clocks spread out that all show the same time and run at the same rate.

You don't have to have real clocks spread out all over the place but you do have to imagine such clocks at each different location or else you aren't following Einstein's convention for setting up a Frame of Reference. Read this from near the end of section 1 of Einstein's paper:

Thus with the help of certain imaginary physical experiments we have settled what is to be understood by synchronous stationary clocks located at different places, and have evidently obtained a definition of “simultaneous,” or “synchronous,” and of “time.” The “time” of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.

We have t as the station time,
and we have t' as Einstein's time.

We have [t,x]s and,
we have [t',x']e

We also have x'=0 as Einstein's location, but what is the location of the station in the station frame? Is it at one location such as where the station-master is at x=17 or is it spread out to include the two-stop clocks between x=13 to x=21 or something else?

You should not think of the events in Einstein's frame as belonging exclusively to Einstein or associated with him. They are merely coordinates in his established Frame of Reference. So any event anywhere and at any time in the station frame also has coordinates in Einstein's frame but that isn't meant to imply that they are located where Einstein is. They can be in front of him or behind him.

So with that, you are saying that when t=14 (the stop-time for the station POV) and x=17 (the distance to the train from the button for the station POV),

[14,17]s = [6.351,x']t for the moment and distance of stop-time at distance x'.

x' = γ(x-tβ) = 1.1547(17-14*0.5) = 1.1547(17-7) = 1.1547(10) = 11.547.
thus;
[14,17]s = [6.351,11.547]e

Now how did the train get up to 11.547 from the button from anyone's POV at press time?

The train isn't at 11.547 when the button is pressed. But there is an imaginary clock located x'=11.547 μls which reads t'=6.351 μs when the button is pressed. Remember, the x' coordinate for the train (where Einstein is located) is always 0. I already showed you how to convert the coordinate time on this imaginary clock to the coordinate time in the station frame at the end of post #10.

The original stipulation was that when the button was pressed, the train was only 10 from the button from the station POV. Now we are talking about a 17 and an 11.547.

In the station frame the button is fixed at x=17 but in Einstein's frame, it is moving and so it has a changing x' coordinate but at the moment the button is pressed, x'=11.547 and t'=6.351.

I accepted that you took it that the stationmaster saw "10" and thus pressed the button, which wasn't the actual stipulation, but we can work with that. So now we have the stop-clock at 14 when the button is pressed.

I specifically asked you:

How does the stationmaster know when to press the stop button?

And you answered:

He doesn't "know when". It was stated that he did press it as 6μls, obviously from the station's POV

I took this to mean he randomly pressed the button for no reason at all but it happened to be "when the train was 6 μls from the first clock". You did also stipulate that the station-master saw 10 μs on the two stop-clocks when he pressed the button so I don't know why you are now saying it wasn't stipulated.

Now the stipulation was also that the button was pressed when the train was 6 from the first clock, or 10 from the button. If again, you assume that the stationmaster must perceive the train to be at 6 from the first clock, then it would take 10 more for him to perceive that (complicating it further) and leading to the assessment that the button was pressed when the train was 5 closer to the button, making it 5 away.

The button is pressed at t=10 and Einstein is only x=10 away (from station POV) or x=5 if you assume the perception issue with the stationmaster. Either way, I am not seeing how you are now coming up with him being 11.547 away from his POV. He cannot be perceiving himself any further way than the stationmaster perceives him.

You seem to be saying that the station sees Einstein 10 away from the button when Einstein sees the button 11.547 away from himself. Those distances can't be different.

If we can clear up your notation enough to see what is what, or if it gets too confusing, I'll just show you my method and we can go from there.

Again, this is not my notation, I didn't invent the concept of a Frame of Reference or the concept of an event or the Lorentz Transform. These are fundamental concepts to learn if you want to understand Einstein's Special Relativity.

Also again, I made no assumption with regard to the stationmaster perceiving the arrival of the train 10 μls away from him. I was just following your statement that he doesn't know when to press the button. So any of your comments regarding timing of events different from what I outlined in post #7 should not be brought up now. You agreed to them.

But you are making a mistake when you say that the button is pressed at t=10. You earlier stated "So now we have the stop-clock at 14 when the button is pressed" so why are you changing your mind now?

If you want to determine how far away Einstein was from the button when it was pressed, one way to do this is to start with the coordinate time of when Einstein arrived at the button. I worked this out near the end of post #10 and it's 29.445. Now you subtract from this the coordinate time of when the button was pressed: 29.445-12.124=17.321. Since Einstein is traveling at 0.5c we can calculate the distance as 17.321*0.5=8.660.

So when you say that the distance the station sees Einstein away from the button cannot be different than the distance Einstein sees himself from the button, you are wrong. This is another fundamental concept of Special Relativity, that distances are relative, specifically, length contracted. The distance of 10 in the station frame becomes 10/γ or 10/1.1547 which equals 8.66.

 

[James S Saint]

So when you say that the distance the station sees Einstein away from the button cannot be different than the distance Einstein sees himself from the button, you are wrong. This is another fundamental concept of Special Relativity, that distances are relative, specifically, length contracted. The distance of 10 in the station frame becomes 10/γ or 10/1.1547 which equals 8.66.

This is a critical part. The rest we can hash out.

It seems that if neither party can know that they are moving, then you have symmetry between their perceptions. There can be no preference involved. All we have (now) is when he pressed the button the train was 6 μls away from the first clock.

If point A sees point B to be 6 μls away, how can point B see point A to be a different distance away? The speed of travel is the same for both of them. One can't have preference such as to be shorter or longer regardless of any clocks. And any calculation that would apply to one frame would equally apply to the other's.

It could have equally been stated that the first stop clock was 6 μls away from Einstein when the button was pressed.

That one number is the only thing actually affixed in the scenario. Everything else is a frame dependent calculation. You have to have something with which to start that is common for both frames. It makes more sense to me to make that first clock x = 0 = x'.

 

[ghwellsjr]

This is a critical part. The rest we can hash out.

It seems that if neither party can know that they are moving, then you have symmetry between their perceptions. There can be no preference involved. All we have (now) is when he pressed the button the train was 6 μls away from the first clock.

If point A sees point B to be 6 μls away, how can point B see point A to be a different distance away? The speed of travel is the same for both of them. One can't have preference such as to be shorter or longer regardless of any clocks. And any calculation that would apply to one frame would equally apply to the other's.

It could have equally been stated that the first stop clock was 6 μls away from Einstein when the button was pressed.

That one number is the only thing actually affixed in the scenario. Everything else is a calculation.

Of course it's symmetrical but you didn't discuss the length of the train or have any other events on the train. If you had, then I could show you how in the station frame, the length of the train is length contracted and from the train frame the length of the station is contracted. If you would have established a symmetrical scenario, then you would see that everything is symmetrical.

 

[James S Saint]

Of course it's symmetrical but you didn't discuss the length of the train or have any other events on the train. If you had, then I could show you how in the station frame, the length of the train is length contracted and from the train frame the length of the station is contracted. If you would have established a symmetrical scenario, then you would see that everything is symmetrical.

I understand that the other lengths are all CONTRACTED, not lengthened. But do we agree that at the button press moment, the train is at 6 μls for both frames?

 

[ghwellsjr]

I understand that the other lengths are all CONTRACTED, not lengthened. But do we agree that at the button press moment, the train is at 6 μls for both frames?

No and I didn't say any length was lengthened. I showed that the length between Einstein and the button at the moment of the press was 10 in the station frame and 8.660 in Einstein's frame.

 

[James S Saint]

No and I didn't say any length was lengthened. I showed that the length between Einstein and the button at the moment of the press was 10 in the station frame and 8.660 in Einstein's frame.

So what are you specifying as common between them so as to make any calculation at all?
I am guessing that you have inadvertently specified that t = 0 at the same moment that t' = 0. But if you do that, the distance 6 μls will not work out properly. In effect, you will be merely converting feet to meters and back again so of course you would get the same results.

At t=0, t' ≠ 0.

Einstein's t'=0 is an irrelevant mark. He merely needs to measure the change in time from the 6 μls mark.

You don't know any history for Einstein's travel, so you can't know nor care what Einstein's clocks might read. I would suggest making them read the same as the station's at that one point of 6 μls (t=14=t') away from the first clock just to clearly see the differences as time moves forward from that point.

[6.351,11.547] Station-master when the stop-button is pressed.

That 11.547 is a problem since the stationmaster, at the same instant, sees the distance as only 10. As you have just now said, it definitely has to be only 8.66.

 

[PeterDonis]

Of course it is expected that the times would be the same.

But I have yet to see that both calculations work out to be the same.

I don't understand why a calculation from the train's frame is even relevant. Read carefully what I posted: you are asking *what time the stop-clocks actually flash once they are stopped*; in other words, you are asking about the actual digits visible on the stop-clocks' LEDs, or which points on the stop-clocks' faces the clock hands point to. That observable depends *only* on the "time" as measured in the stop-clocks' rest frame, since that's what determines the time actually displayed by the stop-clocks; the train frame is irrelevant. The *time coordinates* assigned in the train frame to the events "stop-clock 1 receives a photon and stops" and "stop-clock 2 receives a photon and stops" will be different, but that's not what you are asking for.

 

[darkhorror]

To answer your questions at the start, everyone will see those clocks reading 18 μs. Since there is nothing you need to do relativity wise, as Einstein's frame of reference never comes into play. All you need to know is that you press the button when it's reading 10 μs. And it takes 4 μs for the pulse to get to the clocks so they stop plus it takes the light 4 μs to get from the clock to the station manager so when the manager actually hits the stop button he is 4 μs behind the clocks.

I understand that the other lengths are all CONTRACTED, not lengthened. But do we agree that at the button press moment, the train is at 6 μls for both frames?

That is not possible. I assumed that the information given at the beginning was from the stations frame of reference. In the stations frame of reference you have a point on the track that is 6 μls away from the first stop-clock. This is when the button is pushed. In Einstein's frame of reference that distance is length contracted by .866 so in Einstein's frame of reference that distance is .866*6 = 5.196 μls away. Also in Einstein's frame of reference the distance between the stop button and the clocks is 4*.866 = 3.464 μls. Edit: but since in that frame those points are moving it takes differnet times.

 

[PeterDonis]

Now we use the Lorentz Transform to convert the coordinates of these events into Einstein's rest frame.

This version of the LT is only valid if the two frames have a common origin: in other words, if Einstein is just passing the station-master (assuming the station-master is at the spatial origin, x = 0, in the station frame) when both Einstein's and the station's clocks read 0. I don't think that's true of the OP's scenario, so all the LT formulas become more complicated since they have to include the offset of one frame's origin relative to the other. As far as I can see, all of the calculations in this thread need to be re-done because of this.

 

[James S Saint]

This is the fourth time I have had to rewrite this due to a "500 Internal Server Error" so I'm sure I left something out... grrr...

This version of the LT is only valid if the two frames have a common origin: in other words, if Einstein is just passing the station-master (assuming the station-master is at the spatial origin, x = 0, in the station frame) when both Einstein's and the station's clocks read 0. I don't think that's true of the OP's scenario, so all the LT formulas become more complicated since they have to include the offset of one frame's origin relative to the other. As far as I can see, all of the calculations in this thread need to be re-done because of this.

There you go. The presumption that when t=0, t'=0 corrupts the calculations.

As far as that 6 μls;
If we take it that from the station's frame we have 6 between the clock and the train and thus must have 5.196 from the train's POV, then we have presumed that it is the train that is moving. If we take it that the station is moving and that from the train's POV the distance is 5.196, then we have to say that the station's POV is going to be 5.196*.866, yielding 4.5 μls for the station.

There has to be a defined common state for both frames somewhere. In the typical scenario when [t,x] = [0,0] then [t',x'] = [0,0]. That was Lorenz' formulation presumption. But we have no idea of Einstein's clock readings and thus they cannot be used as a fact for calculation so directly using Lorenz. If we say that when the button is pressed, the train's Clock is at t'=0, we get different results.

In these scenarios we are really only concerned with the change in time readings thus we have to use something else as the common state. It shouldn't matter what Einstein's clocks read at anyone moment as long as they remain consistent from there forward. In this scenario, the only thing common that we can use is that distance and if we don't use it, we have a conundrum of not knowing who is moving such as to indicate which clocks are slower.

 

[darkhorror]

This is the fourth time I have had to rewrite this due to a "500 Internal Server Error" so I'm sure I left something out... grrr...


There you go. The presumption that when t=0, t'=0 corrupts the calculations.

As far as that 6 μls;
If we take it that from the station's frame we have 6 between the clock and the train and thus must have 5.196 from the train's POV, then we have presumed that it is the train that is moving. If we take it that the station is moving and that from the train's POV the distance is 5.196, then we have to say that the station's POV is going to be 5.196*.866, yielding 4.5 μls for the station.

From the stations frame of reference the train IS moving. Just like from the trains frame of reference the station IS moving. In the train's frame of reference it know's that the station is moving, thus it knows that when it looks at the station it's looking at a station that is length contracted. Thus to figure out the length of the station in it's own frame of reference if it's length is 5.196 in the train's. You take 5.196*1/.866 = 6 μls is the distance in the station's frame of reference.

Just as if in the stations frame of reference the train was 10 meter, then in the trains frame of reference the train would be 1/.866 = 11.547 meters.

To make it simple, if something is moving in a frame of reference then in that frame of reference that something is length contracted. If instead you look at a frame of reference where that something is not moving then it won't be length contracted in that frame.

 

[ghwellsjr]

No and I didn't say any length was lengthened. I showed that the length between Einstein and the button at the moment of the press was 10 in the station frame and 8.660 in Einstein's frame.

So what are you specifying as common between them so as to make any calculation at all?
I am guessing that you have inadvertently specified that t = 0 at the same moment that t' = 0.

It wasn't inadvertent--it was purposeful. Of course the origins of the two reference frames have to coincide in order to use the standard configuration of the Lorentz Transformation. And it's not just t=t'=0, it's also x=x'=0 and y=y'=0 and z=z'=0.

But if you do that, the distance 6 μls will not work out properly.

What's not working out properly? I have no idea what your concern is. What are you saying that I'm doing that is not working out properly and what do you say is the right thing to do to make it work out properly?

In effect, you will be merely converting feet to meters and back again so of course you would get the same results.

Please show me the calculations that I'm doing that you don't agree with.

At t=0, t' ≠ 0.

Einstein's t'=0 is an irrelevant mark. He merely needs to measure the change in time from the 6 μls mark.

You don't know any history for Einstein's travel, so you can't know nor care what Einstein's clocks might read. I would suggest making them read the same as the station's at that one point of 6 μls (t=14=t') away from the first clock just to clearly see the differences as time moves forward from that point.

Of course I don't know all these things because you won't tell me. I asked you to provide details of your scenario but you refused. When I provided some, you agreed that they were OK, but now you don't like them.

You should also recognize that the times for the events in Einstein's frame are coordinate times on those imaginary clocks that the real Einstein said we needed in order to establish a frame of reference. I never said that Einstein actually carried a clock but even if he did, I agree that what is important is differences in times on that clock. So do you agree that this is a non-issue?

[6.351,11.547] Station-master when the stop-button is pressed.

That 11.547 is a problem since the stationmaster, at the same instant, sees the distance as only 10. As you have just now said, it definitely has to be only 8.66.

If you go back and look at the listing of events in the station frame (post #7) you will see four of them for different places when the stop-button is pressed. Notice that they all share the same time coordinate of 14. This is a requirement in any Frame of Reference when you want to see how far apart different things are--you have to find events in which the time coordinates are all the same. The fact that when these four events are transformed into Einstein's frame they end up with different time coordinates is demonstrating the relativity of simultaneity--simultaneous events in one frame may not be simultaneous in another frame. That's why I had to do some extra computations to figure out what the various lengths were in Einstein's frame.

 

[ghwellsjr]

This version of the LT is only valid if the two frames have a common origin: in other words, if Einstein is just passing the station-master (assuming the station-master is at the spatial origin, x = 0, in the station frame) when both Einstein's and the station's clocks read 0. I don't think that's true of the OP's scenario, so all the LT formulas become more complicated since they have to include the offset of one frame's origin relative to the other. As far as I can see, all of the calculations in this thread need to be re-done because of this.

There you go. The presumption that when t=0, t'=0 corrupts the calculations.

What are you talking about? Peter just said that this version of the LT (he's talking about the standard configuration) is only valid if the two frames have a common origin which means t=0 and t'=0 (and the same for the x, y, and z coordinates as I mention on the previous thread). He then went on to suggest one way to do this but realized that wouldn't fit your scenario and then said there needed to be an offset between the two frame's origins. Well I accomplished the same thing by providing an offset between the station masters location and the station frame's origin. If you had provide details that you cared about when I asked for them, we could have avoided all these issues that you are now raising.

Please note that Peter is also talking about a station with a clock even though you never described the station, just a station master with no clock and two stop-clocks by the tracks.

As far as that 6 μls;
If we take it that from the station's frame we have 6 between the clock and the train and thus must have 5.196 from the train's POV, then we have presumed that it is the train that is moving. If we take it that the station is moving and that from the train's POV the distance is 5.196, then we have to say that the station's POV is going to be 5.196*.866, yielding 4.5 μls for the station.

No, there is a distance in the station frame that you said was 6 μls. That distance when measured by Einstein is 5.196. It's actual length is frame dependent. In the station frame, it's 6, in Einstein's frame it's 5.196. When Einstein measures it in either frame (or any other frame), he gets 5.196. When the station master measures it in either frame (or any other frame), he gets 6. Nobody measures it as 4.5 although there are frames of reference in which it is 4.5 but not one of the ones we have considered so far.

There has to be a defined common state for both frames somewhere. In the typical scenario when [t,x] = [0,0] then [t',x'] = [0,0]. That was Lorenz' formulation presumption. But we have no idea of Einstein's clock readings and thus they cannot be used as a fact for calculation so directly using Lorenz. If we say that when the button is pressed, the train's Clock is at t'=0, we get different results.

I think this issue has already been dealt with. Just keep in mind that the times for events that got transformed are coordinate times on imaginary clocks and if you want to give Einstein a real clock, you can add or subtract an offset to relate it to the coordinate clocks.

In these scenarios we are really only concerned with the change in time readings thus we have to use something else as the common state. It shouldn't matter what Einstein's clocks read at anyone moment as long as they remain consistent from there forward. In this scenario, the only thing common that we can use is that distance and if we don't use it, we have a conundrum of not knowing who is moving such as to indicate which clocks are slower.

We know who is moving. You told us. But we can transform your description based on the rest frame of the station into any other frame of our own choosing and it will be just as valid as the one you started with.

 

[James S Saint]

No, there is a distance in the station frame that you said was 6 μls. That distance when measured by Einstein is 5.196. It's actual length is frame dependent. In the station frame, it's 6, in Einstein's frame it's 5.196. When Einstein measures it in either frame (or any other frame), he gets 5.196. When the station master measures it in either frame (or any other frame), he gets 6. Nobody measures it as 4.5 although there are frames of reference in which it is 4.5 but not one of the ones we have considered so far.

.
.

We know who is moving. You told us. But we can transform your description based on the rest frame of the station into any other frame of our own choosing and it will be just as valid as the one you started with.

Now I am seeing a common misunderstanding.

There are only 2 frames. The stationmaster can only measure within his own. Einstein can only measure within his own.

In any scenario the travel speed is between the two frames. It never belongs to just one frame. In effect, that is the whole point of relativity. And it is by that thought, that we cannot say that it is "merely" the train moving and thus the station's time is faster. Nor could we have said that it is "merely" the station that is moving and thus the train's time will be faster.

We never know who is moving and who isn't. We choose one or the other for sake of calculations at that moment. But those calculations MUST end up with the same results even if we had picked the other as the one moving, because the actual motion is BETWEEN them.

That is very fundamental in relativity.

I believe that your calculations have translated the original into a different perspective using Lorenz and time, but then merely translated them back using Lorenz and distance, so obviously you will end up with the same answer.

If I had not said that it was a train traveling at .5c, but instead had the "two rockets" scenario and one of their clocks read 14, how would you justified that the other clock was slower or faster? The same is true with a given single distance. If one rocket sees it as 6, the other must also see it as 6.

 

[PeterDonis]

Well I accomplished the same thing by providing an offset between the station masters location and the station frame's origin.

Ah, ok, this would take care of it. So basically you are defining the origin of the station frame (t=0, x=0, y=0, z=0) as the event where Einstein is at some point *other* than the station master's location; and at that event, the station's clock reading *and* Einstein's clock reading are both taken to be zero?

 

[PeterDonis]

We never know who is moving and who isn't. We choose one or the other for sake of calculations at that moment. But those calculations MUST end up with the same results even if we had picked the other as the one moving, because the actual motion is BETWEEN them.

This is true: you should be able to calculate what the stop-clocks' displayed times will be in any frame. Doing it in the station frame is simplest because the clocks are at rest in that frame, so the time they will display corresponds to coordinate time in that frame, and that's easy to calculate from the statement of the problem.

Doing the calculation in any other frame requires you to calculate, from coordinates of events given in *that* frame, what the coordinate time in the station frame is for those same events. In particular, for the events "photon hits stop-clock 1" and "photon hits stop-clock 2". I'm still not entirely clear if that is what you are trying to calculate. If it is, then yes, you should get the same answer doing it this way as you get doing it the simple way, using just the coordinates in the station frame.

Btw, a very useful tool for these types of problems is a spacetime diagram; has anyone tried to draw one for this scenario? I have a feeling that would help clear up a lot of misunderstandings.

 

[PeterDonis]

I believe that your calculations have translated the original into a different perspective using Lorenz and time, but then merely translated them back using Lorenz and distance, so obviously you will end up with the same answer.

Well, since your original statement of the problem only gave coordinates in the station frame, how else are we to proceed? The ultimate answer we want is the coordinate time in the station frame for two specific events, since that's what the stop-clocks will display. Taking the roundabout route of transforming into Einstein's frame, seeing how things look from his perspective, then transforming back, is not necessary to calculate the answer; as I've said before, that can be done entirely within the station frame. But it does illustrate why the answer must be the same regardless of whose frame is used to compute it.

 

[darkhorror]

Now I am seeing a common misunderstanding.

There are only 2 frames. The stationmaster can only measure within his own. Einstein can only measure within his own.

In any scenario the travel speed is between the two frames. It never belongs to just one frame. In effect, that is the whole point of relativity. And it is by that thought, that we cannot say that it is "merely" the train moving and thus the station's time is faster. Nor could we have said that it is "merely" the station that is moving and thus the train's time will be faster.

We never know who is moving and who isn't. We choose one or the other for sake of calculations at that moment. But those calculations MUST end up with the same results even if we had picked the other as the one moving, because the actual motion is BETWEEN them.

But we know exactly who is moving and who isn't moving. In the train's frame of reference the station is moving. In the stations frame of reference the train is moving. Who is moving and who isn't moving just depends on the frame of reference you choose. Both are equally valid.

That is very fundamental in relativity.

I believe that your calculations have translated the original into a different perspective using Lorenz and time, but then merely translated them back using Lorenz and distance, so obviously you will end up with the same answer.

If I had not said that it was a train traveling at .5c, but instead had the "two rockets" scenario and one of their clocks read 14, how would you justified that the other clock was slower or faster? The same is true with a given single distance. If one rocket sees it as 6, the other must also see it as 6.

No that is not right at all, Both rockets would see the other's clock moving slower than their own. Just as the train sees the stations clock moving slower than it's own, and the station see's the train's clock moving slower than it's own.

Now for the distance it's easy to calculate. If one rocket sees a distance of 6 between two points that are not moving in it's frame of reference, then the other rocket sees that length contracted to 5.176.

The problem you are having is that you then you want to say oh well it's length is 5.176 in my frame there for in your frame it's length contracted so it's length is 4.482. The problem with this is that the length you are measuring is moving in your frame of reference. Thus you know that if you switch to the frame of reference where the points aren't moving then you take 5.176 * 1/.866 which gives you 6.

Here is another way to look at this.
Take 2 points, let's name them A and B.
Two frames of reference named 1 and 2.
A and B are stationary with respect to each other.
In frame 1 A and B are not moving.
In frame 2 A and B are moving at .5c.
In frame 2 A and B are 1 light year away from each other.
How far are they away from each other in frame of reference 1?

Well since A and B are moving in frame 2 and are not moving in frame 1. They are 1*(1/.866) away or 1.1547 light years away in frame 1.

Now let's take that 1.1547 distance in frame 1, and you want to know how far they are away in frame 2. Since you know A and B are not moving in frame 1, and are moving at .5c in frame 2. This means you take 1.1547 * .866 which means you get 1 light year in frame 2. Which is what we had started with. So we know that in frame 1 A and B are 1.1547 light years away, and in frame 2 A and B are 1 light year away. They are both true, and both equally valid.

 

[ghwellsjr]

Now I am seeing a common misunderstanding.

There are only 2 frames.

There are an infinite number of frames, all equally valid for describing, illustrating, analyzing, etc any given scenario. You have arbitrarily chosen to focus on two of them and you didn't even specify them, you left that up to me and when I did it, you first accepted them but now you are rejecting them.

The stationmaster can only measure within his own. Einstein can only measure within his own.

No, the stationmaster does not need a frame in which to make measurements, neither does Einstein. We use frames to help us analyze in thought problems what the stationmaster would measure and what Einstein would measure if they were actually carrying out the scenario. But we only need one frame to do this and it doesn't matter which one it is, except of course that the calculations can be very easy using one frame and very difficult using a different frame, as evidenced by my posts on this thread. Since you have described the scenario in a general way from the POV in which the station is at rest, that is the easiest one in which to do a rigorous specification of a frame and events in it and then to do the calculations, because we don't need to do any transformations. We can analyze exactly what the stationmaster would measure (without him using any concept of a frame--he just measures) and exactly what Einstein would measure (without him using any concept of a frame--he just measures).

In any scenario the travel speed is between the two frames.

Not if we don't want it to be.

We start with any frame which we consider to be at rest and in which we describe all observers, objects, clocks, whatever. They can be located anywhere and doing anything in terms of moving and/or accelerating. In the first section of Einstein's 1905 paper he says (emphasis his):

If a material point is at rest relatively to this system of co-ordinates, its position can be defined relatively thereto by the employment of rigid standards of measurement and the methods of Euclidean geometry, and can be expressed in Cartesian co-ordinates.

If we wish to describe the motion of a material point, we give the values of its co-ordinates as functions of the time.

I don't know why you think a single frame doesn't allow motion of objects or observers in it.

We can then pick another frame and if we choose to limit ourselves to the standard convention, we assign an arbitrary speed along the x-axis but we fix the directions of the three axes to be the same and the origins (all four coordinates equal zero) of the two frames to be coincident. We don't have to pick a speed that corresponds to the travel speed of an observer or an object in our original frame but there are certain advantages to doing that, mainly that we don't have to then do another step of calculation to determine the Proper Times on the clocks that were moving at a constant speed along the x-axis in the first frame but are stationary in the second frame.

It never belongs to just one frame.

Nothing belongs to just one frame, not even an observer at rest in the frame. Everything is in all frames...all of the infinity of equally valid frames.

In effect, that is the whole point of relativity. And it is by that thought, that we cannot say that it is "merely" the train moving and thus the station's time is faster. Nor could we have said that it is "merely" the station that is moving and thus the train's time will be faster.

We never know who is moving and who isn't. We choose one or the other for sake of calculations at that moment. But those calculations MUST end up with the same results even if we had picked the other as the one moving, because the actual motion is BETWEEN them.

That is very fundamental in relativity.

What you are describing is the situation prior to Einstein's Special Relativity. All we could know is what each observer could see and measure. But no observer can tell what time is on another moving observer's clock, only what he can see after the image of that clock is transmitted to him at the speed of light. Einstein's Special Relativity allows us to arbitrarily assign coordinate times to distant locations and define what time is on another moving observer's clock at any particular coordinate time. But these coordinate times are dependent on the particular frame we decide to use and will change when we pick a different one in motion to the first one.

I believe that your calculations have translated the original into a different perspective using Lorenz and time, but then merely translated them back using Lorenz and distance, so obviously you will end up with the same answer.

Of course we will get the same answer when we are calculating what each observer sees and measures or what a clock displays when a photon hits it and things like that. But the arbitrary coordinate times and distances are just that--arbitrary and can be different in each frame.

If I had not said that it was a train traveling at .5c, but instead had the "two rockets" scenario and one of their clocks read 14, how would you justified that the other clock was slower or faster? The same is true with a given single distance. If one rocket sees it as 6, the other must also see it as 6.

If you want to talk about two identical rockets traveling toward or away from each other, then we have an exactly symmetrical situation. Each one sees and measures exactly what the other one sees and measures and this is independent of any frame we use to describe and analyze the situation. Each one sees the other ones clock running slower and the other ones rocket as being shorter. But without selecting a frame, neither one can say (nor can we say) what time is on the other ones clock corresponding to any time on their own. Nor can they say how far away either one is at any given time on their own clock. We need SR for that (or some other equivalent theory). And depending on which frame you select, you can get different answers for those coordinates,.especially those that are remote from each observer.

So, if you choose the rest frame for the first rocket, then his clock is defined to be running at the same rate as the coordinate time for the frame and his rocket is defined to be the same length as the coordinate distance for the frame but for the second rocket, his clock is calculated to be running at a slower rate than the coordinate time and his rocket is calculated to be shorter than the coordinate distance. The factor gamma which is based on their relative speed determines that amount of change and since both rockets measure the other one to be going the same speed (but opposite directions), then when we switch to the rest frame of the other rocket, all the calculations switch between the two rockets.

 

[ghwellsjr]

Well I accomplished the same thing by providing an offset between the station masters location and the station frame's origin.

Ah, ok, this would take care of it. So basically you are defining the origin of the station frame (t=0, x=0, y=0, z=0) as the event where Einstein is at some point *other* than the station master's location; and at that event, the station's clock reading *and* Einstein's clock reading are both taken to be zero?

Sort of, but please note, James did not specify a station clock, especially one located with the station master. I did it the way I did so that the coordinate clocks for the station frame would have the same time on them as the two stop-clocks located remotely from the station master.

 

[James S Saint]

This is how I had started analyzing the scenario..

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 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 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.

Does that make sense so far?

 

[ghwellsjr]

Well, since your original statement of the problem only gave coordinates in the station frame, how else are we to proceed? The ultimate answer we want is the coordinate time in the station frame for two specific events, since that's what the stop-clocks will display. Taking the roundabout route of transforming into Einstein's frame, seeing how things look from his perspective, then transforming back, is not necessary to calculate the answer; as I've said before, that can be done entirely within the station frame. But it does illustrate why the answer must be the same regardless of whose frame is used to compute it.

I gave James the same answer in post #2 that you have repeated several times, including pointing out that he included "a lot of extraneous junk" but when he chided me for showing the "obvious and trivial", I concluded that what he really wanted was an explanation of how "The speed of light is the same for all observers" as he opened up his original post, and how the two photons could both be traveling at c in both the station frame and Einstein's frame even though the distances were different, so that what I proceeded to do. I also showed him how to calculate the times on the two stop-clocks in Einstein's rest frame. So now I'm trying to help him understand some of the fundamentals of Special Relativity so that he can understand my answers.

 

[ghwellsjr]

This is how I had started analyzing the scenario..

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 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 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.

Does that make sense so far?

Well you got the same answer I calculated for you in post #7, so without scrutinizing your calculation, I will assume it makes sense so far:

For the first stop-clock, the event of the start of the photon is [6.351,11.547] and the arrival of the photon at the first stop-clock is [13.279,4.619]. If we take the difference between the components of these two events, we get [6.928,-6.928] which shows us that the photon traveled a distance of -6.928 μls in 6.928 μs which means it traveled at c in Einstein's frame.

 

[PeterDonis]

I gave James the same answer in post #2 that you have repeated several times...

Understood; I'm not sure the point has been fully taken yet.

 

[James S Saint]

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

That leaves 4.6984 μs time difference between them at the button press moment.

Still making sense?

 

[PeterDonis]

So from Einstein’s POV for the clocks to show the same 18, they would have to have been out of sink

Wrong. As I said before, what you have calculated is irrelevant to the readings actually *displayed* by the stop-clocks. They display time according to their own state of motion, not Einstein's state of motion. In order to determine, from Einstein's point of view, what the stop-clocks will actually *display*, you have to take what you have calculated and transform it *back* to the station frame, just as ghwellsjr did in previous posts. And if you do that, of course, you get the same answer, 18.

 

[James S Saint]

Wrong. As I said before, what you have calculated is irrelevant to the readings actually *displayed* by the stop-clocks. They display time according to their own state of motion, not Einstein's state of motion.

We are temptorarily accepting that the clocks will end up displaying 18. And yes, that means that by their own time, they will have to be at 18 when the photon strikes.

But Einstein can tell that it is going to take only 2.3096*10^-6 for the light to get to the stop-clock, so the STOP-CLOCK in EINSTEIN's time, must have been 15.690 when the button was pressed, else it cannot end up at 18 in Einstein's time. If it isn't at 18 in Einstein's time when it stops, how could it become 18 later?

I am saying that in EINSTEIN's POV, the clocks would have to be seen out of sink. If that was a confusion.
Basic Lorenz agrees with that.

 

[ghwellsjr]

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

Again, you got the same answer I got so I'll assume you did the correct calculation:

Doing the same thing for the second stop-clock we use the same start event [6.351,11.547] but the arrival of the photon at the second stop-clock is [8.660,13.856] for a difference of [2.309,2.309], again showing that this photon traveled at c in Einstein's frame but has a shorter distance to travel and took less time, as you noted it would in your first post.

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

That leaves 4.6984 μs time difference between them at the button press moment.

Still making sense?

I gave you a way to calculate how Einstein arrives at the correct answer of 18 in post #7:

The second thing to notice is regarding the stopped times on the two stop-clocks. First off, you have to be aware that the times in the events are coordinate times so the events pertaining to the stop-clocks will not show you the times on the stop-clocks. One way to figure out from Einstein's frame what time will be on the stop-clocks, is to see what time would have been on the stop-clocks when Einstein is colocated with them and then determine Einstein's coordinate time deltas from when the photons hit the clocks and then use the time dilation factor to determine the proper time delta on those clocks. It sounds complicated and it is complicated but it will give us the same answer that we got in the station rest frame and that's all that matters.

So for the first stop-clock, we note that in Einstein's frame, he is colocated with it at [22.517,0] when it would have read 26 (the same as the coordinate time in its rest frame). The time at which the the photon hits the stop-clock in Einstein's frame is 13.279. The delta is 9.238. Dividing this by gamma we get 8. Subtracting 8 from 26 we get 18, the flashed time on the first stop-clock.

And for the second stop-clock, we note that in Einstein's frame, he is colocated with it at [36.373,0] when it would have read 42 (the same as the coordinate time in its rest frame). The time at which the the photon hits the stop-clock in Einstein's frame is 8.660. The delta is 27.713. Dividing this by gamma we get 24. Subtracting 24 from 42 we get 18, the flashed time on the second stop-clock.

This is obtained without reverting back to the station frame but it does require Einstein to be colocated with the real clocks in order to provide the correct calculated offsets in a meaningful way, that is, without regard to the arbitrary frame of reference selected.

Please note: there are other ways for Einstein to make the same calculation, I just came up with this one because it seems more directly associated with the actual clocks.

 

[PeterDonis]

But Einstein can tell that it is going to take only 2.3096*10^-6 for the light to get to the stop-clock, so the STOP-CLOCK in EINSTEIN's time, must have been 15.690 when the button was pressed, else it cannot end up at 18 in Einstein's time. If it isn't at 18 in Einstein's time when it stops, how could it become 18 later?

You are still missing the point. The stop-clocks do *not* read "Einstein's time". The stop-clocks are physical objects, and the process that determines what they read is a physical process occurring along a well-defined worldline in spacetime. The clocks' readings are determined by elapsed proper time along the worldlines they follow; the coordinate time assigned to them by someone in relative motion to them, like Einstein, is irrelevant.

Let me suggest another way of looking at it that may make the above clearer. Your original scenario stipulates that the station-master presses the button when he sees the stop-clocks read 10. That means that a light signal must have been emitted from each stop-clock, when they each read 10, and both of those light signals must have arrived at the station-master at the same event (the same instant according to the station-master--he sees them simultaneously).

Pick either stop-clock--for definiteness, we'll use the one closest to Einstein. We can define three definite events from the above:

Event A: The stop-clock emits a light signal showing a reading of 10.

Event B: The station-master receives the light signal from event A and presses the button, emitting a light signal towards the stop-clock.

Event C: The stop-clock receives the light signal from event B, stops, and flashes its final reading.

Now: what determines the final reading of the stop-clock at event C? The answer according to relativity is: the elapsed proper time along the stop-clock's worldline from event A to event C. So the only thing remaining is to show that this elapsed proper time comes out the same when calculated in both the station frame and Einstein's frame.

In the station frame, of course, the calculation is simple: each light signal (A to B and B to C) takes 4 us, so the total time elapsed is 8. Since the stop-clock is at rest in this frame, time elapsed in the station frame is identical to proper time elapsed along the stop-clock's worldline. Hence, the final reading will be 10 + 8 = 18.

In Einstein's frame, you've already done the calculations; it's just a matter of putting them together. You calculated the elapsed time from B to C in this frame in post #37; I will write it (for reasons that will appear in a moment) as 6/.866. You also calculated, in effect, the elapsed time in Einstein's frame from A to B in post #41; the light signal from A to B takes the same amount of time in that frame as the photon whose time you calculated in that post. (If that's not obvious, I can go into detail, but it should be obvious.) I will write that time as 2/.866.

So the total elapsed time between events A and C, in Einstein's frame, is 6/.866 + 2/.866 = 8/.866. But since the stop-clock is moving relative to Einstein, if we want the elapsed *proper time* along the stop-clock's worldline between A and C, we have to apply the time dilation factor of .866. So the elapsed proper time is 8/.866 * .866 = 8. So again, Einstein predicts that the stop-clock will read 10 + 8 = 18 when it stops at event C.

I am saying that in EINSTEIN's POV, the clocks would have to be seen out of sink. If that was a confusion.

You are correct, the two stop-clocks are indeed "out of sync" from Einstein's POV. However, they also receive the photons from the station-master's button press at different times, which exactly compensates for the fact that they are out of sync; the stop-clock closer to Einstein receives the photon (at event C) *later* than the stop-clock further from him by exactly the amount of time, in Einstein's frame, required to allow it to "catch up" to the other clock (which stops, of course, when it receives the photon).

To see this more explicitly, let's now label events on the second stop-clock's worldline. It emits a light signal when it reads 10, at event D, which the station-master receives at event B, the same instant he sees the signal from event A (and he then presses the button). The light signal from event B reaches the second stop-clock at event E. By symmetry, it should be obvious that the elapsed time, in Einstein's frame, from D to B is the same as that from B to C; and the elapsed time in Einstein's frame from B to E is the same as that from A to B. So the total elapsed time from D to E is the same as from A to C, meaning that the elapsed time from each stop-clock reading 10 to each stop-clock receiving the photon from the button press and stopping is the *same*. (And, of course, to convert that elapsed time in Einstein's frame to elapsed proper time for the stop-clock, we have to apply the time dilation factor, as we did above.)

However, it is also true that, in Einstein's frame, event D occurs *before* event A, and event E occurs *before* event B. This is what you are referring to as the clocks being "out of sync". So at any given time in Einstein's frame, *before* the time of event E, the further stop-clock (D to E) will be running *ahead* of the nearer stop-clock (A to B). This must be the case, because the further clock reads 10 at event D, but the nearer stop clock reads 10 at event A, and D is before A in Einstein's frame.

But at event E, the further clock *stops*; so its reading is frozen at 18. At that time, in Einstein's frame, the nearer clock reads something *less* than 18; but it hasn't received its photon yet. By the time it receives its photon from the button-press, and stops, it has just caught up and reads 18, just as predicted.

As I said before, all this would be a lot clearer if you drew a spacetime diagram. If you haven't done that, I recommend doing so.

 

[James S Saint]

Actually I was somewhat hoping that someone would already see the problem. But a little while ago when I started to show the rest of the situation that led to the paradox, I noticed an arithmetic error.
"That leaves 4.6984 μs time difference between them at the button press moment."

That 4.6984 should have been 4.6184.

That error ripples through to create a problem. But once corrected, my method for analyzing this scenario finally works out (and is a lot simpler than others provided).

I think that when I "simplified" the "Stopped Clock Paradox", I messed it up. So I will have to rethink from the original and see if that kind of error is involved in the original "non-simplified".

So thanks for your help guys. (although I still have several issues with some things stated in this thread)

 

[ghwellsjr]

You're welcome.

Does this mean that you no longer believe there is a paradox?

You can still ask for clarification of whatever issues you still have with any concepts stated in this thread.

 

[James S Saint]

You're welcome.

Does this mean that you no longer believe there is a paradox?

You can still ask for clarification of whatever issues you still have with any concepts stated in this thread.

Well, I am pretty sure the original is still a paradox, but I'll have to comb through it when I get time. I didn't like how complicated the original was. It involves 4 clocks. I just mistakenly thought I found a way of stating the same issue in a simpler form...ohhh..wellll..

That issue of the 6 being 6 from both POVs is something that needs to be understood, but I don't want to get into distractive argumentation right now. There were other things, but nothing important.

Thanks again.

 

[ghwellsjr]

That issue of the 6 being 6 from both POVs is something that needs to be understood, but I don't want to get into distractive argumentation right now.

I get the impression that you believe that the only difference between the stationmaster's POV and Einstein's POV is the direction from which they are measuring that distance of 6 μls and so it has to be the same, correct? If so, then you are ignoring the fact that Einstein is traveling at 0.5c with respect to that distance and the stationmaster is not. Let's use Einstein's speed and the time it takes him to traverse that distance as a way to measure that distance from the two POV's.

The events that describes Einstein's motion from the stationmaster's POV are:

[14,7] Einstein when the stop-button is pressed.
[14,13] First stop-clock when the stop-button is pressed.
[26,13] Einstein colocated with first stop-clock (when it would have read 26).

First, we observe that the distance we are talking about is the distance between where Einstein was when the stop-button was pressed and the first stop-clock. Einstein was at the x-coordinate of 7 μls and the stop-clock was at 13 μls for a difference of 6 μls.

But we also note that it took Einstein from the t-coordinate 14 μs to 26 μs to traverse that distance which means it took him 12 μs from the POV of the stationmaster. Since he is traveling at 0.5c, that means the distance is 6 μls.

Now we look at those same events transformed into Einstein's POV:

[12.124,0] Einstein when the stop-button is pressed.
[8.660,6.928] First stop-clock when the stop-button is pressed.
[22.517,0] Einstein colocated with first stop-clock (when it would have read 26).

We note that from Einstein's POV, the time it took him to traverse the distance is the difference between 12.124 μs and 22.517 μs which is 10.393 μs and at a speed of 0.5c, Einstein will measure the distance as 5.196 μls.

The stationmaster will also arrive at this number if he understands relativity because he knows that if Einstein is traveling at 0.5c then his distances will be divided by gamma. Since gamma is 1.1547, he can just divide 6 μls by 1.1547 and get 5.196 μls.

Now I realize that one of the confusing issues for you was the fact that the event corresponding to the first stop-clock when the stop-button was pressed had an x-coordinate of 6.928 μls which is 6 μls times gamma, not divided. But that is because if you want to determine distances in any FoR, you must do it with the time coordinates for the two events equal and in this case they are not equal. You must find an event in the stationmaster's FoR that transforms into an event in Einstein's FoR in which the times are the same for Einstein. It turns out that [17,13] in the stationmaster's FoR transforms into [12.124,5.196] in Einstein's FoR. Note that the times are the same, 12.124, so the distance 5.196 is the correct distance that the 6 transforms into.

There are so many ways to arrive at the same calculation that in Einstein's POV, the 6 μls is 5.196 μls. I would hate to see you spend any time trying to discover a paradox in relativity because it is a waste of your time but if you want to discuss another potential paradox, be my guest.

 

[darkhorror]

Well, I am pretty sure the original is still a paradox, but I'll have to comb through it when I get time. I didn't like how complicated the original was. It involves 4 clocks. I just mistakenly thought I found a way of stating the same issue in a simpler form...ohhh..wellll..

That issue of the 6 being 6 from both POVs is something that needs to be understood, but I don't want to get into distractive argumentation right now. There were other things, but nothing important.

Thanks again.

I already went over the distance multiple times, just go back up and you can see what I said and where you were having a problem.