# Relativity and The Stopped Clock Paradox

1. Feb 23, 2012

### James S Saint

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

"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.

2. Feb 23, 2012

### 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.

Last edited: Feb 23, 2012
3. Feb 23, 2012

### 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?

4. Feb 23, 2012

### Staff: Mentor

How so? Show us your calculations.

5. Feb 23, 2012

### ghwellsjr

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?

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.

6. Feb 23, 2012

### James S Saint

Okay;

7. Feb 24, 2012

### 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.

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?

8. Feb 24, 2012

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

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

(14,13)s + 4 = (14,17)s = button state.
(14,13)s + 8 = (14,21) = second clock state.

I think it will do.
Verified.
The orange parts, I don't believe are relevant, but otherwise okay.

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..

9. Feb 25, 2012

### James S Saint

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

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.
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?

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.

Last edited: Feb 25, 2012
10. Feb 25, 2012

### ghwellsjr

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.
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.
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).
Yes, they are Einstein's distance coordinate and time coordinate not to be confused with the station's coordinate distance and time.
Just remember, they are Einstein's coordinate clocks not the station's coordinate clocks or the stop-clocks.
No, the stop button is pressed when Einstein sees his own coordinate clock reading 12.124.
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?
I hope I have given you the needed help but it's not my notation-it's Einstein's and Lorentz's.

11. Feb 25, 2012

### James S Saint

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.

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.

12. Feb 25, 2012

### ghwellsjr

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

13. Feb 25, 2012

pictures. that is all....

14. Feb 26, 2012

### James S Saint

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

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 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.

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.

15. Feb 26, 2012

### Staff: Mentor

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).

16. Feb 26, 2012

### James S Saint

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.

17. Feb 26, 2012

### ghwellsjr

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:
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.
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.
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 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.

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.

18. Feb 26, 2012

### James S Saint

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'.

Last edited: Feb 26, 2012
19. Feb 26, 2012

### ghwellsjr

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.

20. Feb 26, 2012

### James S Saint

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?