Relativity and The Stopped Clock Paradox

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Discussion Overview

The discussion revolves around the Stopped Clock Paradox within the context of relativity, specifically examining the synchronization of clocks from different frames of reference. Participants explore the implications of light speed constancy and how it affects the readings of clocks as perceived by different observers, including a stationmaster and Einstein on a moving train.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a scenario involving synchronized stop-clocks and a stop-button, questioning the time readings observed by Einstein and the stationmaster.
  • Another participant suggests that the stationmaster can see the clocks reading 10 μs when he presses the button, but later calculations indicate they would actually read 14 μs, leading to further confusion.
  • Some participants challenge the clarity of the scenario, asking for more details about the positions and actions of the stationmaster and Einstein to resolve ambiguities.
  • One participant proposes using the Lorentz Transform to analyze the events from different frames of reference, asserting that this method should yield consistent results without paradoxes.
  • Another participant outlines a detailed frame of reference, defining the positions of Einstein, the stationmaster, and the stop-clocks, and presents calculations regarding the timing of events as perceived by Einstein.
  • There is a discussion about the implications of light travel time and how it affects the perceived readings of the clocks when Einstein observes them.

Areas of Agreement / Disagreement

Participants express differing views on the clarity and completeness of the scenario, with some agreeing on the need for more information while others assert that the existing details are sufficient. There is no consensus on the correct interpretation of the time readings or the calculations involved, leading to ongoing debate.

Contextual Notes

Participants note that the scenario lacks certain details necessary for a complete analysis, such as the exact timing of events from each observer's perspective. The discussion also highlights the complexity of transforming events between different frames of reference using the Lorentz Transform.

  • #31
ghwellsjr said:
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?
 
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  • #32
James S Saint said:
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.
 
  • #33
James S Saint said:
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.
 
  • #34
James S Saint said:
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.
 
  • #35
James S Saint said:
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.
James S Saint said:
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).
James S Saint said:
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.
James S Saint said:
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.
James S Saint said:
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.
James S Saint said:
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.
James S Saint said:
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.
 
  • #36
PeterDonis said:
ghwellsjr said:
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.
 
  • #37
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?
 
  • #38
PeterDonis said:
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.
 
  • #39
James S Saint said:
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:
ghwellsjr said:
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.
 
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  • #40
ghwellsjr said:
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.
 
  • #41
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?
 
  • #42
James S Saint said:
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.
 
  • #43
PeterDonis said:
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.
 
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  • #44
James S Saint said:
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:
ghwellsjr said:
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.
James S Saint said:
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:
ghwellsjr said:
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.
 
  • #45
James S Saint said:
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.

James S Saint said:
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.
 
  • #46
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)
 
  • #47
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.
 
  • #48
ghwellsjr said:
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.
 
  • #49
James S Saint said:
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:
ghwellsjr said:
[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:
ghwellsjr said:
[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.
 
  • #50
James S Saint said:
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.
 
  • #51
What did I say about "distractive argumentation"?

At some point from Einstein's POV;
Clock1 = 11.072
Clock2 = 15.690So the question becomes;
For what time t1 and other clock time t1+4.6184 separated by 6.928 μls would there be an x location from which to observe the difference of 4.6184 μs?

x' = γ(x-tβ)

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

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

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

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

-----------------------------------

The distance of 6 μls was irrelevant.

But you could never convince me that because I said that the train was traveling rather than saying that the station was traveling, the distance between the station and train would be different. Again if it had been 2 rockets coming toward each other, due to symmetry, neither can claim ownership of the distance.

I am curious of one detail though. Is length contraction the same for objects moving away as it is for object moving toward? The equations would imply that it is, but sometimes that sort of thing just gets left out as presumption much like the t=0=t' concern.

{..and I meant that the original had 4 timing paths, not 4 clocks.}
 
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  • #52
James S Saint said:
The distance of 6 μls was irrelevant.

But you could never convince me that because I said that the train was traveling rather than saying that the station was traveling, the distance between the station and train would be different. Again if it had been 2 rockets coming toward each other, due to symmetry, neither can claim ownership of the distance.

I am curious of one detail though. Is length contraction the same for objects moving away as it is for object moving toward? The equations would imply that it is, but sometimes that sort of thing just gets left out as presumption much like the t=0=t' concern.

{..and I meant that the original had 4 timing paths, not 4 clocks.}

Did you read what I wrote?

What you are saying is that you don't believe in length contraction. If the train is 100 meters long in the train's frame of reference, do you think that the train is also 100 meter's in the stations frame of reference if the train is moving at .5c in that frame of reference?
 
  • #53
darkhorror said:
Did you read what I wrote?

What you are saying is that you don't believe in length contraction. If the train is 100 meters long in the train's frame of reference, do you think that the train is also 100 meter's in the stations frame of reference if the train is moving at .5c in that frame of reference?
No. That is a different issue. The length perceived in any frame within that frame will be dilated from how it is perceived from the other frame. But the distance between the moving object of one and the moving object of the other must be equal.
 
  • #54
How about this if on Earth we see a star 100 light years away from Earth in Earth's frame of reference. Then let's say there is a spaceship moving at .5c towards that star, how far is the star and the Earth away from each other in the ships frame of reference?
 
  • #55
And now I see what the simplification error I made was.

If Einstein must see clock1 at 4.6184 at button press time. And the light coming from it telling the stationmaster to press the button had to be 4.6184 - 3.464 = 1.1544.

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

Lorenz would suggest that Einstein’s POV would require that a clock in the station’s frame that reads 10 in that frame, must be reading 8.666 in Einstein’s frame, not 1.544.
 
  • #56
James S Saint said:
No. That is a different issue. The length perceived in any frame within that frame will be dilated from how it is perceived from the other frame. But the distance between the moving object of one and the moving object of the other must be equal.
Are you aware that your claim is in opposition to Special Relativity?
 
  • #57
James S Saint said:
Lorenz would suggest that Einstein’s POV would require that a clock in the station’s frame that reads 10 in that frame, must be reading 8.666 in Einstein’s frame, not 1.544.

No, Lorentz (sp.) would not suggest that. If a clock physically reads "10" at a particular event (meaning the LEDs on its face display the digits "10", or the equivalent depending on the type of clock), then that is a direct physical observable, an invariant which will be the same in *all* frames. Einstein's frame may assign a different *time coordinate* to the event at which the LEDs on the clock's face read 10, but that is a different thing. You continue to miss this point and it appears to me to be a primary source of your confusion.
 
  • #58
James S Saint said:
At some point from Einstein's POV;
Clock1 = 11.072
Clock2 = 15.690So the question becomes;
For what time t1 and other clock time t1+4.6184 separated by 6.928 μls would there be an x location from which to observe the difference of 4.6184 μs?

x' = γ(x-tβ)

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

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

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

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

Can anyone point to the exact error in that post?
 
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  • #59
PeterDonis said:
No, Lorentz (sp.) would not suggest that. If a clock physically reads "10" at a particular event (meaning the LEDs on its face display the digits "10", or the equivalent depending on the type of clock), then that is a direct physical observable, an invariant which will be the same in *all* frames. Einstein's frame may assign a different *time coordinate* to the event at which the LEDs on the clock's face read 10, but that is a different thing. You continue to miss this point and it appears to me to be a primary source of your confusion.
This is just an issue of the language we each are using. Everything is "physical".
 
  • #60
James S Saint said:
This is just an issue of the language we each are using. Everything is "physical".

You appear to be claiming that, at one and the same event, a clock can be reading "10" to one observer but something different to another. That is not an "issue of language": it is a contradiction in terms. The actual observed reading of a clock (the actual digits displayed on its face, or the equivalent) at a particular event is a direct physical observable, and must be the same for everyone.
 

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