Relativity and The Stopped Clock Paradox

[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.

 

[James S Saint]

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.}

 

[darkhorror]

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?

 

[James S Saint]

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.

 

[darkhorror]

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?

 

[James S Saint]

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.

 

[ghwellsjr]

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?

 

[PeterDonis]

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.

 

[James S Saint]

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?

 

[James S Saint]

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".

 

[PeterDonis]

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.

 

[James S Saint]

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.

"seen as reading differently" from Einstein's POV.
That is how we got the relativity of simultaneity.

and must be the same for everyone.

It can't be observed as the same by everyone. That is the whole point to relativity and especially to relativity of simultaneity.

 

[James S Saint]

I can't believe I have been spelling "sync" as "sink" and got somehow talked into spelling Lorentz as "Lorenz"... brain just getting too worn out. :(

 

[ghwellsjr]

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.

"seen as reading differently" from Einstein's POV.
That is how we got the relativity of simultaneity.

and must be the same for everyone.

It can't be observed as the same by everyone. That is the whole point to relativity and especially to relativity of simultaneity.

James, you need to learn something now that I tried to teach you way back in post #12 but you flat out rejected in post #14 and I tried to clarify for you in post #17. There is no point in continuing any discussion with you with regard to Special Relativity until you understand what coordinate time is.

There is no clock that two different people see as having different times on them. Einstein has his own set of clocks and the stationmaster has his own set of clocks. At any given moment at every possible location, there are two clocks colocated, one for Einstein and one for the stationmaster. These two clocks almost always have two different times on them. That is what Relativity of Simultaneity is related to. The Lorentz Transform allows you to pick a clock at any location at any time in one frame (say one of the stationmaster's) and see which clock and what time is on it in another frame (say Einstein's).

Please reread posts #12 and #17 and study the first two sections of Einstein's 1905 paper until you grasp how Einstein constructs a Frame of Reference. If you have any questions about this, please ask--discussing anything else truly will be "distractive argumentation".

 

[PeterDonis]

"seen as reading differently" from Einstein's POV.
That is how we got the relativity of simultaneity.

It can't be observed as the same by everyone. That is the whole point to relativity and especially to relativity of simultaneity.

See ghwellsjr's comment. The actual reading displayed by a particular clock is not "seen as reading differently"; it *is* "observed as the same by everyone". The "time from Einstein's POV" refers to a *different* clock (or set of clocks) moving on a different worldline (or worldlines).

 

[James S Saint]

Well, I have to say that this is getting interesting... in a somewhat depressing way.

I really didn't expect to have to defend fundamental relativity concerns on this forum.

I have asked 2 questions that have been ignored in favor of directing me to go learn to think like Einstein (and thus perhaps make all of the same mistakes) rather than my more direct and simple algebra. I have read many of Einstein's papers but that was years ago when I was more tolerant of having to interpret what someone is trying to say when they wrote something despite the superficial appearances. It is a bit like reading the Bible in that you have to realize what the authors at that time in their situation intended to be saying rather than the more literalist presumptions that ignore writing styles and symbolic language usage of the era.

It seems that there are 2 fundamental complaints preventing me from getting the answers to my simple questions.

PeterDonis seems to not understand that a moving observer at a different point in space will not see the same time on a clock that a non-moving observer would see. He seems to think that the time it takes for light to travel to an observer has no effect on what the observer will see. It surprises me a bit that grwellsjr seems to agree. Einstein has written papers on that issue (not that he should have needed to), most notably his 1920 paper concerning Relativity of Simultaneity (Ref: Relativity: The Special and General Theory. Chap IX. The Relativity of Simultaneity).

Granted if we can't even agree on that issue, there is little point in discussing anything concerning relativity at all. Without that one issue, there would be no relativity concerns at all.

The other complaint seems to be that if I would only workout the math in the far, far more complex and obfuscated way using "coordinate clocks", rather than my simple algebra, I would get the "right" answer. Frankly, to me math is math and the simpler the better. There is nothing magic about relativity. It doesn't need any special worshipful symbols or rituals. And if such is somehow required, then that would be a clear indication of it being fallacious. When math has to be done in only some particular way to get the "right" answer and not done any other way, it could only be because that "right" answer isn't.

All of that is just me ignoring the good probability that you each have misunderstood what Einstein was saying in the first place (especially considering the "observation" issue and the "6 μls" issue).

I have asked 2 questions;
1) can you point out any error in the relatively simple math posts.
2) on a side note, is it commonly understood that length contraction is irrespective of direction of motion?

I don't feel that this is really the thread to be arguing over fundamental relativity issues. I feel like I have proposed a calculus issue to discuss and we are arguing basic arithmetic.

 

[PeterDonis]

PeterDonis seems to not understand that a moving observer at a different point in space will not see the same time on a clock that a non-moving observer would see. He seems to think that the time it takes for light to travel to an observer has no effect on what the observer will see.

Wow. Either you are extremely confused or you are using words in a very different way than ghwellsjr and I are. I think some more precise definition of terms is in order.

In an earlier post I gave labels to a number of events in the scenario in question. Take one of them, say event A, where the stop-clock "closest to Einstein" reads 10, and emits a light signal showing that reading. That light signal then travels to event B, where the station-master sees it and presses the button.

What I have been saying is this:

(1) The statement "the stop-clock at event A reads 10" is an invariant statement; it is true for all observers.

(2) The station-master at event B, when he receives the light signal from event A, "sees" that the stop-clock read 10 when that signal was emitted--i.e., at event A.

Now consider another observer, call him Bob, who is moving relative to the station-master, but who just happens to be passing the station-master precisely at event B--that is, at precisely the instant that the station-master sees the light signal from event A. The following is then also true:

(3) Bob *also* "sees" that the stop-clock read 10 when the light signal was emitted from event A.

Note two things: first, the station-master and the other observer both "see" the *same* time on the stop-clock, even though they are in relative motion; second, both the station-master and the other observer are at a different point in space from the stop-clock, yet they agree on what the stop-clock's reading was at event A.

Now consider still *another* observer, Fred, who is moving at the same speed as Bob relative to the station-master but is separated in space from Bob, such that he passes the station-master some time *after* event B. Fred will receive the light signal from event A at some *other* event than B--so when he receives it, he will be separated in space from Bob, the station-master, *and* the stop-clock. And yet:

(4) Fred *also* "sees" that the stop-clock read 10 when the light signal was emitted from event A.

All of these statements seem to me to be obviously true. Do you agree?

1) can you point out any error in the relatively simple math posts.

I don't see the point, since I'm not even sure how you are using words yet.

2) on a side note, is it commonly understood that length contraction is irrespective of direction of motion?

If you mean "length contraction occurs in the direction of motion, but not transverse to that direction, regardless of which direction the motion is in", then yes.

 

[James S Saint]

What I have been saying is this:

(1) The statement "the stop-clock at event A reads 10" is an invariant statement; it is true for all observers.

That one statement is a "show-stopper". It is defiant of relativity and as long as you believe that statement, there is no point in going further. You are declaring an "absolute truth" frame of reference. If this was a philosophy forum, I could stand on your side and we could possibly go far. But as far as the world of relativity, that statement is heresy.

UP to now our considerations have been referred to a particular body of reference, which we have styled a “railway embankment.” We suppose a very long train traveling along the rails with the constant velocity v and in the direction indicated in Fig. 1. People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment. As a natural consequence, however, the following question arises: 1
Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.

FIG. 1.

2
When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment. But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A —> B on the traveling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M’ in the train did not possesses this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result: 3
Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event. 4
Now before the advent of the theory of relativity it had always tacitly been assumed in physics that the statement of time had an absolute significance, i.e. that it is independent of the state of motion of the body of reference. But we have just seen that this assumption is incompatible with the most natural definition of simultaneity; if we discard this assumption, then the conflict between the law of the propagation of light in vacuo and the principle of relativity (developed in Section VII) disappears.

 

[PeterDonis]

That one statement is a "show-stopper". It is defiant of relativity and as long as you believe that statement, there is no point in going further.

Wow again. Apparently it's option 1: you are extremely confused.

Read my statement again, very carefully:

The statement "the stop-clock at event A reads 10" is an invariant statement; it is true for all observers.

What am I saying? I am saying that "the stop-clock at event A reads 10" is a statement about a direct physical observable. It is a statement similar to the following statement that could be made about the scenario from Einstein that you quote: "Lightning strikes at event A". In other words, the stop-clock reading 10 is a direct observable like a lightning strike. An event, like "event A", is *defined* by what direct physical observable (a particular stop-clock reading, "10", or a lightning strike) happens there.

Note also that my statement above (and in fact all of the statements in my previous post) say absolutely *nothing* about "absolute time". Every single statement referred to direct physical observables at specific events. *None* of them were statements about how "times" at spatially separated events are related. (I never said "what time it was" according to the station-master, or Bob, or Fred, at event B, or any other event; I only said what they "saw" from the light signal that was emitted at event A.)

In summary: direct physical observables *must* be invariants, the same for all observers: otherwise we could not do physics. Saying that different observers in different states of motion can disagree about what the stop-clock reads at event A is like saying that different observers in different states of motion can disagree about whether or not lightning struck at event A.

Different observers in different states of motion *can* (and will) disagree about whether two different events are *simultaneous*; that's what Einstein's scenario is all about. So different observers in different states of motion will disagree about whether the lightning strikes at A and B are simultaneous. Similarly, different observers in different states of motion in your scenario (such as the station-master and Einstein) will disagree about whether, say, the events that I labeled "A" and "D" are simultaneous. But they will *not* disagree about what happens at those events; both will agree that stop-clock #1 reads 10 at event A, and stop-clock #2 reads 10 at event D. For them to disagree about that, once again, would be like the observer on the train disagreeing with the observer on the embankment about whether lightning struck at events A and B.

 

[James S Saint]

What am I saying? I am saying that "the stop-clock at event A reads 10" is a statement about a direct physical observable.

That doesn't exist.
No one "directly observes" anything...ever. There is no such thing. And that is what relativity is all about.

In reference to relativity of simultaneity;
Two clocks that are "in sync" can only be observed to be in sync by someone standing in one particular spot. Anyone else cannot testify by "direct observation" that those clocks are in sync. That is what his paper and that last post was about.

Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa

Clocks in sync are "simultaneous events" with each tic.

My clock1 and clock2 are ONLY simultaneous/"in sync" to the stationmaster.

 

[James S Saint]

Well, I partially take that back.
There is a zy plane intersecting the stationmaster wherein the clocks would be in sync, but not reading the same as what the stationmaster sees at the same moment that the stationmaster sees them. Anyone else in that plane would be out of sync with the stationmaster's purview.