What's wrong with this so-called paradox?

[pavelcherepan]

I don't think that's correct. I think by the equivalence principle you could have the clocks in a uniform gravitational field but at different heights (ie. different gravitational potentials) corresponding to their different x positions on the train. Therefore if you do set up the train so that it accelerates uniformly for some time, you still have the clocks ticking at different rates according to GR and can end up out of sync depending on the rest of the setup. As studiot mentioned, this might not describe the original setup.

 

 

Edit: Just thinking to myself... I understand the instinct to put the clocks at the same height around a massive body in order to get the same field strength, but that's the wrong way to think about it. A uniformly accelerating train is like a uniform gravitational field. It will still take effort to "climb" your way from the back of the accelerating train to the front. The effort is proportional to distance climbed. Equivalently it takes effort proportional to distance climbed, to climb out of a gravity well with a uniform field. The location on the train and in the equivalent gravitational field matters.

 

Thanks for your reply, that's some interesting info. Still, I'm confused about your explanation. You're saying that it's incorrect to picture all three points of the train as being at 'same height' in a uniform gravitational field because it takes energy to move from the back of the train to the front. I totally agree with that. And yet, let's suppose I have two points A and B on the same height in a uniform gravitational field but in different spatial locations. If you want to move from point A to B you'd still require additional energy to counter the gravitational attraction and the amount of this extra energy will be (like in your explanation) proportional to the distance traveled.

 

If all parts of the train start accelerating at the exact same time and at an exact same rate than I really see no reason why clocks in the back of the train would be going any differently from clocks in the middle and in the front.

[xyzt]

 

 

 

If all parts of the train start accelerating at the exact same time and at an exact same rate than I really see no reason why clocks in the back of the train would be going any differently from clocks in the middle and in the front.

if a light signal is sent from the back of the train to the front , in the direction of travel, it will be detected as redshifted because it was sent when the source was traveling at speed [math]v[/math] and it was detected when the receiver was traveling at speed [math]v'>v[/math] AWAY from the signal wavefront.

 

conversely, if a light signal is sent from the front of the train to the back , against the direction of travel, it will be detected as blueshifted because it was sent when the source was traveling at speed [math]v[/math] and it was detected when the receiver was traveling at speed [math]v'>v[/math] TOWARDS from the signal wavefront.

 

THIS IS WHY

Edited March 9, 2015 by xyzt

[pavelcherepan]

if a light signal is sent from the back of the train to the front , in the direction of travel, it will be detected as redshifted because it was sent when the source was traveling at speed [math]v[/math] and it was detected when the receiver was traveling at speed [math]v'>v[/math] AWAY from the signal wavefront.

 

conversely, if a light signal is sent from the front of the train to the back , against the direction of travel, it will be detected as blueshifted because it was sent when the source was traveling at speed [math]v[/math] and it was detected when the receiver was traveling at speed [math]v'>v[/math] TOWARDS from the signal wavefront.

 

THIS IS WHY

 

 

Ok, I get it. In fact I mentioned it in the previous post:

 

 

 

6) If the middle observer had a spectrometer on him/her and analyzed light from both flashes he/she would see that the light coming from the front of the train has been shifted towards blue side of spectrum and the light coming from the back is red-shifted due to him/her moving towards or away from the respective source.

 

But still that doesn't explain that clocks in different parts of the train would be showing a different time. If the flashes at both sides occur simultaneously, they would arrive in the middle at the same time and red/blue shift simply accounts for the observer moving to or from the source, right?

[xyzt]

 

 

 

 

 

But still that doesn't explain that clocks in different parts of the train would be showing a different time. If the flashes at both sides occur simultaneously, they would arrive in the middle at the same time and red/blue shift simply accounts for the observer moving to or from the source, right?

this is a DIFFERENT issue, it was explained to you that , in relativity, simultaneity is frame dependent

[pavelcherepan]

this is a DIFFERENT issue, it was explained to you that , in relativity, simultaneity is frame dependent

 

Yes, I agree again. But aren't all three clocks on the train in the same reference frame?

[xyzt]

 

Yes, I agree again. But aren't all three clocks on the train in the same reference frame?

yes, but they have different x coordinates. so, according to sr, they will be synchronized only in the frame of the train. in all other frames, they will not.

[pavelcherepan]

yes, but they have different x coordinates. so, according to sr, they will be synchronized only in the frame of the train. in all other frames, they will not.

 

Oh, yeah, definitely. I completely agree, I just forgot to correct my earlier post, where I said that assistants will see flashes simultaneously. Now that I've thought about it, it is obvious - we have already determined that in the FoR of train flashes will happen simultaneously and, thus the middle clock will flash, but since the front of the train is moving away from assistants and the back of the train - towards them, then in order for both rays to come to the middle at the same time in assistants' FoR the clock in the front must flash first and the clock in the back will flash later.

Edited March 9, 2015 by pavelcherepan

[md65536]

And yet, let's suppose I have two points A and B on the same height in a uniform gravitational field but in different spatial locations. If you want to move from point A to B you'd still require additional energy to counter the gravitational attraction and the amount of this extra energy will be (like in your explanation) proportional to the distance traveled.

This is no longer completely on-topic but is a misunderstanding of a related idea.

 

No it doesn't take energy proportional to distance, to move sideways in a gravitational field (or along the width of the accelerating train). Suppose you roll a heavy sphere along a smooth frictionless horizontal surface. It will take energy to get it moving, but then the ball will keep rolling due to inertia (not requiring extra energy to keep moving). The energy put into its momentum is kinetic energy. Meanwhile if you lift the ball, the energy goes into gravitational potential energy. It won't keep moving. Each meter you lift it requires additional energy. If you roll a ball up an incline (or forward in an accelerating train), each meter moved will require additional energy.

 

If all parts of the train start accelerating at the exact same time and at an exact same rate than I really see no reason why clocks in the back of the train would be going any differently from clocks in the middle and in the front.

They don't, in the inertial frame where "exact same time" is determined. From the track's perspective, the clocks could be coordinated as you say, and remain in sync. This is not the case being discussed in other recent posts. The accelerating train is not an inertial frame, and "at the exact same time" will not mean the same thing for the differently located accelerating observers (they won't agree on simultaneity of events). After the train has completed accelerating, it has an inertial frame, and the differently located observers would from then agree on simultaneity of events, and could arrange for their clocks to be synchronized.

Edited March 9, 2015 by md65536

[whiskers]

Remember the

 

 

If all parts of the train start accelerating at the exact same time and at an exact same rate than I really see no reason why clocks in the back of the train would be going any differently from clocks in the middle and in the front.

It's weird, alright. If you have two spaceships traveling at the same speed, and they are satisfied that their clocks are synchronized, and then they decide that when their clocks both strike 8 they will give identical puffs of thrust out the back, thus accelerating by the same amount - after 8, the observers in the ships will no longer find that their clocks are synchronized with each other. It seems so odd, since as far as they are concerned they did everything completely symmetrically.

Edited March 12, 2015 by whiskers

[LaurieAG]

Everything so far has been based on Einstein's original thought experiment with a train frame, an embankment frame and an observer so what happens when we move the observer to different locations?

 

If you extend the basic train experiment to use a circular track you can use SR based relativistic rolling wheel solutions that have a wheel, a road and an axle or carriage frame. The points in the wheel frame rotate around the circumference and the road sits stationary as the wheel rolls by. The axle frame is different from the wheel frame in that it doesn't rotate with the wheel but stays at the same location as the axle while the carriage frame is any other frame that is fixed with respect to the axle frame/location and its movement at any time.

 

If the observer moved directly backwards to the center of the circular track it would be equidistant from both the front and the back of the train. If the track has sensors that are triggered by switches built into the front middle and rear of the train as it travels past and these ground switches turn on a light at the side of the track pointed directly at the observer when triggered then how would the pattern of lights appear at the observer?

 

From this location would you see the first image with 2 lights and then a later image with 1 or would you see 1 then 2 then 3 lights?

Or is there a point where it goes from one to the other depending on the distance away?

Extending the train experiment further you can put the observer in an air ship tethered directly above the center of the circular track at a distance of 2 * Pi * R * c/v from the axle or center of rotation/mass. From this location you can capture all the photons emitted during one complete revolution of the train up to v = c knowing that the photons emitted from every light on the track would reach the observer at the same point in time if they had all been emitted at the same time regardless of their point of emission.

This diagram below shows the light paths that would exist in real time between two rotating sources and an observer at various angles to the plane of rotation if the sources continued to rotate and emit for one complete rotation and the photons were not distorted or blocked along the way. Note C, 0 degrees, where the observer is in the same plane as the plane of rotation is a SR based relativistic rolling wheel solution (x and y only, z = 0) and A, 90 degrees, shows where the observer is perpendicular to the plane of rotation of the sources and observes minimal shift.

[LaurieAG]

I have just corrected the image above and swapped the apparent shift colors for paths A and C.

[CasualKilla]

Does this help?

 

 

Atleast it shows the observers on the train disagree with the blokes outside about simultaneity of events.

 

NEW POST!

 

I have being doing some 2D special relativity calculations, and I have come up with some conclusions that may help. I assume a train speed of 0.87c, and I will have the clocks emit a pulses at 2hz, and due to the length of the train, the middle clock will also pulse also at 2hz, but with a slight delay. (with reference to on train observer)

 

Conclusions from calculations:

 

• All clocks experience the same time dilation, and the observer outside the train will see each clock pulsing at 1Hz. (v = 0.87c)
• The outside observer does not see the front and back clocks flash at the same time, though the flash RATES are observed to be the same.

 

• Order of observations (long train):

1. Front clock is observed to flash first.
2. Back clock flashes second.
3. Middle clock flashes last.

However I noticed something very interesting, if the distance between side and middle clocks is small, the time delay as observed by the observer on the train will also be small. If this delay is small enough, the observer outside the train will actually observe the middle clock flash before the back clock. So this changes the order of observation for a short train:

 

• Order of observations(short train):

1. Front clock is observed to flash first.
2. Middle clock flashes second
3. Back clock flashes last

What a truly amazing result.

 

This was a misunderstanding of the maths, since the middle clock is signalled by the front and back clocks with light, I can't just make the delay as short as i want, the delay is directly fixed to the length of the train, but as the length decrease, the time delay between Cb and Cf also decreases, the result in that the middle clock always flashes last. The part at the end of the video about space-time separation should explain this a bit better.

 

This leads to an interesting new conclusion:

assumption: 2 systems at rest within the same reference frame separated by a finite distance L cannot cause an affect on one another faster than t = L/c.

If event A causes event B, then event A occurs before or simultaneously with event B in any reference frame.

 

I believe the rest of my calcs were sound, I used the theory in this lecture as reference.

 

https://www.youtube.com/watch?v=mG-rO9KLpHc

Edited March 27, 2015 by CasualKilla

[xyzt]

 

 

Conclusions from calculations:

 

• All clocks experience the same time dilation, and the observer outside the train will see each clock pulsing at 1Hz. (v = 0.87c)

This is, of course false. An external observer will measure the clock frequency to be larger than 2hz when the clock approaches and less than 2hz when the clock moves away. This is basic Doppler effect. Feel free to continue your practice of giving me negative feedback for correcting your glaring mistakes.

 

This leads to an interesting new conclusion:

assumption: 2 systems at rest within the same reference frame separated by a finite distance L cannot cause an affect on one another faster than t = L/c.

If event A causes event B, then event A occurs before or simultaneously with event B in any reference frame.

 

This is just as false as the rest of your posts.

 

[math]t'=\gamma(t-vx/c^2)[/math]

 

 

so:

 

[math]dt'=\gamma(dt-vdx/c^2)[/math]

 

The above shows that the sign of [math]dt'[/math] may be :

 

-the same

or

-opposite

 

the sign of [math]dt[/math]

 

depending on :

 

-[math]dt>vdx/c^2[/math]

-[math]dt<vdx/c^2[/math]

Edited March 27, 2015 by xyzt

[md65536]

 

This is just as false as the rest of your posts.

 

[math]t'=\gamma(t-vx/c^2)[/math]

 

 

so:

 

[math]dt'=\gamma(dt-vdx/c^2)[/math]

 

The above shows that the sign of [math]dt'[/math] may be :

 

-the same

or

-opposite

 

the sign of [math]dt[/math]

 

depending on :

 

-[math]dt>vdx/c^2[/math]

-[math]dt<vdx/c^2[/math]

The sign can change only for space-like intervals.

Causal relations are time-like or light-like, they can't be space-like.

CasualKilla is correct. If event A causes event B, there is no frame of reference in which event B precedes event A. Do you have a counter-example, of an event that precedes its cause? The only exception I can think of is faster-than-light particles, which are only speculative.

[xyzt]

The sign can change only for space-like intervals.

Causal relations are time-like or light-like, they can't be space-like.

CasualKilla is correct. If event A causes event B, there is no frame of reference in which event B precedes event A. Do you have a counter-example, of an event that precedes its cause? The only exception I can think of is faster-than-light particles, which are only speculative.

So, in the general case, his statement is false. That was the point. In your haste to "correct" me, you obviously missed it. Not the first time you did that. I am amused that you did not tackle my first counter to his incorrect claims. Out of curiosity, do you agree with his claim on clocks being observed to tick at 1hz?

Edited March 27, 2015 by xyzt

[md65536]

So, in the general case, his statement is false. That was the point. In your haste to "correct" me, you obviously missed it. Not the first time you did that. I am amused that you did not tackle my first counter to his incorrect claims. Out of curiosity, do you agree with his claim on clocks being observed to tick at 1hz?

The statement is about causal relations. There is no "general case" in which CasualKilla's statement is false.

[xyzt]

The statement is about causal relations. There is no "general case" in which CasualKilla's statement is false.

This is not what I said. In general, CasualKilla's statement is false. Actually, it can be shown to be false even for causal relations. In order to be true for causal relations, it requires an extra condition. You can stop trolling my posts now.

Edited March 27, 2015 by xyzt

[swansont]

 

[math]t'=\gamma(t-vx/c^2)[/math]

 

 

so:

 

[math]dt'=\gamma(dt-vdx/c^2)[/math]

 

The above shows that the sign of [math]dt'[/math] may be :

 

-the same

or

-opposite

 

the sign of [math]dt[/math]

 

depending on :

 

-[math]dt>vdx/c^2[/math]

-[math]dt<vdx/c^2[/math]

 

 

So for objects separated by L and at a speed of c, it matters whether dt is larger or smaller than L/c.

 

I'm missing where this is different than what was claimed.

[xyzt]

 

 

So for objects separated by L and at a speed of c, it matters whether dt is larger or smaller than L/c.

 

I'm missing where this is different than what was claimed.

For two events A and B separated by a distance [math]dx[/math] in space and a interval [math]dt[/math] in time, one can have:

 

[math]\frac{dx}{dt}>\frac{c^2}{v}[/math]

 

[math]\frac{dx}{dt}=\frac{c^2}{v}[/math]

 

[math]\frac{dx}{dt}<\frac{c^2}{v}[/math]

 

where [math]v[/math] is the relative speed of frame S' wrt. S. The key point is that

[math]\frac{dx}{dt}[/math] is NOT a speed and it is certainly not the speed at which the influence of event A propagates to event B. It is simply the ratio of spatial to the temporal separation. As such, it can have any value (see above) and it can certainly be larger than c.

Edited March 28, 2015 by xyzt

[swansont]

For two events A and B separated by a distance [math]dx[/math] in space and a interval [math]dt[/math] in time, one can have:

 

[math]\frac{dx}{dt}>\frac{c^2}{v}[/math]

 

[math]\frac{dx}{dt}=\frac{c^2}{v}[/math]

 

[math]\frac{dx}{dt}<\frac{c^2}{v}[/math]

 

where [math]\frac{dx}{dt}[/math] is the relative speed of frame S' wrt. S. The key point is that

[math]\frac{dx}{dt}[/math] is NOT a speed and it is certainly not the speed at which the influence of event A propagates to event B. It is simply the ratio of spatial to the temporal separation. As such, it can have any value (see above) and it can certainly be larger than c.

 

OK, now how about actually addressing what I wrote? I didn't say anything about dx/dt. I substituted values for dx and v.

[xyzt]

 

OK, now how about actually addressing what I wrote? I didn't say anything about dx/dt. I substituted values for dx and v.

It is the same thing, I am pointing out that motion between frames can invert the order of events in ALL cases , INCLUDING causal. It is the same math.

[md65536]

It is the same thing, I am pointing out that motion between frames can invert the order of events in ALL cases , INCLUDING causal. It is the same math.

That's wrong. You've written out maths to cover all cases, including space-like intervals which don't have a unique order, but which cannot describe causal relationships. Do you understand that? Or do you have an example where event A causes event B, but B precedes A?

[xyzt]

That's wrong. You've written out maths to cover all cases, including space-like intervals which don't have a unique order, but which cannot describe causal relationships. Do you understand that?

No, it isn't wrong, the math has nothing to do with space-like intervals, despite your insistence. The fact that you can't follow is your problem, not mine.

 

 

 

Or do you have an example where event A causes event B, but B precedes A?

 

Easy. I already gave you a hint to the fact that you are missing a necessary condition. It has to do with clock synchronization.

Edited March 28, 2015 by xyzt

[md65536]

Easy. I already gave you a hint to the fact that you are missing a necessary condition. It has to do with clock synchronization.

Please continue to the example where event A causes event B, but B precedes A. I'm not seeing where you're going (unless you think that the order of events depends on the coordination of clocks, and that an event happens earlier if you set a nearby clock to read earlier).

[xyzt]

Please continue to the example where event A causes event B, but B precedes A.

Has nothing to do with your inability to understand the previous posts, nor is it related in any fashion with this thread. The fact that you insist, amply demonstrates that. Nevertheless, I will explain to you. Clock A shows time "0" when clock B shows T. An event happens at location A causing another event to happen at B after time "t". Clock A shows time t when clock B shows T+t, DESPITE the fact that event A caused event B. You can just as easy have the reverse, clock B showing less elapsed time than clock A, DESPITE event A , again, causing event B. You can also have the two clocks showing the same exact time. You can stop trolling on this totally unrelated subject now.

Edited March 29, 2015 by xyzt