Stopped Clock Paradox - Analysis

[PhysBang]

@James S Saint

The first thing out of every SR defenders mouth concerning the proposed paradox situation is that such a situation cannot ever occur because it requires simultaneity. But as seen, even using SR equations,

Let’s stop at this lie, because you haven’t used SR equations. You have only said that the equations work without using them.

the scenario certainly can exist, one merely must treat the fore and aft parts of the train separately, which can easily be arranged merely by having them as separate engines. The centeredness of the clocks is the only issue, not the lengths.

Except that the length between the different trains changes, too.

And I suspect that the SR defender would then proclaim, “yeah but there can’t be an observer photon”.

It is true that there are no valid reference frames that are co-moving with photons because there is not a valid transformation from a valid reference frame to such a frame and vice versa. This is because SR assumes that all photons move at a set speed relative to any reference frame. Thus if one wants to actually use SR, one can’t have a reference frame co-moving with a photon.

As always, it is important to learn SR before criticizing SR.

Well, there can’t be a train moving at half the speed of light passing a station either. We are not talking about practical scenarios that every house wife has need to be concerned about. We are talking about the very nature of existence and thus are concerned with consistencies, not practicalities.

Again, there difference between these cases is one of mathematical impossibility vs. a toy scenario that exaggerates the values of likely scenarios. If one wants, one can do any train scenario with likely train speeds.

So
A) what would an “almost” observer photon perceive as the scenario played out?
B) what would the other observer photon, from the other direction, perceive?

And then considering both of those,
C) what would the predicted result have to be from the perspective of the photons?

{{finally after two years, getting into the meat of this}}

What scenario? Can you give us some specific values for this one?

[Fixed_Cross]

It’s been a few days, I’m having trouble answering these.
It depends on what you mean - are these almost-photons conventional matter, in that they are influenced in their speed by their source?

In any case they would observe the station and the train heavily compressed so as to be almost flat, it seems.
But if the source does determine the relative speed, then one photon (from the back of the train) perceives a significantly less compressed train than the other.

It’s clear that if we’re not talking about Relativity’s light, the near-lightspeed observer can not be observed to move at the same speed from different reference frames.

[James_S_Saint]

@Fixed Cross

It’s been a few days, I’m having trouble answering these.
It depends on what you mean - are these almost-photons conventional matter, in that they are influenced in their speed by their source?

You are over complicating the issue. The "observer almost-photon"s mass is irrelevant, as was the mass of the station and the train. The issue is merely any changing of distance between the station clock and the train clock. Each are, for example, coming toward the observer photon at infinitesimally close to the speed of light.

Will they reach the observer photon at the same moment according to the observer photon, using SR or not?
Which will reach the observer-photon first?
Which would reach the back observer-photon first?

@Fixed Cross

In any case they would observe the station and the train heavily compressed so as to be almost flat, it seems.

True by SR, but irrelevant. There is still a distance between them that cannot be zero.

@Fixed Cross

But if the source does determine the relative speed, then one photon (from the back of the train) perceives a significantly less compressed train than the other.

“Source” doesn’t determine the observer photon’s light speed and isn’t actually relevant.

We are using an “almost observer photon” merely to examine what the “observer-photon” would “see” if it had the capacity to observe. Something an infinitesimal amount slower, could not see anything much different than an observer-photon would see if it could.

When you think about it, what could be more appropriate to use as your “inertial frame of reference” than the one thing that is declared and proven to never be able to be anything but what it is, the light itself.

[phyllo]

Okay, calculate the velocities of everything (station clock, train clock, front flasher, rear flasher and other photon) with respect to the inertial frame of each photon (front and rear). Calculate the time that each photon reaches each clock.

[James_S_Saint]

A) Both clocks stop - There is no SR frame wherein both clocks can stop
B) Only Station clock stops - SR has only the station clock stopping in the station frame
C) Only the Train clock stops - SR has only the train clock stopping in the train frame
D) Neither clock stops - SR has neither clock stopping in the “Almost Observer Photon” frames.

So SR has three “correct” answers, all of which contradict each other.
It reminds me a bit of college psychology quizzes.

So what would really happen?
Is there another relevant frame?

[PhysBang]

@James S Saint

A) Both clocks stop - There is no SR frame wherein both clocks can stop
B) Only Station clock stops - SR has only the station clock stopping in the station frame
C) Only the Train clock stops - SR has only the train clock stopping in the train frame
D) Neither clock stops - SR has neither clock stopping in the “Almost Observer Photon” frames.

So SR has three “correct” answers, all of which contradict each other.
It reminds me a bit of college psychology quizzes.

So what would really happen?
Is there another relevant frame?

If this is true, you should be able to demonstrate this with actual numbers, right?

So why lie to us?

[James_S_Saint]

Buy a letter.

[PhysBang]

OK, let’s see what SR actually says, not simply what James fantasizes it says.

Let’s say that the trains are going at 0.99 of the speed of light in a reference frame in which the station is at rest.

We will identify a frame co-moving with the station and call that frame, O. We will identify a frame co-moving with the middle train (MT) and call that frame, O’. We will say that the final speed of the trains is, in O, 0.99 (in units where c = 1). We will pick our origins for O and O’ such that O(0,0) is the same spacetime event as O’(0,0). (Coordinates given with ’ indicate a coordinate from the O’ frame.)

We will stipulate that at time t=10, flashers go off on the leading train (LT) and the trailing train (TT). We will stipulate that in O, the trains are 10 always units apart and are going the same speed. This is accomplished because there are sensors at x = 10 and x = -10 such that when the middle of the relevant trains go past the sensors, a flash goes off at these locations. The middle train and the station both have sensors that register when a flash of light hits them and they stop a clock that they carry if they get two flashes of light at the same time.

Let MT(x) = 0.99t – 9.9; LT(x) = 0.99t + 0.1; TT(x) = 0.99t – 19.9; and Station(x) = 0. In this way, the flashers go off when MT is at the origin and LT is at 10 and TT is at -10. The light from the flashers reaches the station at t=20, at which point a sensor in the station, primed to go off if it receives two flashes of light at once, registers the light and stops its clock.

The path of the flash from LT is t = -x + 20. So the sensor on MT receives the flash from LT at about t = 15.03. The path of the flash from TT is t = x + 20. So the sensor on MT receives the flash from LT at about t = 101. Thus the clock on MT does not stop.

If one wants to do the SR calculations (standard Lorentz transforms), then I suggest one go to the following website and work through them. hyperphysics.phy-astr.gsu.edu/hb … trans.html

When the flashers go off, where and when are the trains in O’? (NB: I will limit the numbers to four significant digits.)

We note that γ = 7.089 given our limitation to four significant digits.

MT x’(0,10) = γ (0 – 0.99(10)) = -70.18
MT t’(0,10) = γ (10 – 0/99(0)) = 70.89

LT x’(10,10) = γ (10 – 0.99(10)) = 0.7089
LT t’(10,10) = γ (10 – 0.99(10)) = 0.7089

TT x’(-10,10) = γ (-10 – 0.99(10)) = -141.1
TT t’(-10,10) = γ (10 – 0.99(-10)) = 141.1

Since the trains do not move in O’, we know that, in O’, the distance between MT and LT is 0.7089 - (-70.18) ~ 70.89 and that the distance between MT and TT is -70.18 - (-141.1) ~ 70.92. This is what we expect from length contraction.

When and where does the clock stop in O’?

Stop x’(0,20) = γ (0 – 0.99(20)) = -140.4
Stop t’(0,20) = γ (20 – 0/99(0)) = 141.78

Where is the midpoint of the flashes in O’? It would be at (-141.1 + 0.7089)/2 = -70.20, at the middle train (right where we expect it to be, given the error introduced by limiting digits). And when does the light reach that midpoint? The light from LT reaches this midpoint at 0.7089 + |(0.7089 - (-70.18))| ~ 71.60. The light from TT reaches this midpoint at 141.1 + |(-140.4 - (-70.18))| ~ 211.3. So the light from the two flashers reaches the middle train at different times, not triggering the sensor on that train to stop its clock.

So SR says one thing: the station clock stops and the train clock does not.

[James_S_Saint]

@PhysBang

OK, let’s see what SR actually says, not simply what James fantasizes it says.

Let’s say that the trains are going at 0.99 of the speed of light in a reference frame in which the station is at rest.

We will identify a frame co-moving with the station and call that frame, O. We will identify a frame co-moving with the middle train (MT) and call that frame, O’. We will say that the final speed of the trains is, in O, 0.99 (in units where c = 1). We will pick our origins for O and O’ such that O(0,0) is the same spacetime event as O’(0,0). (Coordinates given with ’ indicate a coordinate from the O’ frame.)

We will stipulate that at time t=10, flashers go off on the leading train (LT) and the trailing train (TT). We will stipulate that in O, the trains are 10 always units apart and are going the same speed. This is accomplished because there are sensors at x = 10 and x = -10 such that when the middle of the relevant trains go past the sensors, a flash goes off at these locations. The middle train and the station both have sensors that register when a flash of light hits them and they stop a clock that they carry if they get two flashes of light at the same time.

Let MT(x) = 0.99t – 9.9; LT(x) = 0.99t + 0.1; TT(x) = 0.99t – 19.9; and Station(x) = 0. In this way, the flashers go off when MT is at the origin and LT is at 10 and TT is at -10. The light from the flashers reaches the station at t=20, at which point a sensor in the station, primed to go off if it receives two flashes of light at once, registers the light and stops its clock.

The path of the flash from LT is t = -x + 20. So the sensor on MT receives the flash from LT at about t = 15.03. The path of the flash from TT is t = x + 20. So the sensor on MT receives the flash from LT at about t = 101. Thus the clock on MT does not stop.

If one wants to do the SR calculations (standard Lorentz transforms), then I suggest one go to the following website and work through them. hyperphysics.phy-astr.gsu.edu/hb … trans.html

When the flashers go off, where and when are the trains in O’? (NB: I will limit the numbers to four significant digits.)

We note that γ = 7.089 given our limitation to four significant digits.

MT x’(0,10) = γ (0 – 0.99(10)) = -70.18
MT t’(0,10) = γ (10 – 0/99(0)) = 70.89

LT x’(10,10) = γ (10 – 0.99(10)) = 0.7089
LT t’(10,10) = γ (10 – 0.99(10)) = 0.7089

TT x’(-10,10) = γ (-10 – 0.99(10)) = -141.1
TT t’(-10,10) = γ (10 – 0.99(-10)) = 141.1

Since the trains do not move in O’, we know that, in O’, the distance between MT and LT is 0.7089 - (-70.18) ~ 70.89 and that the distance between MT and TT is -70.18 - (-141.1) ~ 70.92. This is what we expect from length contraction.

When and where does the clock stop in O’?

Stop x’(0,20) = γ (0 – 0.99(20)) = -140.4
Stop t’(0,20) = γ (20 – 0/99(0)) = 141.78

Where is the midpoint of the flashes in O’? It would be at (-141.1 + 0.7089)/2 = -70.20, at the middle train (right where we expect it to be, given the error introduced by limiting digits). And when does the light reach that midpoint? The light from LT reaches this midpoint at 0.7089 + |(0.7089 - (-70.18))| ~ 71.60. The light from TT reaches this midpoint at 141.1 + |(-140.4 - (-70.18))| ~ 211.3. So the light from the two flashers reaches the middle train at different times, not triggering the sensor on that train to stop its clock.

So SR says one thing: the station clock stops and the train clock does not.

Well, at least you made the attempt (finally).

But you have merely given the perspective from the station. And yes, the station perspective says that only the station clock will stop (option 2).

Now do that exact same thing (perhaps correcting a detail or two) except from the train’s perspective where O is the MTrain, “MT” and it is the station that is moving, thus the station’s frame will be O’. SR doesn’t permit you to declare a privileged frame, so you have to apply the same rules for each frame of motion, each perspective.

So you have half of it (although you seem to want to start at an odd point, no matter).
Of course logic will tell you that you are going to get the exact same calculations, merely reversed, but go ahead and try it.

[PhysBang]

@James S Saint

But you have merely given the perspective from the station.

Are you desperately trying to lie to the other people that might be reading this thread or do you really not realize that well more than half the work in the above demonstrates just what happens in the reference frame in which the trains are not moving?

Now do that exact same thing (perhaps correcting a detail or two) except from the train’s perspective where O is the MTrain, “MT” and it is the station that is moving, thus the station’s frame will be O’.

You seem to not understand that “O” and “O’” are merely names. And they are not the names of objects, but the names of frames of reference. After I give something a name, changing that name doesn’t change the scenario.

You seem not to understand that if what we have laid out what happens in the station frame, then the same events must happen in every frame. That is, since the clock in the station stops in the station frame, then it has to stop in every frame. Because frames are merely ways to describe the same events.

So the above descriptions I gave actually give you the thing that you are still asking for. So go back and read it.

[James_S_Saint]

So you really don’t understand SR at all.
I didn’t think so.

You must go through the same calculations with the station moving, not MT.

MT has no velocity.
X’ refers to the station’s position as it moves passed the train.

[PhysBang]

@James S Saint

So you really don’t understand SR at all.
I didn’t think so.

You must go through the same calculations with the station moving, not MT.

MT has no velocity.
X’ refers to the station’s position as it moves passed the train.

Look up! It’s all there.

In the reference frame in which the trains are not moving, the leading train is at 0.7089, the middle train is at -70.18 and the trailing train is at -141.1. The station is moving towards them at 0.99. The trailing edge of the station passes the leading train at 0.7089. The leading edge of the station passes the trailing train at 141.1.

What do you not understand?

[James_S_Saint]

@PhysBang

@James S Saint

So you really don’t understand SR at all.
I didn’t think so.

You must go through the same calculations with the station moving, not MT.

MT has no velocity.
X’ refers to the station’s position as it moves passed the train.

Look up! It’s all there.

In the reference frame in which the trains are not moving, the leading train is at 0.7089, the middle train is at -70.18 and the trailing train is at -141.1. The station is moving towards them at 0.99. The trailing edge of the station passes the leading train at 0.7089. The leading edge of the station passes the trailing train at 141.1.

What do you not understand?

What I don’t understand is how you could not understand yet think that you do.

If MT is not moving, because you are that observer, “O”, then MT is at x=0 and stays there.

The flashers are going to be Xs distance from the station from the station’s perspective at the moment TM is aligned with the station. If you want to use length contraction, although there actually is none, then you have to calculate how far LT and TT have moved from MT at the moment of flash. That means that you have to length contract the station and Xs into an Xs’ and use that as your distance from MT to LT and also to TT.

Since the station is moving, the time of travel for the light to get to the station will be different for LT than it is for TT, because the station is moving closer to one and further from the other.

[PhysBang]

@James S Saint

If MT is not moving, because you are that observer, “O”, then MT is at x=0 and stays there.

Did you not read what I wrote above? MT is at -70.18 and stays there. What is at 0 is arbitrary, we can easily create a frame where MT is at 0, all we need to to is add 70.18 to every x’ coordinate.

You are free to adjust the coordinates as you wish, but since you didn’t take the time to lay out a physics scenario properly, you missed your chance.

The flashers are going to be Xs distance from the station from the station’s perspective at the moment TM is aligned with the station. If you want to use length contraction, although there actually is none,

Look James, you merely do not want to use SR. That’s fine, you and all other physics deniers are free to use whatever crazy physics you want. I know that nobody is going to allow you to build anything.

However, you can’t simply demand that SR works one way when it clearly doesn’t. If you want to use SR, then you have to use length contraction.

then you have to calculate how far LT and TT have moved from MT at the moment of flash. That means that you have to length contract the station and Xs into an Xs’ and use that as your distance from MT to LT and also to TT.

Since the station is moving, the time of travel for the light to get to the station will be different for LT than it is for TT, because the station is moving closer to one and further from the other.

All those calculations are done above. All of them.

You can either take the time to go through them again and see what you missed, or you can ask me specific questions about what you don’t understand. Or, you can continue to display poor character and lie about SR.

[James_S_Saint]

@PhysBang

@James S Saint

If MT is not moving, because you are that observer, “O”, then MT is at x=0 and stays there.

Did you not read what I wrote above? MT is at -70.18 and stays there. What is at 0 is arbitrary, we can easily create a frame where MT is at 0, all we need to to is add 70.18 to every x’ coordinate.

Well apparently you need to do that, because you have the station remaining evenly between the flashers, and it obviously doesn’t.

[PhysBang]

@James S Saint

Well apparently you need to do that, because you have the station remaining evenly between the flashers, and it obviously doesn’t.

So you apparently went the lying (and poor grammar) route.

There is no reason for the station to not be the midpoint of the flashers. The length contraction is not partial to one side of the station over the other.

Given that in O (the name of the frame that was used in which the station is at rest), the distance between the station and the flashers was 10, the distance in O’ (the name of the frame that was used in which the trains are at rest) is approximately 1.411.

The station movement in O’ is given by x’ = -0.99t’. The motion for the flashers is x’ = -0.99t + 1.411 and x’ = -0.99t - 1.411. One can determine this from looking at the Lorentz transformations of the events where the flashers go off.

What doesn’t remain “even” is the moment when the flashers go off. In O’ there are two different moments.

[Fixed_Cross]

This is where I was struggling to answer.
In relativity, due to the weird nature of the speed of light of being equal in every reference frame, both clocks are actual midpoints, even if compared to each other they relate differently to the light.

We can thus not use light as an inertial reference frame – because to do the calculation, we would need to bring it in relation to itself and this results indeed in ‘‘reality not existing’ (being zero’ in spacetime values).

So as soon as we propose a mass that is moving at .99c, it is no longer behaving as light, and is thus no longer that factor that grounds the Relativity calculations.

I agree with the animations, these are what my answer was going to be, but I don’t see that the .99c near-photon observers are comparable to light.

[PhysBang]

@Fixed Cross

I agree with the animations, these are what my answer was going to be, but I don’t see that the .99c near-photon observers are comparable to light.

If one assumes SR, then the animations are simply wrong: if the flashes are simultaneous in one frame, then they cannot be simultaneous in the other; the animations display the flashes as simultaneous in both frames.

[James_S_Saint]

@Fixed Cross

This is where I was struggling to answer.
In relativity, due to the weird nature of the speed of light of being equal in every reference frame, both clocks are actual midpoints, even if compared to each other they relate differently to the light.

Each is a midpoint in its own frame.

You can only be one observer or the other. So if you are the station, you see yourself being the stationary midpoint between the flashes. If you are the train observer, you see yourself as the stationary midpoint between the flashes.

What PhysBang and others tend to do is look from only one perspective and do the calculations (usually from the station perspective, a Newtonian outlook). They deduce (properly) that the train clock will not stop and their own will stop (although PB’s reasoning was a little messed up). They stop at that point declaring that all is resolved because they got an answer and don’t want to deal with it any more, “first answer is the only answer”. They don’t verify anything, “lazy”.

To verify their deduction, they have to say, "Well, if I was on the train, rather than the station, how would things look and calculate out?". The station and tracks are moving, not the train. The math is even easier when they have only one moving object, the flashers don’t move relative to the train clock. If you can ever get them to do that, against their lazy will, they end up calculating that it must be the train clock that stops and not the station.

They cannot do as PB wants to do and shortcut the verification by saying, “Since we know it is all going to work out the same anyway, and doing all of that thinking will hurt my brain, we can just go ahead and say that we get the same results.”

PB’s reasoning was a bit skewed, but it won’t matter as long as he used the same reasoning in both cases, which he refuses to do. So of course, he can’t verify his conclusion (nor find the errors he made).

@Fixed Cross

We can thus not use light as an inertial reference frame – because to do the calculation, we would need to bring it in relation to itself and this results indeed in ‘‘reality not existing’ (being zero’ in spacetime values).

We didn’t use the light itself, but the “almost observer photon”. If anything is moving at one infinitesimal less that the speed of light (c * .999…), according to SR, nothing is zero but is still merely an infinitesimal different than what the light itself would have to see if it could be an observer.

@Fixed Cross

I agree with the animations, these are what my answer was going to be, but I don’t see that the .99c near-photon observers are comparable to light.

It isn’t “.99c”.
It is “.999999999999999999” times c. Or merely one infinitesimal less than c, “c-”. That is a very different answer than “.99c”.

So using exactly c as the observer speed will get you infinitesimally close to what the “almost observer photon” would “see”. That is what that last animation is showing. The observer photon is seeing the station and train moving toward it at almost light speed, but one less than the other, “c- and c–”.

[PhysBang]

@James S Saint

You can only be one observer or the other. So if you are the station, you see yourself being the stationary midpoint between the flashes. If you are the train observer, you see yourself as the stationary midpoint between the flashes.

What PhysBang and others tend to do is look from only one perspective and do the calculations (usually from the station perspective, a Newtonian outlook). They deduce (properly) that the train clock will not stop and their own will stop (although PB’s reasoning was a little messed up). They stop at that point declaring that all is resolved because they got an answer and don’t want to deal with it any more, “first answer is the only answer”. They don’t verify anything, “lazy”.

This is simply a lie.

Above, I demonstrated what goes on in two different reference frames, ones that we might naturally associate with the two observers. It is not lazy to do something correctly.

To verify their deduction, they have to say, "Well, if I was on the train, rather than the station, how would things look and calculate out?". The station and tracks are moving, not the train. The math is even easier when they have only one moving object, the flashers don’t move relative to the train clock. If you can ever get them to do that, against their lazy will, they end up calculating that it must be the train clock that stops and not the station.

Again, a lie.

The train clock does not stop, because the flashes that go off reach the midpoint at different times. As I have demonstrated above.

Lying does not advance a scientific argument.

It is a good attempt to dodge James’ inability to understand this kind of thing, I suppose.

PB’s reasoning was a bit skewed,

I have never been happier that a liar who cannot follow the relevant physics finds my reasoning “skewed”.

but it won’t matter as long as he used the same reasoning in both cases, which he refuses to do. So of course, he can’t verify his conclusion (nor find the errors he made).

Again, a lie.

I feel in good company, however, since James also lies about what Feynman said.

We didn’t use the light itself, but the “almost observer photon”. If anything is moving at one infinitesimal less that the speed of light (c * .999…), according to SR, nothing is zero but is still merely an infinitesimal different than what the light itself would have to see if it could be an observer.

SR is not about observers, it is about systems of coordinates. One cannot have a system of coordinates that is, roughly speaking, traveling at the speed of light. It just doesn’t work out.