Relative Velocity Measurement – Frame and photon

Who cares how many times it's here again. The real point is: My expression of my point has nothing to do with your literary fumbles and subsequent reiteration. You (all) have been plainly informed of the fatal logical contradiction between Einstein's exposition of ambiguous perceived simultaneity (or lack thereof) versus Postulated constant speed of light anywhere for anybody.

The real point is: A Relativity protagonist is totally unable to prove that either The Rider or The Embankment Observer can show that he observes both constant speed of light and ambiguous simultaneity.

The pundit can choose only one. Not both.

Special Relativity must have both or choke to death.

 

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You plainly "informed" us that you got it completely wrong:

You might want to do your reading again. If the flashes arrive in the middle at different moments, they clearly began at different moments.

The real point is that you have no idea what a "relativity protagonist" thinks.
Neither the rider or the embankment observer "observes" ambiguous simultaneity. In any given reference frame, simultaneity is quite unambiguous. The ambiguity is in the comparison between reference frames.

 

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Just minutes ago, after posting and then trying to LOG OUT, I repeatedly clicked on the log out button but this site would not release my connection. After about a half dozen tries, I was suddenly dumped. When I tried to get back back into the site to verify logout, I was unable to do so. I rebooted my computer and have come back now to report.

Why did your site refuse to log me out?

 

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Pete, I have already noticed that you get confused easy. You have got confused again.

I do not care at all what you write as your opinion of my post or your insulting opinion of me personally.

I have written my serious objections to the logical gaps in Special Relativity. I am not interested in investing any more time or energy in arguing with you about it. You can rebut my utterances or fail. There is no reason for a dialog. And, for my behalf, no shred of a desire for one with you.

So, you can thoroughly satisfy my grievances with your solo posts. Or fail to do so. Many people much smarter than you are are waiting to read your explanations for the points I have raised.

 

Where did I imply 'it is all about [me]'? I'm pointing out that if you actually had something worthwhile you'd not be posting it on a forum, you'd be submitting it to a journal.

I'm not the one making grand claims about turning over a century of science. You are. I just want you to provide some justification/evidence for your claim. I'm sorry to take a request to follow the scientific method as some kind of personal attack but its hardly my fault you can't justify your claims.

So you're not going to provide evidence for your claims because I've been rude to you? You're the one who made a claim, simply repeatedly asking you to justify it because you keep failing to do so isn't a fault of mine. If you'd justified your claim initially we'd not have to keep badgering you.

Do you think everyone who disagrees with you is 'ditzy'? No one here who knows any physics has agreed with you. You have made a claim (about who sees when the light gets to the person's eyes) which is false and I'm not the only one to point it out.

You haven't actually done any SR calculations, you just jump straight to what you assume is the outcome and declare that's what SR says. Attacking SR because you think it says something it doesn't is just silly. You also continue with the "One of the postulates is contradictory", which you haven't justified either, just repeatedly stated.

Tell you what, I'll walk you through it. It's basically the same as the problem I've been correcting Jack_ on so its not difficult.

Person A is in Frame 1, which has the train at rest and coordinates (x,t)
Person B is in Frame 2, which has the embankment at rest and coordinates (x',t')
The train has length 2cT and moves at speed v relative to the embankment.
Person A is in the middle of the train. He fires photons in both directions, up and down the train and waits for them to come back.

So who sees what? Start with person A's point of view.

A is at (0,t) for all t. The light spreads out such that its at \(x=\pm ct\). Note that this is precisely the same swetup as I'm discussing with Jack_ as this is the definition of a 0 dimensional spheres, \(x^{2} = R^{2}\) with \(R=ct\). The photon sphere hits the front and back of the train, at \(x = \pm cT\) simultaneously at t=T and reflects back, with coordinates satisfying \(x^{2} = c^{2}(2T-t)^{2}\) for \(t>T\). Person A sees the light return when x=0 so t=2T. Easy and simple.

Now consider what Person B sees.

In Frame 2 B has coordinates(x',t') = (0,t') for all t'. However, Person A is moving in this frame, with speed v. His location is therefore (x',t') = (vt',t') for all t'. Further more the train has been Lorentz contracted. Its ends satisfied \(x^{2} = (cT)^{2}\) in (x,t) coordinates but a Lorentz boost contracts them by a factor of \(\gamma\) and moves their position in time, ie \(\gamma^{2}(x'-vt')^{2} = (cT)^{2}\).

The photon sphere is produced at the initial position of x'=x=t=t'=0. It spreads such that it satisfies \(x'^{2} = (ct')^{2}\).

Since there's an asymmetry due to the motion we'll consider the two directions in turn, starting with the photons moving towards the back of the train, ie \(x' = -ct'\). Solving the simultaneous equations we get the time they hit the back of the train as \(t' = \sqrt{\frac{c-v}{c+v}}T \equiv T_{0}\). Once again I point you at Jack_'s thread because this is an identical result.

These reflect back. At \(t=T_{0}\) they are at \(x' = -cT_{0}\) and move with velocity c so their location is \(-cT_{0} + (t'-T_{0})c\) for \(t' \geq T_{0}\). When does they catch up with the guy on the train? His coordinates are (x',t') = (vt',t') and so we want \(-cT_{0} + c(t'-T_{0}) = vt'\) or equivalently \(t' =\frac{2c}{c-v}T_{0} = 2T\gamma\).

This is the same result as in the first frame but with the time dilation coming into effect.

Now consider the photons which go towards the front of the train. They are at \((x',t') = (ct',t')\). The front of the train satisfies \(\gamma(x'-vt') = cT\) and the front of the photon sphere satisfies \(x' = ct'\). Putting them together gives \(\sqrt{\frac{c+v}{c-v}}T \equiv T_{1}\). Note that \(T_{0}(v) = T_{1}(-v)\) and the \(\sqrt{\frac{c\pm v}{c\mp v}}\) factor is the usual red/blue shifting, which is essentially what we're doing.

These reflect back and they have location \(x' = cT_{1}\) at \(t'=T_{1}\) and so have location \(x' = cT_{1} - (t'-T_{1})c\) for \(t_{1} \geq T_{1}\). From B's point of view these reach A when \(cT_{1} - (t'-T_{1})c = vt'\), so \(t' = \frac{2c}{c+v}T_{1} = 2T\gamma\).

Lowe and behold, the same time! So the person on the embankment, B, sees the two light pulses return to A at the same moment. He doesn't think they hit the back and front at the same time but because of a symmetry this isn't a problem.

From the B's point of view the time it takes a photon to go from A to the back is the same as it takes a photon to go from the front to A. Likewise, its the same time for a photon to go from A to the front as it is to go from the back to A. It's symmetric and thus the totals are the same.

A contradiction would arise if in Frame 1 both photons get to A at the same time but in Frame 2 they don't. If 3 things meet at a point in space-time in one frame (ie two photons and A) then they must meet at a point in ALL frames. Which is precisely what we see here.

I find it funny that Uno and Jack_ both claim to have whipped special relativity and to have perfect, undeniable logic yet all you need to do to prove them wrong is know how to write down the equation for a sphere!

 

I'm not looking for arguments, I'm just pointing out a very simple error you made. You can correct it or continue to ignore it, depending on how hungry you are for understanding.

 

Thanks, Alpha. Just so you can be confident that someone read and understood it (and to check my understanding as well), I'm checking your post in detail:

1 dimensional

\(t'=T_{0}\)

That last sentence took a minute... it didn't immediately click that you were treating To and T1 as functions of v.

I think they don't speak the language. They see a bunch of equations, and think "that's meaningless rubbish". They are unable to translate equations into meaningful concepts in the same way as they can translate prose into meaningful concepts, so they think that equations are abstract hoo-ha, unrelated to reality. Which is why they rely on prose, and why their logic is so fuzzy. It's because the language they have at their disposal is fuzzy, and our language shapes our thoughts.

 

Not true I'm afraid. A one dimensional sphere is a circle (parametered by only 1 variable, the angle). A zero dimensional sphere is a pair of points.

Generically you have \(S^{n}\) in Cartesian coordinates of \(\mathbb{R}^{n+1}\) as \(\sum_{m=1}^{n+1} x_{m}^{2} = R^{2}\). You have n+1 variables and 1 equation which relates them thus you have an n dimensional space defined in \(\mathbb{R}^{n+1}\). The case of n=0 you have \(x^{2} = R^{2}\). Two solutions, \(x = \pm R\).

Yeah, I thought it was worth pointing out since by \(v \to -v\) you swap from blue to red shift and vice versa since you change from things approaching to things going away.

I agree. They don't understand the terminology and can't do the algebra so they read textbooks or papers are just techno-babble. As a result they think if they babble people will think they are doing science. Unfortunately to those people who do understand the terminology their deception is very obvious.

Yes, the classic "I've done the conceptual work, now you do the boring algebra" crank logic. Heck, Uno's even said that to me in this very thread, saying that I'd just be the calculator monkey for some other creative person's ideas. He sees someone who can actually do the details and so has to come up with a reason to tell himself why he can't do it. And the obvious crank view is learning details treads on creative ability.

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It just illustrates how little he knows about science and research.

 

Ah. So I was confusing the sphere itself with the space in which it is embedded?
Or rather, confusing common and formal terminology, I guess.

It's clear that a sphere in 3-space has a 2-dimensional surface, but it doesn't feel right to call it a 2-dimensional sphere. But then, in common parlance "sphere" also includes the space inside.

 

Well, in my post quoted below, I assumed a rotating, circular system of synchronised clocks. They are arranged around the circumference of a circle, and they are 1.0 light-seconds apart. The diameter is unknown, and the linear/tangential speed of rotation at the perimeter was 0.8c. How far off was I?

 

I can be helpful to your understanding.

Place an O' observer at each lightning strike in the moving frame.


Tell me what you see.

 

You are correct.

Except this is not the light postulate.

The light postulate only say light moves throug spacxe at c and that is all.

There is a hidden third postulate that claims the light path is between the emission point in the frame to the target.

Do not abandon what you are doing.

 

Einstein's Train/Embankment Experiment
Einstein's train/embankment experiment actually refutes Special Relativity.
Einstein wrote the following, "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."
http://www.bartleby.com/173/9.html


The goal is to take M' as stationary and look at the train/embankment experiment. This is completely legitimate by the relativity postulate. Normally, the R of S experiment accepts the view of the embankment observer, M, as absolute truth. However, the view of observer M' as stationary is just as legitimate under the rules of relativity.


Now, the experiment is reworded in an equivalent way.
Points A and B are in the frame of M and are equidistant to M. When M and M' are co-located, lightning strikes both points A and B.

As shown in Figure 5, M' and M are co-located when the simultaneous lightning strikes occur. In addition, by the setup, A and B are equidistant to M at the time of the strikes. Let d be the distance from A to M and from B to M measured in the frame of M.

Now, assume there were M' observers co-located at B and A when the lightning strikes occurred at A and B and call them A' and B'. Therefore, when M and M' are co-located, the distance from M' to B' is d/λ and the distance from M' to A' is d/λ. Obviously, if B' is located at a rest distance d/λ, then M could conclude B' and B are not co-located at the time of the strike since M would view B' at a distance of d/λ^2. However, the views and conclusions of M as stationary are not permitted when evaluating M' as stationary as this would commit the fallacy of frame mixing.

Now, the reductio ad absurdum argument is applied by assuming Einstein’s conclusion that M' sees the light from B prior to A from the view of M' as stationary.

To support Einstein’s conclusion that M' sees the light from B prior to A one of the following two possibilities must be true:
1. M' moves toward the lightning strike at B closing the distance for light to travel relative to the strike at A.
2. The strike at B occurs prior to the strike at A in the time coordinates of M'.

Possibility 1
Since M' is stationary, it is not moving. A, B and M are moving relative to M'. Sure, B closes the distance to M' as the light travels toward M' but this has nothing to do with the distance light traveled in the frame of M'. The distance light traveled would be measured from B' to M' which is the distance from the light emission point in the frame to the strike point in the coordinates of the frame. That distance is d/λ. The same logic applies to A' and M' where that distance is also d/λ. This logic is supported by length contraction under SR. In particular, if a moving rod is of length d and a stationary rod is of length d/λ, then from the view of the stationary frame, the two ends of the rods can be simultaneously co-located. Hence, possibility one is not viable when taking M' as stationary since B' and A' are equidistant from M and M' when they are co-located and the lightning strikes.

Possibility 2
Since, there are the observers B' and A' co-located at the lightning strikes at A and B, it is impossible there is any disagreement between the frames as to whether light is moving along the x-axis or not. Hence, for example, if B' claims lightning just struck, B will make the same claim as well. So, it cannot be claimed the lightning appears for one frame at some location while a co-located observer claims light is not at that location. Therefore, perhaps the time on the clock of B' will show an earlier time than the clock of A' for the light strike and this explains it. In other words, the light emitted from B' before it emitted from A'.

So, let tB' be the time of the lightning strike at B' and tA' be the time of the strike at A'. Therefore, tB' < tA'. By the experiment, B' is located a distance d/λ from M' at the time of the strike and is co-located with B. At that strike at B', as required by the experiment, M and M' are co-located. But, that also implies M moves a distance (tA' - tB')v between the two lightning strikes in the time of M' if they indeed occur at different times in the M' frame. Thus, when the strike at A/A' occurs, M has moved a distance (tA' - tB')v, hence, M and M' are no longer co-located at the time of the strike at A/A'. So, possibility 2 is not viable.

Therefore, using this simple argument, the stationary frame of M' will be struck by two simultaneous equidistant lightning strikes and M will be struck by the lightning from A prior to B.

Given this fact and assuming t = d/c is universally true, it must be the case then that M' will see the strikes simultaneously and also M will see the strikes as simultaneous when M is take as stationary since both were co-located and equidistant to the simultaneous lightning strikes. Yet, this simultaneity for each occurs at two different positions in space, which is contradiction of nature

 

Wrong, although unfortunately not an uncommon misunderstanding. The scenario is first described as it occurs in the rest frame of M, but this does not in any way imply it is absolute, any more than describing every day occurrences on the ground implies that the Earth is absolutely at rest. It's just a convenient reference point.

This is wrong. You're getting mixed up over what distances are contracted in which reference frames.

The M rest frame distances from M to A, M to B, M' to A', and M' to B' are all d.
In that rest frame, M' to A' and M' to B' are moving rulers. In their rest frame, they are longer.
So the M' rest frame distances from M' to A' and M' to B' are dλ, not d/λ.

In the M' rest frame, M to A and M to B are the moving rulers, so they are contracted in that rest frame to a length of d/λ.

 

Wrong you are confused as to what I did.

I put M' observers at A and B.

If the distance is d for A and B to M, then when M' is stationary, B' and A' matching A and B from the view of M are length contracted.

Thus, B' is d/λ to M' and A' is d/λ to M'.

 

Right.
You put A' at A when the lightning strikes at A.
You put B' at B when the lightning strikes at B.

In the rest frame of M, A' is at A, B' is at B, and M' is at M, all at the same time.

So,

...then clearly, the distance is also d for A' and B' to M'. Right?

And this is the length-contracted distance, because A', B', and M' are moving. Right?

...don't necessarily happen at the same time, so A to M to B won't necessarily match A' to M' to B'. In fact, they can't, because...

In the rest frame of M':
M to A and M to B are length contracted to d/λ.
M' to A' and M' to B' are now not length contracted, and measure dλ.

 

No, it is because the embankment, A and B are moving and it is d to them.

So,B' is d/λ to M' and A' is d/λ.

Let's first agree one way or the other here.

Once we get this straight the proof is fine.

 

Please clarify "It is d to them".
Is "It" the distance from A to M and B to M?
Or is "it" the distance from A' to M' and B' to M'?
Or both?

And is "them" A, B, and M?


Also, you said you were "rewording the experiment in an equivalent way". I assumed that this means that the strikes are still simultaneous in the rest frame of M.
Is that right, or are you changing it so that the lightning strikes are simultaneous in the rest frame of M'?

 

d is for the M frame, embankment.

that the strikes are still simultaneous in the rest frame of M
Yes.

Is that right, or are you changing it so that the lightning strikes are simultaneous in the rest frame of M'?

No it is simultaneous in the M frame.

This issue with M' must be proven.

 

So, is "it" the distance from A to M and B to M?
Or is "it" the distance from A' to M' and B' to M'?
Or both?