About "A New Light In Physics"

As you prefer:

Here is a "perfect" experiment thought to show the inconsistency of Relativity Theory.
It’s a new version of the well known twins’ paradox with new features: a symmetrical travel and the possibility of them to take photographs of themselves at a crossover point and send them to the other.
Just to not consider the movement of Hearth we will think in a mother-ship that goes to some “quiet” place at space.
The mother-ship goes there, brakes and stops remaining there. After that, two small space-ships with twins accelerate in opposite directions, travel some time and brake in the same manner making a perfect symmetric travel to stop at some far distance.
After that, they turn their space-ships in the opposite direction and at some time (may be synchronized by the mother-ship that is at equal distance from them) they accelerate and travel back in a second symmetrical travel deviating a negligible little (to not collide) just to pass very near of them and the mother-ship at the same instant but they don't brake.
The intention is to capture the movement as they are travelling at some considerable velocity to detect some relativistic effects.
We must consider that the state of both twins at the crossover point can be directly observed by them and by the people in the mother-ship. For example the twins can take photographs of their own face at the instant of crossing and be sent to everybody, even to us, to analyze the phenomenon.
We could observe for example that if one twin aged more he would have him with a long beard while the other would not.
We are going to determine the different observations (relativistic predictions) in the three possible different referentials (one in the mother-ship and one for each twin) and compare them.
Now the situation is:
Both are travelling at some velocity v in relation to the mother-ship but in opposite directions just in front of it.
For simplicity we will consider this instant as time zero and apply Lorentz Transform to the twins to see how they are aging.
First we will consider the problem as seen by the twins themselves.
Choosing the directions of the referentials of the twins as the directions of their velocities they see each other travelling at a velocity w (classically is 2v but with the relativistic addition of velocities is something different).
For them we must consider:
k = (1-w^2/c^2)^-1/2
As with the considered assumptions the twins verify x=wt then for both twins we will have the same times’ relation:
t' = t/k
This means that for each twin the other one age less. They both would see themselves with a long beard and the other one without it.
This means opposite contradictory results.
Now from the referential at the mother-ship both twins are observed aging the same and in a different amount since for it they travel at velocity v and not w. The mother-ship would see both with a “half beard”.
Then two inconsistencies were found because of contradictory results.

NOTE:
If we want to use clocks to compute times elapsed we can synchronize them simultaneously choosing the instant of take off of the space-ships out of the mother-ship.
This means that Relativity Theory is a wrong theory. A right theory cannot be inconsistent; the same phenomenon must have consistent predictions in different referentials of observation.

 

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Note that symmetrical travel by the twins means their beards will be the same throughout.

The presence of the mothership is taken to be at a "fixed" point relative to the journeys that the twins take, and that is the point to which each twin's journey is relative to. Do I have that correct in your estimation?

Ok, no difference in beards at this point, right?

Still no difference in beards yet, since there is no time dilation due to the fact that they have had identical journeys relative to each other and relative to the mothership.

The velocity isn't the determine factor in this case since it is the same, and since at the cross over point they have had identical journeys relative to each other and the mothership.

OK, I got my copies, and the beards where the same in all photos.

No, they have both had a net zero relative motion from start to finish, and so no difference in beards.

I don't think so.

The mother ship would see them both with the same beard, true. But the reference to it being a half beard relative to what the twins would observe is incorrect, since regardless of if you are a traveling twin, or at "rest" in the mothership, there has been no time dilation, if I understand correctly.

No, the contradiction here is that your determination that the twins would see each other to have the longer beard is inconsistent with the effect expected from their symmetrical journeys.

 

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Your análisis is wrong since you tried to find the right solution to the problem all the time while the problem is about the three different predictions of Relativity Theory in each frame. The three predictions are different so giving contradictory results for the same phenomenon. There is the inconsistency.

 

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I think that the first prediction you mention is that the twins could each perceive themselves to be at rest relative to the other twin. In that case they would each conclude that the other twin had travelled at relativistic velocities relative to their rest position, and so the contradiction would be that they wouldn't agee on each others beard length? That boils down to the absence of a common rest position, so neither could know which one was actually in motion.

But you gave us a common rest location with the introduction of the mothership. So each twin would know that they were in motion relative to the common rest position of the mothership. That makes the symmetry 0f their travels override their misinterpretations of who was at rest and who was in motion in the first prediction. Knowing that, they would expect to see the other have the same length beard.

 

How could you call each twin's frame prediction as "misinterpretations"? They are relativistic predictions according to the Lorentz Transforms! I mean you have three frames in the problem and their predictions related by the Lorentz Transforms (that`s for what the transforms exist) and the inconsistency is that they give different contradictory results for the same phenomenon. That's what you refuse to see I think. You continue with the aim to choose the right solution from the three ones while that is not the case. The inconsistency is the existence of three contradictory relativistic predictions.

 

I understand. But there has always been that "contradiction" in SR, ie, the twins paradox. You are right, and I am seeing the solution to the contradiction being in the knowledge of which twin is at rest, and at rest relative to what frame. Your example provides the needed knowledge to know that both twins are in relative motion on symmetrical paths, and so they will experience no time dilation between themselves.

As for the observation from the mothership, their beards will be observed to be the same length, as we agreed.

I didn't read any further into your paper, but does the contradiction you point out make any difference to the rest of your conclusions?

 

Makes no difference at all. The rest has nothing to do with the twins paradox.
Actually I would prefer to consider other topics. I would prefer you to consider topics of the theory itself rather than being stuck in my considerations against Relativity Theory, something very few people have done I think. Who knows it could inspire something good to do in Physics...

 

I did suggest earlier that if you truly believe you have invalidated SR or in any way found fault, then you need to submit a properly peer reviewed paper on the nature of your findings.
But of the many that have gone before you, claiming similar, SR still remains viable and validated.
I'm no professional, but I'm sure one will be along shortly, and if he finds your claim as what I believe it to be, he should show you the error of your ways....
In the meantime.....
http://www.phys.vt.edu/~takeuchi/relativity/notes/section12.html

 

Maybe I'll read on some then, but do you agree that the twins paradox is resolved by knowing for sure which twin is actually in motion and which is at rest?

 

In Relativity Theory always exist the observer "at rest" but in relation to its own frame of reference. In the other frame always exist the moving one always moving in relation to the observer frame. If you change the frame of reference (frame of observation) all this change and you will have a new observer "at rest" in relation to his new reference frame while that first one will appear now moving. As there's no privileged frame you cannot decide which could be really at rest or really moving.

 

Sorry, that's the way a physicist would do but I'm an Engineer and I believe in another way which begun writing an appropriated book to compile all my findings (which are many you know). Now is time to find some ones that could be interested in my work and be interested to do something about.
As for your link I can only say that time dilation and space contraction are both represented simultaneously in the Lorentz Transforms "embedded" in its equations. I do know enough about them.

 

Then if I understand you, in the thought experiment you set up three frames, and used the assumption that the observer is always at rest to then point out the inconsistencies. That is not remarkable, given that the twins paradox is already recognized, which we agree on. Once you introduce a preferred frame, like you do when you establish the rest location of the mothership, you have departed from the original postulates of SR, I would think, but I have no argument with you pointing to the twins paradox as evidence that SR is not a complete physical analysis. Instead, it is a mathematical explanation for concepts of length contraction and time dilation that are consequences of relativity theory.

 

He can be at rest or accelerating for the purposes of observation. They result in different observations.

He can be motionless, moving at a steady speed or accelerating.

Correct - there is no privileged frame.

 

I haven't "departed from the original postulates of SR, you cannot say that. I didn't say the mothership frame would be an absolute rest frame, not at all. It is a special frame in the problem because it allows the setup of a perfect symmetrical travel of the twins and the best point to suit a reference clock to synchronize the travels and even the final point to analyze the situation of the twins but it is just another relativistic frame of reference, nothing else. In the manuscript I say:
"Just to not consider the movement of Hearth we will think in a mother-ship that goes to some “quiet” place at space.
The mother-ship goes there, brakes and stops remaining there."
So the intention is to determine some place far from any other kind of external influence in the problem. That's why I wrote "quite" between quotes. It is needed to determine some place, for example outside the galaxy and yet far from other galaxies which we would identify and set as the destination of the mothership which must reach there and stop there (at that predetermined location turning any propulsion motor off before the twins begin their travel until their back to the initial point) allowing the perfect symmetrical travel of the twins.
I stay within the relativistic assumptions all the time.

 

No, they'd see each other at about the same age. They switch planes of simultaneity as they decelerate at the halfway point of the trip; this effectively 'resets' the perceived differences between the two twins (as they observe the other.)

Draw the Minkowski diagram of your described trip; it will show you what's going on.

 

No, that "halfway point of the trip" as you say could be set as the initial points of the trips to analyze (clocks=zero) if wanted.

 

Yes, you could. And both would still observe the other as having the same age when they passed each other. As they accelerated towards each other, the planes of simultaneity would again shift. (Remember, it is only INERTIAL frames that are locally invariant; once you start accelerating you can no longer make that assumption.)

Again, draw the Minkowski diagram.

 

No way. Don't need Minkowsky diagram. Each one sees the other one at some velocity so aging less.

 

Well, you seem to not understand the results of your thought experiment. If you could draw the Minkowsky diagram, it would help you understand.

 

Although not an expert on Minkowsky diagram I know enough about it.
By the way I was editing the post while you posted.