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A Disproof of the Principle and Theory of Relativity


lidal

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 A Disproof of the Principle and Theory of Relativity

Galileo’s ship thought experiment:

 Consider a light source emitting a light pulse from some point in the Earth's frame, at t=0.  At the instant of light emission, an observer/detector is at distance D from the source and is moving away from the source with velocity v, in the Earth's frame. 

 We know that the light will catch up with the observer/detector at t = D/ ( c - v ) . This is a well-known and accepted fact even in the Special Relativity Theory SRT and has been confirmed by experiments. Now I will use this in my argument against the principle of relativity.

 Consider Galileo's ship thought experiment. A physicist in a closed room of the ship is doing a physics experiment. There are two light sources S1 and S2, with the distance between them equal to2D. The line connecting the sources is parallel to the longitudinal axis of the ship, and hence to the velocity of the ship. S2 is in front of S1.  A detector is placed at the midpoint between the sources, at distance D from each of the sources. The light sources each emit a short light pulse simultaneously every second.  The detector detects the time difference between the pulses.

The observer in the closed room first has to synchronize the clocks at S1 and S2. For this, a short light pulse is emitted from S1 towards S2. Suppose that S1 emits the light pulse at t=0. The physicist in the closed room synchronizes the clocks based on the principle of isotropy of the speed of light, because according to SRT the speed of light is isotropic in Galileo’s ship! However, unknown to him/her, we know that the clocks synchronized by this procedure will be out of synch by an amount:

                                                 (  2D/ ( c - v )  )   -   2D/c    =  2D  v / v(c-v)  

The clock at S2 will be behind the clock at S1 by this amount.

 It should be noted that, according to special relativity, the clocks synchronized by this procedure will be in synch. However, from experience we know that the clocks will be out of synch. Therefore, we know that the relativistic procedure is wrong, based on experience. Therefore we analyze the experiment classically as follows.

 

   
     
 

image.png.5d957ce8611bf2338103ba2019b25fb7.png

 

The sources each emit a short light 'simultaneously' (quoted because the clocks are not actually in synch), every second. The physicist expects the pulses to arrive simultaneously, which they do not, as we will see.

Let S1 emit the light pulse at t = t0. Then S2 will emit 'simultaneously' at time,

                                            t0  +  2D  v / v(c-v)

The light from S1 arrives at the detector at time,

                                       t1 =     t0  +  D/(c -v)

The light from S2 arrives at the detector at time,

                                   t2 =    [    t0  +  2D  v / v(c-v)  ]  D/(c+v)

The difference in the time of arrival of the two pulses at the detector will be:

                                    t2 -  t1  =  (2D/c) β2 /(1-β2 )  

where β  = v/c

The physicist synchronized the clocks by assuming isotropy of the speed of light, placed the detector at the midpoint between the sources, and the sources emitted light pulses 'simultaneously'. He/she would expect the light pulses to arrive simultaneously at the detector, which they don't. The light pulses always arrive with a time difference of Δ that depends on velocity. The observer would have no way to explain this other than abandoning the principle of isotropy of the speed of light. To anyone rejecting this argument, my response is this: let an actual experiment be done to test it. We know that the origin of the problem lies in the observer assuming isotropy of the speed of light while synchronizing the clocks. This disproves both the principle and theory of relativity.

 

 

 

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An immediate problem I see is that you are using v to refer to the relative speed of the detector with respect to the emitter, and also the speed of the rocket ship. Since these are different things, you should use different variables, so you don’t use the wrong one.

For instance, in 2D/ ( c - v ) used in your clock synchronization, v=0 because the emitter and detector are at rest with respect to each other. The rocket speed doesn’t show up in the equation.

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4 hours ago, lidal said:

At the instant of light emission, an observer/detector is at distance D from the source and is moving away from the source with velocity v, in the Earth's frame. 

A a distance D in which frame ?

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Just now, swansont said:

An immediate problem I see is that you are using v to refer to the relative speed of the detector with respect to the emitter, and also the speed of the rocket ship. Since these are different things, you should use different variables, so you don’t use the wrong one.

For instance, in 2D/ ( c - v ) used in your clock synchronization, v=0 because the emitter and detector are at rest with respect to each other. The rocket speed doesn’t show up in the equation.

1. The sources and the detector are fixed to the ship. V is the velocity of the ship.

2. It is not a space ship, it is Galileo's ship steadily sailing on the sea. So it is a terresterial experiment. 

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1 hour ago, lidal said:

1. The sources and the detector are fixed to the ship. V is the velocity of the ship.

But in your original equation. v is the relative velocity. That’s the problem. The source and detector are stationary with respect to each other, so there is no difference in the time.

1 hour ago, lidal said:

2. It is not a space ship, it is Galileo's ship steadily sailing on the sea. So it is a terresterial experiment. 

Whatever. 

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11 hours ago, studiot said:

A a distance D in which frame ?

No, this is a classical analysis. I am claiming that if a real experiment was done, by following the principle of light speed isotropy to synchronize the clocks, the physicist inside the closed room could detect some time difference between the two pulses, which disproves relativity. My argument is this: 1. If you say that no time difference will be detected, then let an actual experiment be done. 2. If you accept that a difference in time of arrival will be detected, then this disproves relativity. 

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9 hours ago, swansont said:

What “experience” is this? 

 

For example, GPS Sagnac correction.

11 hours ago, swansont said:

But in your original equation. v is the relative velocity. That’s the problem. The source and detector are stationary with respect to each other, so there is no difference in the time.

 

In the first part, I was only referring to our experience regarding moving observer, for example, GPS Sagnac correction. So you can start reading from the third paragraph talking about Galileo's ship.

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It seems that you're groping your way towards a theory of absolute orientation...

Mmmm...

I'm looking forward to your Earth-shattering predictions.

Otherwise, you've just misinterpreted the results of experiments, which sounds like what's really happened.

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5 hours ago, lidal said:

For example, GPS Sagnac correction.

In the first part, I was only referring to our experience regarding moving observer, for example, GPS Sagnac correction. So you can start reading from the third paragraph talking about Galileo's ship.

Sagnac correction happens because the clocks are in an accelerated (rotating) coordinate system. SR applies to inertial systems.

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32 minutes ago, swansont said:

Sagnac correction happens because the clocks are in an accelerated (rotating) coordinate system. SR applies to inertial systems.

I think the GPS receiver is approximated to be in inertial motion during the very short transit time of the satellite signals. But there is no need for us to argue over whether light takes more time to catch up with an observer moving away from the light source. My argument is simple: in the Galileo's ship thought experiment described in the OP, will t2 - t1 be zero or different from zero ? 

Edited by lidal
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25 minutes ago, lidal said:

I think the GPS receiver is approximated to be in inertial motion during the very short transit time of the satellite signals. But there is no need for us to argue over whether light takes more time to catch up with an observer moving away from the light source. My argument is simple: in the Galileo's ship thought experiment described in the OP, will t2 - t1 be zero or different from zero ? 

If we’re in an inertial system, since v=0 (the relative speed), the times will be the same.

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7 hours ago, lidal said:

No, this is a classical analysis.

And you are making the classic(al) mistake.

That of mixing values in two different frames.

 

When you specify or calculate a value or ask a question about a quantity that result is meaningless unless you also specify the frame you are working in.

This is true in both galilean and einstinian relativity.

 

So I ask  the question again

In which frame are you measuring D.

 

Note you also need to do this for time quantities.

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18 minutes ago, swansont said:

If we’re in an inertial system, since v=0 (the relative speed), the times will be the same.

So you are saying that t2 - t1 = 0, that is the pulses will arrive simultaneously.

But the experiment can be approximated to be inertial to a high degree of precision (of course, if you agree on this). Supposing you agree, I would say that the two pulses will not arrive at the detector simultaneously. 

Since you can't convince me (and I can't convince you) by arguments alone, the only way to settle this would be to carry out a real version of the thought experiment described above.

But I also wonder how many mainstream physicists would agree with you when you said the two pulses will arrive simultaneously.

7 minutes ago, studiot said:

And you are making the classic(al) mistake.

That of mixing values in two different frames.

 

When you specify or calculate a value or ask a question about a quantity that result is meaningless unless you also specify the frame you are working in.

This is true in both galilean and einstinian relativity.

 

So I ask  the question again

In which frame are you measuring D.

 

Note you also need to do this for time quantities.

 

Let us not mix-up two things. The proposed experiment and my analysis of the experiment, and therefore my prediction of the outcome (the calculations). 

I could have omitted the analysis and gone directly to the thought experiment. The experiment is as described already. A physicist is inside a closed room in a steadily sailing ship. He/she is doing a physics experiment. He places a detector at the mid- point between two light sources. The clocks at the sources are synchronized by the relativistic procedure, that is by assuming isotropy of the speed of light. The sources emit short light pulses simultaneously, say, every second. Will the two pulses arrive at the detector simultaneously or not ? 

 

Edited by lidal
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2 minutes ago, lidal said:

But I also wonder how many mainstream physicists would agree with you when you said the two pulses arrive simultaneously.

Swansont, of course, didn't say such silly thing. "Simultaneous" is a relative concept, as you should know by now. In what frame?

Both @studiot and @swansont have pointed out or implied... In what frame? at several points. Almost every other claim of inconsistency of SR has a flaw of this kind. Yours has it too.

I confess I'm taking all this very lightly, trying not to stray too deep into your personal rabbit's hole, choosing to fix on one particular inalienable reason why your argument cannot be true.

But I must be doing fine, as you chose not to address any of my concerns. Clear indication that you have no answer to them. 

Simultaneous? In what frame? Non-isotropic space within a Galilean ship? How could that be?

 

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1 hour ago, joigus said:

Swansont, of course, didn't say such silly thing. "Simultaneous" is a relative concept, as you should know by now. In what frame?

Both @studiot and @swansont have pointed out or implied... In what frame? at several points. Almost every other claim of inconsistency of SR has a flaw of this kind. Yours has it too.

I confess I'm taking all this very lightly, trying not to stray too deep into your personal rabbit's hole, choosing to fix on one particular inalienable reason why your argument cannot be true.

But I must be doing fine, as you chose not to address any of my concerns. Clear indication that you have no answer to them. 

Simultaneous? In what frame? Non-isotropic space within a Galilean ship? How could that be?

 

 I didn't mention the frame because it was too obvious. Since we are talking about time difference as measured in the ship's frame, I meant simultaneous in the reference frame of the ship. 

And which of your comments did I fail to address? In the one comment you already gave, it seemed you didn't mean it to be replied to because it was more general. To address your concern, my argument is not like such and such experiment disproves relativity. I started from a fact that has been a source of confusions and debates: that light takes more time to catch-up with a moving observer. This fact is often used by 'anti-relativists' to refute special relativity, and mainstream accepts it to be conforming to relativity. However, despite countless efforts, I have seen no argument that has made effective use of this fact against SRT. The arguments usually gave room for (not always consistent) relativistic counter-arguments. The present argument leaves no room for such counter-arguments, for example, reference frame arguments, etc. Note again that I am not directly arguing that moving observer experiments disprove relativity. I am using this fact effectively in a way that refutes relativity.

I proposed a simple thought experiment. In the thought experiment a simple question was asked: is t2 - t1 = 0 .  ( well, in the reference frame of the ship, as measured by the detector). That is, what time difference will the detector record? The time difference could then be printed on paper to avoid "in what reference frame?" arguments.  

Edited by lidal
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28 minutes ago, lidal said:

I didn't mention the frame because it was too obvious. Since we are talking about time difference as measured in the ship's frame, I meant simultaneous in the reference frame of the ship. 

There is no v in the problem you're setting up. The whole point of relativity is that S2 is not receding. No observer attached to the ship can measure any such v. There is no dragging of the speed of light. The measured speed for light inside the ship is c, not c-v, as you claim. You got everything wrong. This v is only in your mind. So the very first time you wrote an equation and you said you were going to hold it against relativity. Well... You wrote the wrong equation. S2 is not receding from the POV of anybody anchored to the ship --at rest relative to it.

Nobody, repeat, nobody stuck to the ship has any right to even start talking about such v. What v? What are you talking about?

What you are doing is using a quantity that only makes sense in the frame attached to a certain "rest observer" you're telling us nothing about in your calculations about another frame.

And yes, according to your main line of reasoning, space within a ship is non-isotropic, which is a ludicrous claim, of course.

On 10/30/2023 at 1:43 PM, lidal said:

We know that the light will catch up with the observer/detector at t = D/ ( c - v ) . This is a well-known and accepted fact even in the Special Relativity Theory SRT and has been confirmed by experiments. Now I will use this in my argument against the principle of relativity.

(My emphasis.)

What?! This is false. In the ship's inertial system such time is t=D/c, not what you say.

The whole thing is ridiculous.

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Good stuff Joigus.  +1

The observer is at rest in his own frame.

But it is also worth pointing out that in his additional papers and books, Einstein took great pains to address the issue of simultaneity.

That is how does the observer know (measure) this ?

 

 

 

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50 minutes ago, studiot said:

Good stuff Joigus.  +1

Thank you. But it was both you and Swanson who set me on the right track when you said,

3 hours ago, studiot said:

And you are making the classic(al) mistake.

That of mixing values in two different frames.

 

3 hours ago, swansont said:

If we’re in an inertial system, since v=0 (the relative speed), the times will be the same.

I've credited accordingly.

You see, it sometimes takes some time to see how deep in shit* an OP's argument is.

You have to, in a way, accept an unacceptable logic.

* Pardon my French.

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2 hours ago, lidal said:

I started from a fact that has been a source of confusions and debates: that light takes more time to catch-up with a moving observer.

You keep doing this - moving with respect to what? It’s a critical detail.

Answer: with respect to the source. If the source and observer aren’t moving with respect to each other, it doesn’t take more time.

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Just now, joigus said:

 The whole point of relativity is that S2 is not receding. 

S2 is not receding from the POV of anybody anchored to the ship --at rest relative to it.

Can you put this in another way? I don't understand it when you say "S2 is not receding"

And did you mean S1 when you said S2?

 

Just now, joigus said:

 

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18 minutes ago, lidal said:

Can you put this in another way? I don't understand it when you say "S2 is not receding"

And did you mean S1 when you said S2?

 

You have mentioned two things without explanation.

1)

In the title you refer to The Principle of Relativity.

2) You refer to the isotropy of space.

 

Can you enlighten us as to what you think these two phrases mean please ?

Where , for instance, did they come from ?

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15 minutes ago, lidal said:

Can you put this in another way? I don't understand it when you say "S2 is not receding"

And did you mean S1 when you said S2?

It doesn't matter.  The point is both of the light pulses S1 and S2 will move at c relative to the observer on the ship, v never is considered.

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5 minutes ago, lidal said:

Can you put this in another way? I don't understand it when you say "S2 is not receding"

 

Ok. Riddle me this: S2 recedes with respect to what --or whom--?

Not the detector --or its associate observer--, since those see no motion.

Only with respect to a fiducial --but, mind you, unmentioned by you-- observer sitting at the port's pier, let's say.

See the flaw?

If you go over Einstein's classic papers, you will notice that all the c-v or c+v terms come from observers calculating distances or delay times from their "rest" frames! According to them, the wall (mirror, or whatever) has moved. Not according to the co-moving observer!

x-posted with @Bufofrog

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