A possible proof that negative ageing doesn't occur in special relativity?

[Ssssssss]

You can look at my paper, I just use classic physics for analysis. Now James gives a formula for calculating speed. I just apply this formula to the calculation of classical physics. Isn't that okay?

Using Newtonian physics to try to describe light is not okay and neither is mixing James' correct relativistic formula with Newtonian physics so the problems are all with your incompetent application of physics.

Can't James' speed be used for things on earth?

Of course but you have to use it correctly. You are mixing it with Newtonian physics and then trying to pretend it's surprising that the result is nonsense.

It seems that it is not me who get nonsense.

It really is.

If the speed formula given by James is not suitable for simple classical physics calculations, please explain.

You are mixing Newtonian and relativistic formulas as I have already said which is not a problem with either theory but a case of you applying them incompetently.

Please use your SR theory to calculate the elapsed time of light in different directions in Morley's experiment.

Try not to be disingenuous and call it "my" SR theory as if it were some piece of nonsense I'd dreamed up instead of a thoroughly tested and coherent theory that underlies all of modern physics.

You still haven't said which frame you are working in so I am going to choose to work in the local rest frame of the Earth where the apparatus and medium are at rest which means that the total path length along each arm is 20m and the speed of light is isotropic and hence that the answer to both your questions is \(20/(3\times 10^8)\)s. If you want an answer in any other frame you'll have to specify which one you want.

 

[Ssssssss]

Let's just do this because honestly there's zero chance that someone who thinks dimensional analysis is a sign of a poor grasp of physics is going to correctly analyse a relativistic scenario.

Initially we'll work in the rest frame of an MM apparatus immersed in a medium which is also at rest and has refractive index \(n\) in its rest frame. The apparatus is placed with its beam splitter at the origin of coordinates and is rotated so its arms point along directions rotated \(\theta\) anticlockwise from the x any y axes. A pulse of light leaves the beam splitter and enters the two arms at time zero. Clearly they return at the same time from symmetry so there are four interesting events which are the emission event \(E_0\) the reflection event in one arm \(E_{r1}\) the reflection event in the other arm \(E_{r2}\) and the final return event \(E_f\). We can easily write down the coordinates of these events in this frame\[\begin{array}{rrcl}
E_0:&x_0&=&0\\
&y_0&=&0\\
&t_0&=&0\\
E_{r1}:&x_{r1}&=&L\cos\theta\\
&y_{r1}&=&L\sin\theta\\
&t_{r1}&=&n\frac Lc\\
E_{r2}:&x_{r2}&=&-L\sin\theta\\
&y_{r2}&=&L\cos\theta\\
&t_{r2}&=&n\frac Lc\\
E_f:&x_f&=&0\\
&y_f&=&0\\
&t_f&=&2n\frac Lc
\end{array}
\]Then we can use the Lorentz transforms\[\begin{eqnarray*}
x'&=&\gamma(x-vt)\\
y'&=&y\\
t'&=&\gamma\left(t-\frac{v}{c^2}x\right)\\
\gamma&=&\frac{1}{\sqrt{1-v^2/c^2}}
\end{eqnarray*}\]to calculate the coordinates of the events in another frame where the apparatus is moving at speed \(v\) in the \(+x\) direction.\[\begin{array}{rrcl}
E_0:&x'_0&=&0\\
&y'_0&=&0\\
&t'_0&=&0\\
E_{r1}:&x'_{r1}&=&\gamma L\left(\cos\theta-\frac{nv}c\right)\\
&y'_{r1}&=&L\sin\theta\\
&t'_{r1}&=&\gamma\frac Lc\left(n-\frac vc\cos\theta\right)\\
E_{r2}:&x'_{r2}&=&-\gamma L\left(\sin\theta+\frac{nv}c\right)\\
&y'_{r2}&=&L\cos\theta\\
&t'_{r2}&=&\gamma\frac Lc\left(n+\frac vc\sin\theta\right)\\
E_f:&x'_f&=&-2\gamma Ln\frac vc\\
&y'_f&=&0\\
&t'_f&=&2\gamma n\frac Lc
\end{array}
\]So far we haven't used velocity transforms at all and we really don't need to but we can derive them at this point if we want because \(\vec u=(\Delta x/\Delta t,\Delta y/\Delta t)\) and \(\vec u'=(\Delta x'/\Delta t',\Delta y'/\Delta t')\).\[\begin{eqnarray*}
u_x&=&\frac{x_{r1}-x_0}{t_{r1}-t_0}\\
&=&c\frac{\cos\theta}{n}\\
u_y&=&\frac{y_{r1}-y_0}{t_{r1}-t_0}\\
&=&c\frac{\sin\theta}{n}\\
u'_x&=&\frac{x'_{r1}-x'_0}{t'_{r1}-t'_0}\\
&=&c\frac{c\cos\theta-nv}{nc-v\cos\theta}\\
u'_y&=&\frac{y'_{r1}-y'_0}{t'_{r1}-t'_0}\\
&=&c\frac{c\sin\theta}{\gamma\left(nc-v\cos\theta\right)}
\end{eqnarray*}\]All that's left to do is to use the first two results to eliminate \(\theta\) from the second two\[\begin{eqnarray*}
u'_x&=&\frac{u_x-v}{\left(1-\frac{vu_x}{c^2}\right)}\\
u'_y&=&\frac{u_y}{\gamma\left(1-\frac{vu_x}{c^2}\right)}
\end{eqnarray*}\]The transform for the \(x\) velocity is obviously the same as what James wrote but the transformation for the \(y\) velocity is different. Presumably James assumed TonyYuan was talking about the one dimensional case where all velocities are colinear. In this more general situation note that in the case that \(u_x=0\) and \(u_y\neq 0\) that \(u'_x\neq 0\) and \(u'_y\neq u_y\). This is why TonyYuan's calculation for the term he calls v3 is incorrect if he wants to work in a frame where the air and apparatus are moving. The terms he calls v1 and v2 are wrong because he's mixing Newtonian and relativistic maths there. Note that here we've derived the velocity transforms starting from a requirement that there be no time difference in the return time of the pulses so the velocity transforms are obviously consistent with there being no fringes. You are welcome to repeat the derivation using the time and position differences between \(E_0\) and \(E_{r2}\) or between either of the reflection events and \(E_f\) if you want and it would be a good exercise if you've never done it.

TonyYuan: this is how you write a scenario. You state what the experiment is and clearly where all the components are and how they are moving and what frame you intend to work in then you write down the coordinates of various events of interest and then you apply the Lorentz transforms and derive any quantities you want to know. If you keep to this simple recipe you will not end up making embarrassingly mistaken claims about relativity being wrong.

I'm gonna leave this here at this point. That's the correct special relativistic analysis of the Michelson Morley experiment in its rest frame and in a moving frame even in the case that the apparatus is at rest in a medium with a refractive index. You will not see fringes unless the apparatus is moving with respect to a medium which is what M&M thought was going on when they did the experiment. TonyYuan's incoherent mixture of Newtonian and relativistic maths is simply wrong but if he actually lays his scenario out formally like this instead of asking leading questions and failing to understand the answers we might be able to help him do better in the future.

 

[TonyYuan]

You still haven't said which frame you are working in so I am going to choose to work in the local rest frame of the Earth where the apparatus and medium are at rest which means that the total path length along each arm is 20m and the speed of light is isotropic and hence that the answer to both your questions is 20/(3×108)20/(3×108)20/(3\times 10^8)s. If you want an answer in any other frame you'll have to specify which one you want.

OK, now there's a car with a speed 20m/s relative to the ground, parallel to the direction of the earth's revolution. Now take the car as a reference.
Please use your SR theory to calculate the elapsed time of light in different directions in Morley's experiment.
Initial conditions:
1. len = 10m
2. The earth's revolution speed is 29.783km/s.
3. The speed of light in the air is 300000 km/s.
(If you think the len and these speeds are derived from classical physics, please give the len and the speed of light and the speed of the earth under SR.)

1. Parallel to the direction of the earth's revolution speed
Your ans:


2. Perpendicular to the direction of the earth's revolution speed
Your ans:

 

[TonyYuan]

$$v_1=\frac{v_0+100}{1+\frac{100v_0}{c^2}}$$

Suppose v0 = 0.5c, the speed of the medium is 0.5c,
If the light velocity and the medium velocity are in the same direction, then
v1 = (0.5c + 0.5c) / (1 + 0.5c * 0.5c / c / c) = 0.8c, relative to the ground light speed is 0.8c.

If it's in the opposite direction, then
v1 = (0.5c-0.5c) / (1-0.5c * 0.5c / c / c) = 0, relative to the ground light speed is 0.

James, is that right?

 

[Ssssssss]

OK, now there's a car with a speed 20m/s relative to the ground, parallel to the direction of the earth's revolution. Now take the car as a reference.

This is wholly inadequate Tony and you need to go back and read the second and last but one paragraphs of #62. This quality of experimental description would get a failing mark in a child's school science project never mind from someone who thinks he's finding errors in relativity. You need to say whether the MM apparatus is on the ground or in the car and if it's in the car whether or not the car is enclosed. And you should say what len is supposed to be so I don't have to guess whether it's the arm length of the interferometer or the round trip length and you should say what frame your speeds are measured in.

your SR theory

I repeat that this is you attempting to pretend that there's some equivalence between your fantasies and a theory with literally millions of successful experiments supporting it. I called that disingenuous before but you're still doing it so let's call it what it is: a lie. Please stop trying to pretend that you not understanding relativity means it's some incoherent made up rubbish like you peddle.

Another thing why don't you analyse this scenario using relativity once you have actually specified it? If your lie that SR is "my" theory were true then you'd have an excuse for not being able to do it but SR is a well documented theory known to thousands of people and there are a great many resources from which you could learn so you could at least try to demonstrate that you've learned it correctly. I don't expect you to take up this challenge because you think you're a master physicist being all Socratic and challenging the dogma or whatever but you're really not and you'd have a better chance of learning something useful if you at least tried to demonstrate you know what you're doing.

James, is that right?

Yes assuming all velocities are colinear but it's not really needed for anything in this topic as I pointed out in #62.

 

[James R]

Obviously v1, v2 and v3 are not equal, so Morley should be able to observe the obvious movement of interference fringes, but why didn't he observe it?

Because relativity tells us that the speed of light in a vacuum does not depend on the speed of the observer.

 

[TonyYuan]

This is wholly inadequate Tony and you need to go back and read the second and last but one paragraphs of #62. This quality of experimental description would get a failing mark in a child's school science project never mind from someone who thinks he's finding errors in relativity. You need to say whether the MM apparatus is on the ground or in the car and if it's in the car whether or not the car is enclosed. And you should say what len is supposed to be so I don't have to guess whether it's the arm length of the interferometer or the round trip length and you should say what frame your speeds are measured in.

I repeat that this is you attempting to pretend that there's some equivalence between your fantasies and a theory with literally millions of successful experiments supporting it. I called that disingenuous before but you're still doing it so let's call it what it is: a lie. Please stop trying to pretend that you not understanding relativity means it's some incoherent made up rubbish like you peddle.

Another thing why don't you analyse this scenario using relativity once you have actually specified it? If your lie that SR is "my" theory were true then you'd have an excuse for not being able to do it but SR is a well documented theory known to thousands of people and there are a great many resources from which you could learn so you could at least try to demonstrate that you've learned it correctly. I don't expect you to take up this challenge because you think you're a master physicist being all Socratic and challenging the dogma or whatever but you're really not and you'd have a better chance of learning something useful if you at least tried to demonstrate you know what you're doing.

Yes assuming all velocities are colinear but it's not really needed for anything in this topic as I pointed out in #62.

You should be very clear about my question. Did you reply to the post#61? Don't avoid my question, please use your SR knowledge to give the answer.
OK, now there's a car with a speed 20m/s relative to the ground, parallel to the direction of the earth's revolution. Now take the car as a reference.
Please use your SR theory to calculate the elapsed time of light in different directions in Morley's experiment.
Initial conditions:
1. len = 10m (You already know what this is at post#61)
2. The earth's revolution speed is 29.783km/s.(Please Google this speed, you should be very clear about this speed)
3. The speed of light in the air is 300000 km/s.(I think you should know the speed.)
(If you think the len and these speeds are derived from classical physics, please give the len and the speed of light and the speed of the earth under SR.)

1. Parallel to the direction of the earth's revolution speed
Your ans:


2. Perpendicular to the direction of the earth's revolution speed
Your ans:

 

[TonyYuan]

Yes assuming all velocities are colinear but it's not really needed for anything in this topic as I pointed out in #62.

The person I asked is James, and I am waiting for his answer. Thank you.

 

[TonyYuan]

Because relativity tells us that the speed of light in a vacuum does not depend on the speed of the observer.

James, #post64, please give a clear answer instead of telling me that the great Einstein assumed "the speed of light is constant".In my opinion, Einstein is no different from an ordinary physics enthusiast.

 

[James R]

Tony: the vacuum is not a medium.

 

[Ssssssss]

You should be very clear about my question

I cannot be. I told you exactly what you needed to tell me:

You need to say whether the MM apparatus is on the ground or in the car and if it's in the car whether or not the car is enclosed.

You did not do that so I have no idea where you think the apparatus is and what conditions it is operating in. Is it at rest with respect to the air? Is it at rest with respect to the observer? I don't know because you didn't tell me and that's why I said you deserved a failing grade. I made and openly stated an assumption when I answered you in #61 but I have no idea what assumption you intend here so you need to tell me.

your SR theory

You continue to lie I see.

You already know what this is at post#61

As I said I guessed and I shouldn't have to guess about your experiment.

Please Google this speed, you should be very clear about this speed

The Earth's radius is 6,400km making its circumference 40,200,000m and its rotation period is 86400s making the speed at the equator 465m/s. Presumably your 29m/s is the speed at some latitude but why would I care? I don't need it for anything. More to the point why should I have to search for the parameters of your experiment? This is the kind of basic stuff you should explain anyway if you think it's relevant and leave out if not.

The person I asked is James, and I am waiting for his answer.

So? Do you think his answer will be any different?

In my opinion, Einstein is no different from an ordinary physics enthusiast.

Right. I mean the ordinary physics enthusiast develops the basics of atomic power and weapons and contributes fundamental advances to cosmology gravity and quantum theory right?

Actually your v1 v2 and v3 are different. As I said it doesn't matter since it's the travel times that matter and your time1 and time2 would be equal if you hadn't completely messed up the calculation (and then ignored me when I pointed it out if we're going to talk about "avoiding the question").

 

[TonyYuan]

Tony: the vacuum is not a medium.

What is a vacuum, is it vacuum if you can’t see it?
What is a magnetic field?
What is the gravitational field?
We can't see, don't understand, it doesn't mean there is no existence, it doesn't mean there is no matter. "The speed of light is constant" because human beings were ignorant of science at that time and put forward this hypothesis. It is not much different from humans’ proposing the existence of God.

The light is dragged by the gravitational field, it's as simple as that. WE DONT NEED THE "The speed of light is constant".

 

[TonyYuan]

The earth's revolution speed is 29.783km/s.
English is not my native language, and I don't know what words you use to describe the speed of the earth's revolution around the sun.

I cannot be. I told you exactly what you needed to tell me:

You can continue to be ignorant.

 

[Ssssssss]

"The speed of light is constant" because human beings were ignorant of science at that time and put forward this hypothesis.

You are aware that we've directly tested the statement aren't you?

The light is dragged by the gravitational field, it's as simple as that.

This has all the same problems as ether dragging hypotheses. Partial dragging doesn't explain terrestrial experiments and full dragging doesn't explain astronomical ones.

The earth's revolution speed is 29.783km/s.
English is not my native language, and I don't know what words you use to describe the speed of the earth's revolution around the sun.

Orbital speed would be the correct term. Again this is not relevant to anything.

You can continue to be ignorant.

Seriously? You're going to stop talking because I am not a mind reader and have to ask you to clarify your experimental design when you can't be bothered to state it clearly? And you have the nerve to accuse me of avoiding questions? Wow.

 

[TonyYuan]

You are aware that we've directly tested the statement aren't you?

This has all the same problems as ether dragging hypotheses. Partial dragging doesn't explain terrestrial experiments and full dragging doesn't explain astronomical ones.

Orbital speed would be the correct term. Again this is not relevant to anything.

Seriously? You're going to stop talking because I am not a mind reader and have to ask you to clarify your experimental design when you can't be bothered to state it clearly? And you have the nerve to accuse me of avoiding questions? Wow.

You should be very clear about my question. Did you reply to the post#61? Don't avoid my question, please use your SR knowledge to give the answer.
OK, now there's a car with a speed 20m/s relative to the ground, parallel to the direction of the earth's revolution. Now take the car as a reference.
Please use your SR theory to calculate the elapsed time of light in different directions in Morley's experiment.
Initial conditions:
1. len = 10m (You already know what this is at post#61)
2. The earth's revolution speed is 29.783km/s.(Please Google this speed, you should be very clear about this speed)
3. The speed of light in the air is 300000 km/s.(I think you should know the speed.)
(If you think the len and these speeds are derived from classical physics, please give the len and the speed of light and the speed of the earth under SR.)

1. Parallel to the direction of the earth's revolution speed
Your ans:


2. Perpendicular to the direction of the earth's revolution speed
Your ans:

Please do the questions carefully and don't say so many useless words.

 

[Ssssssss]

You should be very clear about my question.

But you still haven't answered my clarifying question about your setup. I cannot answer your question until you answer mine and I do not understand why this is difficult for you. Here is the question: Where is the interferometer? Is it:
(a) on the ground?
(b) inside the car and protected against the wind?
(c) on top of the car not protected against the wind?
You have not answered this question despite this being at least the third time I have asked and now you really only have to give a one letter answer. You must know the answer because it's your experiment but if you still won't tell me what experiment you are thinking of then I cannot analyse it.

your SR theory

Still lying I see.

don't say so many useless words.

Despite me asking specific clarifying questions you have simply repeated the exact same (edit: I suppose you did agree that my guess that len was the arm length rather than the round trip distance was a correct guess so not quite exactly the same) unclear experimental design three times now and you want to lecture me about useless words?

 

[TonyYuan]

But you still haven't answered my clarifying question about your setup. I cannot answer your question until you answer mine and I do not understand why this is difficult for you. Here is the question: Where is the interferometer? Is it:
(a) on the ground?
(b) inside the car and protected against the wind?
(c) on top of the car not protected against the wind?
You have not answered this question despite this being at least the third time I have asked and now you really only have to give a one letter answer. You must know the answer because it's your experiment but if you still won't tell me what experiment you are thinking of then I cannot analyse it.

Still lying I see.

Despite me asking specific clarifying questions you have simply repeated the exact same (edit: I suppose you did agree that my guess that len was the arm length rather than the round trip distance was a correct guess so not quite exactly the same) unclear experimental design three times now and you want to lecture me about useless words?

If you understand Morley’s experiment, you should know that it cannot be done on a car. This experiment can only be done on the ground.
Since you mentioned the wind, by the way, the speed of the wind is 0, please don't ask me whether it is relative to the car or relative to the ground.
Hope you have some thoughts on your own.

 

[Ssssssss]

If you understand Morley’s experiment, you should know that it cannot be done on a car. This experiment can only be done on the ground.

It would be an engineering challenge but there is no theoretical obstacle to doing the experiment in a car and engineering problems are frequently ignored in thought experiments.

So all you want to know is the time for pulse return along each arm as measured by an observer moving with respect to an interferometer that is at rest on the Earth's surface. So the answer to both your questions is \(t'_f=2\gamma nL/c\) as derived in #62 which is just \(\gamma\) times the answer I gave in #61 where \(\gamma\) is the Lorentz gamma factor of the car in the ground frame which is \(1/\sqrt{1-20^2/(3\times 10^8)^2}\).

 

[Ssssssss]

So the answer to both your questions is \(t'_f=2\gamma nL/c\) as derived in #62 which is just \(\gamma\) times the answer I gave in #61

This answer is obvious of course because each arm of the interferometer is just a light clock with the beam splitter replacing one of the mirrors so time dilation says that viewed from a frame where the interferometer is moving both arms take \(\gamma\) times longer than they do in the interferometer's rest frame.

You cannot prove SR wrong with this kind of thought experiment because it is a simple self consistent theory so will always give consistent results to any thought experiment unless you mess up the maths. If you want to prove it wrong you would have to do an actual experiment and show that the results are different from the predictions of SR and we've been doing experiments to test SR for over a century now and never seen any problems that weren't operator error.

 

[phyti]

Neddy;

No matter what times we choose, this signal method only tells us what we already should have known just from the td of 1/2

That was Mike's opinion. IF the experiment goes according to plan, the predictions will be in agreement. It's a conditional statement, depending on a successful experiment, i.e. speculation. Real world future events don't have a guarantee.

No, there is no universal now while the twins are in relative motion, because each one says the other is younger by 1/2 at any given time.

Never said there is. The finite speed of light rules that out.
There is a relative 'now' for each observer by forming a simulated clock synchronization.
When each makes measurements and calculations, it's for the purpose of simultaneity. What does the other clock read when my clock reads t? Each has their own description of the same events which will differ from that of everyone else. That is true even when no relative motion, no two people can occupy the same space simultaneously.
Adding a 3rd person is redundant, and only good for one location!
It also doesn't matter what time is selected to use the signals. The procedure is the same. The verification is always after the event.

The bottom line is always the same. There is no instant knowledge.
Definition-know: be aware of, have facts or evidence of