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

Discussion in 'Alternative Theories' started by Mike_Fontenot, Jun 13, 2021.

1. ### SsssssssRegistered Senior Member

Messages:
302
And since we've already done explicit application of the Lorentz transforms and a shortcut with time dilation let's do this a third way.

Light travelling at speed $\pm c/n$ in the x direction in the rest frame of the interferometer moves at speed $(\pm c/n-v)/(1\mp v/(nc))$ in the frame of the car and in the +x direction it moves a distance $x'_{r1}-x'_0=\gamma L(1-nv/c)$ and in the -x direction it moves a distance $x'_f-x'_{r1}=-2\gamma Lnv/c-\gamma L(1-nv/c)=-\gamma L(1+nv/c)$ where the $x'$s are as defined in #62 with $\theta=0$ so that the interferometer arms are aligned with the +x and +y axes. The total elapsed time is given by distance over speed in the outward direction plus distance over speed in the inward direction which is $\frac{\gamma L(1-nv/c)(1-v/nc)}{c/n-v}-\frac{\gamma L(1+nv/c)(1+v/nc)}{-c/n-v}$which simplifies to $2\gamma L n/c$ as expected.

Now let's look at the light in the other arm of the interferometer. Light travelling at speed $\pm c/n$ in the y direction in the rest frame of the interferometer has x-velocity $-v$ and y-velocity $\pm c/\gamma n$ which makes a speed of $\sqrt{v^2+(c^2-v^2)/n^2}$ and it moves an x distance of $x'_{r2}-x'_0=-\gamma Lnv/c$ and a y distance of $y'_{r2}-y'_0=L$ which means a total distance of $\gamma L\sqrt{1+(n^2-1)v^2/c^2}$ in the upwards direction and the same in the downward direction from symmetry. Again elapsed time is distance over speed for the two legs so it is $\frac{\gamma L\sqrt{1+(n^2-1)v^2/c^2}}{\sqrt{v^2+(c^2-v^2)/n^2}}+\frac{\gamma L\sqrt{1+(n^2-1)v^2/c^2}}{\sqrt{v^2+(c^2-v^2)/n^2}}$which again simplifies to $2\gamma Ln/c$ as expected.

TonyYuan take note that this is the right way to do the calculation you messed up in #56 and #58. Comparing the first calculation to yours you had the velocities (your v1 and v2) correct until you added Earth's orbital velocity for no reason but you had the distances travelled incorrect because you failed to allow for the fact that the interferometer is moving in a frame where it is not at rest. In the second calculation you had the velocity (your v3) wrong because you failed to do the transform properly and you had the distances wrong because you again failed to allow for the fact that the interferometer is moving when it's not at rest. So your belief that there ought to be phase differences comes from your failure to do the maths correctly.

Last edited: Jul 9, 2021

3. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
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)
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
Ssssssss's ans:

elapsed time of light is $1/\sqrt{1-20^2/(3\times 10^8)^2}* [20/(3\times 10^8)]$

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

elapsed time of light is $1/\sqrt{1-20^2/(3\times 10^8)^2}*[20/(3\times 10^8)]$

Last edited: Jul 10, 2021

5. ### SsssssssRegistered Senior Member

Messages:
302
I find it interesting how dishonest you are willing to be. You keep repeating the "your" SR theory even though I've pointed out the false equivalence and repeated it several times. And you keep repeating the "don't avoid the question" thing even after you acknowledged that your experimental description was deficient and finally answered my simple questions. I realise that you have to have a certain blind arrogance to pretend to yourself that you could have overturned GR when you can't even do the maths to correctly describe the simple physical situation we're discussing here but you could at least try to pretend you're thinking about what I'm telling you.
Seriously? After I've proved something three times in three different ways you are asking if it's the right answer? On the other hand this is you copying my expression so let's just check.
I love that you've added brackets when you combined my two expressions but you've done it incompetently so that the result is still ambiguous. It's so typical of your skills. What you've written can be interpreted as$\frac{1}{\sqrt{1-20^2/(3\times 10^8)^2}\times 20/(3\times 10^8)}$ or $\frac{1}{\sqrt{1-20^2/(3\times 10^8)^2}}\frac{20}{3\times 10^8}$depending on whether you interpret $1/ab$ as $1/(ab)$ or $(1/a)b$ which is a matter of personal style so what you wrote is maybe the right answer. It's the second interpretation that is correct at least if you are happy to use $3\times 10^8\mathrm{m/s}$ as the speed of light so it would be much easier to say that the result is what I actually wrote: $2\gamma Ln/c$.

Last edited: Jul 10, 2021

7. ### SsssssssRegistered Senior Member

Messages:
302
And while we're on this topic you first asked the question badly in #63 and finally clarified it in #77 but as I said in my answer in #78 the answer you were asking for was $t'_f$ which I defined in #62. So while you were accusing me of avoiding the question and whining about me daring to ask for clarification I had already answered it (something I could not know due to your poor question but you could) so it is clear that you are either not reading the answers you are getting or failing to understand them. Which is it?

8. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
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)
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
Ssssssss's ans:

elapsed time of light is $[1/\sqrt{1-20^2/(3\times 10^8)^2}] \times [20/(3\times 10^8)]$

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

elapsed time of light is $[1/\sqrt{1-20^2/(3\times 10^8)^2}] \times [20/(3\times 10^8)]$

Ssssssss, is this your last answer? Cut the crap and I'll put the brackets on it for you.

9. ### SsssssssRegistered Senior Member

Messages:
302
You mean stop pointing out that you keep lying and you clearly either don't read or don't understand the responses you get? Then stop lying (all you need to do is say "use SR theory" instead of "use your SR theory") and acknowledge that I did not avoid your question and in fact had answered it even before you asked it. Or contest the accusation with a better explanation for your behaviour.
As I said it is clear that you either do not read or do not understand the responses. From #83: It's the second interpretation that is correct at least if you are happy to use $3\times 10^8\mathrm{m/s}$ as the speed of light. You need to put units on it though as in #61.
Behave with integrity and I won't have grounds to criticise.
Why do you think the brackets are for me? They should have been there because of basic mathematical literacy.

10. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
Don't you think you talk a lot of nonsense?

11. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
1. Parallel to the direction of the earth's revolution speed
Ssssssss's ans:

elapsed time of light is $[1/\sqrt{1-20^2/(3\times 10^8)^2}] \times [20/(3\times 10^8)]$

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

elapsed time of light is $[1/\sqrt{1-20^2/(3\times 10^8)^2}] \times [20/(3\times 10^8)]$

Ssssssss, you just tell us if your answer is like the above. Thanks very much.

12. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
Ssssssss, I will reply to you tomorrow, hoping not to be so nonsense all the time.

13. ### SsssssssRegistered Senior Member

Messages:
302
No. I think everything I say about you can be backed up with evidence from this thread and your "gravitational waves" one.
I already told you twice. What didn't you understand about It's the second interpretation that is correct at least if you are happy to use $3\times 10^8\mathrm{m/s}$ as the speed of light?

14. ### phytiRegistered Senior Member

Messages:
627
Tony;
This pdf is equivalent to the MMX by sending signals 360 deg simultaneously.
Either you didn't understand it or ignored it.
It gets tiresome watching someone argue a case that was settled long ago.

File size:
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1
15. ### exchemistValued Senior Member

Messages:
11,261
Some hope! You have proven to be nonsense all the time so far and I see no sign of change.

16. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
1. Parallel to the direction of the earth's revolution speed
Ssssssss's ans:

elapsed time of light is ________________.

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

elapsed time of light is _________________.

17. ### SsssssssRegistered Senior Member

Messages:
302
No. I have already answered three times and you are supposed to be an intelligent person not a 1980s OCR program that cannot understand anything not written in specified boxes.

I'll make it easy for you and quote exactly what I wrote the first of the three times I already answered you. From post #83:
What you've written can be interpreted as$\frac{1}{\sqrt{1-20^2/(3\times 10^8)^2}\times 20/(3\times 10^8)}$ or $\frac{1}{\sqrt{1-20^2/(3\times 10^8)^2}}\frac{20}{3\times 10^8}$depending on whether you interpret $1/ab$ as $1/(ab)$ or $(1/a)b$ which is a matter of personal style so what you wrote is maybe the right answer. It's the second interpretation that is correct at least if you are happy to use $3\times 10^8\mathrm{m/s}$ as the speed of light so it would be much easier to say that the result is what I actually wrote: $2\gamma Ln/c$.

In the bit I've put in bold which formula do I mean by "the second"?
Is that formula the same as the one you wrote in #85 and #88?

If you are unable to answer these questions then act like an intelligent human being and say where you get stuck instead of repeating your INPUT ERROR COMPUTER DOES NOT UNDERSTAND behaviour.

18. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
1. Parallel to the direction of the earth's revolution speed
Ssssssss's ans:

elapsed time of light is ________________.

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

elapsed time of light is _________________.

Don't worry about the answer I gave anymore, just assume that I have never given any answer.

19. ### SsssssssRegistered Senior Member

Messages:
302
No. For the fifth and final time the correct answer to both your questions is my second formula in post #83 and as repeated in post #94. If you are not confident in your ability to count to two I suggest you enrol in an infant school or if you have some other problem understanding what I wrote then say what you do understand and we can go from there.

Until then I am done with this line of conversation.

Last edited: Jul 11, 2021
20. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
652
I don't know why you don't have the courage to fill in the answer directly above.

21. ### SsssssssRegistered Senior Member

Messages:
302
You are an even worse psychological manipulator than you are physicist which I didn't think was possible.

You have been told my answer to both your questions five times and if you won't accept it because it is not written in the place you want then your problem is your own petty bureaucratic mentality.

Last edited: Jul 11, 2021
22. ### exchemistValued Senior Member

Messages:
11,261
I suspect it is just an increasingly desperate stratagem on his part to be able to claim, at least in his own mind, that you have somehow been defeated by his question.

This fool has been hawking the same shit round numerous science forums, most of which he has now been banned from, of course. So he winds up here.....

Ssssssss likes this.
23. ### SsssssssRegistered Senior Member

Messages:
302
I came back to read this because I was wondering if I was being too obstructionist by just repeating my answer rather than rephrasing given that TonyYuan is working in a second language but then I remembered that this is a guy who claims to have overturned relativity and says that to him Einstein is "just another physics enthusiast". I can see four different ways to tell which of my two formulas is correct:
1. Count to two and take the second one as I've told him
2. Put the units in and see which one has units of time
3. See which one is the same as the algebraic answer $2\gamma Ln/c$ that I've given several times
4. See which one is the product of the velocity I stated in #61 and the gamma factor I stated in #78 when I said that the right answer was the product of the two
Any of these would work even if a reader was struggling a bit with my language. Now I'm a smart guy but I know Einstein was a lot smarter than me and I would expect him to see at least these four ways to verify which formula I meant if he needed to. TonyYuan seems strangely uncertain about these fairly simple concepts for someone who has such a low opinion of Einstein.
I see what you did there.

exchemist likes this.