A contradiction in special relativity?

[lidal]

 

It is a basic requirement of special relativity theory (SRT)  that all relatively moving inertial observers agree on an observable (interference fringe shift, for example). It is shown that SRT trivially leads to a disagreement on the observables ( interference fringe shift) in two relatively moving inertial reference frames.

Suppose that the interference fringe shift of a light speed experiment predicted in frame S is N, and the fringe shift predicted in frame S’ for the same experiment is N’. It is a requirement of relativity theory that there should be an agreement on the observable (the fringe shift) in both frames, i.e.  N = N’. Let us see if this is actually the case.

We know that,

                       N  = c ∆t / λ      and      N'  =  c ∆t' / λ'

But,

                    ∆t'  =  γ  ( ∆t  -  v ∆x/c² )          and    λ'   =   γ λ

 

Therefore,

                N'  =  c ∆t' / λ'   =   c ( γ  ( ∆t  -  v ∆x/c² ) ) / (γ λ) =   (c ∆t / λ  ) -  ( v ∆x/(cλ)  =  N -  ( v ∆x/(cλ) ≠ N 

 

Galilean relativity, however, does not lead to such disagreement, as shown below.

                                    ∆x'  =  ∆x  -  v ∆t

                                   ∆t'  =  ∆t

                                        c'  =  c ± V  ⟹  c' / c = 1± v/c 

                                         f' = f   ⟹ c' / λ'  =  c / λ  ⟹  λ'  = λ (1± v/c)

 

Therefore,

                                 N'  = ( c' ∆t' ) / λ'  =  ( c ± V ) (∆t) / λ (1± v/c)  = c ∆t / λ = N 

 

 

 

 

 

trivial disproof of SRT 3 FINAL.pdf

[Bufofrog]

2 hours ago, lidal said:

N  = c ∆t / λ

Could you give a source for that equation?   N = 7.5 x 10^14 for a time of 1 sec.  I don't really know what that is supposed to tell me.  Thanks

[lidal]

10 hours ago, Bufofrog said:

Could you give a source for that equation?   N = 7.5 x 10^14 for a time of 1 sec.  I don't really know what that is supposed to tell me.  Thanks

 

You may read " Michelson-Morley experiment", Wikipedia 

∆t is the (change in ) difference in arrival times of the two interfering light beams, caused by observer motion (for example, Sagnac effect) or possibly absolute motion.

Also, " Sagnac effect" , Wikipedia

       ∆ф = 2Πc∆t/λ

[Sensei]

22 minutes ago, Bufofrog said:

Could you give a source for that equation?   N = 7.5 x 10^14 for a time of 1 sec.  I don't really know what that is supposed to tell me.  Thanks

c has unit m/s

t has unit s

λ has unit m

so ( m/s * s ) / m is unitless scalar..

22 minutes ago, Bufofrog said:

N = 7.5 x 10^14 for a time of 1 sec.

How did you get such value? For visible light?

N = 299792458 m/s * 1s / 400*10^-9 m = 0.00074948114 for 400 nm wavelength

N' = 299792458 m/s * 1s / 700*10^-9 m = 0.00042827494 for 700 nm wavelength

 

[Bufofrog]

 

9 minutes ago, Sensei said:

N = 299792458 m/s * 1s / 400*10^-9 m = 0.00074948114

I think you made a math error, you're dividing by 10^-9.

15 minutes ago, lidal said:

You may read " Michelson-Morley experiment", Wikipedia

That didn't help, I could not find that equation in the article

[Sensei]

13 minutes ago, Bufofrog said:

I think you made a math error

..you're right.. I just put the equation to the Google calculator:

https://www.google.com/search?q=299792458%2F+400*10^-9

but it should be:

https://www.google.com/search?q=299792458%2F(400*10^-9)

or

https://www.google.com/search?q=299792458%2F+400e-9

 

[lidal]

10 hours ago, Bufofrog said:

 

 

That didn't help, I could not find that equation in the article

 Go to the formula for the fringe shift, n :

               n = ( ∆λ₁  -  ∆λ₂ )/ λ

where ∆λ₁ and ∆λ₁ are the difference in path lengths of the two beams for two absolute velocities.

Also read the Wikipedia article " Sagnac effect ". 

[swansont]

The Sagnac effect is for rotational systems, which are not inertial frames.

Any article that mentions absolute velocities isn’t a reliable source of relativity information ; it is not mentioned in the body of the wikipedia article (which you should link to if you’re going to reference it)

1 hour ago, lidal said:

 Go to the formula for the fringe shift, n :

               n = ( ∆λ₁  -  ∆λ₂ )/ λ

where ∆λ₁ and ∆λ₁ are the difference in path lengths of the two beams for two absolute velocities.

Also read the Wikipedia article " Sagnac effect ". 

 

What the article actually says is “the fringes of the interference pattern should shift when rotating it by 90° as the two beams have exchanged roles. To find the fringe shift, subtract the path difference in first orientation by the path difference in the second, then divide by the wavelength, λ, of light”

and rather importantly this is an equation that assumes an aether, and was found to be wrong. It’s not derived from SR. So how does this show anything about SR?

[Bufofrog]

2 hours ago, Sensei said:

..you're right.. I just put the equation to the Google calculator:

https://www.google.com/search?q=299792458%2F+400*10^-9

but it should be:

https://www.google.com/search?q=299792458%2F(400*10^-9)

or

https://www.google.com/search?q=299792458%2F+400e-9

I could easily be just loosing my mind but isn't the equation this:

= c ∆t / λ = 299792458 m/s X 1sec / 400*10^-9 m = 299792458 m / 400*10^-9 m = 74,931,145,000,000?

400 nm X  (10^-9 m/1 nm) = 400 x 10^-9 m

3 hours ago, lidal said:

Go to the formula for the fringe shift, n :

               n = ( ∆λ₁  -  ∆λ₂ )/ λ

where ∆λ₁ and ∆λ₁ are the difference in path lengths of the two beams for two absolute velocities.

I asked you for a source for N  = c ∆t / λ, so are you saying ( ∆λ₁  -  ∆λ₂ )/ λ = c ∆t / λ?

[lidal]

Just now, swansont said:

The Sagnac effect is for rotational systems, which are not inertial frames.

Any article that mentions absolute velocities isn’t a reliable source of relativity information ; it is not mentioned in the body of the wikipedia article (which you should link to if you’re going to reference it)

 

What the article actually says is “the fringes of the interference pattern should shift when rotating it by 90° as the two beams have exchanged roles. To find the fringe shift, subtract the path difference in first orientation by the path difference in the second, then divide by the wavelength, λ, of light”

and rather importantly this is an equation that assumes an aether, and was found to be wrong. It’s not derived from SR. So how does this show anything about SR?

In this case I did not mention in support for my argument. The formula N  = c ∆t / λ, is not specific to the Sagnac effect; it applies also to the Michelson-Morley experiment. 

 

[swansont]

31 minutes ago, lidal said:

In this case I did not mention in support for my argument. The formula N  = c ∆t / λ, is not specific to the Sagnac effect; it applies also to the Michelson-Morley experiment. 

The M-M experiment failed. There was no fringe shift. Do you have an equation that works, i.e. is consistent with relativity, to discuss?

[lidal]

Just now, swansont said:

The M-M experiment failed. There was no fringe shift. Do you have an equation that works, i.e. is consistent with relativity, to discuss?

 

Just now, swansont said:

The M-M experiment failed. There was no fringe shift. Do you have an equation that works, i.e. is consistent with relativity, to discuss?

The Michelson-Morley (MM ) experiment did not detect a fringe shift because                 N' = c ∆t'/ λ' =  c ( γ  ( ∆t  -  v ∆x/c² ) ) / (γ λ) =   (c ∆t / λ  ) -  ( v ∆x/(cλ)  = 0

, because ∆x = 0 and ∆t = 0.

Therefore, in the case of MM, N= N' = 0 .

The Michelson-Morley experiment is perhaps the only experiment to which special relativity applies correctly.

[lidal]

On 6/8/2022 at 10:07 AM, Bufofrog said:

 

I asked you for a source for N  = c ∆t / λ, so are you saying ( ∆λ₁  -  ∆λ₂ )/ λ = c ∆t / λ?

Introduction to Special Relativity, Robert Resnick, page 23

Einstein's Space Time  An Introduction to Special and General Relativity, Rafael Ferraro, page 39

Note that ∆t in the formula N  = c ∆t / λ , is the change in time difference the two light beams, induced by motion.

 

[swansont]

43 minutes ago, lidal said:

Introduction to Special Relativity, Robert Resnick, page 23

Einstein's Space Time  An Introduction to Special and General Relativity, Rafael Ferraro, page 39

Are they citing the equation based on an aether, or one based on special relativity? Since you cite pages early in the book, it seems the former is likely.

14 hours ago, lidal said:

 

The Michelson-Morley (MM ) experiment did not detect a fringe shift because                 N' = c ∆t'/ λ' =  c ( γ  ( ∆t  -  v ∆x/c² ) ) / (γ λ) =   (c ∆t / λ  ) -  ( v ∆x/(cλ)  = 0

, because ∆x = 0 and ∆t = 0.

Therefore, in the case of MM, N= N' = 0 .

The Michelson-Morley experiment is perhaps the only experiment to which special relativity applies correctly.

You can’t use SR applied to an equation derived assuming the existence of an aether. The assumption is invalid, so the equation can’t be assumed to hold.

Saying ∆x = 0 and ∆t = 0 is an assertion that time dilation and length contraction don’t happen, which is a circular argument. And we have experimental evidence that SR is correct.

 

[lidal]

On 6/9/2022 at 2:39 AM, swansont said:

 

You can’t use SR applied to an equation derived assuming the existence of an aether. The assumption is invalid, so the equation can’t be assumed to hold.

 

The formula for fringe shift applies to any inertial reference frame!

[swansont]

1 hour ago, lidal said:

The formula for fringe shift applies to any inertial reference frame!

Please show the derivation