Three Experiments Challenging SRT

Possibly if you could show that you understood the "relativity of simultaneity" thoroughly we may make progress. It is not a matter of whether it is right or wrong, it is merely a matter of knowing what you are working with.

 

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So many people have been stuck on this pseudo paradox of two contradictory words... "relative" and "simultaneity" when combined indicate a paradox yes?
I agree with this assessment that it is indeed indicative of a failed theory as science attempts to deal with a premise that forces it into this situation.
However it is the current state of play and virtually impossible to over turn regardless of how counter intuitive [ paradoxical] it may seem.
I can assure you that every arguement you put up will fail unless you deal with the fundamental reasons for the need for SRT to begin with.

Challenge:

Solve the problem as if, no one has solved it.
Imagine you are confronted with this giant puzzle

• You have light speed invariance.
• You have proven time dilation phenonema.
• you state that Light speed invariance exists for all observers regardless of relative velocity.

The key criteria is:
That all observers will measure 'c' to be the same value regardless of relative velocity.


Now work out a theory that accommodates all aspects totally and fully with out exception across all phenonema observed.

you will end up with SRT as no other solution is available.
and if you are really into it you will derive E=mc^2 as well.

If you do the above exercise you will indeed end up with SRT as you must do so to accommodate the "belief" not "fact" that light travels across a vacuumous void of space at the speed of 'c' in a way that is invariant to all observers. A belief started by 'Ole Rommer's observation of the moons of Jupiter in 1644.


In other words create a SRT as if it was your own....and see why it can't be rendered false.
Discover the sheer beauty of the Lorentz Transforms and work out why there is no other way for your self.

 

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Little change:

1. Our screen are one and only.
2. This screen has an electronic device with a electric light (bulb).
3. If the rate of sequence of picture areas of the first observer more: the bulb lights up.
Otherwise: the bulb not lights up.

The first observer believes that the light must shine, and the second observer believes that the light should not shine.

Question: Will the light shine?
Which of observers to be right, and - why?

Verbiage about the relativity of certainty does not solve the problem of contradiction SRT to common sense, which arise because of time dilation.
The principle of simultaneity does not resolve this conflict.

What makes you deceive yourself?

 

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Slowing down time is not a necessary condition for the invariance of the speed of light.
The speed of light can be invariant without slowing down time.
Master Theory has the invariant speed of light and the absolute time.

 

How does this theory account for each observer measuring 'c' regardless of relative velocity?
How does it maintain the integrity of dopler effects regardless of relative velocity?
I must admit I am a bit perplexed.
If light speed is invariant to all observers regardless of relative velocity how can it be measured as such if there is no time dilation [ at the least]?

 

If tell us briefly:

SRT transverse scale is fixed.
Einstein (without explanation) made the transverse scale absolute.

I take away this absoluteness from the transverse scale, and give away this absoluteness to the time.
The invariance of the speed of light remains.

Look Master Teory.

 

False:

From On the Electrodynamics of Moving Bodies, near the end of section 3.

This is more or less the same thing I explained way back in [POST=2957966]post #63[/POST]: there are many possible velocity dependent transformations with an invariant c, but most of them don't form a valid symmetry group. Masterov's in particular does not. His claim that he has absolute time is also false, as I explained in the same post:

 

I am wrong in this place possible.
Einstein gave an explanation of the reasons that led him to transverse scale absolute.
I have shown (in Master Theory) that there is no reason to make the transverse scale of the absolute.

 

You've ignored the reason for making the transverse scale absolute. That's not the same as showing there is no reason.

The reason is that the physical significance of a coordinate transformation, like the Lorentz transformation, comes from the fact that it forms a symmetry group so it can be incorporated as a symmetry of physical theories. In general you can't just assert that clocks are dilating by whatever factor you want, or that objects are contracting in length however you want. You would normally have to give a detailed explanation of how and why that happens in terms of the object's components and how they are affected by velocity. Instead you just give arbitrary equations. The fact that the Lorentz group is a symmetry group is the reason we don't have to worry about these complications in relativity: clocks will all dilate by the same factor of \(1/\gamma\), regardless of the details of their construction, if their structure and evolution is governed by Lorentz-symmetric physics. Without symmetry, you just have arbitrary equations and no reason they should be followed by real physical clocks.

Like I said before, you can multiply the Lorentz transformation by any factor \(\lambda\) and still have an invariant speed of light. For this sort of problem, it is convenient to express the Lorentz transformation in terms of the rapidity \(\phi\), defined by \(\tanh(\phi) \,=\, v/c\). With this definition, \(\gamma \,=\, \cosh(\phi)\) and \(\gamma v/c \,=\, \sinh(\phi)\). With this parameter, the generalised Lorentz boost along the x axis can be written in matrix form as

\( \begin{bmatrix} ct' \\ x' \end{bmatrix} \,=\, \lambda(\phi) \begin{bmatrix} \cosh(\phi) & -\sinh(\phi) \\ -\sinh(\phi) & \cosh(\phi) \end{bmatrix} \begin{bmatrix} ct \\ x \end{bmatrix} \,, \)​

where in general the scaling factor \(\lambda = \lambda(\phi)\) is allowed to depend on the rapidity (and therefore the velocity).

If you perform two successive boosts \((ct,\, x) \,\rightarrow\, (ct',\, x')\) and \((ct',\, x') \,\rightarrow\, (ct'',\, x'')\) of rapidities \(\phi_{1}\) and \(\phi_{2}\), you get:

\( \begin{eqnarray} \begin{bmatrix} ct'' \\ x'' \end{bmatrix} &=& \lambda(\phi_{2}) \begin{bmatrix} \cosh(\phi_{2}) & -\sinh(\phi_{2}) \\ -\sinh(\phi_{2}) & \cosh(\phi_{2}) \end{bmatrix} \lambda(\phi_{1}) \begin{bmatrix} \cosh(\phi_{1}) & -\sinh(\phi_{1}) \\ -\sinh(\phi_{1}) & \cosh(\phi_{1}) \end{bmatrix} \begin{bmatrix} ct \\ x \end{bmatrix} \\ &=& \lambda(\phi_{1}) \lambda(\phi_{2}) \begin{bmatrix} \cosh(\phi_{1} \,+\, \phi_{2}) & -\sinh(\phi_{1} \,+\, \phi_{2}) \\ -\sinh(\phi_{1} \,+\, \phi_{2}) & \cosh(\phi_{1} \,+\, \phi_{2}) \end{bmatrix} \begin{bmatrix} ct \\ x \end{bmatrix} \,. \end{eqnarray} \)​

The symmetry group condition says this has to take the same form as a single boost:

\( \begin{bmatrix} ct'' \\ x'' \end{bmatrix} \,=\, \lambda(\tilde{\phi}) \begin{bmatrix} \cosh(\tilde{\phi}) & -\sinh(\tilde{\phi}) \\ -\sinh(\tilde{\phi}) & \cosh(\tilde{\phi}) \end{bmatrix} \begin{bmatrix} ct \\ x \end{bmatrix} \,, \)​

for some rapidity \(\tilde{\phi} \,=\, \tilde{\phi}(\phi_{1},\, \phi_{2})\). This imposes the constraints

\( \begin{eqnarray} \lambda(\tilde{\phi}) \cosh(\tilde{\phi}) &=& \lambda(\phi_{1}) \lambda(\phi_{2}) \cosh(\phi_{1} \,+\, \phi_{2}) \,, \\ \lambda(\tilde{\phi}) \sinh(\tilde{\phi}) &=& \lambda(\phi_{1}) \lambda(\phi_{2}) \sinh(\phi_{1} \,+\, \phi_{2}) \,. \end{eqnarray} \)​

It is easy to show that \(\tilde{\phi} \,=\, \phi_{1} \,+\, \phi_{2}\) and \(\lambda(\phi_{1} \,+\, \phi_{2}) \,=\, \lambda(\phi_{1}) \lambda(\phi_{2})\), which only works if \(\lambda(\phi) \,=\, 1\) or \(\lambda(\phi)\) is exponential, and that's only considering successive boosts in the same direction. I would expect that in 1+3 dimensions the exponential \(\lambda\) also gets ruled out.

Either way, your own result does not conform to these constraints. Like I have said before, you do not have a symmetry group.

 

Are you trolling or do you actually believe this? The GPS system is hard evidence that Einstein was correct. You should have all the reference material available to you to prove this to yourself.

Do you care to try to prove what you're claiming? And then, when I prove you are wrong, will you admit it? (chinglu said he would, but then ran away). All I am asking is whether you will admit you were wrong when proven wrong, otherwise you're a troll and I won't waste my time, nor will I feed your need for cynical self-gratification.

By the way I'll bet even a novice can tell Masterov why satellites will vary in their amount of time dilation.

 

Look Lorentz's error, post 219, page 11 of this thema.

 

???

Call the number and subject of your post.

 

*sigh*

You still haven't meaningfully replied to any of these posts.

Put simply:

The solution to this problem is not given by Minkowski's equation, just like it is not given by \(E \,=\, mc^{2}\) or \(\bar{F} \,=\, m \bar{a}\) or \(f(t) \,=\, \frac{1}{\sqrt{2\pi}} \int \mathrm{d} \omega \, \tilde{f}(\omega) \, e^{-i \omega t}\) or most other equations. I suppose you're going to tell us all these equations are wrong?

 

I've already answered all these questions.

 

The question relates to the topic, which I did discussion the relative and absolute spatial and temporal scales.

This topic is not discussed in the relativistic mass and energy, and certainly not discussed in the Fourier transform.

 

It also has nothing to do with Lorentz transformations or the equation \((ct')^{2} \,-\, x'^{2} \,=\, (ct)^{2} \,-\, x^{2}\). The problem, as you describe it [POST=2928447]here[/POST] and repeat [POST=2964266]here[/POST] does not require the use of a coordinate transformation as you are only considering one reference frame. It is a simple kinematics problem that is solved with simple kinematics. It doesn't require relativity at all. I don't see how I can explain this more clearly. If you really believe the times you're looking for should be given by Minkowski's equation the way you're trying to apply it, then you don't understand relativity. It's as simple as that.

 

I am ready to discuss the post Lorentz's error, post 219, page 11 of this thema or the thema Master Theory.

I do not understand your problems.

Do I understand you so that: you think that the transit time of a photon (in terms of a moving observer) will be the same in both directions?

 

No, the transit times are different in both directions. That is not the problem. The problem is: relativity does not predict otherwise. The transit times are not supposed to be roots of \((ct')^{2} \,-\, L^{2} \,=\, 0\). That is not what the equation means or how you are supposed to apply it.

 

Can you reconcile your statement with an expression?:
\((ct')^2-x'^2=(ct)^2-x^2\)

 

I already did:

I can illustrate how this equation works in more detail, by considering more than one reference frame.

Start with the mirrors at rest a distance L from one another. Then \(\Delta x_{1} \,=\, -L\) and \(\Delta x_{2} \,=\, +L\), and \(\Delta t_{1} \,=\, -L/-c \,=\, L/c\) and \(\Delta t_{2} \,=\, L/c\).

Because the pulse is moving at the speed of light in both directions, \((c \Delta t_{1})^{2} \,-\, (\Delta x_{1})^{2} \,=\, 0\) and \((c \Delta t_{2})^{2} \,-\, (\Delta x_{2})^{2} \,=\, 0\).

You can transfer to the case where the mirrors are moving with velocity +v via a Lorentz transform:

\( \begin{eqnarray} \Delta t' &=& \gamma (\Delta t \,+\, \frac{v}{c^{2}} \Delta x) \\ \Delta x' &=& \gamma (\Delta x \,+\, v \Delta t) \,, \end{eqnarray} [/indent]\)

with \(\gamma \,=\, \gamma(v) \,=\, \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}\).

You get different transit times in both directions because \(\Delta x_{1} \,=\, -\Delta x_{2}\):

\( \begin{eqnarray} \Delta t'_{1} &=& \gamma (\Delta t_{1} \,+\, \frac{v}{c^{2}} \Delta x_{1}) \\ &=& \gamma (L/c \,-\, \frac{v}{c^{2}} L) \\ &=& (L/c) \gamma (1 \,-\, \frac{v}{c}) \\ &=& (L/c) \frac{1}{\gamma (1 \,+\, \frac{v}{c})} \\ &=& \frac{L'}{c \,+\, v} \\ \Delta t'_{2} &=& \gamma (\Delta t_{2} \,+\, \frac{v}{c^{2}} \Delta x_{2}) \\ &=& \gamma (L/c \,+\, \frac{v}{c^{2}} L) \\ &=& \frac{L'}{c \,-\, v} \,, \end{eqnarray} \)​

with \(L' \,=\, L/\gamma\), reflecting relativistic length contraction. To make the expressions more recognisable I used that \(\gamma \,=\, \frac{1}{\sqrt{1 \,+\, \frac{v}{c}} \sqrt{1 \,-\, \frac{v}{c}}}\) to work out that \(\gamma (1 \,\mp\, \frac{v}{c}) \,=\, \frac{1}{\gamma (1 \,\pm\, \frac{v}{c})}\).

So you get the different transit times like you expected: \(\Delta t'_{1} \,=\, L'/(c \,+\, v)\) and \(\Delta t'_{2} \,=\, L'/(c \,-\, v)\), and \(\Delta t'_{1} \,\neq\, \Delta t'_{2}\).

According to Minkowski's equation,

\( \begin{eqnarray} (c \Delta t_{1})^{2} \,-\, (\Delta x_{1})^{2} \,=\, 0 &\Rightarrow& (c \Delta t'_{1})^{2} \,-\, (\Delta x'_{1})^{2} \,=\, 0 \,, \\ (c \Delta t_{2})^{2} \,-\, (\Delta x_{2})^{2} \,=\, 0 &\Rightarrow& (c \Delta t'_{2})^{2} \,-\, (\Delta x'_{2})^{2} \,=\, 0 \,. \end{eqnarray} \)​

If you want to verify that this still holds, you need to work out \(\Delta x'_{1}\) and \(\Delta x'_{2}\). The transit distances are not \(\pm L'\), because the mirrors are moving while the pulse moves between them. You can work them out with the Lorentz transformation again:

\( \begin{eqnarray} \Delta x'_{1} &=& \gamma (\Delta x_{1} \,+\, \Delta t_{1}) \\ &=& \gamma (-L \,+\, v L/c) \\ &=& - L \gamma (1 \,+\, \frac{v}{c}) \\ &=& - \frac{L'}{1 \,+\, \frac{v}{c}} \\ \Delta x'_{2} &=& \gamma (\Delta x_{2} \,+\, \Delta t_{2}) \\ &=& \gamma (L \,+\, v L/c) \\ &=& \frac{L'}{1 \,-\, \frac{v}{c}} \,. \end{eqnarray} \)​

Now you can see that \(\Delta x'_{1} \,=\, - \frac{L'}{1 \,+\, \frac{v}{c}} \,=\, - \frac{c L'}{c \,+\, v} \,=\, - c \Delta t'_{1}\) and \(\Delta x'_{2} \,=\, c \Delta t'_{2}\), so Minkowski's equation holds just fine:

\( \begin{eqnarray} (c \Delta t'_{1})^{2} \,-\, (\Delta x'_{1})^{2} \,=& (c \Delta t'_{1})^{2} \,-\, (- c \Delta t'_{1})^{2} &=\, 0 \,, \\ (c \Delta t'_{2})^{2} \,-\, (\Delta x'_{2})^{2} \,=& (c \Delta t'_{2})^{2} \,-\, (c \Delta t'_{2})^{2} &=\, 0 \,. \end{eqnarray} \)​

Hopefully this rather long post will illustrate how the Lorentz transformation and Minkowski's equation apply to your problem.​