# Thread: Three Experiments Challenging SRT

1. Originally Posted by przyk
$\begin{eqnarray}(c \Delta t'_{1})^{2} \,-\, (\Delta x'_{1})^{2} \,=& (c \Delta t'_{1})^{2} \,-\, (- c \Delta t'_{1})^{2} &=\, 0 \,, \\
(c \Delta t'_{2})^{2} \,-\, (\Delta x'_{2})^{2} \,=& (c \Delta t'_{2})^{2} \,-\, (c \Delta t'_{2})^{2} &=\, 0 \,.
\end{eqnarray}$
It's no correct, because you have:
$|x_1|\neq |x_2|$
$|x'_1|\neq |x'_2|$

Correct so:
$x_1=-x_2=L$
$x'_1=-x'_2=L'$

2. Originally Posted by Masterov
[tex]Correct so:
$x_1=-x_2=L$
Correct (if you like; I used opposite signs so $\Delta x_{1} \,=\, -L$ and $\Delta x_{2} \,=\, +L$).

$x'_1=-x'_2=L'$
Not correct. With the sign conventions I used before, $\Delta x'_{1} \,=\, - \frac{L'}{1 \,+\, v/c}$ and $\Delta x'_{2} \,=\, \frac{L'}{1 \,-\, v/c}$. Obviously, $| \Delta x'_{1} | \,\neq\, | \Delta x'_{2} |$.

Example for $\Delta x'_{1}$:

Trajectory of mirror: $x'_{\mathrm{m}}(t') \,=\, - L' \,+\, vt'$.

Trajectory of light pulse: $x'_{\mathrm{p}}(t') \,=\, - ct'$.

They coincide where $x'_{\mathrm{m}}(t'_{1}) \,=\, x'_{\mathrm{p}}(t'_{1})$:

$
\begin{eqnarray}
&& -L' \,+\, vt'_{1} \,=\, - ct'_{1} \\
&\Rightarrow& t'_{1} \,=\, \frac{L'}{c \,+\, v} \,.
\end{eqnarray}
$

The position of the light pulse at that time is $x'_{\mathrm{p}}(t'_{1}) \,=\, x'_{\mathrm{m}}(t'_{1}) \,=\, - \frac{c L'}{c \,+\, v} \,=\, - \frac{L'}{1 \,+\, v/c}$.

The transit displacement is $\Delta x'_{1} \,=\, x'_{\mathrm{p}}(t'_{1}) \,-\, x'_{\mathrm{p}}(0) \,=\, - \frac{L'}{1 \,+\, v/c}$.

3. Originally Posted by przyk
Obviously, $| \Delta x'_{1} | \,\neq\, | \Delta x'_{2} |$.
It's error.

Correct so:
$| \Delta x'_{1} | \,=\, | \Delta x'_{2} |=L'$

$t'_1=L'/(c+v)$
$t'_2=L'/(c-v)$
$\Delta x'_1=-\Delta x'_2=L'$

4. Originally Posted by Masterov
It's error.

Correct so:
$| \Delta x'_{1} | \,=\, | \Delta x'_{2} |=L'$
No. $| \Delta x'_{1} | \,=\, \frac{L'}{1 \,+\, v/c}$ and $| \Delta x'_{2} | \,=\, \frac{L'}{1 \,-\, v/c}$. Example derivation for $\Delta x'_{1}$ here.

5. Continue the debate is pointless.

You need to go back to school.

6. This really isn't complicated. For $\Delta x'_{2}$:

Key: lp: light pulse; m1, m2: mirrors.

As anyone with a working pair of eyes can see, $| \Delta x'_{2} | \,\neq\, L'$.

7. Originally Posted by przyk

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}$

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.
Hi guys. Trying to follow your discussion in the context of 'motion' per se. To that end, would you (przyk) clarify for me what your "at rest" mirrors and your "moving with velocity +v" mirrors are being compared to? I only want to make sure there is no 'absolute' rest/motion frames/assumptions 'built into' any of the equations/scenario being put by either of you in your discussion.

I just want to make sure I am not misunderstanding either of you, that's all. So if you can clarify the foregoing aspect for me I would greatly appreciate it. Thanks.

8. Originally Posted by RealityCheck
Hi guys. Trying to follow your discussion in the context of 'motion' per se. To that end, would you (przyk) clarify for me what your "at rest" mirrors and your "moving with velocity +v" mirrors are being compared to?
The mirrors are at rest and moving in suitably chosen reference frames (or compared with suitably positioned imaginary observers, if you prefer).

There's no possibility of introducing any sort of "absolute" frame in this problem because it only concerns basic kinematics (where and when trajectories will coincide, and so on). The relativity principle is about the laws of physics taking the same form in all inertial reference frames. This is a non issue here because I'm not applying any physical laws.

9. Originally Posted by przyk
As anyone with a working pair of eyes can see, $| \Delta x'_{2} | \,\neq\, L'$.
Ok.

You have contraction:
$\Delta x'_1\equiv c\Delta t'_1$
$\Delta x'_2\equiv c\Delta t'_2$
it's identity by definition.
You have a definitions:
$\Delta x'^2_1-(c\Delta t'_1)^2\equiv 0$
$\Delta x'^2_2-(c\Delta t'_2)^2\equiv 0$

It is a vast difference with:
$\Delta x'^2_1-(c\Delta t'_1)^2=0$
$\Delta x'^2_2-(c\Delta t'_2)^2=0$

10. Originally Posted by Masterov
Ok.

You have contraction:
$\Delta x'_1\equiv c\Delta t'_1$
$\Delta x'_2\equiv c\Delta t'_2$
No, I have

$
\begin{eqnarray}
\Delta x'_{1} &=& - c \Delta t'_{1} \\
\Delta x'_{2} &=& + c \Delta t'_{2} \,.
\end{eqnarray}
$

it's identity by definition.
The light pulse is moving at the speed of light. So yes, $| \Delta x' | \,=\, | c \Delta t' |$ by definition, and that is all the equation $(c \Delta t')^{2} \,-\, (\Delta x')^{2} \,=\, 0$ is saying. That's what makes your argument so absurd.

The equation $(c \Delta t')^{2} \,-\, (\Delta x')^{2} \,=\, (c \Delta t)^{2} \,-\, (\Delta x)^{2}$ is just a constraint on the Lorentz transfomation:

$
\begin{eqnarray}
c \Delta t' &=& \gamma (c \Delta t \,-\, \frac{v}{c} \Delta x ) \\
\Delta x' &=& \gamma ( \Delta x \,-\, \frac{v}{c} c \Delta t ) \,.
\end{eqnarray}
$

Check:

$
\begin{eqnarray}
(c \Delta t')^{2} \,-\, (\Delta x')^{2} &=& \gamma^{2} (c \Delta t \,-\, \frac{v}{c} \Delta x )^{2} \,-\, \gamma^{2} ( \Delta x \,-\, \frac{v}{c} c \Delta t )^{2} \\
&=& \gamma^{2} \bigl[ \, (c \Delta t)^{2} \,-\, 2 \frac{v}{c} (c \Delta t) (\Delta x) \,+\, \frac{v^{2}}{c^{2}} (\Delta x)^{2} \,-\, (\Delta x)^{2} \,+\, 2 \frac{v}{c} (\Delta x) (c \Delta t) \,-\, \frac{v^{2}}{c^{2}} (c \Delta t)^{2} \, \bigr] \\
&=& \gamma^{2} \bigl[ \, \bigl(1 \,-\, \frac{v^{2}}{c^{2}} \bigr) (c \Delta t)^{2} \,-\, \bigl(1 \,-\, \frac{v^{2}}{c^{2}} \bigr) (\Delta x)^{2} \, \bigr] \\
&=& \frac{\gamma^{2}}{\gamma^{2}} \bigl[ \, (c \Delta t)^{2} \,-\, (\Delta x)^{2} \, \bigr] \\
&=& (c \Delta t)^{2} \,-\, (\Delta x)^{2}\,.
\end{eqnarray}
$

So it is no surprise that when I started with $(c \Delta t_{i})^{2} \,-\, (\Delta x_{i})^{2} \,=\, 0$ in post #280 and then applied a Lorentz transformation, I got a result with $(c \Delta t'_{i})^{2} \,-\, (\Delta x'_{i})^{2} \,=\, 0$. That's what happens when you apply relativity correctly.

In any case, post #286 makes it clear that $| \Delta x'_{1} | \,\neq\, L' \,\neq\, | \Delta x'_{2} |$, unless $v \,=\, 0$.

11. I have doubt that I am right.
przyk is right apparently.
Expression:
$\Delta x'_1=-c\Delta t'_1$
$\Delta x'_2=c\Delta t'_2$
denotes the invariance of speed of light.
This is relevant to the frame of reference, which is connected to the moving observer.
Lorenz was not mistaken possible.
I'll be thinking.

But even if I'm wrong on this issue, it will not affect the validity of Master Theory, because MT is a consequence of the transfer of the absoluteness (of the transverse scale) to the time. It's a first.

Second: Lorentz transformations for visual coordinates. Physical coordinates obey Galilean transformations and can be calculated by double-integration (of the time) of acceleration.

12. Expression:
$\Delta x'_1=-c\Delta t'_1$
$\Delta x'_2=c\Delta t'_2$
denotes the invariance of speed of light.
It's a fact.

$\Delta x'^2_i=(c\Delta t'_i)^2$ - It's a fact too.

As it is consistent with my expression?:
$(\Delta x'-v\Delta t')^2=(c\Delta t')^2$

My $\Delta x'$ is distance between the mirrors.
$|\Delta x'|=L'$

przyk's $\Delta x'$ it's pass of photon which it run between the mirrors in the reference frame of the mobile observer for $\Delta t'$.

I did not do a calculations of this pass.

Consequently:

Expression: $\Delta x'^2_i=(c\Delta t'_i)^2$ which means the invariance of the speed of light, it's a right for Master Theory too.
(If $\Delta x'$ is pass of photon in the reference frame of the mobile observer.)

Question:

Is LT a right for Master Theory too?

No.

Physical coordinates are take to Galilean transformations.

Time dilation does not exist.
Time is absolute.

13. Originally Posted by Masterov
Time is absolute.
According to whose frame of reference?

One interpretation of SR is that time is in fact a completely subjective assessment of the order of events. That confines time to a completely relative perspective. Though this holds true for all frames of reference, SR begins with the assertion that it holds for all inertial frames of reference, likely because the variables become far more complex in accelerating frames.

So again, since time is in fact an observation of the rate of change, and both the rate that our clocks tick at and the rate that change appears to occur at, are both frame dependent.., time is absolute according to whose frame of reference?

This discussion assumes that there is some frame of reference for time that does not require or include an observer. Even if such a frame does exist, it is outside our ability to observe and measure. Time for us is relative to the frame of reference from which we observe the underlying change. That excludes for us any frame of reference in which time could be considered absolute. We are limited to seeing the world from where we stand at the moment, and from where we stand, in any one moment, time of light delays define what appears to be simutaneous.

The only thing absolute about time is that no two observers can ever agree on the sequence of events sufficiently removed from their physical locations, without an understanding of the relativity of simultaneity.

14. All relativistic effects are visual and is sequent behaviors of electromagnetic wave.
Time and accelerations in all inertial reference systems are identical.
Time is determined by the indication of clocks.

15. Originally Posted by Masterov
All relativistic effects are visual and is sequent behaviors of electromagnetic wave.
Time and accelerations in all inertial reference systems are identical.
Time is determined by the indication of clocks.
This can only be even thought of as accurate in the flat spacetime of SR and where there is no acceleration or gravity to be considered.

Two observers whose motion is inertial relative to one another, and yet whose proximity to a gravitational mass (location in a gravity well) is different, will hold clocks whose measurement of time will not agree. The same is true for any two observers who are, either moving at different relative velocities or under different accelerations.

While it is true that each observer will experience the sequence of events relative to their location and the time of light delays involved, the different rates at which their clocks tick will add an additional variable to their accounts of what they experience.

Time and acceleration cannot in fact be equivalent in all inertial reference systems. Only the laws that govern them remain equal. Consider two inertial systems one with a relative velocity of 0.5 c with respect to the other. They both being of equal rest mass exert an identical force in an attempt to accelerate along a common x axis. The observer already in motion relative to the other cannot experience an equal acceleration, since as the velocity of an object approaches c the force required to further accelerate it also increases. Though both observers may agree that a given force will result in a specific acceleration, they will not experience and equivalent acceleration, as a result of the aplication of an equal force.

As soon as you add accelerations or gravity to any inertial system, SR can no longer be applied in exactly the same manner. Both acceleration and gravity affect clocks differently. When one observer is in motion relative to the other, and they both exert the same force to accelerate, the change in velocity experienced by each will be different. Knowing their initial relative velocities, you can determine the difference in acceleration, resulting for identical forces. (The only way this can be untrue, is if the determination of the force applied is based on measurements of changes in inertia. However, in that case the observer already in motion would experience an equal inertial resistance to a change in motion, from the addition of a smaller excellerating force.)

16. Originally Posted by OnlyMe
This can only be even thought of as accurate in the flat spacetime of SR and where there is no acceleration or gravity to be considered.

Two observers whose motion is inertial relative to one another, and yet whose proximity to a gravitational mass (location in a gravity well) is different, will hold clocks whose measurement of time will not agree. The same is true for any two observers who are, either moving at different relative velocities or under different accelerations.

While it is true that each observer will experience the sequence of events relative to their location and the time of light delays involved, the different rates at which their clocks tick will add an additional variable to their accounts of what they experience.

Time and acceleration cannot in fact be equivalent in all inertial reference systems. Only the laws that govern them remain equal. Consider two inertial systems one with a relative velocity of 0.5 c with respect to the other. They both being of equal rest mass exert an identical force in an attempt to accelerate along a common x axis. The observer already in motion relative to the other cannot experience an equal acceleration, since as the velocity of an object approaches c the force required to further accelerate it also increases. Though both observers may agree that a given force will result in a specific acceleration, they will not experience and equivalent acceleration, as a result of the aplication of an equal force.

As soon as you add accelerations or gravity to any inertial system, SR can no longer be applied in exactly the same manner. Both acceleration and gravity affect clocks differently. When one observer is in motion relative to the other, and they both exert the same force to accelerate, the change in velocity experienced by each will be different. Knowing their initial relative velocities, you can determine the difference in acceleration, resulting for identical forces. (The only way this can be untrue, is if the determination of the force applied is based on measurements of changes in inertia. However, in that case the observer already in motion would experience an equal inertial resistance to a change in motion, from the addition of a smaller excellerating force.)
I agree.. this issue is more about how arguing against SRT using a "obscured" universal RF yet denying it simutaneously.

There can be absolutely NO universal reference frame.. if one wishes to understand SRT [ even the use of zero implied or actual - is relative ]
New thread on the subject of using zero as a universal reference frame...here

17. Acceleration can be measured by a weight+spring in a confined space where there is no windows.
Acceleration - a separate physical unit, which can be measured without calculations (without analyzing the changes of coordinates).

18. Originally Posted by Masterov
Acceleration can be measured by a weight+spring in a confined space where there is no windows.
Acceleration - a separate physical unit, which can be measured without calculations (without analyzing the changes of coordinates).
Just for the record, constant proper acceleration (e.g. as measured by a weight and spring) in relativity means a trajectory of the form

$
\begin{eqnarray}
ct &=& \frac{c^{2}}{a} \, \sinh(a \tau / c)
x &=& \frac{c^{2}}{a} \, \cosh(a \tau / c) \,,
\end{eqnarray}
$

or $x(t) \,=\, c \sqrt{t^{2} \,+\, c^{2}/a^{2}}$, if the weight and spring is always measuring acceleration $a$. See here for an explanation.

19. It's hard to believe.

20. Originally Posted by Masterov
It's hard to believe.
It is hard to believe.., that you did not already understand this.

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