One way speed of light experiment proposal

[DanMP]

Yes. You can't spin the rotating apparatus up without it distorting as different parts are moving at different speeds. ...
You can find a way of correcting for all this stuff, but the correction process itself will end up with the assumption that the one-way speed is equal to the two-way speed hidden somewhere.

Thank you. I reached a similar conclusion during the weekend :)

This is the well - known method, it is so - called Double Fizeau Toothed wheel
...

I knew it but forgot the name ... and didn't find it on Google at "speed of light experiment toothed wheels". Thank you.

 

[kkris1]

"I don't think I understood the @kkris1's problem last week. I think the point of his experiment (or the way I would phrase it if I were in his shoes) was that you could measure a value of the speed of light as a function of the velocity of the rod - call it c ′ ( v ) c′(v) - and then observe that this is asymptotic to some value as v → 0 v→0. And this asymptote would indeed be c c. One way speed of light, right? Since he's getting c for the one-way speed, clocks synchronised by this method must agree with Einstein synchronised clocks (DanMP's addition of two clocks disguised as cogs makes this explicit). The un-expressed question is: what decision did kkris1 make that led to the Einstein convention and not any other? "

Let me rephrase the experiment:
Let's have opaque rod with narrow slits close to both ends A and B. Let's position two lasers so the light is getting through both slits to the photo sensors at A' and B' respectively. Now we move the rod out, accelerate it to the speed v (around 10m/s, so the length contraction is negligible ~10^-14m ;besides, we can't really tell what is being contracted, the rod or the distance between the lasers, since we can assume that the rod is at rest and the lasers are moving with constant speed -v).
When the light from laser A will be transmitted through the leading slit, the light from laser B will pass through the trailing slit. The clocks at A' and B' will be synchronized, since if leading slit is at A, the trailing slit is simultaneously at B. Having both clocks synchronized, we can now measure one way speed of light.
There is no dependency of the speed of the rod and measured value of one way speed of light.

Using this experiment, we do not even need synchronized clocks. Instead we can reflect the light from A' and B' to the photo sensor somewhere in front and using digital oscilloscope measure the difference in arrival time of the two signals.

 

[Ibix]

There is no dependency of the speed of the rod and measured value of one way speed of light.

This is false, as I pointed out in the part of my post that you quoted. And several other posters pointed out.

Also - accelerate the whole apparatus to 0.6c (not just the rod) and repeat. Are your "simultaneous" events simultaneous?

And what value of c are you putting into your length contraction calculation? And, for that matter, how do you justify using the length contraction equation at all, since it is based on assuming Einstein's simultaneity convention?

 

[kkris1]

"This is false, as I pointed out in the part of my post that you quoted. And several other posters pointed out. Also - accelerate the whole apparatus to 0.6c (not just the rod) and repeat. Are your "simultaneous" events simultaneous? And what value of c are you putting into your length contraction calculation? And, for that matter, how do you justify using the length contraction equation at all, since it is based on assuming Einstein's simultaneity convention?"

How can you practically accelerate anything to 0.6c? I'm talking about real , not just "thought" experiment. But even if you did, it would have changed nothing, since everything in the experiment would be affected the same way.(e.g length contraction, time dilation)

We can talk only about relative length contraction; our 10m rod gliding 10m/s would be contracted according to formula:
L'=L(sqrt(1-v^2/c^2) which will yield ~(10^-13)m . I'm aiming to position the lasers with ~(10^-9)m (1nm ) accuracy.
Regarding dependency of the measured value of one way speed of light on v: there is no such dependency.
If the rod glides with lower speed (say 1m/s) past the lasers, the signals from photo sensors will be longer and less sharp, but still we should be able find their peaks, or middle points and their physical separation will be 10m.
Just imagine- when leading slit of the rod triggers the signal from A', at the same time 10m behind another signal is created at B'. This two signals separated by the distance of 10m are going with the speed of light to the sensor somewhere in front and you can measure the time between these two signals. Your distance is 10m, divided by the time between these two signals will give you one way speed of light.

 

[Dale]

@kkris1 we have already asked you multiple times to work out the math. What you are saying is simply false.

You assume that the length contraction is negligible at 10 m/s, but that assumption already assumes that the one way speed of light is c. If the one way speed of light is not c then such assumptions cannot be made and you need to work out the math, as has been requested multiple times.

There is simply no way to measure the one way speed of light which does not assume the very thing it purports to measure

 

[Ibix]

"This is false, as I pointed out in the part of my post that you quoted. And several other posters pointed out. Also - accelerate the whole apparatus to 0.6c (not just the rod) and repeat. Are your "simultaneous" events simultaneous? And what value of c are you putting into your length contraction calculation? And, for that matter, how do you justify using the length contraction equation at all, since it is based on assuming Einstein's simultaneity convention?"

How can you practically accelerate anything to 0.6c? I'm talking about real , not just "thought" experiment.

There are two obvious answers to this. First, impractical and impossible are not the same thing - so all you are doing here is saying that you refuse to think about your theory in a regime where it will fail. Fine if you admit that you're approximating something, but you claim you're not.

Second, back of the envelope, if your whole apparatus is moving at $u$ ($u <<c$) then the flight time of light (your rod is length l; Einstein synchronisation assumed) is $l/c$. The error from ignoring the relativity of simultaneity is approximately $\gamma ul/c^2$. That makes the fractional timing error approximately $u/c$, treating $\gamma\simeq 1$. Modern atomic clocks are accurate to one part in 10¹⁴, so can potentially detect the synchronisation errors when the apparatus moves at a speed of three microns per second. That's probably optimistic, but I have a lot of orders of magnitude in hand before generating velocities is even remotely challenging. 0.6c just makes it more dramatic.

But even if you did, it would have changed nothing, since everything in the experiment would be affected the same way.(e.g length contraction, time dilation)

Incorrect, as shown above.

We can talk only about relative length contraction; our 10m rod gliding 10m/s would be contracted according to formula:
L'=L(sqrt(1-v^2/c^2) which will yield ~(10^-13)m .

You are here explicitly assuming the Einstein synchronisation convention. You need to derive the length contraction equation without assuming the speed of light is the same in both direction. Or, indeed making any assumption about the speed of light. Good luck...

I'm aiming to position the lasers with ~(10^-9)m (1nm ) accuracy.
Regarding dependency of the measured value of one way speed of light on v: there is no such dependency.
If the rod glides with lower speed (say 1m/s) past the lasers, the signals from photo sensors will be longer and less sharp, but still we should be able find their peaks, or middle points and their physical separation will be 10m.

Nonsense. Your entire purpose in keeping $v$ low was to attempt to duck length contraction. If there's no effect from traveling quickly, why are you specifying low speed?

Just imagine- when leading slit of the rod triggers the signal from A', at the same time 10m behind another signal is created at B'. This two signals separated by the distance of 10m are going with the speed of light to the sensor somewhere in front and you can measure the time between these two signals. Your distance is 10m, divided by the time between these two signals will give you one way speed of light.

...except that you assumed the speed of light when you used the length contraction equation. So you will get the answer you put in.

 

[Mister T]

;besides, we can't really tell what is being contracted, the rod or the distance between the lasers, since we can assume that the rod is at rest and the lasers are moving with constant speed -v).

This claim alone reveals a misconception. Length contraction is real for both, not just one or the other. Perhaps if you can understand the role simultaneity takes in making this arrangement (symmetry of length contraction) possible you would be better able to understand the objections voiced by the other posters.

 

[Ibix]

You need to derive the length contraction equation without assuming the speed of light is the same in both direction.

Just for fun...

If the round trip speed of light is to be $c$, it is easy to see that the one-way speeds ($c_+$ in the +x direction and $c_-$ in the -x direction) must satisfy $2/c=1/c_++1/c_-$, where $c$ is the usual invariant two-way speed of light.

Imagine a radar set moving inertially. It emits a radar pulse in the +x direction at some time $t_e$ and receives the reflection at time $t_r$. The event the pulse reflects off is assigned coordinates $(x_a,t_a)$ (a for asymmetric):$$\begin{eqnarray*}
x_a&=&c(t_r-t_e)/2\\
t_a&=&t_e+(t_r-t_e)\frac c{2c_+}
\end{eqnarray*}$$If we assume the Einstein synchronisation convention and let $c_+=c$ then this reduces to:$$\begin{eqnarray*}
x&=&c(t_r-t_e)/2\\
t&=&(t_r+t_e)/2
\end{eqnarray*}$$These are the usual Einstein coordinates (for more discussion see Dolby and Gull - https://arxiv.org/abs/gr-qc/0104077). We can eliminate $t_e$ and $t_r$ to get a transformation between the asymmetric speed coordinates and the familiar Einstein coordinates. Writing this in matrix form:$$\begin{eqnarray*}
\left(\begin{array}{c}x_a\\t_a\end{array}\right)&=&
\left(\begin{array}{cc}1&0\\\frac{c-c_+}{cc_+}&1\end{array}\right)
\left(\begin{array}{c}x\\t\end{array}\right)\\
\vec{x_a}&=&\mathbf m \vec {x}
\end{eqnarray*}$$Writing in matrix form makes it easy to chain together coordinate transforms, so we can write a generalisation of the Lorentz transform that transforms between frames using asymmetric light speeds as $\vec{x'_a}=\mathbf {m\Lambda m}^{-1}\vec{x_a}$, where $\mathbf\Lambda$ is the usual Lorentz transform matrix:$$
\mathbf\Lambda=\left(\begin{array}{cc}\gamma&-v\gamma\\-\frac{v}{c^2}\gamma&\gamma\end{array}\right)$$This is basically switching from our asymmetric coordinates to the usual Einstein coordinates, switching to Einstein coordinates in another frame, then switching back to our asymmetric coordinates that correspond to that frame. The algebra is tedious, but we eventually find:$$\begin{eqnarray*}
x'_a&=&\gamma\left(\left(1-\left(1-\frac c{c_+}\right)\frac vc\right)x_a-vt_a\right)\\
t'_a&=&\gamma\left(\left(1-\frac v{c_+}+\frac vc\right)t_a-\left(\frac 2{c_+}-\frac c{c_+^2}\right)\frac vcx_a\right)
\end{eqnarray*}$$Note that $v$ is the velocity of the primed Einstein frame as measured by the unprimed Einstein frame; I should probably carry the calculation through to get the velocity as measured by the unprimed asymmetric frame (since it has a different notion of time, it has a different notion of velocity) but it's more convenient here not to. Note also that $\gamma$ has its usual relativistic meaning.

Now we can derive the correct length contraction formula for the case where the speed of light is asymmetric. Imagine a rod of length $l$ at rest in the unprimed frame. One end is at $x_a=0$ and the other at $x_a=l$. In the primed frame at $t'_a=0$ one end is also at the origin and the other is at $x'_a=l'$. Substituting $x_a=l$, $x'_a=l'$ and $t'_a=0$ into our transformation equations we get$$l'=\left(\frac {c_+}{c_+-(1-c_+/c)v}\right)\frac l\gamma$$Now we have everything we need to analyse kkris1's experiment. Assuming that the one-way speed of light is $c_+$ the expected flight time of the light along a rod of length $l$ is $T=l/c_+$. The systematic timing error from ignoring length contraction is $\delta T=(l-l')/v$ so the fractional error is $$\frac{\delta T}T=\frac{c_+}{\gamma v} \left(\gamma-\frac {c_+}{c_+-(1-c_+/c)v}\right)$$Substituting l=10m, v=1m/s we can plot the fractional error as a function of $c_+$ (note that the possible range of $c_+$ is $c/2\leq c_+$).

Clearly, length contraction is only negligible for $c\simeq c_+$ - so neglecting length contraction as kkris1 does is equivalent to assuming the Einstein synchronisation convention. Note that this isn't the same choice and error I mentioned in my earlier posts; there appear to be many choices implicit in the experimental design. In other words, simultaneity convention remains a convention.

I hope I typeset all the above correctly (Edit: I didn't - minor corrections above). A Maxima .mac file to do the algebra is:

/* Maxima batch file to derive length contraction in coordinates where */
/* the speed of light is not isotropic                                 */

/* Define Einstein x,t coordinates in terms of the emission and reception */
/* times (te and tr) of a radar set at rest.                              */
x:c*(tr-te)/2;
t:(tr+te)/2;

/* Ditto for the anisotropic coordinates */
xa:x;
ta:te+(tr-te)*(c/(2*cp));

/* Eliminate tr and te...*/
linsolve([xx=x, tt=t], [tr,te]);
/* ...and get the expression for ta in terms of x and t (xa is trivial) */
ratsimp(subst(rhs(%[2]),te,subst(rhs(%[1]),tr,ta)));

/* Transcribe the results into matrix form and define the Lorentz transform */
m:matrix([1,0],[(c-cp)/(c*cp),1]);
L:matrix([gamma,-v*gamma],[-v*gamma/c^2,gamma]);

/* Express the complete transformation matrix and apply it to a vector */
La:m.L.invert(m);
Laxa:ratsimp(La.matrix([xxa],[tta]));

/* Now get the length contraction formula by insisting that ta'=0... */
solve(Laxa[2][1],tta);
/* ...and xa=l. This expression is l', the length contraction formula. */
ratsimp(subst(l,xxa,subst(rhs(%[1]),tta,Laxa[1][1])));
/* Simplified it manually - just check we didn't mess up - this should be 0 */
ratsimp(%+cp*l*gamma*(1-v^2/c^2)/((1-cp/c)*v-cp));

 

[Dale]

Just for fun...

Excellent post. Very well written as well as a good analysis.

 

[Nugatory]

If the round trip speed of light is to be $c$, it is easy to see that the one-way speeds ($c_+$ in the +x direction and $c_-$ in the -x direction) must satisfy $2/c=1/c_++1/c_-$, where $c$ is the usual invariant two-way speed of light.

That constraint is one of the things that make the postulate so compelling; only a very strange and bizarre anistropy could satisfy that constraint everywhere. But we're still talking postulate here - if the alternatives to a postulate are strange and bizarre... that's probably why we chose the postulate in the first place.

 

[Ibix]

Excellent post. Very well written as well as a good analysis.

Thank you.

That constraint is one of the things that make the postulate so compelling; only a very strange and bizarre anistropy could satisfy that constraint everywhere. But we're still talking postulate here - if the alternatives to a postulate are strange and bizarre... that's probably why we chose the postulate in the first place.

It's also a choice to use orthogonal coordinates; the anisotropic ones share a time axis with Einstein coordinates but the x axes are skewed with respect to one another. That's why the algebra gets so much nastier.

 

[DrGreg]

One other thing to mention about using a non-standard synchronisation: it doesn't just affect the one-way speed of light, it affects all (one-way) coordinate speeds.

For example, using Einstein synchronisation, we know that if two equal masses collide with equal but opposite velocities, inelastically so that they stick together, the resulting mass will have zero velocity. Using a different synchronisation, those masses would not have equal coordinate speeds before collision, so in such a coordinate system $\gamma m\textbf{v}$ is not conserved. You'd need a different formula to get a conserved momentum.