One-way speed of light experiment, no clocks or mirrors (with simulation)

Question

As can be seen in the many creative questions here, it is hotly disputed whether you can measure the speed of light in only one direction to the point where a infamous postulate was made that the speed of light might be instantaneous in one direction; you can never know.

So here is yet another idea that seems to be capable of falsifying above. The question is, what seems to be wrong here / how might the universe conspire?

This is a variation of the original measurement, close to what is described here: Measuring the one-way speed of light?

Instead of a cogwheel we just rotate a path and shine a collimated beam of photons through it. At the end is a flux integrator. The predictions are then:

1. there should be some rotational speed where no light can make it through the path, if the speed of light is finite.

2. if the speed of light is finite, the integrator will show a particular shape (amplitude over time) for a particular angular frequency.

expanding on 2, at any given speed you should be able to measure the deviation between an instantaneous assumed speed of light (that will always have the same shape) and the experimentally measured value (which will have start and ends increasingly cut off), thus be able to derive the speed of light.

Of course, if you are able to rotate the device at close to relativistic speeds, you can also derive the speed from 1). But again, the idea is any rotational speed should show a deviation.

Goes without saying the apperatus can be rotated on any axis.

Here's a simulation where the speed of light is slowly increased.

https://www.youtube.com/watch?v=7y-GldVtuVU

Here are two pictures that sort of tell the story from the simulation.

Answer 1

Unfortunately, there is simply no possible way to measure the speed of light independently of your synchronization convention. In this case, if you use the standard isotropic synchronization convention then length contraction is also isotropic and so is the one way speed of light. If you use an anisotropic synchronization convention then length contraction is anisotropic and so is the one way speed of light.

In this case the anisotropic length contraction leads to the wall of the device becoming curved. Because of this curvature the light will pass through at the same rotational velocity regardless of the one way speed of light.

Answer 2

Let's simplify the experiment slightly by replacing the rotation of the shutter with linear movement perpendicular to the light path. Say the width of the gap, in the rest frame measured perpendicular to the light path, is $y$.

Since the shutter is moving at relativistic speed $v$, the contracted width $y'$ becomes

$$y'=y\sqrt{1-v^2/c_y^2}$$

where $c_y$ is the speed of light along the axis perpendicular to the light path. This value is needed in order to calculate the original value $c_x$ which is the one-way speed of light along the horizontal x-axis:

$$c_x=L/t = L v / y'$$

where $L$ is the horizontal length of the shutter passage the light beam moves through and $t$ is the time it takes for the light to barely pass from the opening edge to the opposite closing edge of the moving passage as measured in the rest frame (we can experimentally select a speed $v$ for this to be the case.)

But now you have shifted the problem to the perpendicular axis - how do you measure $c_y$?

Answer 3

The way we determine something's shape ultimately boils down to interacting with it by bouncing a c-propagating signal off of it. If the apparatus looks like a circle with a straight-line aperture through it to us, interrogating it with c-propagating signals, then the apparatus looks like a circle with a straight-line aperture through the middle of it to the c-propagating signal you're trying to shoot through the apparatus.

If the apparatus looks like a circle with a wavy line (like a Pepsi logo or a yin-yang) through the middle of it (because it's spinning fast enough to deform), then it will look like a circle with a wavy line through the middle of it to the laser you're trying to shoot through the apparatus.

You get Dale's answer if we let the apparatus be perfectly rigid, that is, the speed at which forces propagate through the medium is the speed of light, so the apparatus deforms as the speed of light vs the rotational velocity.

You can run an argument that the speed of sound in any given medium scales linearly with the speed of light, but it's unnecessary. If the photons bouncing off of the apparatus and into our retinas tell us there's a straight path from one end to another, then there's a path from one end to the other that's straight for photons.

Answer 4

I assume this question was inspired by the same Veritasium video as all the others.

As explained in answers to earlier questions, all that he does in that video is define "velocity" differently, so that the same physical phenomena are assigned different velocity numbers, some of which are $\infty$ instead of $c$. It can't be falsified because it's just a word game.

See, e.g., this answer. To quote from it:

What we mean when we say that the speed of light is constant is that there exist coordinates with respect to which it's constant. In a Newtonian corpuscular world, no such coordinates would exist, so the fact that they do exist in the real world is physically meaningful. [...]

There also exist coordinates with respect to which the speed of light isn't constant. This is not physically meaningful, because no theory could ever avoid them; you can always do a formal substitution of variables [...]

If $(x,t)$ are standard inertial coordinates, then with respect to coordinates $(x,t')$ where $t'=t-x$, the speed of light $|dx/dt'|$ ranges from $c/2$ to $\infty$ depending on direction. Why don't we see this as an anisotropy in the sky? Because the universe in different directions has aged by different amounts, and their ages differ by just the right amount to compensate for the different light travel times.

So, let's apply that here. You wrote:

there should be some rotational speed where no light can make it through the path, if the speed of light is finite.

If the light is allowed to change direction, then it can always make it through, because the speed of the apparatus is everywhere less than the speed of light, so light can make radial progress while also moving transversely to avoid the sides.

If the light is not allowed to change direction, then it can't make it through at high enough rotational speeds even if its "velocity" is infinite, because with respect to the $t'$ used to define the "infinite" velocity, the path is curved enough that there is no straight-line way through. Why is it curved? Because (supposing the $(x,t)$ coordinates we started with were the center-of-momentum frame of the path), $t' = t-x$ cuts through different isotropic times $t$ at different $x$ positions, catching, in effect, different times in the path's rotation.

The path isn't consistently curved in these coordinates. It's most curved when it's roughly parallel to the $x$ axis, and straight when it's in the $yz$ plane. But you can't get the light through when it's in the $yz$ plane because the speed of light in that direction isn't $\infty$, but $c$.

This directional anisotropy isn't real. The problem is intrinsically symmetric. It appears anisotropic in these coordinates because the coordinates themselves are anisotropic for no reason.