Measuring the One Way Speed of Light

[Ibix]

So, if the horses I referred to above took longer to run their course in one direction to the other, are you saying the interval in their arrival times at the finish line would always be the same?

As long as their two-way speed remains constant, and the rule that if they arrived togther they must always arrive together is obeyed (i.e. that it's possible to assume isotropy), then yes.

 

[weirdoguy]

Do you guys recall how the scientific mainstream simply accepted as read that there was no way to test quantum entanglement?

Can you provide a reference that states that there is no way to test entanglement by its very definition? If not then I don't see any connection to the one way speed of light.

 

[Dale]

IF there is isotropy in both the calibration and two tests then the arrival time intervals (for the two tests) would be the same.

Only if you ASSUME that the one way speed is equal to the two way speed. You are calibrating by measuring the two way speed.

The assumption in the calibration is that the two-way trip delivers a result being the average speed in both directions*. The two test runs are then compared to 'that' singular average speed result. Asymmetry in the results of the two tests would support an anisotropy hypothesis.

You don’t need to go there. That is already refuted by measurement. The entire discussion is already predicated on the experimentally established fact that the two way speed of light is isotropic. That is uncontroversial and already established.

Knowing that the two way speed of light is isotropic does not imply that the one way speed of light is also isotropic

 

[Dale]

Do you guys recall how the scientific mainstream simply accepted as read that there was no way to test quantum entanglement?

This is not that. This is a simple matter of definition. You have been utterly unwilling to address that.

The one way speed requires two synchronized clocks and a straight unidirectional path, by definition. It is $v=(t_1-t_0)/d$ where $t_0$ is the time of departure from clock 0, $t_1$ is the time of arrival at clock 1, and $d$ is the straight line distance between clock 0 and clock 1. This is the definition of a one way speed.

You are focusing on irrelevant experimental details. They don’t matter at all. There is nothing in the experiments that change the definition. If you don’t have two synchronized clocks then you are not measuring a one way speed.

 

[Vanadium 50]

Do you guys recall how the scientific mainstream simply accepted as read that there was no way to test quantum entanglement?

Remember when the mathematical mainstream said there simply accepted as read that there was no way to solve one equation in two unknowns?

 

[vanhees71]

One can formulate the special theory of relativity also in terms of a symmetry assumption on spacetime. It all starts with the special principle of relativity, which also holds in relativistic physics:

(1) There exists a class of reference frames, called inertial reference frames, where Newton's 1st Law holds, i.e., wrt. to such a reference frame anybody not being interacting with anything will move with constant velocity.

In order to be able to define what velocity is you need first a model for the kinematics, i.e., a way to measure distances and describe locations of the body wrt. the reference frame. In Newtonian as well as special relativistic physics one postulates

(2) For any inertial observer there exists a standard clock (you can just take the definition of the second from the SI and build an atomic clock establishing the standard Cs frequency $\nu_{\text{Cs}}$). The laws of Nature are time-translation invariant (i.e., the outcome of any experiment by an inertial observer won't depend on the time it is made) and that observer describes space as a 3D affine Euclidean manifold (implying the symmetries of this manifold, i.e., translation invariance (homogeneity) and rotational invariance (isotopy)).

From this it follows that an inertial reference frame may be defined by a family of synchronized clocks at each point of space, and a spatial position at any given time is determined by an arbitrary (for convenience right-handed Cartesian coordinate systems).

From these assumptions alone you can derive the form of the transformations of time and spatial coordinates from one inertial reference frame to any other inertial frame moving with some constant velocity relative to it. As it turns out there are, up to isomorphy, only two symmetry groups fitting these assumptions:

(a) Newtonian spacetime with an absolute time, implying that any standard clock once synchronized at one place (defining the origin of the inertial reference frame) and then transported to any other point in space the time shown by this transported clock stays synchronized.

(b) Minkowski spacetime with a universal limiting speed, which empirically, with very high precision) turns out to be the speed electromagnetic signals travel in vacuo. We can thus use light signals to establish standards of lengths measurements. By the assumed symmetries (homogeneity of time and space as well isotropy and Euclidicity of space) the one-way speed of light is the same in all directions in any inertial reference frame, and the standard clocks by definition are synchronized such that if you send a light signal from the origin at time $t$ to any other place at distance $r$ from the origin the arrival time of the light signal, as shown by the clock at this place, is $t+r/c$.

As you see, in this formulation of the postulates the isotropy of the one-way speed of light is an assumption and the clock synchronization a la Einstein is a useful convention to measure space and time intervals, and as it turns out, indeed this leads to the Lorentz transformations and taking the space-time translations under consideration the Poincare transformations building the symmetry group of spacetime (to be precise from the symmetries we can only argue about the proper orthochronous Poincare group, which is that part of the symmetry group that is continuously connected to the identity) keeping Minkowski products between four-vectors invariant.

As with any set of postulates of a physical theory you can also test special relativity only for consistency with observations, i.e., you can test the consequences this theory makes for measurable phenomena, among them for special relativity time dilation/length contraction, the null result of the Michelson-Morley experiment (testing the isotropy of the two-way speed of light), etc. AFAIK you cannot empirically independently test the isotropy of the one-way speed of light, which (as was stressed many times by many people in this thread) is rather a postulate, enabling the synchronization of clocks at rest in one inertial reference frame using Einstein's clock-synchronization. Particularly it follows that the set of clocks synchronized in one inertial reference frame are not synchronized with the clocks, which are synchronized within another inertial reference frame. So there's neither an absolute meaning of time nor an absolute meaning of simultaneousness for observations of observers not being at rest in one common inertial reference frame.

In GR it's even more involved, because there you really can only objectively establish inertial reference frames only locally, and all there is objectively measurable are local observations of a observer as defined in his or her local inertial reference frame. The only thing that's left, and this is the mathematical formulation of the strong equivalence principle, is that at any spacetime point there's a local inertial reference frame.

 

[cmb]

This is not that. This is a simple matter of definition. You have been utterly unwilling to address that.

The one way speed requires two synchronized clocks and a straight unidirectional path, by definition. It is $v=(t_1-t_0)/d$ where $t_0$ is the time of departure from clock 0, $t_1$ is the time of arrival at clock 1, and $d$ is the straight line distance between clock 0 and clock 1. This is the definition of a one way speed.

You are focusing on irrelevant experimental details. They don’t matter at all. There is nothing in the experiments that change the definition. If you don’t have two synchronized clocks then you are not measuring a one way speed.

If you could clarify which two clocks I have referred to it might help?

My calibration and tests are done with a single [one] clock that is not synchronised with any time reference, and does not move during any testing/measurements.

Is there some clock in my description I did not notice?

The diagrams in post #143 do not appear to represent what I have tried to illustrate in words. Maybe I am being completely opaque in words and a diagram might help?

 

[Dale]

If you could clarify which two clocks I have referred to it might help?

My calibration and tests are done with a single [one] clock that is not synchronised with any time reference, and does not move during any testing/measurements.

Yes, therefore it is not a one way measurement, by definition.

 

[Ibix]

The diagrams in post #143 do not appear to represent what I have tried to illustrate in words. Maybe I am being completely opaque in words and a diagram might help?

You can add a diagram, but make sure you understand mine first.

This is a displacement-time graph of what I think you were describing. Time is left-to-right, displacement vertical. Velocity is the slope of a line. The two horses leave the start line at the same time. The red horse is going faster, so it gets to the finish line first. The experiment is shown twice, once starting from each end.

The only difference between this and the first diagram in #143 is that the red horse doesn't do the extra trips that prove that the difference in arrivals is invariant.

 

[cmb]

Yes, therefore it is not a one way measurement, by definition.

I don't understand. The horses only run one way. Which way is 'the other' way?

 

[Ibix]

I don't understand. The horses only run one way. Which way is 'the other' way?

When you turn the experiment around.

 

[cmb]

You can add a diagram, but make sure you understand mine first.

View attachment 275096
This is a displacement-time graph of what I think you were describing. Time is left-to-right, displacement vertical. Velocity is the slope of a line. The two horses leave the start line at the same time. The red horse is going faster, so it gets to the finish line first. The experiment is shown twice, once starting from each end.

The only difference between this and the first diagram in #143 is that the red horse doesn't do the extra trips that prove that the difference in arrivals is invariant.

Sure.

Now show the horses going 'to' the finish line both at x1.5 their 'normal' [isotropic] speed and on the return path x.5 their 'normal' speed, as if their 'normal speed' varied and they didn't realize it.

Average there and back is 'normal speed' for both horses, but the gap between the red and blue lines on the top line is smaller than the gap between red and blue on the lower line.

 

[cmb]

When you turn the experiment around.

You don't need to turn the experiment around. You can determine their speed one way, just from the interval at the finish line.

The 'synchronisation' to the first horse is the second horse, but if both are affected by a common cause effect (say, wind speed) then their variation of speeds can be noticed against each other.

 

[Ibix]

Now show the horses going 'to' the finish line both at x1.5 their 'normal' [isotropic] speed and on the return path x.5 their 'normal' speed, as if their 'normal speed' varied and they didn't realize it.

Third diagram in #143 is the correct way to do that. And 1.5x speed and 0.5x speed averages to 0.75x speed, by the way.

 

[Ibix]

You don't need to turn the experiment around. You can determine their speed one way, just from the interval at the finish line.

That's not what you said in #140. There, you said you didn't need to turn the experiment around if you knew the velocity ratio. I already asked how you were going to determine the velocity ratio, and you didn't answer. You can't do it without a one-way speed measure, which relies on clock synchronisation. You also said that if you didn't know the velocity ratio you needed to turn the experiment around.

 

[cmb]

Third diagram in #143 is the correct way to do that.

The third diagram does not show what I have said.

In the third diagram, the red horse runs at infinite speed. What is half of infinite speed? It's still infinite speed. So the red and blue lines going 'out' should both be vertical.

There is never an interval between red and blue horses passing the finish line.

If you have an optical cable and it is 0.6c propagation speed and you time light transmitted in vacuum and a second path via optical cable, then; if light transmission was instantaneous in some direction then the two signals would arrive together in that direction.

 

[cmb]

That's not what you said in #140. There, you said you didn't need to turn the experiment around if you knew the velocity ratio. I already asked how you were going to determine the velocity ratio, and you didn't answer.

I have repeated several times, by sending the light around the cable medium in a loop. The two directions, if anisotropic, will cancel each other out and as mentioned is already an experimentally proven fact that there is no detectable effect.

The calibration CAN be averaged as a two-way test, there is no issue there. This determines the average two-way speed for light whether or not it is isotropic. It is the subsequent test which is one-way.

 

[Ibix]

In the third diagram, the red horse runs at infinite speed. What is half of infinite speed? It's still infinite speed. So the red and blue lines going 'out' should both be vertical.

As noted in the post you quoted (although I edited the second sentence in, so you may not have seen), 1.5x speed and 0.5x speed averages to 0.75x speed. So your numbers would not work. Mine would. You can use a finite one-way speed if you like - as I noted in #143, all it does is shear the diagram.

If you have an optical cable and it is 0.6c propagation speed and you time light transmitted in vacuum and a second path via optical cable, then; if light transmission was instantaneous in some direction then the two signals would arrive together in that direction.

That is incorrect, largely because velocities don't transform the way you appear to think they do if you are using non-isotropic light speed. Thus what an Einstein frame would call 0.6c, your non-isotropic coordinates would not call 0.6c (even if we used a finite one-way propagation speed for light).

 

[Ibix]

I have repeated several times, by sending the light around the cable medium in a loop.

...which measures the ratio of two-way velocities, which is not the same as the ratio of one-way velocities. It's the latter you need to measure.

 

[Dale]

Ok @cmb we are done here. This is such a fragmented mess now it is irredeemable.

IF you choose to open a new thread on this topic then include a good diagram or more detailing the entire process. Describe the measurement procedure in detail. Make sure to explicitly label all experimental devices and measured quantities. And write down all calculated quantities including especially the formula used to calculate the final speed result. Any round-trip measurements of the speed of light in vacuum are assumed to be isotropic and equal to c in accordance with experiments.

Remember, if the light travels in anything other than a single straight line then it is not a one way measurement. Especially a loop (I am still not sure why you don’t seem to get that) is obviously not a one way measurement.

 

[astrodummy]

Hello,

My name is Steve, I am a retired IT guy from Canberra, Australia. I study physics as much as I can. I have done the AstroX series of astrophysics courses on edX, Special Relativity on Coursera plus a bunch of self-pacing study using a bunch on textbooks. I'd like to say 'hello' by kicking around some ideas I found in a video I just stumbled across. I look forward to a discussion with you learned people.

This video ( ) by Veritasium discusses how the speed of light can only be determined by bouncing it off a mirror. It opens the possibility that the speed of light may not be the same outbound as it is on return.

In one specific example it suggests the speed of light may be c/2 out bound, and 'infinite' on return. The only plausible argument against this I can come up with is the galactic redshift. We would not see any relativistic redshift in receding galaxies since their velocity would be insignificant compared to the now infinite light speed. The Lorentz factor would be precisely equal to one. So no redshift, no Hubble Constant, no clue the universe is expanding, no dark energy. Also the entire universe would be observable, not just limited to the objects that have had enough time for their light to get us since the Big Bang.

The fact we DO see redshift in receding galaxies indicates the speed of light in the return direction cannot be infinite. The speed of light on return could still be different to the outbound speed, however. But as the Veritasium states, we may never know.

Discussion?

 

[PeterDonis]

It opens the possibility that the speed of light may not be the same outbound as it is on return.

More precisely, it means that the isotropy or lack thereof of the speed of light is a matter of one's choice of coordinates, not physics. As has already been discussed in multiple previous posts in this thread. Please review the thread.