Measuring Light Speed with MMX in Water

[grav-universe]

For the MMX, the speed of light is measured isotropically at c in free space regardless of the frame, so gives a null result. To another frame of observation, light still travels isotropically at c while the apparatus is length contracted by sqrt(1 - (v/c)^2) in the line of motion, whereby a null result is still achieved. If we place the MMX underwater, the index of refraction is 1.333, so the speed of light is measured at 3 c / 4, although still isotropically to a frame at rest to the water, so still gives a null result. But what is the result when the apparatus is moving through the water? Certainly the MMX apparatus is still length contracted by sqrt(1 - (v/c)^2) according to the rest frame even though light is measured to travel at 3 c / 4, isn't it?

 

[Dale]

This is related to the Fizeau experiment, one of the experiments which pre-dates SR and is explained by SR. Essentially, light traveling through a moving medium uses the velocity addition formula.

http://en.wikipedia.org/wiki/Fizeau_experiment

 

[grav-universe]

This is related to the Fizeau experiment, one of the experiments which pre-dates SR and is explained by SR. Essentially, light traveling through a moving medium uses the velocity addition formula.

http://en.wikipedia.org/wiki/Fizeau_experiment

Right, light traveling through moving water uses Fizeau/Fresnel's equations, but we would just be taking the point of view of the rest frame, where the water is also at rest, so measure light to travel at 3 c / 4 isotropically, so we don't really need them for this. But what about the length contraction (and time dilation) of a body moving through the water as measured by the rest frame? Surely it would still be sqrt(1 - (v/c)^2) in the line of motion rather than using the speed of light as measured by the rest frame underwater, but I'm not sure how to prove that.

 

[Dale]

Surely it would still be sqrt(1 - (v/c)^2) in the line of motion rather than using the speed of light as measured by the rest frame underwater, but I'm not sure how to prove that.

Start in the rest frame of the device where it is uncontracted and the speed of light is anisotropic. Then boost it to the frame where the water is at rest. You will get length contraction and the same phase shift.

 

[grav-universe]

Start in the rest frame of the device where it is uncontracted and the speed of light is anisotropic. Then boost it to the frame where the water is at rest. You will get length contraction and the same phase shift.

Well, it would depend upon the contraction of the device to find the anisotropic speed as measured by the device to begin with, which would differ from the Fizeau equations if the device were to contract differently from that of ordinary SR due to the different speed of light underwater, although I don't think it would, but I don't want to just assume it. The closest thing I can think of that is related to what you are saying would be to employ a tube filled with moving water with the device inside. The lab frame is now at rest with the device and measures c outside of the experiment and the device and water are moving past each other. So the question now becomes, if I drop an object into moving water, will the lab frame measure that the length of the object has now changed along the line of motion of the moving water? It doesn't seem that it should, but I'm looking for some principle that would ensure it. The difference wouldn't be noticable unless the water is flowing at relativistic speeds. Without such a principle, though, at this point I suppose I would just have to assume it remains the same unless some experiment were to show otherwise, which I doubt. Thanks for your time, Dalespam. :)

 

[Dale]

So the question now becomes, if I drop an object into moving water, will the lab frame measure that the length of the object has now changed along the line of motion of the moving water? It doesn't seem that it should, but I'm looking for some principle that would ensure it.

I already gave that to you. We know how the laws of physics work in the rest frame of the device, we know that the laws of physics are invariant under the Lorentz transform, so Lorentz transform from the rest frame of the device to the rest frame of the water.

 

[grav-universe]

I already gave that to you. We know how the laws of physics work in the rest frame of the device, we know that the laws of physics are invariant under the Lorentz transform, so Lorentz transform from the rest frame of the device to the rest frame of the water.

Thanks Dalespam, but that's just it. We don't know how the laws of physics work in the rest frame of a device moving through water. We only know how the laws of physics work in the lab frame, outside of the tube of water with light traveling at c and using Fizeau's equations to find the speed of light in the moving water.

We have two main possibilities. The laws of physics for the length contraction of the device are the same as in the lab frame, outside of the tube of water, where observers underwater with the device at rest with the water still measure a length contraction of sqrt(1 - (v/c)^2) of the device moving through the water, giving a null result only for the device at rest with the water but a non-null result for the MMX performed by the moving device since light travels at 3 c / 4 according to the frame at rest with the water, or that the laws of physics underwater are the same as in the lab frame by applying 3 c / 4 instead of c, so both devices still give a null result, but that would require a length contraction measured underwater of sqrt(1 - (v / (3 c / 4))^2), and the device moving through the water would measure light isotropically as well at 3 c / 4. In other words, we know what happens with the speed of light as it travels through the water according to the lab frame, but we don't know what physically happens to the devices themselves. I'm thinking the first scenario but as far as I can tell, we couldn't know for sure without performing an experiment to find out.

 

[Dale]

Thanks Dalespam, but that's just it. We don't know how the laws of physics work in the rest frame of a device moving through water.

Yes, we do know. The only law of physics involved here is Maxwell's equations, and Maxwell's equations are Lorentz invariant.

 

[grav-universe]

Yes, we do know. The only law of physics involved here is Maxwell's equations, and Maxwell's equations are Lorentz invariant.

Maxwell's equations apply only to the speed of light in vacuum, whereas MMX performed inertially in a vacuum will always produce null results. Due to the lesser speed of light underwater, however, MMX performed in moving water will produce non-null results, only null at rest with the water, so the physics changes somewhat within the medium, at least in terms of the results gained from such experiments. That is, unless objects underwater contract to sqrt(1 - (v / (3 c / 4)^2) as one frame observes another instead of sqrt(1 - (v/c)^2) as they would in vacuum, whereby the result of MMX in any inertial frame underwater would still be null. Of course the former seems much more likely, but I'm looking for a principle that would tell one way or the other. I think I have an idea about a principle that could be applied to this to find out, though, so I'll work on it some more. Thanks again.

 

[Dale]

Maxwell's equations apply only to the speed of light in vacuum

That is incorrect. Maxwell's equations govern all non-quantum aspects of electromagnetism.

 

[grav-universe]

Well, the principle I was trying to employ doesn't seem to be working out as I had hoped. The principle is just that we cannot find any absolute speed as told by the contraction of the device, but it seems that no matter how the device contracts, it would only be telling its relative speed to the water, so I'll have to find something else to use.

 

[grav-universe]

That is incorrect. Maxwell's equations govern all non-quantum aspects of electromagnetism.

Well, I'm not quite sure what you are trying to say at this point, so let me ask you this. Do you think the MMX performed in moving water would still produce a null result or no? If the device contracts by sqrt(1 - (v / c)^2) using light's speed in a vacuum, as observed by a frame at rest with the water, then the result will be non-null. If it contracts by sqrt(1 - (v / (3 c / 4))^2) using light's underwater speed, then the result will still be null in any frame.

 

[Dale]

Do you think the MMX performed in moving water would still produce a null result or no? If the device contracts by sqrt(1 - (v / c)^2) using light's speed in a vacuum, then the result will be non-null.

The result will be non null.

 

[JesseM]

Do you think the MMX performed in moving water would still produce a null result or no?

There is some ambiguity about the phrase "moving water", if you just mean the water is moving relative to the MMX apparatus then you can go with DaleSpam's answer of a non-null result, but if you mean that the water is "moving" in some absolute sense (relative to the ether in a Lorentz ether theory, say), then as long as the MMX apparatus is at rest relative to the water and Maxwell's laws apply in the frame where both at rest, the answer will be null despite this absolute motion.

 

[grav-universe]

The result will be non null.

Yes, that is what I figure too. I felt kind of silly asking, but I can't find a principle that would indicate that definitely, since a different contraction of the device which gives a null result shouldn't violate SR in any way, I don't think, but I also don't see why the dimensions of an object should change simply by submerging it in a medium, moving or not, except of course by the normal pressures that would be applied, but anyway, I agree. Thank you again, Dalespam.

 

[Dale]

I can't find a principle that would indicate that definitely

the LORENTZ TRANSFORM!

 

[grav-universe]

Okay, wait, one more quick question. :) If we have a light clock underwater, it will tick at 3/4 the rate as in a vacuum in a frame at rest with the water. For a light clock moving through the water at a speed v, the tick rate that the rest frame measures compared to its own, t = d / (3 c / 4), would be

((3 c / 4) t')^2 = d^2 + (v t')^2

t' = d / sqrt((3 c / 4)^2 - v^2)

so the rest frame measures a time dilation of the moving light clock to its own of

t / t' = sqrt(1 - (v / (3 c / 4)^2))

However, a mechanical watch traveling with the moving light clock would still tick with a time dilation of sqrt(1 - (v/c)^2), right? This would be because c isn't really light speed at all, but a universal speed that works the same underwater as in a vacuum even if light itself doesn't, correct?

 

[Dale]

A light clock with water instead of vacuum will undergo time dilation the same as a mechanical clock. Remember, the speed of light through water is not invariant.

 

[grav-universe]

A light clock with water instead of vacuum will undergo time dilation the same as a mechanical clock. Remember, the speed of light through water is not invariant.

I doubt you mean that the mechanical clock will have a time dilation of sqrt(1 - (v / (3 c / 4))^2), since of course that is not the same as the Lorentz transform in a vacuum, which it seems you've been driving at. :) But if you mean that the moving light clock will tick with a time dilation of sqrt(1 - (v/c)^2), then that does not account for the speed of light underwater. Could you clarify please?

 

[Dale]

If you fill a light clock with water then it will tick with a lower frequency. That is NOT time dilation. That is simply the tick rate of this new clock design.

If you then boost this clock it will time dilate just like any other clock.

 

[grav-universe]

If you fill a light clock with water then it will tick with a lower frequency. That is NOT time dilation. That is simply the tick rate of this new clock design.

If you then boost this clock it will time dilate just like any other clock.

Right, yes, two enclosed light clocks will have the time dilation sqrt(1 - (v/c)^2) because they are carrying the medium with them. That is where we would apply Fizeau's equations to derive sqrt(1 - (v/c)^2) (or vice versa) for the moving clock with the water moving with it since it would be ticking at the same rate in its own frame as measured for the the light clock at rest in the rest frame. I was referring to the case where we just have a single large open body of water with two mirrors at rest with the water and two mirrors moving through the water at v. The light clock at rest will tick at a rate t = d / (3 c / 4) while light will bounce between the two moving mirrors at a rate of t' = d / sqrt((3 c / 4)^2 - v^2), giving a time dilation of the moving light clock as measured in the frame of the light clock at rest of sqrt(1 - (v / (3 c / 4))^2). But regardless, the time dilation of a mechanical clock traveling with the moving light clock as measured by the frame at rest would still be sqrt(1 - (v/c)^2), right?

 

[Dale]

I was referring to the case where we just have a single large open body of water with two mirrors at rest with the water and two mirrors moving through the water at v. The light clock at rest will tick at a rate t = d / (3 c / 4) while light will bounce between the two moving mirrors at a rate of t' = d / sqrt((3 c / 4)^2 - v^2), giving a time dilation of the moving light clock as measured in the frame of the light clock at rest of sqrt(1 - (v / (3 c / 4))^2).

That is not time dilation. Time dilation is between two identically constructed clocks moving relative to each other, or (more clearly) for a single clock considered in different reference frames. The two clocks you are describing are not identically constructed, for instance if a collimated light source is used then the angle must be different. You would not call this time dilation any more than if you constructed two pendulum clocks, one with a 100 cm pendulum and one with a 50 cm pendulum.

ALL clocks undergo the same time dilation by the standard Lorentz factor, whether the clock is mechanical or optical or nuclear or whatever. This is required by the first postulate. If you get something that says otherwise then you know from first principles that it is either incorrectly calculated or does not represent time dilation.

 

[grav-universe]

Ah yes, of course. Since the medium physically affects light's speed, it would be considered part of the light clock as well, so the light clock constitutes more than just mirrors and photons in this case. Therefore a light clock with the medium at rest and one with the medium flowing through it are not considered identical and the ratio of tick rates are not considered time dilation. Even more so since simply changing the angle between the mirrors in the moving frame changes the tick rate if MMX produces a non-null result. Makes sense. Thank you, Dalespam.