High School One-Way Speed of Light

[Orodruin]

If Snell's law wasn't isotropic, then all the optics in the world would need constant adjustment as the world spins around. Since this is not necessary, Snell's law is isotropic.

It is isotropic only if you have orthogonal temporal and spatial axes. In essence, you are making the assumption of using coordinates where the speed of light is isotropic because that is what your version of Snell’s law states. Only in such coordinates, where the one-way speed of light is the same as the two-way speed of light, does your conclusion make sense.

 

[shawnhcorey]

Again, you aren't understanding. If you can measure the one-way speed of light then you can synchronise remote clocks without assumptions. You seem to understand that this synchronisation is impossible, but then contradict yourself by claiming a one-way speed measure that would allow it.

I said synchronization is impossible using two clocks. That does not imply it is impossible using one clock.

 

[shawnhcorey]

It is isotropic only if you have orthogonal temporal and spatial axes. In essence, you are making the assumption of using coordinates where the speed of light is isotropic because that is what your version of Snell’s law states. Only in such coordinates, where the one-way speed of light is the same as the two-way speed of light, does your conclusion make sense.

That is also how they measure the two-way speed of light. They ignore any acceleration and therefore assume orthogonal axes. If they do it for the two-way speed of light, it should be done the same for the one-way speed of light.

 

[Orodruin]

That is also how they measure the two-way speed of light. They ignore any acceleration and therefore assume orthogonal axes. If they do it for the two-way speed of light, it should be done the same for the one-way speed of light.

That’s just ignorant. Further discussion is useless until you have actually learned relativity and are ready to accept that you may be wrong.

 

[Ibix]

That does not imply it is impossible using one clock.

What are you synchronising if you only have one clock..?

 

[PeterDonis]

I said synchronization is impossible using two clocks. That does not imply it is impossible using one clock.

As I said in post #27, if your experiment only has one clock, it can't be done the way you describe it--you can't put the transparent medium in only one leg of the light beam's travel. You can only do that if there are two separate clocks.

 

[PeterDonis]

That is also how they measure the two-way speed of light. They ignore any acceleration and therefore assume orthogonal axes. If they do it for the two-way speed of light, it should be done the same for the one-way speed of light.

This is simply wrong. No "axes" have to be assumed at all to measure the round-trip speed of light. All you need is a clock and a mirror at rest relative to it, with a ruler extended between the clock and the mirror to read off the distance between them. You don't even need coordinates; everything in the experiment is a direct observable.

 

[javisot]

If you are using only one clock, as you say, that means the light has to return to the same point that it started from, since it has to come back to the same clock. Which means you can't place a transparent medium in "one side" of its path and not the other. Both legs of the light's path--outbound and return--cover the same path in space. So if the medium is there on one leg, it's there on the other as well.

If, on the other hand, you insist on having the light hit the mirror at an angle so that the two legs of its path are in different parts of space, then you can't measure the light's flight time using just one clock. You need two.

Either way, your claim to have discovered a way to make an invariant (independent of any conventions) measurement of the one-way speed of light is not valid.

If I understand you correctly, Peter, you're saying that "measuring the one-way speed of light" means that we have emitted a photon, overtaken it, and received it ourselves without any intermediate interaction.

Since the speed of light is the maximum speed in the universe, it is impossible to overtake a photon that we have emitted ourselves, and this is equivalent to "not being able to measure the one-way speed of light."

Is this correct?

 

[PeterDonis]

If I understand you correctly, Peter, you're saying that "measuring the one-way speed of light" means that we have emitted a photon, overtaken it, and received it ourselves without any intermediate interaction.

No. I have no idea where you got any of this from.

Also, please do not use the term "photon" in this context. This is the relativity forum, not the quantum physics forum. (Even in a quantum context, the term "photon" doesn't mean what you think it means--but that's a discussion for some other thread, not this one.)

 

[javisot]

If you are using only one clock, as you say, that means the light has to return to the same point that it started from, since it has to come back to the same clock. Which means you can't place a transparent medium in "one side" of its path and not the other. Both legs of the light's path--outbound and return--cover the same path in space. So if the medium is there on one leg, it's there on the other as well.

If, on the other hand, you insist on having the light hit the mirror at an angle so that the two legs of its path are in different parts of space, then you can't measure the light's flight time using just one clock. You need two.

Either way, your claim to have discovered a way to make an invariant (independent of any conventions) measurement of the one-way speed of light is not valid.

I could say light instead of photon (like you) if that's more appropriate, ok. I'm trying to understand this. Could you explain more?

 

[PeroK]

That is also how they measure the two-way speed of light. They ignore any acceleration and therefore assume orthogonal axes. If they do it for the two-way speed of light, it should be done the same for the one-way speed of light.

When you say "they" I guess you mean professional physicists. Given that @Orodruin is a physics professor, you're now actually telling him how he does things!

 

[Nugatory]

That is also how they measure the two-way speed of light. They ignore any acceleration and therefore assume orthogonal axes.

The two way measurement uses the proper time along the clock's worldline between the emission event and the reception event; no coordinates and coordinate axes enter into the computation at any time.

 

[Nugatory]

If I understand you correctly, Peter, you're saying that "measuring the one-way speed of light" means that we have emitted a photon, overtaken it, and received it ourselves without any intermediate interaction.

A one way measurement is a measurement between two different points in space, with a clock at each point. We say the flash of light (not photon! photons are not what you think they are and light is not photons moving from one point to another) left point A when the clock there read zero, arrived at point B when the clock there read $T$, if the two points are separated by distance $D$ then the one-way speed of light is $D/T$.
It's the same logic as if leave my house at noon, arrive at my destination 30 miles away when the clock there reads 1:00, therefore my average speed was 30 miles per hour.

There's no catching up or overtaking involved.

 

[Ibix]

If I understand you correctly, Peter, you're saying that "measuring the one-way speed of light" means that we have emitted a photon, overtaken it, and received it ourselves without any intermediate interaction.

I think Peter's saying that you can only measure a one-way speed with one clock if you can overtake the light. You can do one-way measures using two clocks. The setup under discussion here is really a two-way measure - the OP only thinks it's a one-way measure because he's failing to use the refractive index correctly in his anisotropic system.

 

[javisot]

I think Peter's saying that you can only measure a one-way speed with one clock if you can overtake the light.

That's what I said, but Peter comments that's not correct. I have asked him to explain more about this.

 

[Ibix]

Hold on that's what I said, but Peter comments that's not correct. I have asked him to explain more about this.

Not quite - you omitted the "with one clock" bit. You can measure the one-way speed with two clocks without travelling faster than light (obviously the answer is assumption dependent).

Or at least, that's what I think Peter's point is.

 

[PeterDonis]

I think Peter's saying that you can only measure a one-way speed with one clock if you can overtake the light.

That's not what I was saying, no. @Nugatory in post #43 correctly described what I was saying.

 

[PeterDonis]

That's what I said, but Peter comments that's not correct. I have asked him to explain more about this.

@Nugatory in post #43 correctly explained what I was saying. However, there's also an additional point.

Since a clock can't move faster than light, consider trying to measure the speed of a bullet instead. Does it even make sense to send a clock along the bullet's path faster than the bullet in order to measure the bullet's speed? The clock is moving. That means, whatever you think you're measuring, it isn't the speed of the bullet relative to clocks at rest, which is what a one-way speed measurement is supposed to be measuring.

Add to that the fact that clocks can't move faster than light, or even at the speed of light, to begin with, and what does that leave?

 

[javisot]

Add to that the fact that clocks can't move faster than light, or even at the speed of light, to begin with, and what does that leave?

That we can't measure the one-way speed of light "with a single clock".

We can bounce light off a mirror, we can bend geometry to make it back, we can use two clocks and send a signal back when the light reaches point B, but that's not measuring the one-way speed of light with a single clock (at rest).

 

[Dale]

TL;DR Summary: The one-way speed of light can be measure with the use of a transparent medium.

To measure the two-way speed of light, a pulse of light is sent thru a vacuum to bounce off a mirror and return. By knowing the distance to the mirror and how long the pulse takes to travel, the two-way speed of light can be calculated. To measure the one-way speed of light, place a transparent medium in one side of the path and measure the time. Then repeat for the other side. The difference in the times can be use to calculate the difference in the speed of light (if any) in both directions.

Example

Set up the mirror so that it is 29.9792458 cm from the emitter and the detector. It will take 2.00 ns for the pulse to travel to the mirror and back. Place a transparent medium in one side that slows light by 10%.

Same Speed

If light travels at the same speed in both directions, it takes 1.00 ns for one direction. By slowing the light by 10% on one side, it will take 1.10 ns. Same for the other side. Total travel time is 2.10 ns.

Different Speed

For the sake of argument, assume light takes 0.50 ns in one direction and 1.5 ns in the other. Total round-trip time is 2.00 ns.

Place the medium in the side of 0.50 ns. This slows the speed to 0.55 ns and has a total round-trip time of 2.05 ns. Placing it in the other side slows light to 1.65 ns for a round-trip time of 2.15 ns.

If the times for the medium in each side is the same, the light travels the same speed in both directions. But what if the amount the medium slows light down is dependent on the direction?

Snell's Law

Snell's law states that the ratio of the sine of angle of incident to the sine of the angle of refraction is equal to the ratio of the corresponding speed of light in the mediums. Set up an apparatus so that the light in the vacuum is in the same direction as one direction of the pulse of light and then in the other direction. If the angles remain the same, then the medium slow light by the same factor.

In order to measure the one-way speed of light you cannot assume it. You have to use a framework where the one-way speed of light is a variable, and then show how your measurement depends on that variable.

So, we can use Anderson's approach where the one way speed of light is determined by a variable $\kappa$. The transform between an Einstein synchronized frame and an Anderson frame is: $$T=t-\kappa x/c$$$$X=x$$$$Y=y$$$$Z=z$$ where the capitalized coordinates are Anderson’s and the lower-case coordinates are Einstein’s. From this you can calculate that $$V=\frac{dX}{dT}=v\frac{1}{1-\kappa v/c}$$ where $v=dx/dt$.

For this problem the one way time for each leg is $$\Delta T = \frac{\Delta X}{V}=\frac{\Delta x}{v/(1-\kappa v/c)}$$ For the forward path $\Delta x = \Delta X = 29.9792458 \mathrm{\ cm} = 1 \mathrm{\ ns}\ c$ and for the reverse path $\Delta x = \Delta X = -29.9792458 \mathrm{\ cm}= 1 \mathrm{\ ns}\ c$.

When the forward path is empty the velocity is $v=c$ which is $V=c/(1-\kappa)$.

When the forward path has the medium the velocity is $v=0.9 c$ which is $V=0.9 c/(1-0.9 \kappa)$.

When the reverse path has the medium the velocity is $v=-0.9 c$ which is $V=-0.9c/(1+0.9\kappa)$.

When the reverse path is empty the velocity is $v=-c$ which is $V=-c/(1+\kappa)$.

Each individual leg depends on the one-way speed of light parameter $\kappa$. Now, let's sum the times to find the result of the experiment. For the medium in the forward path we get $$\Delta T = \frac{1 \mathrm{\ ns}\ c}{0.9 c/(1-0.9 \kappa)}+\frac{-1 \mathrm{\ ns}\ c}{-c/(1+\kappa)} = 2.11 \mathrm{\ ns}$$ And for the medium in the reverse path we get $$\Delta T = \frac{1 \mathrm{\ ns}\ c}{ c/(1- \kappa)}+\frac{-1 \mathrm{\ ns}\ c}{-0.9 c/(1+0.9 \kappa)} = 2.11 \mathrm{\ ns}$$

So although each leg does depend on $\kappa$, the overall experiment does not. It turns out that there is no clever way around this. It is not possible for any experimental measurement to depend on $\kappa$ without assuming it first, typically through clock synchronization or some assumption of isotropy.

Explicitly, as was mentioned before, assuming that Snell's law is isotropic is not consistent with assuming that the one way speed of light is anisotropic. The correct formula for determining one way speeds is given above. It is not just the one way speed of light that is affected, as the math clearly shows.

 

[PeterDonis]

That we can't measure the one-way speed of light "with a single clock".

Yes, that's correct. But it's not what you said in the post of yours that I responded to earlier.

 

[pervect]

While this isn't strictly relevant to the question of the one-way speed of light, for ANYTHING ELSE, anything that moves slower than light, one can define a proper velocity that does not require any syncrhonization convention.

The procedure is straightforward and easy to explain without detailed math. Instead of using two clocks, one at the source and one at the destination, one measures the trip time with a clock on the moving object. For instance, one might put a clock on an airplane (or a boat, or a train, or whatever), starting the clock when it files over the start line of some course, and stopping it when it crosses the "finish line".

Note that one cannot use this technique with light, light does not have a proper time. One can take an appropriate limit by imagining an object moving faster and faster - in this limit, the trip time always approaches zero for a high enough velocity. So, while light doesn't have a proper time, one can say in the limit as the speedd of a particle appraoches the speed of light, the proper time of a trip approaches zero.

To make this observation a bit relevant to light, one can point out that the speed of light can be taken as the limiting velocity of an object with a lot of energy in any theory compatible with relativity. Light is just very convenient, as it moves at the limiting velocity.

The underlying physical principle here was mentioned by Einstein in his 1905 paper - it is called isotropy.

The assumption that if two objects moving in opposite directions have the same elapsed time to cover the course should also be measured to take the same time to finish the course by any "fair" measurement is equivalent to the asumption of isotropy and Einstein's scheme for synchronizing clocks.

 

[PeterDonis]

one can define a proper velocity that does not require any syncrhonization convention.

What distance does one use to do this? Yes, a clock records proper time along its worldline--but along its worldline, the distance it travels is zero. Everything else is moving, the clock is at rest.

in the limit as the speedd of a particle appraoches the speed of light, the proper time of a trip approaches zero.

How do you take this limit? Note that, again, in the particle's rest frame, its speed is zero, not "approaching the speed of light".

 

[pervect]

What distance does one use to do this? Yes, a clock records proper time along its worldline--but along its worldline, the distance it travels is zero. Everything else is moving, the clock is at rest.

The distance is defined in some particular frame - it's a frame dependent quantity. The proper time is, of course, frame independent.

How do you take this limit? Note that, again, in the particle's rest frame, its speed is zero, not "approaching the speed of light".

One picks a spatial frame of reference (which defines the distance, as above, and also and a notion of "rest". It also gives rise to the meaning of velocity, though if we insist on using the standard defnition of velocity we have to introduce a simultaneity convention. The point is that the closely related notion of velocity called "proper velocity" doesn't need a synchronization convention, and one can do a lot of physics with it. In fact, I think it is conceptually simpler - IMO it's really mostly experimental constraints that make stationary clocks more popular, but using those to measure velocity does require some notion of synchronization.

To recap and explain further, In this frame we can compute the proper velocity, by dividing the distance by the proper time elapsed along the worldline. As we increase the velocity, as defined above by our choice of frame of reference, the proper time approaches zero. The proper velocity numerically approaches infinity as we do this, since the distance is constant and the proper time approaches zero. The ordinary velocity will approach the same velocity as light in SR - essentially we are not creating a new theory, we're just providing an alternate way of thinking about SR.

Essentially this is just special relativity, but we've made a point of thinking of and presenting "c" as "the ultimate speed", as per the title of a video I often mention, rather than assuming a priori that light is somehow special. The nature of light really isn't all that important to the underlying theory of relativity - this is just a way of attempting to communicate that. It probably won't short-circuit the usual seemingly endless discussion of one-way vs two-way speeds of light, but I can always hope this alternate approach will be useful to a few readers.

Another way of saying this - it is the standard choice of "velocity" using the whole setup of two clocks that requires a notion of synchronization. Using "proper velocity" rather than the usual formulation allows us to downplay the signnificane of that and illustrate it's true nature as a convenience, rather than anything fundamentally important.

 

[PeterDonis]

The distance is defined in some particular frame - it's a frame dependent quantity. The proper time is, of course, frame independent.

Which means that the "proper velocity", as you are calling it, is not an invariant (because it's calculated from a frame-dependent quantity), so it's (a) misnamed (the adjective "proper" is usually reserved for invariants) and (b) not physically meaningful.

the closely related notion of velocity called "proper velocity" doesn't need a synchronization convention

I don't see how that can be true if it's a frame-dependent quantity, per the above.

Essentially this is just special relativity, but we've made a point of thinking of and presenting "c" as "the ultimate speed", as per the title of a video I often mention, rather than assuming a priori that light is somehow special.

Another way of saying this - it is the standard choice of "velocity" using the whole setup of two clocks that requires a notion of synchronization. Using "proper velocity" rather than the usual formulation allows us to downplay the signnificane of that and illustrate it's true nature as a convenience, rather than anything fundamentally important.

Do you have a reference to a textbook or peer-reviewed paper (not a video) that expounds this approach? I'm quite skeptical that it actually helps.

 

[javisot]

I'm confused by these last comments, I understood that the proper time of light is zero, it has no reference frame at rest and its proper velocity is not defined in SR. Right?

 

[cianfa72]

In order to measure the one-way speed of light you cannot assume it. You have to use a framework where the one-way speed of light is a variable, and then show how your measurement depends on that variable.

So, we can use Anderson's approach where the one way speed of light is determined by a variable $\kappa$. The transform between an Einstein synchronized frame and an Anderson frame is: $$T=t-\kappa x/c$$$$X=x$$$$Y=y$$$$Z=z$$ where the capitalized coordinates are Anderson’s and the lower-case coordinates are Einstein’s. From this you can calculate that $$V=\frac{dX}{dT}=v\frac{1}{1-\kappa v/c}$$ where $v=dx/dt$.

Btw, I'd add the following (we are assuming the realm of SR).

Just to keep it simple let's drop two dimensions and consider the frame representation of inertial motion. W.r.t. the Einstein's synchronized frame it has the form $x= a + bt$. Transforming to Anderson's frame we get also a linear form $X= A +BT$. Since w.r.t. Anderson's frame inertial motion occurs at constant coordinate velocity then, by definition, it is an inertial frame as well.

Now, the very fact that both frames result to be inertial, allows us to take seriuosly both round-trip measurements about the fact that the two-way speed of light is $c$.

 

[PeterDonis]

I understood that the proper time of light is zero

The spacetime interval along the worldline of a light ray is zero. The term "proper time" doesn't even apply to light because that term only applies to timelike intervals, not lightlike intervals.

it has no reference frame at rest

That's correct.

its proper velocity is not defined in SR.

The definition @pervect gave for "proper velocity" would be undefined for light, yes.

 

[cianfa72]

The definition @pervect gave for "proper velocity" would be undefined for light, yes.

Sorry, maybe I missed the definition of "proper velocity" given by @pervect. Is it basically related to the "spacetime angle" between a test object w.r.t. a reference body at the point where the two worldlines cross ?

 

[PeterDonis]

maybe I missed the definition of "proper velocity" given by @pervect.

It's given in post #54.