Speed of light for a Rindler observer

[Gaussian97]

TL;DR

I've read that speed of light for a Rindler observer is not constant.

I've read that speed of light for a Rindler observer is not constant. I wasn't sure why and I tried to do this as an exercise for myself, I think I came with the correct answer, but I'm not sure, is the following argument correct?

Let's describe the (flat) space-time using Rindler coordinates $\tau, \rho$ and let's assume a (non-inertial) observer moving by the worldline $\rho=\rho_0$. I want to compute the speed of light for that observer.
First of all, to define the velocity we need to talk about distances and time. In SR the time measured by any observer is the "length" of its worldline, $s$. Therefore because our observer's worldline is defined as $d\rho=0$ the Rindler metric gives us $ds=\rho_0 d\tau$. Now the distance that this observed sees is the "negative length" between two points at equal time, i.e., equal $\tau$. So again the Rindler metric gives $dl=-ds=d\rho$. Finally, the velocity is the change in space divided by the change on time, so
$$v=\frac{1}{\rho_0}\frac{d\rho}{d\tau}$$
Because a lightray has $ds=0$ we find that
$$v=\frac{1}{\rho_0}\frac{d\rho}{d\tau}=\frac{\rho}{\rho_0}$$
Which depends on the $\rho$ coordinate.

 

[Dale]

What surprises me is that, not only is not constant but also can be greater than 1.

Why does that surprise you? The speed of light is c in an inertial frame. The Rindler frame is non-inertial.

 

[Gaussian97]

I don't know, many years hearing the phrase "the speed of light is the absolute maximum speed" that, the first time you see a violation of this, I think it must surprise you. Anyway, my question was not about how much surprise should this provoke, but rather about if the reasoning there is correct or not.

 

[Nugatory]

Which depends on the $\rho$ coordinate. What surprises me is that, not only is not constant but also can be greater than 1.

That's a coordinate velocity; it just means that you've chosen position and time coordinates in such a way that the difference between the position coordinate of the light flash at the same time (note the slipperiness of that phrase!) that a clock on the accelerating object reads $\tau_1$ and the position coordinate of the light flash at the same time that a clock on the accelerating object reads $\tau_2$ is greater than $c(\tau_2-\tau_1)$. Clearly that has more to do with the choice of coordinates than anything else.

The speed of light limit is still there though. Consider any timelike wordline that intersects the null worldline of the flash of light; that timelike worldline is the worldline of some hypothetical observer. How does their speed compare with the speed of the flash of light in the neighborhood of the intersection? This calculation may be easier to do if you transform back to Minkowski coordinates first.

Informally: no matter who does the calcuation and what speeds they calculate, no one is ever going to win a race with a flash of light.

 

[A.T.]

I don't know, many years hearing the phrase "the speed of light is the absolute maximum speed" that, the first time you see a violation of this,

Go outside, and look at the stars. In the non-inertial rotating rest frame of the Earth they move faster than c (except our Sun).

 

[Dale]

Anyway, my question was not about how much surprise should this provoke, but rather about if the reasoning there is correct or not.

Yes, your reasoning is fine. In non-inertial coordinates there may be multiple ways to get a velocity. What you did seems fine to me.

I probably would have calculated a parametric expression for the worldline and then used it to calculate $d\rho/d\tau$. But again what you did seems good to me.

 

[Gaussian97]

Yes, your reasoning is fine. In non-inertial coordinates there may be multiple ways to get a velocity. What you did seems fine to me.

I probably would have calculated a parametric expression for the worldline and then used it to calculate $d\rho/d\tau$. But again what you did seems good to me.

Ok thanks :D

 

[cianfa72]

That's a coordinate velocity; it just means that you've chosen position and time coordinates in such a way that the difference between the position coordinate of the light flash at the same time (note the slipperiness of that phrase!) that a clock on the accelerating object reads $\tau_1$ and the position coordinate of the light flash at the same time that a clock on the accelerating object reads $\tau_2$ is greater than $c(\tau_2-\tau_1)$. Clearly that has more to do with the choice of coordinates than anything else.

I believe $\tau_1$ and $\tau_2$ readings of the two clocks "bolted" on the accelerating object are really the "coordinate clock time".

Is the Rindler coordinate time equal to the proper time elapsed for a Rindler observer (basically the time elapsed for an observer with fixed coordinate position in Rindler coordinates) ?

 

[PeterDonis]

I believe $\tau_1$ and $\tau_2$ readings of the two clocks "bolted" on the accelerating object are really the "coordinate clock time".

For an appropriately chosen Rindler observer, the two are the same. If you are only dealing with one Rindler observer, obviously that observer is going to be the one for whom the two are the same.

Is the Rindler coordinate time equal to the proper time elapsed for a Rindler observer (basically the time elapsed for an observer with fixed coordinate position in Rindler coordinates) ?

It is for one particular Rindler observer at one particular value of the Rindler space coordinates. And, as above, if you are only dealing with one Rindler observer, that observer will be at those space coordinates.

 

[cianfa72]

To measure/define the (coordinate) speed of the light propagation process I believe two Rindler observers are actually needed (in fact the light path in spacetime is null-like).

To me it is not entirely clear the reasoning of the OP to get the result.

 

[Nugatory]

To measure/define the (coordinate) speed of the light propagation process for a Rindler observer I believe two Rindler observers are actually needed (in fact the light path in spacetime is null-like).

Strictly speaking we don’t need any Rindler observers at all. We just need the $x$ and $t$ coordinates of two events: flash of light emitted and flash of light received, and we can calculate these from observations made by anyone, no matter their state of motion. But no matter how we come up with these values (and strategically placed Rindler observers may be the easiest way) by definition then the coordinate speed of light is $\Delta x$, the difference between the $x$ coordinates, divided by $\Delta t$, the difference between the $t$ coordinates. The light path is null but neither $\Delta x$ nor $\Delta t$ will be zero.

 

[cianfa72]

Strictly speaking we don’t need any Rindler observers at all. We just need the $x$ and $t$ coordinates of two events: flash of light emitted and flash of light received, and we can calculate these from observations made by anyone, no matter their state of motion.

Sure, my point was that no same 'physical' body can be at the emission and reception of a flash of light thus really a 'family' of observers is actually needed (in this case the family of Rindler observers). This family of observers account for the $x$ and $t$ coordinates

 

[PeterDonis]

To measure/define the (coordinate) speed of the light propagation process I believe two Rindler observers are actually needed

One observer and a mirror would be enough; you don't need a clock at the mirror, you just need to know its Rindler space coordinate. The point is that you would calculate the speed using Rindler coordinate values.

 

[Gaussian97]

Sure, my point was that no same 'physical' body can be at the emission and reception of a flash of light thus really a 'family' of observers is actually needed (in this case the family of Rindler observers). This family of observers account for the $x$ and $t$ coordinates

Well, maybe I'm being very innocent, but I think that it is similar to Special Relativity, in there you can talk about the velocity of some object, imagine that it starts at some space-time point A, and it ends at B. Then the velocity is the spatial distance between the two points divided with the time distance between them. You just need one observer, to specify the coordinates of A and B, of course, that has nothing to do with the observer being at both space-time points (as long as he/she can determine the coordinates of both points). But maybe you are saying something deeper than what I said and I need to think it more carefully.

 

[Nugatory]

Well, maybe I'm being very innocent, but I think that it is similar to Special Relativity, in there you can talk about the velocity of some object, imagine that it starts at some space-time point A, and it ends at B. Then the velocity is the spatial distance between the two points divided with the time distance between them. You just need one observer, to specify the coordinates of A and B, of course, that has nothing to do with the observer being at both space-time points (as long as he/she can determine the coordinates of both points).

You’ve got it right... if all speed measurements required an observer present at one of the endpoints being measured, we wouldn’t get much of an answer when we googled for “orbital speed of moon”. Many textbook examples, especially in relativity, include observers strategically placed at these endpoints but that’s done to simplify determining the coordinates of the relevant events, not because it’s necessary.

 

[Halc]

I agree that one point of measurement is enough. It kind of defines the frame.
The moon is an interesting case since the speed is typically specified relative to Earth, which is isn't nearly as inertial as the Earth/moon barycenter. But nobody specifies the orbital speeds of the two objects relative to that. Why not?

Also, the speed is very much frame dependent under GR. An observer on the moon is going to measure a slower speed than the Earth observer simply because his clock runs faster in the lower gravitational potential up there.

OK, this is a Rindler thread. We're not talking about inertial frames now. We have a rindler coordinate system. The x coordinate is conveniently painted on the walls of our accelerating craft, so that's the easy part. Problem is, clocks placed at different x locations don't stay in sync, so you can't use multiple synced clocks to run any measurement. So speed of light depends on where the light goes.

I place a mirror 100m up (in the direction of acceleration) and another 100m down. I flash a light pulse and the once from ahead comes back before the rear one. Speed of light (c +- a bit respectively) is direction dependent in this non-local test. It is c locally, but my mirrors are placed at different potentials and they 'experience' different g forces. Of course the timing one way isn't expected to be the same as the other.

 

[cianfa72]

Then the velocity is the spatial distance between the two points divided with the time distance between them. You just need one observer, to specify the coordinates of A and B, of course, that has nothing to do with the observer being at both space-time points (as long as he/she can determine the coordinates of both points).

Right, thus coming back to first post the coordinate speed of light should result just in $\rho$ and not $\frac{\rho}{\rho_0}$ I believe...

One observer and a mirror would be enough; you don't need a clock at the mirror, you just need to know its Rindler space coordinate. The point is that you would calculate the speed using Rindler coordinate values.

Thus what you're actually considering is the measurement of two-way (round-trip) velocity of a light pulse bouncing back at the mirror. From mirror's Rindler space coordinate and $ds=0$ (light propagation equation) you can actually calculate both (one way and two way coordinate speed of light propagation process)

 

[cianfa72]

Right, thus coming back to first post the coordinate speed of light should result just in $\rho$ and not $\frac{\rho}{\rho_0}$ I believe...

Are you agree ? What do you think about ?

 

[Gaussian97]

Are you agree ? What do you think about ?

Well, that was part of my question, so as far as I know, I think it should be as I originally said $\frac{\rho}{\rho_0}$. Now we should clarify what you mean by coordinate velocity if you mean the change in coordinates, i.e. how $\rho$ changes when $\tau$ changes, then yes for sure
$$\frac{d\rho}{d\tau}=\rho$$
What I mean is the physical velocity i.e. how many meters would the light travel every second, measured with respect to an observer moving along $\rho_0$.

 

[PeterDonis]

What I mean is the physical velocity i.e. how many meters would the light travel every second, measured with respect to an observer moving along $\rho_0$.

That observer can only measure the "physical velocity" of light rays that are passing his spatial location. He will measure their "physical velocity" to be $c$. He cannot measure the "physical velocity" of light rays that are spatially distant from him.

 

[Gaussian97]

That observer can only measure the "physical velocity" of light rays that are passing his spatial location. He will measure their "physical velocity" to be $c$. He cannot measure the "physical velocity" of light rays that are spatially distant from him.

Mmm... Okay so this sounds to me very similar to what cianfa said in #10. But I don't actually see why is this, let me put some example, to simplify things let assume that I'm a Rindler observer with $\rho_0 = 0.5$. Now suppose someone emits a lightray at $t=0, x=1$ (Minkowskian) and it's absorbed at $t=1, x=2$. Why I can't do the following argument:
I observe the emission at $\tau=0, \rho=1$, and the absorption at $\tau\approx0.5493, \rho=\sqrt{3}$, then I measure the time distance, $\Delta T$ between these two events, because I know that time is related to $\tau$ as $\Delta T=\frac{\Delta \tau}{2}\approx 0.2747$ and the physical distance is exactly $\Delta L = \Delta \rho \approx 0.7321$, so the average velocity of the lightray is
$$\frac{\Delta L}{\Delta T} = 2.665$$

Is this argument incorrect?

 

[PeterDonis]

I observe the emission at $\tau=0, \rho=1$, and the absorption at $\tau\approx0.5493, \rho=\sqrt{3}$

No, you don't. You calculate these values as coordinate values for the emission and absorption events. So your calculation is a calculation of a coordinate speed, not a "physical speed".

 

[Gaussian97]

Well, I don't observe them, of course, but I know that they are occurring at those coordinates. Doesn't that allow me to say that the lightray has traveled 0.7 meters in 0.3 seconds? (Well, of course, I'm using $c=1$ so my units wouldn't be meters and seconds, but that's not the important here)

 

[PeterDonis]

I don't observe them, of course, but I know that they are occurring at those coordinates.

Yes. Note the word "coordinates", that you agree is the correct one to use.

Doesn't that allow me to say that the lightray has traveled 0.7 meters in 0.3 seconds?

As a coordinate speed. Not as a "physical speed". The two are not the same. And you already agreed--see above--that "coordinates" is the correct term to use for this case, which means the speed you are giving here is a coordinate speed.

 

[Dale]

Well, I don't observe them, of course, but I know that they are occurring at those coordinates. Doesn't that allow me to say that the lightray has traveled 0.7 meters in 0.3 seconds?

Sure, you can do that. But most people will call that a coordinate speed, not a physical speed. Physical speed would usually be a coordinate-independent quantity, which would only ever be c.

 

[Gaussian97]

Ok, then we're talking about the same thing, sorry for confusing you using the word "physical", I was referring to what you call the coordinate speed. I thought that coordinate speed means $\frac{\Delta \rho}{\Delta \tau}$. But if with coordinate speed we already account for the $\rho_0$ factor I'm ok with that.

 

[PeterDonis]

I thought that coordinate speed means $\frac{\Delta \rho}{\Delta \tau}$.

It does. In post #21 you described a light ray with $\Delta \rho \approx 0.7321$ and $\Delta \tau \approx 0.5493$. So the coordinate speed of that light ray will be $\approx 1.3328$.

I don't understand how you are getting what you call $\Delta T$ in post #21, or what you think it means, or what you think $\Delta L / \Delta T$ in that post means. It certainly isn't either a coordinate speed or a "physical speed".

 

[Gaussian97]

Well, then definitely there's something I'm understanding wrong, my reasoning is precisely the one I describe in #1; basically, the main goal of this threat that, until #17 nobody seem to have any problem with . Anyway, I repeat it here:
Because I'm an observer with worldline defined as (in Rindler coordinates) $\rho=\rho_0=\text{const.}$, what I perceived as "time" is not the coordinate $\tau$, but $\rho_0 \tau$. This is obtained, of course from the metric:
$$ds^2 = \rho_0^2 d\tau^2$$
Therefore if I observe two events separated by $\Delta \tau$, the time I observe between them is not $\Delta \tau$ but $\rho_0\Delta \tau$.
In #21 I consider as example $\rho_0=0.5$ so the time between the two events that I describe, that are separated by $\Delta \tau \approx 0.5$, is precisely $\Delta T \approx 0.3$.

At least I think this gives more useful information than the coordinate velocity, because if now I use different coordinates defined as $\tau' = 2\tau$ you coordinate velocity changes by a factor 2, but my velocity stays the same, because there's also a factor 2 in the relation between "time" and $\tau'$ that cancels.
Actually, I would say that as I defined, my concept of "physical velocity" is invariant under general coordinate transformations...

 

[PeterDonis]

Because I'm an observer with worldline defined as (in Rindler coordinates) $\rho=\rho_0=\text{const.}$, what I perceived as "time" is not the coordinate $\tau$, but $\rho_0 \tau$.

"What you perceive as time" is vague. The correct statement is that your proper time as a function of coordinate time along your worldline is $\rho_0 \tau$.

Therefore if I observe two events separated by $\Delta \tau$

You can only observe such events if they are on your worldline. The events you are trying to analyze are not, so you cannot observe them. You can only calculate their coordinates.

At least I think this gives more useful information than the coordinate velocity

No, it doesn't. It gives less useful information, because it throws together quantities that have no actual relationship: they aren't both coordinate quantities in the same coordinate chart, and they aren't both directly observable invariants.

Actually, I would say that as I defined, my concept of "physical velocity" is invariant under general coordinate transformations...

I challenge you to prove this. You will lose the challenge.

 

[PeterDonis]

Well, then definitely there's something I'm understanding wrong, my reasoning is precisely the one I describe in #1; basically, the main goal of this threat that, until #17 nobody seem to have any problem with .

Everybody all along has been telling you that the things you are calculating are coordinate speeds, not physical speeds. Apparently you haven't been paying attention.