Speed of light for a Rindler observer

In summary: No, there is a difference between coordinate time and proper time. coordinate time is the time measured by an observer in Rindler coordinates, while proper time is the time measured by an observer in any frame.
  • #1
Gaussian97
Homework Helper
683
412
TL;DR Summary
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.
 
Last edited:
Physics news on Phys.org
  • #2
Gaussian97 said:
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.
 
  • #3
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.
 
  • #4
Gaussian97 said:
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.
 
  • Like
Likes cianfa72, PeroK, Ibix and 1 other person
  • #5
Gaussian97 said:
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).
 
  • Like
Likes etotheipi and Nugatory
  • #6
Gaussian97 said:
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.
 
  • Like
Likes Gaussian97
  • #7
Dale said:
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
 
  • Like
Likes Dale
  • #8
Nugatory said:
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) ?
 
  • #9
cianfa72 said:
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.

cianfa72 said:
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.
 
  • #10
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. :rolleyes:
 
Last edited:
  • #11
cianfa72 said:
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.
 
  • #12
Nugatory said:
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
 
  • #13
cianfa72 said:
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.
 
  • #14
cianfa72 said:
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.
 
  • #15
Gaussian97 said:
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.
 
  • Like
Likes Dale
  • #16
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.
 
  • #17
Gaussian97 said:
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...

PeterDonis said:
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)
 
Last edited:
  • #18
cianfa72 said:
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 ?
 
  • #19
cianfa72 said:
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##.
 
  • #20
Gaussian97 said:
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.
 
  • #21
PeterDonis said:
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?
 
  • #22
Gaussian97 said:
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".
 
  • #23
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)
 
  • #24
Gaussian97 said:
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.

Gaussian97 said:
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.
 
  • #25
Gaussian97 said:
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.
 
  • #26
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.
 
  • #27
Gaussian97 said:
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".
 
  • Like
Likes cianfa72
  • #28
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...
 
  • #29
Gaussian97 said:
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##.

Gaussian97 said:
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.

Gaussian97 said:
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.

Gaussian97 said:
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.
 
  • Like
Likes cianfa72
  • #30
Gaussian97 said:
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.
 
  • #31
Gaussian97 said:
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.

You have it backwards. If you want to remove the factor of 2 between your "time" and ##\tau##, you need to define ##\tau' = 0.5 \tau##. In other words, you want the ##\tau## intervals along your chosen worldline, the one with ##\rho = 0.5##, to be marked off the same as the proper time intervals along that worldline; that means marking them off with the same scaling as ##\rho_0##.

It is true that this rescaling of ##\tau## will change the coordinate speed by a factor of ##2##, without changing your "speed", since you have defined your "speed" using the proper time scaling along your chosen worldline. Now try rescaling ##\rho##.
 
  • #32
PeterDonis said:
No, this is wrong. If you just rescale ##\tau## but don't change ##\rho##, then your observer is still at ##\rho_0 = 0.5## and his proper time is still ##\tau' / 2##. So, in these new coordinates, ##\Delta \rho## stays the same and ##\Delta \tau## gets multiplied by ##2##, so coordinate speed is half what it was in the old coordinates. And the "speed" for your observer by your definition is still twice the coordinate speed, so it also is half what it was in the old coordinates.

In other words, in these new coordinates, the coordinate speed will be ##\approx 0.6664## and the "speed" for your observer by your definition will be ##\approx 1.3328##.
Let me do it in detail, I change from ##(\tau, \rho)## to ##(\tau', \rho)##, therefore my new metric is ##ds^2=\frac{1}{4}\rho^2d\tau'^2 - d\rho^2##. And I'm still at ##\rho=0.5## so my proper time is related to ##\tau'## as follows:
$$ds^2=\frac{1}{4}\rho_0^2d\tau'^2 \Longrightarrow \Delta T = \frac{\rho_0}{2}\Delta \tau'$$
Now before the two event were separated by ##\Delta \tau = 0.55 \Longrightarrow \Delta \tau'=1.1##. Therefore the time between the two events is still ##\Delta T = \frac{0.5 \cdot 1.1}{2}=0.27## as before, I haven't change the way I define distance, so velocity is still ##2.7##.
However, as you say, coordinate velocity has changed from ##1.33## to ##0.66##, just because we have changed how we describe mathematically the real world.

Now if I rescale ##\rho'=2\rho## now the metric becomes ##ds^2=\frac{{\rho'}^{2}}{4}d\tau^2-\frac{1}{4}d\rho'^2##. Now of course I'm at ##\rho'_0=2\rho_0=1## and the proper time is
$$ds^2=\frac{{\rho'_0}^2}{4}d\tau^2 \Longrightarrow \Delta T = \frac{\rho'_0}{2}\Delta \tau = \frac{\Delta \tau}{2}$$. So again, the time between the two events is still ##0.27##. But now we have also modify how we measure distance, because at constant times
##-ds^2=\frac{1}{4}d\rho'^2 \Longrightarrow \Delta L = \frac{\Delta \rho'}{2}##. So the two event are separated by ##\Delta \rho' = 1.46##, which means a distance of ##\Delta L = 0.73##, the same as before, if I compute the velocity I get, again ##2.7##. And you're right now the coordinate velocity is in fact ##2.7## so both are equal, but I don't understand why you say that coordinate velocity stays the same while mine changes...

PeterDonis said:
I challenge you to prove this.
Well, then probably I'm wrong, I haven't proved rigorously, but my intuition was something like this:
The proper time between two points of my worldline is just the relativistic distance between them, this is a scalar quantity. I also define the physical distance as the relativistic distance between two points at the same time (because we have a scalar notion of time, there no problem there, and because the relativistic distance is invariant, we should have an invariant "physical" distance). Therefore the velocity as I compute it would be invariant. As I said, this was my reasoning and probably it's wrong. But surely, for the easy cases that I've done before, I still don't understand why the coordinate velocity is better.

PeterDonis said:
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.
Well, I'm sorry, English is not my main language, and I really thought that everyone before message #17 was ok with my calculations.
 
  • #33
Gaussian97 said:
until #17 nobody seem to have any problem with
I still don’t have a problem with it, but I would not call it a physical speed.

I would be fine calling it a coordinate speed even though it is not the standard Rindler coordinate speed because you can simply rescale the time coordinate to get coordinates where it is the standard coordinate speed.
 
  • Like
Likes cianfa72
  • #34
Gaussian97 said:
Let me do it in detail

I agree with your metrics and with the coordinates you obtain for both cases.

Gaussian97 said:
now we have also modify how we measure distance

This, however, is a new element. Before you had ##\Delta L = \Delta \rho##. Now you have ##\Delta L = \Delta \rho' / 2##.

So it appears that what you are actually trying to define is something like: ##ds## along a spacelike curve of constant time / ##ds## along a timelike curve of constant spatial location. And you are picking the curves and intervals as follows:

The spacelike curve is a curve of constant time between the event of emission of the light ray, and the spatial location where the light ray is received (but the event you end up at will not be the same as the reception event).

The timelike curve is the segment of your chosen observer's worldline between the event that is simultaneous, in your chosen coordinates, with the emission of the light ray, and the event that is simultaneous, in your chosen coordinates, with the reception of the light ray.

If you define things as above, then you are correct that the ratio of these two ##ds## values is an invariant, since both of them are arc lengths along well-defined segments of well-defined curves in spacetime, and arc lengths along specific curves in spacetime are invariants.

However, your interpretation of this ratio as a "physical speed" is still not correct, because it does not correspond to any direct observation that any observer would make. It is still a calculation that your observer does about events none of which take place on his worldline, meaning that he cannot observe any of them directly. Strictly speaking, I suppose, since your ratio, if defined as above, is a ratio of invariants, there would be some way of constructing a measurement that gave that ratio directly; but such a measurement would be obviously gerrymandered and would not meet any reasonable definition of a direct measurement of "physical speed".

Note also that the conditions under which it is even possible to find curves that meet your specifications is extremely limited. Notice the limitations in how your curves are defined: the observer whose worldline we are using has to be at rest in your chosen coordinates; and both of the spatial locations for the light ray (emission and reception) also have to be at rest in your chosen coordinates. Also, the "physical distance" between the two spatial locations, of emission and reception, has to be constant (since you're only calculating it at one time, the time of emission, so you are assuming that it doesn't change). That means there must be three timelike worldlines that can all be viewed as at rest relative to each other in suitable coordinates, and physical distances along spacelike curves of constant time in those coordinates must be constant in time. That is an extremely restrictive condition (the technical way of stating it is that all three curves must be integral curves of the same timelike Killing vector field). In most spacetimes, no such set of curves even exists.
 
  • Like
Likes cianfa72
  • #35
PeterDonis said:
So it appears that what you are actually trying to define is something like: ##ds## along a spacelike curve of constant time / ##ds## along a timelike curve of constant spatial location. And you are picking the curves and intervals as follows:

The spacelike curve is a curve of constant time between the event of emission of the light ray, and the spatial location where the light ray is received (but the event you end up at will not be the same as the reception event).

The timelike curve is the segment of your chosen observer's worldline between the event that is simultaneous, in your chosen coordinates, with the emission of the light ray, and the event that is simultaneous, in your chosen coordinates, with the reception of the light ray.
Just to point out that Peter is talking about the 'coordinate time' and 'coordinate spatial location' of events in the chosen coordinate system (Rindler coordinate system for flat spacetime).

Thus for instance in the sententence "The spacelike curve is a curve of constant time between the event of emission of the light ray, and the spatial location where the light ray is received" constant time is really constant coordinate time.
 

Similar threads

  • Special and General Relativity
Replies
11
Views
518
  • Special and General Relativity
Replies
17
Views
366
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
7
Views
1K
  • Special and General Relativity
Replies
0
Views
2K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
17
Views
2K
  • Special and General Relativity
Replies
12
Views
1K
  • Special and General Relativity
Replies
15
Views
2K
  • Special and General Relativity
2
Replies
38
Views
3K
Back
Top