Speed of light not an invariant in GR

[PeroK]

@PeroK as per your post in the other thread, the measured velocity $ds/d\tau$ of a light beam should vary according to its propagation direction?

Not at all. That was one calculation for a radial light ray, where the locally measured speed was $ds/d\tau = 1$.

You could do that calculation for any other light path. It's a simple matter to show that $ds/d\tau = 1$, generally, for any null geodesic as measured by someone at rest in Schwarzschild coordinates.

 

[PeterDonis]

it's a simple matter to show that $ds/d\tau = 1$, generally, for any null geodesic as measured by someone at rest in Schwarzschild coordinates.

This needs to be clarified a bit. What the calculation you made in the other thread actually shows is that, in the limit in which the change in radial coordinate involved in the measurement of the speed of the radial light ray is negligible, $ds / d\tau = 1$. Your calculation assumes $r$ is constant, but it's impossible for $r$ to be exactly constant during a measurement of the speed of a radial light ray. There has to be some change in $r$. If the change in $r$ is not negligible, you no longer have $c = 1$.

If you do the calculation in local coordinates (i.e., by treating a local patch of Schwarzschild spacetime as a patch of Minkowski spacetime, ignoring spacetime curvature), then the coordinates in which the observer is at rest are Rindler coordinates, and you have the same issue: to get $c = 1$ you have to take the limit in which the change in the Rindler spatial coordinate is negligible. If the change in Rindler spatial coordinate is not negligible, you no longer have $c = 1$.

 

[PeroK]

This needs to be clarified a bit. What the calculation you made in the other thread actually shows is that, in the limit in which the change in radial coordinate involved in the measurement of the speed of the radial light ray is negligible, $ds / d\tau = 1$. Your calculation assumes $r$ is constant, but it's impossible for $r$ to be exactly constant during a measurement of the speed of a radial light ray. There has to be some change in $r$. If the change in $r$ is not negligible, you no longer have $c = 1$.

If you do the calculation in local coordinates (i.e., by treating a local patch of Schwarzschild spacetime as a patch of Minkowski spacetime, ignoring spacetime curvature), then the coordinates in which the observer is at rest are Rindler coordinates, and you have the same issue: to get $c = 1$ you have to take the limit in which the change in the Rindler spatial coordinate is negligible. If the change in Rindler spatial coordinate is not negligible, you no longer have $c = 1$.

Okay, but $$\frac{ds}{d\tau} = \lim \frac{\Delta s}{\Delta \tau}$$ And, I guess, you need to bounce the light back off a mirror to get the limit of a two-way measurement.

 

[PeterDonis]

This needs to be clarified a bit.

Another way of putting the issue I'm describing here is that, if the observer is not in free fall, i.e., the observer's worldline is not a geodesic, then you have two different kinds of curvature to worry about: spacetime curvature and the path curvature of the observer's worldline. We take care of spacetime curvature by assuming that we make the measurement of $c$ in a small enough patch of spacetime that spacetime curvature can be ignored. But that in itself does not take care of path curvature; you have to separately impose whatever condition is required to make that negligible in your particular scenario if you want to derive the prediction that you will measure $c = 1$.

 

[PeterDonis]

Okay, but $$\frac{ds}{d\tau} = \lim \frac{\Delta s}{\Delta \tau}$$

Yes, but that just rephrases the issue: you don't measure $ds / d\tau$, you measure $\Delta s / \Delta \tau$. Your measurement will involve some finite change in $s$ and $\tau$.

And, I guess, you need to bounce the light back off a mirror to get the limit of a two-way measurement.

No, you bounce the light back off a mirror to avoid all the issues with simultaneity that come into play in a one-way measurement. But bouncing the light back off a mirror doesn't allow you to measure over an infinitesimal interval of $s$ or $\tau$; you still have a finite change in both.

 

[PeroK]

Yes, but that just rephrases the issue: you don't measure $ds / d\tau$, you measure $\Delta s / \Delta \tau$. Your measurement will involve some finite change in $s$ and $\tau$.

It's never been otherwise. Measuring simple acceleration or any rate of change, we are always using the derivative as the limit or approximation of a physical measurement.

 

[Ibix]

Is it helpful to say this:

An ideal, infinitely small, light speed measuring device will always measure $c$. No exceptions.

A real device will always measure $c$ as long as "flat spacetime" is a reasonable approximation across it (in all four directions!) and as long as any proper acceleration is kept low enough that $c^2/a$ is much larger than the size of the apparatus.

As others have noted, a sufficiently large Michelson interferometer stood on edge would show lightspeed anisotropy, and a Michelson interferometer under sufficiently brutal acceleration parallel to one arm would show lightspeed anisotropy.

 

[anuttarasammyak]

So there's no way to measure c < 1 or even > 1 in any frame, IFR or not?

Back to the Einstein paper case where $g_0i=0$ synchronicity is shared among all observers and its parameter is coordinate time t. Different Local clock times readings may have no meanings to others but its transformation to coordinate time, e.g. every observer has not only his natural clock but also coordinate time clock as we have local time clock and GMT clock on the wall, tells his event is in past, now or future for her in distant place also holding the two type clocks. If divided by coordinate time not local clock time light speed varies with location and bending light front is commonly recognized with observers worldwide.

 

[Pyter]

You could do that calculation for any other light path. It's a simple matter to show that $ds/d\tau = 1$, generally, for any null geodesic as measured by someone at rest in Schwarzschild coordinates.

Makes sense. I didn't do the math for every direction of propagation in the Schwarzschild metric, but some authors define $d\tau \triangleq \pm ds$, so $|ds/d\tau|$ must be 1 in every FR by definition. In practice, as many of you remarked, what you measure instrumentally may differ.

 

[PeterDonis]

some authors define $d\tau \triangleq \pm ds$, so $|ds/d\tau|$ must be 1 in every FR by definition.

No, this is not correct; you are confusing two different meanings of the notation $d \tau$.

When authors define $d\tau \triangleq \pm ds$, they are using $d\tau$ to refer to the same differential line element as $ds$--the definition is just a matter of which metric signature convention the author wants to use.

When we talk about $ds / d\tau$ in terms of the speed of light, $ds$ and $d\tau$ refer to two different differential line elements, one spacelike (the distance the light travels) and one timelike (the time the light takes to travel that distance).

 

[Pyter]

No, this is not correct; you are confusing two different meanings of the notation $d \tau$.

When authors define $d\tau \triangleq \pm ds$, they are using $d\tau$ to refer to the same differential line element as $ds$--the definition is just a matter of which metric signature convention the author wants to use.

When we talk about $ds / d\tau$ in terms of the speed of light, $ds$ and $d\tau$ refer to two different differential line elements, one spacelike (the distance the light travels) and one timelike (the time the light takes to travel that distance).

Right. Actually I've only seen $d\tau$ defined that way in the SR context. In GR, it should be, at the FR origin, $$ds^2 = g^{44} dx^4$$ (or $g^{00} dx^0$ in the convention where $t = x^0$) $$ \implies ds^2 = g^{44} d\tau^2.$$
Is this compatible with $ds/d\tau \equiv 1$ in every direction, every FR (also non-Lorentz), as per @PeroK post?

 

[vanhees71]

For local distance measures of space wrt. an observer one can use the constancy of the (two-way!) speed of light to define the distance of a infinitesimally distant point from him by sending a light signal and measures the time when he receives the reflected light signal. For the light signals you have
$$g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} = g_00 (\mathrm{d} x^0)^2 + 2 g_{0k} \mathrm{d} x^0 \mathrm{d} x^k + g_{jk} \mathrm{d} x^j \mathrm{d} x^k=0.$$
For this you get two solutions for $\mathrm{d} x^0$ describing the coordinate times needed for the light signal to move forth and back
$$(\mathrm{d} x^0)_{\pm}=\frac{1}{g_{00}} \left (-g_{0k} \mathrm{d} x^{k} \pm \sqrt{(g_{0j} g_{0k}-g_{00} g_{jk})\mathrm{d} x^j \mathrm{d} x^k} \right).$$
The coordinate-time difference thus is
$$\mathrm{d} \Delta x^0=(\mathrm{d} x^0)_+ -(\mathrm{d} x^0)_- = \frac{2}{g_00} \sqrt{(g_{0j} g_{0k}-g_{00} g_{jk})\mathrm{d} x^j \mathrm{d} x^k}.$$
The time measured by the observer between the sending the light signal and receiving the reflected one is
$$\mathrm{d} \tau = \sqrt{g_{00}} \mathrm{d} \Delta x^0.$$
So the distance as measured by the observer is half that. So we finally have
$$\mathrm{d} l = \sqrt{\frac{g_{0j} g_{0k}-g_{00} g_{jk}}{g_{00}} \mathrm{d} x^j \mathrm{d} x^k}.$$
In this way, for sufficiently small distances, one sees that in GR as in SR the (two-way) velocity of light is just 1 by convention.

 

[Pyter]

For local distance measures of space wrt. an observer one can use the constancy of the (two-way!) speed of light to define the distance of a infinitesimally distant point from him by sending a light signal and measures the time when he receives the reflected light signal. For the light signals you have
$$g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} = g_00 (\mathrm{d} x^0)^2 + 2 g_{0k} \mathrm{d} x^0 \mathrm{d} x^k + g_{jk} \mathrm{d} x^j \mathrm{d} x^k=0.$$
For this you get two solutions for $\mathrm{d} x^0$ describing the coordinate times needed for the light signal to move forth and back
$$(\mathrm{d} x^0)_{\pm}=\frac{1}{g_{00}} \left (-g_{0k} \mathrm{d} x^{k} \pm \sqrt{(g_{0j} g_{0k}-g_{00} g_{jk})\mathrm{d} x^j \mathrm{d} x^k} \right).$$

Why is there a negative solution for $\mathrm{d}x^0$?

The coordinate-time difference thus is
$$\mathrm{d} \Delta x^0=(\mathrm{d} x^0)_+ -(\mathrm{d} x^0)_- = \frac{2}{g_00} \sqrt{(g_{0j} g_{0k}-g_{00} g_{jk})\mathrm{d} x^j \mathrm{d} x^k}.$$

I don't get the physical meaning of this difference. Why should it be the round-trip duration in coordinate time?

The time measured by the observer between the sending the light signal and receiving the reflected one is
$$\mathrm{d} \tau = \sqrt{g_{00}} \mathrm{d} \Delta x^0.$$
So the distance as measured by the observer is half that.

How do you compute the distance from the proper time $\mathrm{d}\tau$? Do you multiply it by $c \triangleq 1$?

So we finally have
$$\mathrm{d} l = \sqrt{\frac{g_{0j} g_{0k}-g_{00} g_{jk}}{g_{00}} \mathrm{d} x^j \mathrm{d} x^k}.$$

$\mathrm{d}l = \mathrm{d} \tau /2$? I don't follow.

In this way, for sufficiently small distances, one sees that in GR as in SR the (two-way) velocity of light is just 1 by convention.

 

[PeterDonis]

Right. Actually I've only seen $d\tau$ defined that way in the SR context. In GR, it should be, at the FR origin, $$ds^2 = g^{44} dx^4$$ (or $g^{00} dx^0$ in the convention where $t = x^0$) $$ \implies ds^2 = g^{44} d\tau^2.$$

Whether we are in flat spacetime (SR) or curved spacetime (GR), "it should be" whatever is correct for the particular curve that is being considered. $ds^2 = g_{00} \left( dx^0 \right)^2$ (note the index placements and the squaring of the coordinate differential) would only be correct for a timelike curve that is "at rest" (i.e., its spatial coordinates are not changing) in the chosen coordinates at the chosen point. It would not be correct for a spacelike curve (which is what you need for a "distance") or a null curve (the worldline of a light ray).

Is this compatible with $ds/d\tau \equiv 1$ in every direction, every FR (also non-Lorentz), as per @PeroK post?

What you're saying is not even correct as you state it, so it's not compatible with anything. See above.

 

[vanhees71]

Why is there a negative solution for $\mathrm{d}x^0$?

I don't get the physical meaning of this difference. Why should it be the round-trip duration in coordinate time?

How do you compute the distance from the proper time $\mathrm{d}\tau$? Do you multiply it by $c \triangleq 1$?

$\mathrm{d}l = \mathrm{d} \tau /2$? I don't follow.

Here's an infinitesimal Minkowski diagram defined by the observer's tetrad. The derivation is from Landau&Lifshitz vol. 2. There you find a very clear discussion of spacetime geometry/tensor analysis.

 

[Pyter]

I guess my main issue is grasping the physical meaning of $\mathrm{d}s$ and $\mathrm{d}s^2$.

How do you compute $\mathrm{d}s$ from $\mathrm{d}s^2$?
If it was simply $\mathrm{d}s = \sqrt {\mathrm{d}s^2}$, then for a light beam it would be $\mathrm{d}s \equiv 0$, because $\mathrm{d}s^2 \equiv 0$. Then the light speed $\mathrm{d}s/\mathrm{d}\tau$ would be $\equiv 0 \equiv$ constant. And for a space-like worldline if would be an imaginary number because $\mathrm{d}s^2 < 0$.

On the LHS of the equation, you have $\mathrm{d}s^2$, on the RHS you have $\langle {\mathbf {\mathrm{dx}}, \mathbf{\mathrm{dx}}} \rangle$ (the inner product of $dx^\mu$ by itself in the metric defined by $g_{**}$ ). Thus $\mathrm{d}s^2$ might be considered as $\| \mathbf {\mathrm{dx}} \|^2$, i.e. the square norm of $\mathrm{d}x^\mu$, but not in the Euclidean metric.

$\mathrm{d}s^2$ and $\mathrm{d}s$ are not Euclidean lengths, so why do they appear at the numerator of the light's (or any body's) speed equation: $\mathrm{d}s/\mathrm{d}\tau$?

 

[Pyter]

$ds^2 = g_{00} \left( dx^0 \right)^2$ (note the index placements and the squaring of the coordinate differential)

Yes sorry, I wrote them wrong, I can't no longer edit my post.

 

[PeterDonis]

I guess my main issue is grasping the physical meaning of $\mathrm{d}s$ and $\mathrm{d}s^2$.

$\mathrm{d}s^2$ is the infinitesimal squared interval from some chosen event to a neighboring event in spacetime. The line element gives a formula for this squared interval in terms of the coordinate differentials between the two events, i.e., the infinitesimal changes in the coordinate values from one event to the other. If $\mathrm{d}s^2$ is timelike, then $\mathrm{d}s$ represents the proper time elapsed on a clock traveling from one event to the other; if $\mathrm{d}s^2$ is spacelike, then $\mathrm{d}s$ represents the proper distance between the two events, as measured by a ruler whose motion is such that both events happen simultaneously for an observer at rest with respect to the ruler. If $\mathrm{d}s^2 = 0$, then $\mathrm{d}s$ represents an infinitesimal segment of a light ray from one event to the other.

How do you compute $\mathrm{d}s$ from $\mathrm{d}s^2$?

$\mathrm{d}s$ is just the square root of $\mathrm{d}s^2$, ignoring the sign (so the value is always real), plus the specification of whether the interval is timelike or spacelike (which you get from the sign of $\mathrm{d}s^2$ and the signature convention being used). If $\mathrm{d}s^2 = 0$, then $\mathrm{d}s = 0$ and the interval is null.

$\mathrm{d}s^2$ and $\mathrm{d}s$ are not Euclidean lengths, so why do they appear at the numerator of the light's (or any body's) speed equation: $\mathrm{d}s/\mathrm{d}\tau$?

Because "speed" in no way requires Euclidean lengths. It only requires spacetime intervals. That's what $\mathrm{d}s$ and $\mathrm{d}\tau$ are. They're spacetime intervals along two different infinitesimal segments, one spacelike and one timelike, as I explained in an earlier post. Those segments correspond to (the limit of) the distance the light travels and (the limit of) the time it takes to travel that distance.

 

[Ibix]

I haven't been following this thread too closely, so I may be wrong, but perhaps an insight you seem to be missing, @Pyter, is that $ds$ is associated with a chosen path. It's analogous to length along a path in Euclidean space. If there is more than one path between two points, there can be more than one length along those paths. Similarly, if there is more than one path between two events, there can be different $ds$ (or $\int ds$, to be precise) along those paths.

A two way speed measurement is a laser and light detector (with a timelike path through spacetime) which emits a light pulse that ounces off a mirror and returns. The laser pulse has a null path through spacetime. So there are two separate relevant quantities that you could denote $ds$. One is $cd\tau$, where $d\tau$ is the time elapsed on a clock attached to the laser/light detector. The other is the interval along the light pulse's path, which is zero. Both are "lengths" and could be called $ds$, but they are different things.

As I say, I haven't read this thread too carefully, so this could be an irrelevance - apologies if so.

 

[Pyter]

If $\mathrm{d}s^2$ is timelike, then $\mathrm{d}s$ represents the proper time elapsed on a clock traveling from one event to the other;

So in this case $ds^2 = d\tau^2$, as I stated in one of my previous posts.

If $\mathrm{d}s^2 = 0$, then $\mathrm{d}s$ represents an infinitesimal segment of a light ray from one event to the other.
[...]
$\mathrm{d}s$ is just the square root of $\mathrm{d}s^2$
[...]
If $\mathrm{d}s^2 = 0$, then $\mathrm{d}s = 0$ and the interval is null.

Then since in the other thread @PeroK gives the measured light speed as $ds/d\tau$, it should be constantly 0, because $ds \equiv 0$.

Because "speed" in no way requires Euclidean lengths.

I was under the impression that measuring any speed requires dividing a measured (Euclidean) length by a measured time interval.

It only requires spacetime intervals. That's what $\mathrm{d}s$ and $\mathrm{d}\tau$ are. They're spacetime intervals along two different infinitesimal segments, one spacelike and one timelike, as I explained in an earlier post. Those segments correspond to (the limit of) the distance the light travels and (the limit of) the time it takes to travel that distance.

They would be "intervals" in the intuitive sense if they were the (Euclidean) length of a difference between two Euclidean vectors, but they're not.

 

[Pyter]

So there are two separate relevant quantities that you could denote $ds$. One is $cd\tau$, where $d\tau$ is the time elapsed on a clock attached to the laser/light detector. The other is the interval along the light pulse's path, which is zero. Both are "lengths" and could be called $ds$, but they are different things.

So the two events are: A - the light departs from the source, B - the light bounces from the mirror, and the first $ds$ is the "path length" between these two events along the clock's wordline, and the second $ds$ the "path length" between the same events along the light beam's wordline?
But what is the "path length"?

 

[PeterDonis]

So in this case $ds^2 = d\tau^2$, as I stated in one of my previous posts.

If you are interpreting $d\tau^2$ as the (square of) the proper time it takes for the light to travel, yes. But there is also a spacelike $ds^2$ that describes the distance the light travels. That $ds$ is not the same as $d\tau$; it's a different spacetime interval. See below.

Then since in the other thread @PeroK gives the measured light speed as $ds/d\tau$, it should be constantly 0, because $ds \equiv 0$.

No. I already pointed out this confusion in an earlier post. If you are interpreting $ds$ as the interval along the light ray's worldline, then $ds = 0$, but there is no $d\tau$ at all, and $ds / d\tau$ makes no sense.

The only way $ds / d\tau$ makes sense is if $ds$ is the spacelike interval describing the distance the light travels, and $d\tau$ is the timelike interval describing the time it takes for the light to travel. And then $ds / d\tau$ does give you the measured speed of the light.

I was under the impression that measuring any speed requires dividing a measured (Euclidean) length by a measured time interval.

And that's just what I described above, except that the length doesn't have to be "Euclidean" because the geometry of space doesn't have to be Euclidean. The length just has to be a spacelike interval describing the distance the light travels.

 

[Dale]

So in this case $ds^2 = d\tau^2$, as I stated in one of my previous posts.

This is a little bit of an aside, but my preferred convention is to have $ds^2=-c^2 d\tau ^2$ with $ds^2>0$ for spacelike intervals. This gives $ds$ in units of length for proper distances and $d\tau$ in units of time for proper times.

 

[PeroK]

Then since in the other thread @PeroK gives the measured light speed as $ds/d\tau$, it should be constantly 0, because $ds \equiv 0$.

As explained, I used $ds$ to be the spacelike interval traveled by the light. To avoid confusion, we could use $dl$ for this. The calculation in flat spacetime (in some inertial frame) would look like:

1) Light travels from $(t_0, x_0, y_0, z_0)$ to $(t_1, x_1, y_1, z_1)$.

2) As the light path is null, we have $(t_1 - t_0)^2 = (x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2$.

3) The spatial distance traveled by the light (as measured in the IRF) is $\Delta l^2 = (x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2$.

4) The time that the light traveled (as measured in the IRF) is $\Delta t = (t_1 - t_0)$

5) The speed of light (as measured in the IRF) is $\frac{\Delta l}{\Delta t} = 1$.

Note that you can elaborate this by having the light bounce off a mirror back to the starting point, so that the time interval may be measured by the proper time of a single clock at rest in the IRF. In which case you measure the speed of light as $\frac{2\Delta l}{\Delta \tau} = \frac{2\Delta l}{2 \Delta t} = 1$.

In general coordinates, or curved spacetime, you would have to to the calculation done by @vanhees71 where the calculations involve the metric tensor components and a small/infinitesimal spatial interval over which the light travels - i.e. the theoretical limit of actual measurements. And you should get the measured speed of light $\frac{2dl}{d\tau} = 1$.

@Pyter a good exercise for you would be to extend the above for the case of an arbitrary diagonal metric.

 

[vanhees71]

I guess my main issue is grasping the physical meaning of $\mathrm{d}s$ and $\mathrm{d}s^2$.

How do you compute $\mathrm{d}s$ from $\mathrm{d}s^2$?
If it was simply $\mathrm{d}s = \sqrt {\mathrm{d}s^2}$, then for a light beam it would be $\mathrm{d}s \equiv 0$, because $\mathrm{d}s^2 \equiv 0$. Then the light speed $\mathrm{d}s/\mathrm{d}\tau$ would be $\equiv 0 \equiv$ constant. And for a space-like worldline if would be an imaginary number because $\mathrm{d}s^2 < 0$.

On the LHS of the equation, you have $\mathrm{d}s^2$, on the RHS you have $\langle {\mathbf {\mathrm{dx}}, \mathbf{\mathrm{dx}}} \rangle$ (the inner product of $dx^\mu$ by itself in the metric defined by $g_{**}$ ). Thus $\mathrm{d}s^2$ might be considered as $\| \mathbf {\mathrm{dx}} \|^2$, i.e. the square norm of $\mathrm{d}x^\mu$, but not in the Euclidean metric.

$\mathrm{d}s^2$ and $\mathrm{d}s$ are not Euclidean lengths, so why do they appear at the numerator of the light's (or any body's) speed equation: $\mathrm{d}s/\mathrm{d}\tau$?

$\mathrm{d} s^2$ is defined as
$$\mathrm{d} s^2=g_{\mu \nu} \mathrm{d} x^{\mu} x^{\nu}$$
along a worldline, described by a function $x^{\mu}(\lambda)$ of the coordinates with an arbitary parameter $\lambda$. Parametrized in this way you can also write
$$\mathrm{d} s^2 = \mathrm{d} \lambda^2 g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
In my postings above (#47 and #50) in the argument how to measure infinitesimal spatial distances between an observer (the left vertical worldline in the picture of #50) and an arbitrary point (right vertical worldline in the picture of #50) by "radar timing", which is one way to define infinitesimal spatial distances in GR using that in a local inertial frame the two-way speed of light is $c$, I used (a) the worldlines of the light signals, which are necessarilly "null worldlines", characterized by $\mathrm{d} s^2=0$, and (b) the worldline of an observer who is defining a local inertial reference frame by using his four-velocity as time-like basis vector of a tetrad. He is using his clock to measure the time from sending out his signal and receiving the signal reflected on the object the distance he wants to measure. For his wordline, defined by being momentarily at rest in the sense of the coordinates $x^{\mu}$, i.e., $\mathrm{d} x^j=0$ (where $j \in \{1,2,3\}$). The time measured by his clock is his "proper time" and given in the used coordinates by
$$\mathrm{d} \tau^2=\mathrm{d} s^2=g_{00} (\mathrm{d} x^0)^2.$$
I used a natural system of units where $c=1$.

 

[Pyter]

As explained, I used $ds$ to be the spacelike interval traveled by the light. To avoid confusion, we could use $dl$ for this. The calculation in flat spacetime (in some inertial frame) would look like:

[...]

3) The spatial distance traveled by the light (as measured in the IRF) is $\Delta l^2 = (x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2$.

At last. You do need a measured Euclidean length at the numerator to compute a velocity.

@vanhees71 it's all clear (more or less) except this point:

The time measured by his clock is his "proper time" and given in the used coordinates by
$$\mathrm{d} \tau^2=\mathrm{d} s^2=g_{00} (\mathrm{d} x^0)^2.$$
I used a natural system of units where $c=1$.

As you say, the time measured by his clock is his "proper time", which as we all know is also designated by the symbol $\tau$. But this time (difference) is $\mathrm{d}x^0$ (his local time coordinate), and in general $g_{00} \neq 0$, so how's possible that $\mathrm{d}\tau^2 =( \mathrm{d}x^0)^2$?

 

[PeterDonis]

You do need a measured Euclidean length at the numerator to compute a velocity.

No. You need a measured length. But, as I have already pointed out several times now, the length does not have to be "Euclidean". It only happens to be in the particular case @PeroK gave because he is assuming flat spacetime. In a general curved spacetime the length would not be "Euclidean" and its functional dependence on the coordinate differences would be more complicated.

 

[PeterDonis]

this time (difference) is $\mathrm{d}x^0$ (his local time coordinate)

Only if $g_{00} = 1$. But as you note, in general that is not the case. If $g_{00} \neq 1$ then $\mathrm{d}\tau$ is not the same as $\mathrm{d}x^0$. (You and @vanhees71 are also assuming that the observer is at rest in the given coordinates, so only $\mathrm{d}x^0$ is nonzero.)

how's possible that $\mathrm{d}\tau^2 =( \mathrm{d}x^0)^2$?

It doesn't in general. Read the formula @vanhees71 actually wrote more carefully.

 

[Pyter]

Only if g00=1. But as you note, in general that is not the case. If g00≠1 then dτ is not the same as dx0. (You and @vanhees71 are also assuming that the observer is at rest in the given coordinates, so only dx0 is nonzero.)

It doesn't in general. Read the formula @vanhees71 actually wrote more carefully.

That's what I meant. How can it be that: $$\mathrm{d}\tau^ \triangleq \mathrm{d}x^0$$ (by definition) and: $$\mathrm{d} \tau^2=\mathrm{d} s^2=g_{00} (\mathrm{d} x^0)^2$$ at the same time, if $g_{00} \neq 1$?

 

[PeroK]

That's what I meant. How can it be that: $$\mathrm{d}\tau^ \triangleq \mathrm{d}x^0$$ (by definition).

You could say that $dt = dx^0$, in the sense that $t = x^0$ is the "time" coordinate. And, of course, you have $\tau = t$ in a local inertial frame.

 

[Pyter]

@PeroK isn't the the proper time $\tau$, by definition, the time $x^0$ measured by an observer at the origin of the coordinate system where he's stationary?

 

[PeroK]

@PeroK isn't the the proper time $\tau$, by definition, the time $x^0$ measured by an observer at the origin of the coordinate system where he's stationary?

No.

 

[vanhees71]

At last. You do need a measured Euclidean length at the numerator to compute a velocity.

@vanhees71 it's all clear (more or less) except this point:As you say, the time measured by his clock is his "proper time", which as we all know is also designated by the symbol $\tau$. But this time (difference) is $\mathrm{d}x^0$ (his local time coordinate), and in general $g_{00} \neq 0$, so how's possible that $\mathrm{d}\tau^2 =( \mathrm{d}x^0)^2$?

Again you have to be careful, about which world line we are talking. For the world line of the light signal ("photon") you have $\mathrm{d} s^2=0$. For the world line of the observer at rest in the reference frame defined by the coordinates $x^{\mu}$ it's $\mathrm{d}s^2=g_{00} (\mathrm{d} x^0)^2$, and that's the proper time measured by the observer's clock, i.e., $\mathrm{d} \tau=\sqrt{g_{00}} \mathrm{d} x^0$.

One should be aware that all that's measurable by some device is an invariant quantity at a point, i.e., time intervals shown by a clock are proper times of some observer.

 

[Pyter]

I've checked out the proper time definition on Wikipedia and there are indeed two different definitions for SR and GR. What do you know, the definition for GR:
$$\Delta\tau = \int_P \, d\tau = \int_P \frac{1}{c}\sqrt{g_{\mu\nu} \; dx^\mu \; dx^\nu}$$
is claimed to be invariant and also contains c.
So we may safely assume, according to Wikipedia, that the postulate of c invariance also holds for GR.

 

[vanhees71]

As the name suggests SR is a special case of GR (describing situations, where gravitational fields can be neglected). The definition of proper time is the same in SR and GR. It's a functional of a time-like worldline and defined as written in Wikipedia for both SR and GR. The distinction is that in SR you can always introduce a global (!) inertial reference frame with a constant tetrad everywhere as reference frame. In this frame, globally you have $g_{\mu \nu}=\eta_{\mu \nu}$. If a gravitational field is present, you can only introduce coordinates at one point, such that $g_{\mu \nu}=\eta_{\mu \nu}$ at this one point but not globally!

You can understand GR as arising from SR by "gauging Lorentz symmetry", i.e., making Lorentz invariance a local symmetry. This is the modern definition of what's called "equivalence principle", i.e., it says that at any space-time point there is a local (!) inertial reference frame. If the gravitational field cannot be neglected, i.e., if the Riemann curvature tensor doesn't vanish, in a spacetime region, you never have a global inertial reference frame.