Velocity measurement by a stationary observer in GR

[sergiokapone]

It would be better if you stop constructing it in such a confuse way. The way you construct an equation should be for it to be more understandable, not the opposite.

May be it is more understandable, but I want to understand it in such form.

When you say so, you seem to assume that the observer dictates how the math should look like.

Oh, sorry, I forgot to put "sarcasm" sign here.

Now I realize the $\sqrt{g_{00}}dx^0$ is just the rate of the stationary observer's clocks compared to the rate of the coordinate clocks, and the $\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)$ is just the diference in time by observer's clock between two events "observer see particle in the distance $dl$ (marked by $x^i + dx^i$) relative to him" and "observer meet particle at $x^0$ in $x^i$" as shown in the picture #22
I.e., if particle turns out at place $dx^i$ in the moment of coordinate time $dx^0$, the stationary observer at a $x^i$ will see it not at the moment of time $\sqrt{g_{00}}dx^0$, but little bit later by term $\sqrt{g_{00}}\frac{g_{0i}}{g_{00}}dx^i$

Only for metrics, where $g_{0i} = g_{i0} = 0$ such "times" looks the same.

Correct me, if I wrong.

 

[PeterDonis]

as one can check this formula leads to

If you put in terms that cancel each other, and then separate them out, yes. But first you have to justify why you would do such a thing. It doesn't seem like any of the sources you are consulting give any such justification. I think there is one, but they shouldn't just take it for granted, since there are a number of complexities involved. See below.

As mentioned in Physical time and physical space in general relativity by Richard J. Cook

The terms "physical time" and "physical space" are extremely misleading, I think. Let me try to restate what this paper is attempting to do in terms which are IMO less misleading.

Suppose we have a collection of observers (the paper calls them "fiducial observers") with the following property: if any two neighboring observers exchange round-trip light signals, the round-trip travel time of those signals, as measured by either observer's clock, remains constant. In this sense, any two neighboring observers remain at a constant "distance" apart (where "distance" is defined as half the round-trip light travel time, in units where the speed of light is 1--see below). Each observer carries a set of three measuring rods of some standard length, which are oriented to be mutually perpendicular, and each of which points towards some particular neighboring observer (and which one does not change with time); and the measuring rods of neighboring observers are all "lined up" so that they point in the same directions.

This collection of observers can then be treated as a "reference frame" in much the same sense as that term is used in special relativity; however, unlike the SR case, in general the metric tensor for a system of coordinates in which each observer is at rest will not be the Minkowski metric.

Now consider the "distance" again, as defined above: half the round-trip light travel time, in units where the speed of light is 1. Cook's paper derives a formula for this distance, which he calls $d \ell$:

$$
d \ell^2 = \gamma_{ij} dx^i dx^j = \left( g_{ij} - \frac{g_{0i} g_{0j}}{g_{00}} \right) dx^i dx^j
$$

Notice that this formula is not quite the same as the one you wrote down (there is a sign change, plus the fraction is different).

Cook calls $\gamma_{ij}$ the "metric of physical space", but this is, as I said, highly misleading. Why? Because the "space" in question does not, in general, correspond to any spacelike hypersurface in the spacetime. In other words, there is no way to view spacetime as being composed of an infinite number of "spaces" of this form, parameterized by "time". Instead, this "space" is a quotient space, i.e., heurisitically, it is what you get if you consider each of the worldlines of the fiducial observers to be a single "point" and use the "radar distance" as defined above (half the round-trip light travel time in units where c = 1) to define the "distance" between the "points". (The mathematical notion of "quotient space" makes all of this rigorous.)

Similarly, Cook uses the term "physical time" for the piece of the full metric that you get when you subtract out this "metric of physical space" from the full metric of spacetime. But this is also misleading, because this "time" does not (as you have discovered) correspond to actual proper time for anything except the fiducial observers themselves, for which it reduces to $d\tilde{t} = \sqrt{g_{00}} dx^0$ (in units where c = 1). The idea, of course, is that, with the metric split up into $d\tilde{t}$ and $d\ell$ in this way, things look very similar to the way they look in SR: the spacetime interval between two nearby events is just $d\tau^2 = d\tilde{t}^2 - d\ell^2$ (which agrees with what you obtained for proper time for a moving object--I was not clear about that before because I didn't understand the notation you were using, but your reference to the Cook paper has cleared that up). So $\tilde{t}$ and $\ell$ act, locally, just like the $t$ and $x$ of SR in standard inertial coordinates. (But only locally.) That is the justification for splitting things up the way Cook does (which involves, as I said, inserting two terms that cancel each other in the metric, and then splitting them apart, one becoming part of $\tilde{t}$ and one becoming part of $\ell$), but it does not, IMO, justify the misleading terms "physical time" and "physical space" for $\tilde{t}$ and $\ell$, since the whole point is that they behave like coordinate time and space in SR, and "coordinate" is not the same as "physical".

if observer use this time, he measure the actual speed of anything moving past him, as well as spedd of light c=1c=1c = 1. How does he know when to use this time, and when to use other one $\sqrt{g_{00}}dx^0$ ?

For a fiducial observer, which is the only kind of observer we are talking about for this purpose, the two are the same--see above. For any observer who is not a fiducial observer, $\tilde{t}$ is not his proper time--see above--and if such an observer wants to measure the actual speed of anything moving past him, he has to use his own proper time, not $\tilde{t}$.

 

[sergiokapone]

Notice that this formula is not quite the same as the one you wrote down (there is a sign change, plus the fraction is different).

Singn is due to agreement on signature(I prefer +---), but "fraction is different" this is because I systematically forgot about $g_{0j}$ (!sorry, mea culpa!)

Did you agree with two statements:

1. The $\sqrt{g_{00}}dx^0$ is rate of the stationary observer's clocks compared to the rate of the coordinate clocks,
2. $\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)$ is just the diference in time by observer's clock between two events $\Delta\tau =$ "time, when observer see particle in the distance $dl$ (marked by $x^i + dx^i$) relative to him" minus"time, when observer meet particle at $x^0$ in $x^i$" as shown in the picture #22

which I quoted above in #36?

 

[PeterDonis]

Did you agree with two statements

Statement #1 is correct, yes.

Statement #2 is not. The difference $d\tilde{t}$ between two events is analogous to the coordinate time difference in SR, in an inertial frame in which the "fiducial" observer is at rest, between two events. That is not the same as the difference in time, by the fiducial observer's clock, between when the fiducial observer sees the particle pass him, and when he sees (as in, receives light from) the particle at a distance $d\ell$ from him. The latter difference must also take into account the light travel time over the distance $d\ell$.

 

[sergiokapone]

Statement #2 is not. The difference $d\tilde{t}$ between two events is analogous to the coordinate time difference in SR, in an inertial frame in which the "fiducial" observer is at rest, between two events. That is not the same as the difference in time, by the fiducial observer's clock, between when the fiducial observer sees the particle pass him, and when he sees (as in, receives light from) the particle at a distance $d\ell$ from him. The latter difference must also take into account the light travel time over the distance $d\ell$.

Oh, yes, you right, I forgot take into account the travel time of light over the distance back to observer.

But, what about pictures in #22? It is correct? (in picture: gray line --- is the light signal, red lines --- is the lines of simultaneity). The lines of simultaneity was drawn according to Einstein's synchronization procedure.

I had redraw the picture slightly

 

[PeterDonis]

The lines of simultaneity was drawn according to Einstein's synchronization procedure.

No, it isn't. You are failing to take into account that the geometry of spacetime, locally, is not Euclidean--it's Lorentzian. Lines of simultaneity are not perpendicular to the observer's worldline in the Euclidean sense, which is how you have drawn them. They are "perpendicular" (orthogonal) in the Lorentzian sense.

Furthermore, since you have adopted coordinates in which the metric is not Minkowski, even the Lorentzian concept of orthogonality will not be represented in your diagram the way it would be represented in a standard spacetime diagram of the kind used in special relativity. In fact, without some specific coordinate chart and metric in mind, there is no way to even know how orthogonality would be represented in your diagram.

The way I would proceed would be to draw an SR-style spacetime diagram with $\tilde{t}$ as the vertical (time) axis and $\ell$ as the horizontal (space) axis--since we know that, locally, the metric takes the Minkowski form in terms of those quantities, i.e., $d\tau^2 = d\tilde{t}^2 - d\ell^2$ (in units where c = 1). In that diagram, lines of simultaneity in the sense you mean (i.e., lines of constant $\tilde{t}$) are horizontal, so it's easy to draw them, and it's also easy to see how they relate to the round-trip light paths, since those are 45 degree lines in terms of $\tilde{t}$ and $\ell$ in units where c = 1, by definition (note that in your diagram, you can't even assume that--it's quite possible that in the general coordinates of your diagram, light rays travel on curves that are not 45 degree lines).

Once you have draw that diagram, you can then try to draw what the coordinate "grid lines" in some general chart $x^0, x^i$ would look like. But again, that will depend, in general, on the coordinates.

 

[sergiokapone]

No, it isn't. You are failing to take into account that the geometry of spacetime, locally, is not Euclidean--it's Lorentzian. Lines of simultaneity are not perpendicular to the observer's worldline in the Euclidean sense, which is how you have drawn them. They are "perpendicular" (orthogonal) in the Lorentzian sense.

I did not think about perpendicularity of lines of simultaneity, I just draw the light signals according to equation $ds^2 = 0$. In my case I took $g_{00} = 1$, $g_{01} = \cos\alpha$, $g_{11} = 1$ for drawing. I get the equation of signal lines $x^0 = 1/(\cos\alpha \pm \sqrt{\cos^2\alpha + 1} )x$, sign $+$ for forward light, sign $-$ for backward. Red line was drawn as distance divided by 2 between intersection of $x^0$ axis by forward and backward signals. And it so happened that they became perpendicular, it was not done on purpose.

If you are interested, I attach a code (LaTeX/TikZ)

May be for correct result, it need to be putted hyperbolyc $g_{01} = \cosh\alpha$ instead of "circular" $\cos$, but I'm not sure about this.

 

[PeterDonis]

May be for correct result, it need to be putted hyperbolyc $g_{01} = \cosh\alpha$ instead of "circular" coscos\cos, but I'm not sure about this.

I'm not sure about either one. In general the metric coefficients will be functions of the coordinates; they won't be constants, which is what you appear to be assuming.

 

[sergiokapone]

I'm not sure about either one. In general the metric coefficients will be functions of the coordinates; they won't be constants, which is what you appear to be assuming.

Yes, they could be functions of the coordinates. But for drawing I neglected of their dependence. We can assume that we have considered a small chart.

 

[sergiokapone]

The way I would proceed would be to draw an SR-style spacetime diagram with $\tilde{t}$ as the vertical (time) axis and $\ell$ as the horizontal (space) axis--since we know that, locally, the metric takes the Minkowski form in terms of those quantities, ...

I had constructed it. I just drew the perpendicular to the $x^0$ the $\ell$ axis and define coordinates on it as $\sqrt{1 - \cos^2\alpha} \cdot (x\, coordinate\,of\,particle)$ (because of $d\ell^2 = (1 - g_{01}^2)(dx^1)^2$ for my drawing) and connected a points on $v^{\nu}$ with poins on $\ell$, so I get pretty perpendiculares to the $\ell$ axis. It is look like magic.

 

[PeterDonis]

I had constructed it.

I don't think you have. See below.

I just drew the perpendicular to the $x^0$ the $\ell$ axis

The $\ell$ axis might not be perpendicular to the $x^0$ axis. You need to prove that, if it's true, not just assume it.

More generally, I don't understand how you are constructing your diagram. Whatever you're doing, it's not what I suggested in post #41. If you were doing that, your axes would be labeled $\tilde{t}$ and $\ell$ and would be perpendicular (one vertical and one horizontal), and light rays would be 45 degree lines.

 

[sergiokapone]

Firstly the $x^0$ axis coincide in direction with $\tilde{t}$ axis. I will try to proof this.

As we know, the $x^0$ axis is the time axis of an observer (which has the 4-velocity $u^{\mu} = \left(\frac{dx^0}{d\tau},0,0,0\right)$.

How can we measure the proper time, i.e., the real time elabsed between two coordinatex $x^0$ and $x^0 + dx^0$?

Let's calculate the value (norm) of $u^{\mu}$.
\begin{equation}
\mathbf{g}(u^{\mu},u^{\mu}) = 1
\end{equation}
where $\mathbf{g}(\cdot,\cdot)$ --- it is a metric tensor.

Remark: $\mathbf{g}(u^{\mu},v^{\nu}) = \cosh\beta$ --- projection of $v^{\nu}$ on direction $u^{\mu}$ is just a $\cosh\beta$ between this two vectors (as we know from SR, and also, SR is valid in small region of space-time because of equivalence principle).​

So,
$\mathbf{g}(u^{\mu},u^{\mu}) = g_{00}u^0u^0 = g_{00}\left(\frac{dx^0}{d\tau}\right)^2 = 1$, thus
\begin{equation}
d\tau = \sqrt{g_{00}}dx^0
\end{equation}
here $d\tau$ is a proper time of an observer
How can we measure the proper time, betwen two points (observer meet particle) and (particle at a distance $d\ell$ from observer)?
We can calculate the projection $v^{\nu}$ on direction $u^{\mu}$:
\begin{equation}
\mathbf{g}(u^{\mu},v^{\nu}) = g_{00}u^0v^0 + g_{0i}u^0v^i = \sqrt{g_{00}}v^0 +{g_{0i}}\frac{1}{\sqrt{g_{00}}}v^i = \sqrt{g_{00}}\frac{dx^0}{ds} +\frac{g_{0i}}{\sqrt{g_{00}}}\frac{dx^i}{ds}
\end{equation}
here $ds$ is a proper time of an particle.
The observer's proper time, betwen two points (observer meet particle) and (particle at a distance $d\ell$ from observer)

\begin{equation}
d\tau = \mathbf{g}(u^{\mu},v^{\nu})ds =\sqrt{g_{00}}\left(dx^0 +\frac{g_{0i}}{{g_{00}}}{dx^i}\right) = \sqrt{g_{00}}d\tilde{t}
\end{equation}

Thus, the prope time $d\tau$ differ from $d\tilde{t}$ just in $\sqrt{g_{00}}$ times, but not in direction, and $\tau$ coinside in direction with $x^0$ as well as with $\tilde{t}$.