# I Velocity measurement by a stationary observer in GR

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1. Nov 5, 2017

### Staff: Mentor

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.

2. Nov 5, 2017

### sergiokapone

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.

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3. Nov 5, 2017

### Staff: Mentor

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.

4. Nov 6, 2017

### sergiokapone

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.

5. Nov 6, 2017

### sergiokapone

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.

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6. Nov 6, 2017

### Staff: Mentor

I don't think you have. See below.

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.

7. Nov 7, 2017

### 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 $u^{\mu}$.

\mathbf{g}(u^{\mu},u^{\mu}) = 1

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

d\tau = \sqrt{g_{00}}dx^0

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 $u^{\mu}$:

\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}

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)

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}

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 $\tilde{t}$.

Last edited: Nov 7, 2017