# I Velocity measurement by a stationary observer in GR

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1. Oct 25, 2017

### sergiokapone

In almost general case, the space-time metrics looks like:

ds^2 = g_{00}(dx^0)^2 + 2g_{0i}dx^0dx^i + g_{ik}dx^idx^k,

where $i,k = 1 \ldots 3$ - are spatial indeces.

The spatial distance between points (as determined, for example, by the stationary observer):

dl^2 = \left( -g_{ik} + \frac{g_{0i}g_{0j}}{g_{00}}\right)dx^idx^j = \gamma_{ik}dx^idx^j,

where

\gamma_{ik} = -g_{ik} + \frac{g_{0i}g_{0j}}{g_{00}} ,

And we can rewrite merics in a form:

ds^2 = g_{00}(dx^0 - g_idx^i)^2 - dl^2,

in the last expression $g_i$ is:

g_i = -\frac{g_{0i}}{g_{00}}.

The proper time in such metrics (when $dx^i = 0$):

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

How do the absolute value of velocity of a certain particle are determined by stationary observer?
I thaught that is to be $v = \frac{proper distance}{proper time}$

v = \frac{dl}{\sqrt{g_{00}}dx^0},

The correct answer should to be

v = \frac{dl}{\sqrt{g_{00}}(dx^0 - g_idx^i)},

and I don't undertand why.

2. Oct 26, 2017

### Tio Barnabe

The thing is that the observer, as you defined him, will see the particle at rest, because, as you say, he's stationary with respect to the particle. The other observer will see it moving with velocity modulo $v$.

3. Oct 26, 2017

### sergiokapone

I thaught the observer is stationary respect to the some milestoun (which is marked by spatial coordinate $x^i$), but particle moving from one $x^i$ to another $x^i + dx^i$. Am I wrong?

4. Oct 26, 2017

### Tio Barnabe

Yes, but that doesn't matter when you are considering the system to be (particle + observer). It doesn't matter where the observer is located in his own reference frame, he could easily set his coordinate position to be $0$.

5. Oct 26, 2017

### sergiokapone

Why do think I'm considering system "particle + observer"?

6. Oct 26, 2017

### Orodruin

Staff Emeritus
The easiest way of finding the relative velocity between two objects is to consider their normalised 4-velocities. The inner product of the normalised 4-velocities is the relative gamma factor, solve for $v$.

7. Oct 26, 2017

### sergiokapone

Yes, this is the good idea. But I try to understand what is velocity in tetms of proper time and proper distance.

For example, in Schwarzschild metric for the stationary observer, which is at rest at some coordinate $r = const$ the $proper\,time = \sqrt{1-\frac{2M}{r}}dx^0$, and the proper distance $dl = \frac{dr}{\sqrt{1-\frac{2M}{r}}}$. So, the velocity of particle, that is move past him and determined by him $v = \frac{proper\,distanse}{proper\,time} = \frac{dr}{dx^0}\frac{1}{1-\frac{2M}{r}}$.

Last edited: Oct 26, 2017
8. Oct 27, 2017

### Staff: Mentor

What @Orodruin described is the velocity in terms of proper time and proper distance--once you've picked the observer whose proper time and proper distance you are using. Proper time and proper distance are not absolute properties of spacetime alone; they are properties of particular observers following particular worldlines through spacetime.

For example, in the Schwarzschild spacetime case you describe, the proper time you are using is the proper time of an observer who is "hovering" at a constant altitude above the black hole, and the proper distance you are using is the radial proper distance in the "hovering" observer's immediate vicinity, as measured by rulers at rest with respect to the "hovering" observer. And the velocity you are getting is specifically for an object moving only in the radial direction, past the "hovering" observer. Without all of those specific qualifications I just gave, any statement about proper time or proper distance is meaningless.

9. Oct 27, 2017

### Staff: Mentor

This is one way of defining "stationary", but it has the strong disadvantage of depending on the coordinates that you choose. Two different people who make different choices of coordinates will disagree on which objects are stationary by this definition.

In certain spacetimes (Schwarzschild spacetime is one of them), there are choices of coordinates in which observers who are stationary in those coordinates are also stationary in a different sense, one that is picked out by some physical property. In the Schwarzschild case, an observer who is "hovering" at a constant altitude above the black hole's horizon is stationary in this different sense--he is stationary not just relative to a choice of coordinates but relative to an invariant feature of the spacetime geometry, the hole's horizon. So this definition of "stationary" does not depend on any choice of coordinates and all observers will agree on which objects are stationary by this definition.

When you see the term "stationary" used in the scientific literature on GR (as opposed to informal pop science discussions), the latter definition is the way in which the term is meant. (And there is a more technical definition that makes the heuristic description I gave above precise.)

10. Oct 27, 2017

### Orodruin

Staff Emeritus
To expand on this. Consider the 4-velocities of the two observers in the Schwarzschild space-time. The stationary observer only has a 4-velocity component in the $t$ direction and so
$$(1-\phi) (U^0)^2 = 1 \quad \Longrightarrow \quad U^0 = \frac{1}{\sqrt{1-\phi}}.$$
The radially moving observer parametrised by the coordinate time $t$ has a 4-velocity $V = (k,k\dot r,0,0)$ in Schwarzschild coordinates and so
$$1 = V^2 = k^2\left[1-\phi - \frac{\dot r^2}{1-\phi}\right] = k^2 \frac{(1-\phi)^2 - \dot r^2}{1-\phi} \quad \Longrightarrow \quad k = \sqrt{\frac{1-\phi}{(1-\phi)^2 - \dot r^2}}.$$
The gamma factor between the observers is therefore given by
$$\gamma^2 = (U\cdot V)^2 = (1-\phi)^2 (U^0 k)^2 = \frac{(1-\phi)^2}{(1-\phi)^2 - \dot r^2}.$$
Now, since $\gamma^2 (1-v^2) = 1$, it follows that
$$v^2 = 1 - \frac{1}{\gamma^2} = 1 - \frac{(1-\phi)^2 - \dot r^2}{(1-\phi)^2} = \frac{\dot r^2}{(1-\phi)^2}$$
or, equivalently,
$$v = \frac{|\dot r|}{1-\phi}.$$
This is exactly the same expression as that in post #7.

(Note, everywhere I have used $\phi = r_s/r$.)

11. Oct 28, 2017

### sergiokapone

Ok, how one can proof the formula $\gamma = U \cdot V$?

12. Oct 28, 2017

### Orodruin

Staff Emeritus
It is true in SR and it is a local measurement. (Go to local normal coordinates.)

13. Oct 28, 2017

### sergiokapone

$g_{\mu\nu}u^{\mu}v^{\nu} = \eta_{\mu\nu}U^{\mu}V^{\nu} = \gamma$

14. Oct 28, 2017

### sergiokapone

Now I want to obtain velocity which is determined by an observer standing at position $x^i = 0$ form the metric (fomr post #1)

ds^2 = g_{00}(dx^0)^2 + 2g_{0i}dx^0dx^i + g_{ik}dx^idx^k,

where $i,k = 1 \ldots 3$ - are spatial indeces.

velocity of an observr is $u^{\mu} = (u^0,0,0,0) = (1,0,0,0)$ and velocity of an particleis $v^{\nu}$ so

g_{00}(v^0 - g_i v^i) = \gamma

Let remember the proper time of an observer $d\tau = \sqrt{g_{00}^o}dx^0$,

g_{00} - g_i \frac{dx^i}{dx^0} =\sqrt{g_{00}^o} \gamma

but now I don't know where I can get $\frac{dx^i}{dx^0}$ for determining $v$.

Last edited: Oct 28, 2017
15. Oct 28, 2017

### Orodruin

Staff Emeritus
This is not correct. It holds only in local coordinates where the time direction has been normalised. If you square the 4-velocity you should get 1. The same essentially goes for your other 4-velocity.

Edit: For example, see how I normalised the 4-velocity of the stationary observer in post #10 in the Schwarzschild metric.

16. Oct 28, 2017

### sergiokapone

Now I want to obtain velocity which is determined by an observer standing at position $x^i = const$ form the metric (fomr post #1)

velocity of an observr is $u^{\mu} = (u^0,0,0,0) = (\frac{1}{\sqrt{g_{00}^o}},0,0,0)$ and velocity of an particleis $v^{\nu}$ so

\frac{g_{00}}{\sqrt{g_{00}^o}}(v^0 -2 g_i v^i) = \gamma

and for $v^{\mu}$ we have a condition:

g_{00}(v^0)^2 + 2g_{0i}(v^i)^2 + g_{ik}v^iv^k = 1

Last edited: Oct 28, 2017
17. Oct 28, 2017

### Orodruin

Staff Emeritus
What is the difference between $g_{00}^o$ and $g_{00}$? You will also have to relate the $v^i$ to some parametrisation of the world-line of the object.

18. Oct 28, 2017

### sergiokapone

It is a $g_{00}^o$ in place where the observer is, $g_{00}$ is the function of coordinates.

19. Oct 28, 2017

### Orodruin

Staff Emeritus
I suspected as much. This is a misconception that you have to dispel immediately. The measurement of relative velocity can only be made locally, i.e., where the observer is. Any other method in a curved space-time is going to be convention dependent.

20. Oct 28, 2017

### sergiokapone

Ok, $g_{00}$ belong to the observer, then

\sqrt{g_{00}}(v^0 -2 g_i v^i) = \gamma

for parametrization worldline I want to choose particle proper time $\tau$

g_{00} ( v^0 - g_{0i} v^i)^2 + \frac{dl^2}{d\tau^2}=1

[Note: edited by a moderator to correct a formatting problem]

Last edited by a moderator: Oct 30, 2017
21. Oct 29, 2017

### sergiokapone

Correct formula should be

\sqrt{g_{00}}(v^0 - g_i v^i) = \gamma

because $g_{i0} u^iv^0 = 0$ , but $g_{0i} u^0v^i \neq 0$.

for parametrization worldline I want to choose particle proper time $\tau$, so for the norm of $v^{\nu}$

g_{00} ( v^0 - g_{0i} v^i)^2 + \frac{dl^2}{d\tau^2}=1

and get definition of $\gamma$

\gamma^2 + \frac{dl^2}{d\tau^2} = 1

The statement, that I do not uderstand (Landau L.D., Lifshitz E.M. Vol. 2. The classical theory of fields (4ed., Butterworth-Heinemann, 1994), page 270)

"It is easy to show that the expression (88.9) (for energy $E_0 = \frac{mc^2\sqrt{g_{00}}}{\sqrt{1-\frac{v^2}{c^2}}}$)remains valid also for a stationary field, if
only the velocity $v$ is measured in terms of the proper time, as determined by clocks
synchronized along the trajectory of the particle
. If the particle departs from point A at the
moment of world time x° and arrives at the infinitesimally distant point B at the moment $x^0 + dx^0$, then to determine the velocity we must now take, not the time interval $(x^0 + dx^0) - x^0 = dx^0$, but rather the difference between $(x^0 + dx^0)$ and the moment $(x^0 - (g_{0\alpha}/g_{00})dx^{\alpha})$ which issimultaneous at the point $B$ with the moment $x^0$ at the point $A$:

(x^0 + dx^0) - (x^0 - (g_{0\alpha}/g_{00})dx^{\alpha}) = dx^0 + g_{0\alpha}/g_{00})dx^{\alpha}

"
I thaught the velocity shoud be mesured by an observer at $x^i = const$ in terms of the proper time, determined by the observer's clock, but not with clocks synchronized along the trajectory of the particle!!! Who does measure the time by clocks synchronized along the trajectory of the particle!? An observer who moves with a particle?

Last edited: Oct 29, 2017
22. Oct 30, 2017

### sergiokapone

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23. Nov 2, 2017

### Yukterez

The most general case for the velocity components is given in the Wikipedia-article of the Kerr-Newman-metric, this velocity v is measured relative to a locally stationary observer, also called ZAMO, and is 1 for the speed of light. If your velocity is differentiated by the proper time of the test particle the speed of light is no longer 1 but infinite (v·γ). The equation with μ=0 for photons and μ=-1 for particles below the speed of light should be then

${\rm \dot x^i} = {\rm v^i / \sqrt{1+ \mu \ v^2}}/ \sqrt{{|g_{\rm i i|}} - \dot {\rm t} \ g_{\rm t i}/g_{ i i}}$

where the overdot is differentiation by proper time τ (here it's not the velocity which is differentiated by the proper time, but the coordinates themselves) with μ=-1 and by the spatial affine parameter with μ=0. The total velocity v is simply

$\rm v=\sqrt{ \sum _{i=1}^3 (v^i)^2}$

Last edited: Nov 2, 2017
24. Nov 4, 2017

### sergiokapone

Thank you for article.

I think every observer (ZAMO or other ones) measure (of course, locally) speed of light equal to 1, and thus, all measurements of the velocity should be division distance by proper time of an observer. That is my understanding of velocity.

The proper time equal to $\sqrt{g_{00}}dx^0$ in general metric, as well as for Schwarzschild case (where $g_{0i} = 0$). But when the $g_{0i} \neq 0$, for velocity measurements one should be divide distance by not proper time, but synchronized time of an observer, only in this case we can obtain physical reasonable result. Mathematically, the procedure of velocity measurement is a projection of particle 4-velocity on the observer 4-velocity, but it is not obvious for beginner level, I think. Now I try to find a way (using space-time diagrams) to explain for some imaginary student, why for velocity measurement we should be divide distance by synchronized time.

25. Nov 4, 2017

### Staff: Mentor

This works for locally measured velocity, so your position amounts to saying that the term "velocity" is properly used only to describe such local measurements of velocity (i.e., measurements carried out within a single local inertial frame, in which the laws of SR hold).

This does not seem to be consistent with what you said earlier in the same post (quoted above). Also, it's not clear what you mean by "synchronized time", how it is different from proper time, or why you think it must be used instead of proper time to obtain a "physically reasonable" result.

Which can only be done within a single local inertial frame, i.e., where the particle's worldline crosses the observer's worldline, so the 4-velocities can be directly compared. This is consistent with saying that "velocity" properly refers only to locally measured velocity. But it does not appear to be consistent with your statement about using "sychronized time".