Velocity measurement by a stationary observer in GR

[sergiokapone]

In almost general case, the space-time metrics looks like:
\begin{equation}
ds^2 = g_{00}(dx^0)^2 + 2g_{0i}dx^0dx^i + g_{ik}dx^idx^k,
\end{equation}
where ##i,k = 1 \ldots 3## - are spatial indeces.

The spatial distance between points (as determined, for example, by the stationary observer):
\begin{equation}
dl^2 = \left( -g_{ik} + \frac{g_{0i}g_{0j}}{g_{00}}\right)dx^idx^j = \gamma_{ik}dx^idx^j,
\end{equation}
where
\begin{equation}
\gamma_{ik} = -g_{ik} + \frac{g_{0i}g_{0j}}{g_{00}} ,
\end{equation}

And we can rewrite merics in a form:
\begin{equation}
ds^2 = g_{00}(dx^0 - g_idx^i)^2 - dl^2,
\end{equation}
in the last expression ##g_i## is:
\begin{equation}
g_i = -\frac{g_{0i}}{g_{00}}.
\end{equation}

The proper time in such metrics (when ##dx^i = 0##):
\begin{equation}
d\tau = \sqrt{g_{00}}dx^0.
\end{equation}

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}##
\begin{equation}
v = \frac{dl}{\sqrt{g_{00}}dx^0},
\end{equation}
but it it wrong answer.

The correct answer should to be
\begin{equation}
v = \frac{dl}{\sqrt{g_{00}}(dx^0 - g_idx^i)},
\end{equation}
and I don't undertand why.

 

[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##.

 

[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?

 

[Tio Barnabe]

I thaught the observer is stationary respect to the some milestoun (which is marked by spatial coordinate ##x^i##)

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##.

 

[sergiokapone]

Yes, but that doesn't matter when you are considering the system to be (particle + observer).

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

 

[Orodruin]

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##.

 

[sergiokapone]

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##.

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}}##.

 

[PeterDonis]

I try to understand what is velocity in tetms of proper time and proper distance.

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.

 

[PeterDonis]

I thaught the observer is stationary respect to the some milestoun (which is marked by spatial coordinate ##x^i##)

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.)

 

[Orodruin]

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.

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##.)

 

[sergiokapone]

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

 

[Orodruin]

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

 

[sergiokapone]

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

 

[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)

\begin{equation}
ds^2 = g_{00}(dx^0)^2 + 2g_{0i}dx^0dx^i + g_{ik}dx^idx^k,
\end{equation}
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
\begin{equation}
g_{00}(v^0 - g_i v^i) = \gamma
\end{equation}

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

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

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

 

[Orodruin]

velocity of an observr is ##u^{\mu} = (u^0,0,0,0) = (1,0,0,0)##

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.

 

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

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

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

 

[Orodruin]

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.

 

[sergiokapone]

What is the difference between ##g_{00}^o## and ##g_{00}##?

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

 

[Orodruin]

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

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.

 

[sergiokapone]

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.

Ok, ##g_{00}## belong to the observer, then
\begin{equation}
\sqrt{g_{00}}(v^0 -2 g_i v^i) = \gamma
\end{equation}
for parametrization worldline I want to choose particle proper time ##\tau##
\begin{equation}
g_{00} ( v^0 - g_{0i} v^i)^2 + \frac{dl^2}{d\tau^2}=1
\end{equation}

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

 

[sergiokapone]

Correct formula should be
\begin{equation}
\sqrt{g_{00}}(v^0 - g_i v^i) = \gamma
\end{equation}
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}##
\begin{equation}
g_{00} ( v^0 - g_{0i} v^i)^2 + \frac{dl^2}{d\tau^2}=1
\end{equation}

and get definition of ##\gamma##
\begin{equation}
\gamma^2 + \frac{dl^2}{d\tau^2} = 1
\end{equation}


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##:
\begin{equation}
(x^0 + dx^0) - (x^0 - (g_{0\alpha}/g_{00})dx^{\alpha}) = dx^0 + g_{0\alpha}/g_{00})dx^{\alpha}
\end{equation}
"
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?

 

[sergiokapone]

Is it correct?

Velocity ##v## is measured in terms of the proper time of an observer, as determined by his clocks, but the clocks should be synchronized along the trajectory of the particle.

http://i95.fastpic.ru/big/2017/1030/23/dad89b2556464763fbacaaa601e59723.gif

 

[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}##

 

[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.

 

[PeterDonis]

all measurements of the velocity should be division distance by proper time of an observer. That is my understanding of velocity.

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).

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.

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.

the procedure of velocity measurement is a projection of particle 4-velocity on the observer 4-velocity

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".

 

[sergiokapone]

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.

What is "synchronized time" is from book of Landau (is a proper time, as determined by clocks
synchronized along the trajectory of the particle), I quoted the above in post #21
and I just do not understand why one need to divide by synchronized time, as Landau mentioned, but only this procedure leads to reasonable results.

And we have two types of proper times:
1. "simple proper time" ##\sqrt{g_{00}}dx^0## (then it is not clear what it is)
and
2."proper time, as determined by clocks synchronized along the trajectory of the particle" ##\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)##
this is confusing

 

[PeterDonis]

What is "synchronized time" is from book of Landau (is a proper time, as determined by clocks
synchronized along the trajectory of the particle)

Ok, then the usual terminology would just be "the proper time of the particle", since that's what the clocks in question are recording. You confused me because you said that "synchronized time" is different from proper time--your phrase in bold in the quote above clearly shows they are the same, not different.

we have two types of proper times

No, we don't. Proper time is the time recorded by a clock that is moving along with the particle. That's the only kind of proper time. But different particles can have different states of motion.

"simple proper time" ##\sqrt{g_{00}}dx^0## (then it is not clear what it is)

It's the proper time of an object (or observer or clock) that remains stationary at constant ##(x^1, x^2, x^3)##, i.e., ##dx^i = 0## for ##i = 1, 2, 3##.

"proper time, as determined by clocks synchronized along the trajectory of the particle" ##\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)##

This is the proper time of an object (or observer or clock) that is moving along with the particle, which is not stationary in the chosen coordinates but has a nonzero coordinate velocity, i.e., ##dx^i \neq 0## for ##i = 1, 2, 3##.

Notice that if you plug ##dx^i = 0## into this formula you get the formula you gave above; in other words, the two formulas are really the same, just for two different objects, one with zero ##dx^i## and the other with nonzero ##dx^i##.

this is confusing

Hopefully the above helps to clarify things.

 

[sergiokapone]

##\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)##

This is the proper time of an object (or observer or clock) that is moving along with the particle, which is not stationary in the chosen coordinates but has a nonzero coordinate velocity, i.e., dxi≠0dx^i \neq 0 for i=1,2,3i = 1, 2, 3.

Strange, I thaught the proper time of an object (or observer or clock) that is moving along with the particle should be ##d\tau^2 = g_{00}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)^2 + dl^2##.

 

[PeterDonis]

I thaught the proper time of an object (or observer or clock) that is moving along with the particle should be ##d\tau^2 = g_{00}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)^2 + dl^2##.

What is ##dl^2##?

 

[sergiokapone]

What is ##dl^2##?

##dl = \sqrt{\left(-g_{ik} + \frac{g_{0i}}{g_{00}}\right)dx^idx^k}## --- spatial distance between ##x^i## and ##x^i + dx^i## as measured by stationary observer at ##x^i##

 

[PeterDonis]

##dl = \sqrt{\left(-g_{ik} + \frac{g_{0i}}{g_{00}}\right)dx^idx^k}## --- spatial distance between ##x^i## and ##x^i + dx^i## as measured by stationary observer at ##x^i##

Why would you expect that to affect the proper time? That's a formula for distance, not proper time.

 

[PeterDonis]

Why would you expect that to affect the proper time?

Or, to put it another way: instead of waving your hands with various formulas pulled from thin air, why not go back to the standard metric formula, ##d\tau^2 = g_{\alpha \beta} dx^\alpha dx^\beta##, where ##\alpha, \beta## range from 0 to 3 (i.e., the "time" and the three "space" coordinates), then go from there?

 

[PeterDonis]

instead of waving your hands with various formulas

Note, btw, that I have not checked your sources for these formulas, nor have I checked them against other sources. I'm simply assuming that you are quoting the correct formulas. If you're not sure about that, you should check them yourself, using the method I suggested in my last post.

 

[sergiokapone]

Note, btw, that I have not checked your sources for these formulas, nor have I checked them against other sources. I'm simply assuming that you are quoting the correct formulas.

My sources for formulas is the book of Landau L.D., Lifshitz E.M. Vol. 2. The classical theory of fields (4ed., Butterworth-Heinemann, 1994, and I assume it is correct.

Or, to put it another way: instead of waving your hands with various formulas pulled from thin air, why not go back to the standard metric formula, ##d\tau^2 = g_{\alpha \beta} dx^\alpha dx^\beta##, where ##\alpha, \beta## range from 0 to 3 (i.e., the "time" and the three "space" coordinates), then go from there?

Ok, let's go back to , ##d\tau^2 = g_{\alpha \beta} dx^\alpha dx^\beta##, as one can check this formula leads to ##d\tau^2 = g_{00}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)^2 - \left(-g_{ik} + \frac{g_{0i}}{g_{00}}\right)dx^idx^j##, where ##i,j = 1\ldots 3## --- spatial indices, thus I conclude, the proper time of an moving particle is ##d\tau^2 = g_{00}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)^2 - dl^2##, but not just ##\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)##. But I want to undestand nature of ##\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)##, which is caled in literature as "physical time" (see Physical time and physical space in general relativity by Richard J. Cook, for example )

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

The all important point of this discussion is that, as a temporal coordinate, the physical time ##\tilde t## is a very special one. It is the time actually used by the fiducial observer for measurements in his local reference frame. If it were not so, this observer would not measure the local light speed ##c##.
Thus the notion that all fiducial observers measure the same speed ##c## for light (Einstein’s postulate) is equivalent to the notion that all fiducial observers use the physical time ##\tilde t## in their local reference frames for the measurement of that speed.

(Allocating by bold is mine)

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

 

[Tio Barnabe]

Ok, let's go back to , ##d\tau^2 = g_{\alpha \beta} dx^\alpha dx^\beta##, as one can check this formula leads to ##d\tau^2 = g_{00}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)^2 - \left(-g_{ik} + \frac{g_{0i}}{g_{00}}\right)dx^idx^j##, where ##i,j = 1\ldots 3## --- spatial indices, thus I conclude, the proper time of an moving particle is ##d\tau^2 = g_{00}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)^2 - dl^2##, but not just ##\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)##. But I want to undestand nature of ##\sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}}dx^i\right)##, which is caled in literature as "physical time"

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.

Proper time is given by ##d \tau^2 = g_{\alpha \beta}dx^\alpha dx^\beta = - dt^2 + \sum (dx^i)^2## for a Minkowski metric of signature (-1,1,1,1). We can consider this specific metric in our analysis, because it's the metric one will use in most of the applications.

How does he know when to use this time, and when to use other one ##\sqrt{g_{00}}dx^0## ?

When you say so, you seem to assume that the observer dictates how the math should look like. That's not correct. The physical situation is imposed on us. The time the observer will measure is the time measured by any working clock in his frame, and this is independent of him wanting so or not.