I 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.

 

[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 $\tilde{t}$.