# Can g_00 of the metric tensor depend on time

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1. Nov 16, 2014

### birulami

In SRT, the line element is $c^2ds^2 = c^2dt^2 - dx^2 -dy^2-dz^2$ and $g_{00} = 1$ (or $-1$ depending on sign conventions). In the Schwarzschild metric we have
$$g_{00}=(c^2-\frac{2 GM}{r}) .$$ So in the first example, $g_{00}$ is constant, in the second it depends on another coordinate ($r$).

Are there examples in GRT where $g_{00}$ depends on $t$ or can it be proven that this cannot be the case?

2. Nov 16, 2014

### Orodruin

Staff Emeritus
This depends completely on the coordinates you pick. Note that, for the Schwarzschild solution, the r coordinate is the time coordinate for r smaller than the Schwarzschild radius.

3. Nov 16, 2014

### birulami

Do you have an example?

Not sure whether this really is an example. Just because r gets smaller than the Schwarzschild radius does not change the Schwarzschild metric, so $g_{00}$ still does not depend on $t$.

4. Nov 16, 2014

### Orodruin

Staff Emeritus
What I am trying to say is that t is just an arbitrary coordinate. You could just as well have called the r coordinate t, which would make perfect sense inside of the Schwarzschild radius. It is only a matter of what is convenient to use, you could also define a new coordinate $w$ as $w = t + t^{21}$, which certainly would have a factor in front of the $dw^2$ in the metric. I suggest you do not try to interpret too much out of the coordinate being named $t$.

5. Nov 16, 2014

### WannabeNewton

FRW space-time, PP-wave space-time, Vaidya space-time...

If there were no solutions to Einstein's equations that weren't dynamical then GR would be quite a useless theory wouldn't it, given that all the interesting phenomenon in nature are dynamical?

Also, as Orodruin stated, we can always find a coordinate system in which $g_{00}$ depends on some time coordinate $t$ even if $g_{00}$ was independent of any time coordinate in the previous coordinate system. Except for special cases (stationary asymptotically flat space-times, space-times generated by a time-like congruence etc.), $t$ has no physical significance in general; we just call it a time coordinate for technical reasons.

6. Nov 16, 2014

### Staff: Mentor

Just a pedantic note: the $g_{00}$ metric coefficient in the FRW metric, in standard coordinates, does not depend on $t$. Other metric coefficients do, but $g_{00}$ does not.

7. Nov 16, 2014

### Staff: Mentor

But the title of this thread is "depends on time", not "depends on the $t$ coordinate". Which are you interested in?

If you're interested in the $t$ coordinate, that's fine, but inside the horizon in the Schwarzschild metric, the $t$ coordinate is spacelike, not timelike, so not depending on $t$ is not the same as "not depending on time".

If you're interested in "time", in the sense of "along a timelike path", then inside the horizon in the Schwarzschild metric, there are no timelike paths along which the $g_{00}$ metric coefficient is constant (this is a more general way of saying what Orodruin said, that the $r$ coordinate is a "time coordinate" inside the horizon). So inside the horizon, $g_{00}$ does depend on "time".

8. Nov 16, 2014

### Orodruin

Staff Emeritus
Just changing spatial coordinates to co-moving coordinates should remedy this if I am not mistaken (it is late) ... Again an example of this statement being coordinate dependent.

9. Nov 16, 2014

### Staff: Mentor

I think you mean "from", not "to". ;) The standard FRW coordinates, in which $g_{00} = -1$, are comoving coordinates. But there are certainly other charts in which $g_{00}$ depends on the time coordinate, yes. (I've never actually seen those other charts used, but I'm certainly not intimately familiar with the literature in cosmology.)

Yes, definitely.

10. Nov 16, 2014

### Orodruin

Staff Emeritus
Definitely, as I said, it is late - I was bound to write something stupid ;)

I mean $g_{00} = +1$, but that is a convention discussion for another time

11. Nov 16, 2014

### Staff: Mentor

Ah, you're right, the OP was using that convention.

12. Nov 17, 2014

### stevendaryl

Staff Emeritus
Well, $t$ is just a letter. You can certainly write the Schwarzschild metric in the form:

$\frac{c^2}{\frac{2GM}{c^3 t} - 1} dt^2 - (\frac{2GM}{c^3 t} - 1) dr^2 - c^2 t^2 d\Omega^2$

which is valid inside the event horizon of a black hole.

13. Nov 17, 2014

### ChrisVer

Last edited: Nov 17, 2014
14. Nov 17, 2014

### Staff: Mentor

Careful. The "static" is often included, even by highly reputable sources, but it's not actually entirely correct. The only source I'm aware of that gets this exactly right is MTW. Even good textbooks like Wald, or good online lecture notes like Sean Carroll's, use the word "static" too broadly in this connection.

Here's the entirely correct statement: the Schwarzschild geometry is the unique solution of the spherically symmetric vacuum Einstein Field Equation. (Note the "vacuum", and that "static" is not there.) This statement is a theorem, known as "Birkhoff's Theorem".

It turns out that this unique solution is static outside the horizon. And since the region outside the horizon is often the only region that needs to be discussed, you very often see sources saying that the Schwarzschild geometry is "static", without bothering to mention that that's only true outside the horizon. Many sources even say that Birkhoff's Theorem says that the Schwarzschild geometry is static (which is not what the theorem says, though the fact that the Schwarzschild geometry is static outside the horizon is indeed an immediate consequence of the theorem--consequence, not premise).

Inside the horizon, the Schwarzschild geometry is not static. (If you want the gory technical details, I can provide them, but I don't want to make this post too long.) But it is still the unique solution of the spherically symmetric vacuum EFE. So the fact that this solution is not static inside the horizon shows that, as I said in post #7, there are no timelike paths inside the horizon along which the metric coefficient $g_{00}$ (or indeed any metric coefficient) is constant. So inside the horizon, the metric does "depend on time".

15. Nov 18, 2014

### JorisL

I don't believe Carroll's notes have been updated since he released them as a book.
I can say that he doesn't make this mistake in the book following are exact quotes.

Introducing the metric in section 5.1 and derive it afterwards
Birkhoff's theorem in section 5.2
The part in parentheses brings us back to the issue at hand here. He started out well, but added this static as a sort of corollary.
He does mention however how the timelike coordinate becomes spacelike in the interior.
I cannot say if he goes into detail however. I'll check that out later today if I find the time.