# Schwarz's metric gamma

1. Aug 27, 2014

### ChrisVer

Can the above logic be applied to Schw. Metric as well?
Suppose I have an object moving with a radial velocity $v=const$, then can I do the same to derive the Schwarchild time dilation as in the Minkowski?

$dr = v ~ dt$

$ds^{2} = [K - \frac{v^2}{K} ] dt^2$

So $\gamma ^{-1} = \sqrt{K} [1 - v^2/K^2]^{1/2}$?

In that case the relativistic factor $\gamma$ doesn't seem to show a limit of the velocity $v=1$ but $v_{max}= K$ (is it correct to say that that's coordinate dependent?). In this case, when the metric gets flat (far away from the gravitational source), the velocity approaches 1. However it tells us that the maximum speed an object can have in the act of a gravitational force can't reach the speed of light.
Is this correct?

2. Aug 27, 2014

### Orodruin

Staff Emeritus
In the case of GR you have to distinguish coordinate velocities from actual velocities that would be measured by observers. In the case of the Schwarzschild metric, t is global time and not the local time for an observer (there is a prefactor before the $dt^2$ in the metric. However, you can use the metric directly to find out some time dilation relations. For example, the time dilation between an observer at rest at infinity and an observer at a fixed position $r$ (note that such an observer is accelerating).

3. Aug 27, 2014

### wil

I don't see any logics in these computations.

This is probably the well known exercise of the derivation of time dilation, or even the whole LT, from the spacetime interval, but these both are equivalent.

There is no possibility to derive the interval invariance without using LT-Lorentz Transform, and vice versa.

4. Aug 27, 2014

### Ich

No. It tells us that the coordinate speed of light is dr/dt = K<1, and that this is also the maximum velocity for any object.
As Orodruin explained: In Minkowski spacetime, dt and dx have well defined meanings that allow the interpretation of dx/dt as a local relative velocity. In Schwarzschild coordinates, neither dt nor dr correspond directly to locally measurable times or distances. Their ratio is therefore not a local relative velocity, but something different.
You can still learn something from this coordinate velocity: as it goes to zero near the event horizon, it tells you that infalling objects come to a halt in Schwarzschild coordinates and don't reach the horizon in finite Schwarzschild time. That's why black holes sometimes have been referred to as "frozen stars".

5. Aug 27, 2014

### wil

On the horizon the speed must be equal to 1, means c, not less, for a distant observer.

Thus for the local obsrver this is infinity already: c -> oo.

And this is the well known situation during free falling to the horizon we can see the end of the world, because the incoming light isn't limited to 1, but c/0+ -> inf.

And the freezing at the horizon is due to a signal lost - light can't escape out of there, so this means we get information, about the state of a falling body, after infinite time span, what means - never!

6. Aug 27, 2014

### pervect

Staff Emeritus
I haven't checked your calculations, but you can solve directly for the coordinate velocity of light by setting ds=0. You should find the maximum coordinate velocity an object can have is the coordinate velocity of light.

You'll also notice that the velocity of light in standard Schwarzschild coordinates depends on direction, the radial coodinate velocity is different than the tangential coordinate velocity (multiplying $d\phi/dt$ by r). The maximum velocity of an object would also then depend on direction.

Isotropic Schwarzschild coordinates have the property that the speed of light is the same in all directions.

Hopefully this will give you some idea of what is going on. The main point is that coordinate velocities depend on what coordinates you use, and shouldn't be confused with the velocity a local obserer would measure.

As a corollary, the choice of coordinates is arbitrary, so(( you can make a coordinate velocity have any value you like.

((Sometimes I tell this to people, and they don't accept it, believing apparaently that Schwarzschild coordinates are the One True coordinates, and imparting a significance to them they don't actually have. Hopefullly this isn't the case here, but I thought I should mention it since my previous argument depends on accepting that the choice of coordinates IS arbitrary)).

7. Aug 28, 2014

### wil

Such frivolous freedom does not exist in geometry.

I think you're talking about the pure mathematics, which usually do not define the units of measurement, but simply accept: unit = 1.

The consequences of failure to set the units of measure in the correct way are quite devastating.

For example Schwarzschild explicitly and correctly determined the units of measurement in his calculations.

8. Aug 28, 2014

### ChrisVer

In fact I don't understand why r and t don't correspond to locally measurable distances/times. I mean they are the coordinates $x^{0,1}$. If you meant that the 0,1 coordinate are not what we call time or space, then this could be an explanation. However I'd question that...
In general the scwarchzild metric has some symmetry (for this example the rotational). That's also why you have in fact a sphere for the other coords 2,3 (they are the $d \Omega$).
What would then be r if not the radial distance, and t the time to complete spacetime? maybe some combination of them two?

9. Aug 28, 2014

### ChrisVer

(Because I think this came out to be a big reply, I bolded the questions)

The way you state how to find the speed of light, is the same as what I used. In your case setting $ds^{2}=0$ (which, now, I actually find more correct) is going to give you $v^{2}=K^{2}$ ... the way I set the maximum velocity was for $\gamma$ to be well defined and neither be imaginary or $\frac{1}{0}$. But as you pointed out $K(r)$ is coordinate dependent and that's why I said my velocity came out to be coordinate dependent.
As for the angular velocity (tangential coordinate velocity) I considered it to be zero (an object moving only on the r-direction) in order to make $ds^{2}$ look simpler. If I'd accept some $\dot{x}^{2,3}$ then I would have a different $ds^{2}$ I guess because I'd need (in your case) to change the $d \phi^{2}$ in the metric as well.
I think your other statement about the isotropic Schw. is simply reflected to the fact that $K \equiv K(r)$ is a function only of r. It also appears in the metric as such when you ask for isotropy.
How are (or are they) the coordinate velocities connected to the velocities measured by a local observer? By choosing a comoving coord system?

As for the last, yes it's "arbitrary" since in GR you have the reparametrization invariance and you can choose some other coordinates $y^{\mu}(x^{\nu})$ and get the same things. At least that's how I get your statement.

10. Aug 28, 2014

### Staff: Mentor

The point is that the coordinates are not what defines the geometry. The geometry is defined by the metric, the coordinates are just labels.

Please see Chapter 2 of Carroll's lecture notes here: http://preposterousuniverse.com/grnotes/

11. Aug 28, 2014

### Staff: Mentor

The reason is fundamentally that the coordinate basis is not normal in Schwarzschild coordinates.

In other words, the quantity $g_{\mu\nu}dr^{\mu}dr^{\nu} \ne 1$. Each point in the manifold the basis vector dr has a different physically measurable length. That means that some $\Delta r$ measured on the surface of a planet is a different distance than the same $\Delta r$ measured in deep space.

Similarly with dt.

12. Aug 28, 2014

### stevendaryl

Staff Emeritus
I think you misunderstand the role of coordinates in physics. Let me give an example from classical physics: If you use rectangular coordinates $x$ and $y$ to describe points on a 2-D plane, then you can compute distances this way:

$\delta s = \sqrt{\delta x^2 + \delta y^2}$

On the other hand, if you use polar coordinates $r$ and $\theta$, then you compute distances this way:

$\delta s = \sqrt{\delta r^2 + r^2 \delta \theta^2}$

So just knowing "Coordinate $x_1$ changed by an amount $\delta x_1$ and coordinate $x_2$ changed by an amount $\delta x_2$" doesn't tell you how much far you have traveled. What does that is the metric tensor. The most general metric tensor for a two-dimensional space has 4 components: $g_{11}, g_{12}, g_{21}, g_{22}$ and then distances are computed by:

$\delta s = \sqrt{g_{11} (\delta x_1)^2 + g_{12} (\delta x_1)(\delta x_2) + g_{21}(\delta x_2)(\delta x_1) + g_{22} (\delta x_2)^2}$

There is nothing special about Schwarzschild coordinates, and there is nothing special about the components of the tensor. Both together determine distances and proper times. You can use any other coordinate system you like, as long as you change the metric tensor components accordingly.

The expression $\delta s^2 = \sum_{\alpha} \sum_{\beta} g_{\alpha \beta} \delta x^\alpha \delta x^\beta$ has the same value in any coordinate system whatsoever. That's what's physically meaningful, not the coordinates themselves.