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Another question about the Schwarzschild solution of General Relativity:

In the derivation (shown below) of the Schwarzschild metric from the vacuum Einstein Equation, at the step marked "HERE," we have the freedom to rescale the time coordinate from $dt \rightarrow e^{-g(t)}dt$. However, later in the derivation, when we pick up an integration constant at the step marked "HERE2," we are unable to absorb that constant into the metric. Actually this constant becomes a fundamental part of the metric itself. Why can't we incorporate this constant into the differentials of the metric as well?

Here's the derivation:

$$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} = 0$$

Assume the most general spherically symmetric metric solution to the above:

$$g_{\mu\nu} = \begin{pmatrix} -e^{2\alpha(r,t)} & 0 & 0 & 0\\ 0 & e^{2\beta(r,t)}&0&0\\0&0&r^2 & 0\\0 & 0 &0 & r^2\sin^2(\theta) \end{pmatrix}$$

Take the trace of both sides:

$$R + 2R = 0\\ \Rightarrow R = 0\\ R_{\mu\nu} - \frac{1}{2}(0)= 0\\ R_{\mu\nu}=0$$

We are free to say:

$$R_{01} = \frac{2}{r}\dot{\beta}=0\\ \Rightarrow \dot{\beta} = 0\\ \therefore \beta = \beta(r) $$

Now, we can take a time derivative of $R_{22}$,

$$\dot{R_{22}} = \frac{d\left(e^{-2\beta}[r(\beta'-\alpha') - 1] + 1\right)}{dt} = 0 = e^{2\beta}r\dot{\alpha}'$$ (Because $\dot{\beta} = 0$, $r$ is just a spacetime coordinate (and is therefore independent of other spacetime coordinates), and $\dot{\beta}' = \partial_r\partial_t\beta = \partial_r(0)$.) $$\dot{\alpha}' = 0\\ \therefore \alpha = f(r) + g(t)$$ because this is the only functional form for which $\partial_r\partial_t\alpha = \partial_t\partial_r\alpha = 0$. Now, the metric becomes:

$$g_{\mu\nu} = \begin{pmatrix} -e^{2f(r)}e^{2g(t)} & 0 & 0 & 0\\ 0 & e^{2\beta(r)}&0&0\\0&0&r^2 & 0\\0 & 0 &0 & r^2\sin^2(\theta) \end{pmatrix}$$

HERE, at this point, we can rescale time: $dt \rightarrow e^{-g(t)}dt$, which makes the metric:

$$g_{\mu\nu} = \begin{pmatrix} -e^{2f(r)} & 0 & 0 & 0\\ 0 & e^{2\beta(r)}&0&0\\0&0&r^2 & 0\\0 & 0 &0 & r^2\sin^2(\theta) \end{pmatrix}$$

Setting $R_{11} = R_{00}$, we find:

$$e^{2(\beta-\alpha)}R_{00} + R_{11} = 0\\ \frac{2}{r}\alpha' + \frac{2}{r}\beta' = 0\\ \Rightarrow \alpha = -\beta + const.$$

This constant is absorbed into $dr^2$.

HERE2, Now comes the crucial part, which I don't understand: looking back at $R_{22} = 0$, we have

$$(re^{2\alpha})' = 1\\ \int (re^{2\alpha})'= \int 1\\ re^{2\alpha} = r + \underbrace{\mu}_{\text{Constant of integration}}\\ e^{2\alpha} = 1 + \frac{\mu}{r}$$

So the metric finally ends up being:

$$ds^2 = -(1 + \frac{\mu}{r})dt^2 + (1 + \frac{\mu}{r})^{-1}dr^2 + r^2 d\Omega^2.$$

Why can't $\frac{\mu}{r}$ be absorbed into $dt^2$ and $dr^2$ like $g(t)$ and the other integration constant in this derivation were?? I don't understand the distinction at all, any help would be greatly appreciated!!

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Here is another Phys.SE post about deriving Birkhoff's Theorem. –  Qmechanic Mar 4 '13 at 9:11

2 Answers 2

up vote 2 down vote accepted

Let's start with the metric

$$\mathrm{d}s^2 = -\left( 1- \frac{2 G M}{r} \right) \mathrm{d}t^2 + \left( 1- \frac{2 G M}{r} \right)^{-1} \mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2 .$$

"Absorbing" a metric coefficient really means defining a new coordinate so that in terms of the new coordinates that coefficient disappears (or becomes one). Then you probably give the new coordinate the same name as the old one, but that's a seperate step.

So let's try to define a new time coordinate $\tau(t,r)$ such that

$$ \mathrm{d}\tau = \frac{\partial \tau}{\partial t} \mathrm{d}t + \frac{\partial \tau}{\partial r} \mathrm{d}r= \sqrt{1- \frac{2 G M}{r}} \mathrm{d}t. $$

If this works then the metric will be $\mathrm{d}s^2 = -\mathrm{d}\tau^2 + \cdots$, which is what I think you're after. From the previous equation we have $\partial \tau/\partial r = 0$, so $\tau(t,r)=\tau(t)$, but then we need $\partial \tau/\partial t$ to be a function of $r$, which is clearly impossible if $\tau$ itself is not a function of $r$. So there is no $\tau$ coordinate we could introduce with the desired property (this is actually a consequence of the nontrivial curvature). Mathematically $\sqrt{1- \frac{2 G M}{r}} \mathrm{d}t$ is not an exact differential. You can get around this if you are willing to introduce cross terms $\mathrm{d}\tau \mathrm{d}r$ etc.

It's different for the $r$ coordinate. Introduce $\rho(t,r)$ such that:

$$ \mathrm{d}\rho = \frac{\partial \rho}{\partial t} \mathrm{d}t + \frac{\partial \rho}{\partial r} \mathrm{d}r= \left(1- \frac{2 G M}{r}\right)^{-1/2} \mathrm{d}r. $$

The first term gives $\rho(t,r) = \rho(r)$ and the second gives

$$\frac{\mathrm{d}\rho}{\mathrm{d} r} = \left(1- \frac{2 G M}{r}\right)^{-1/2}, $$

which integrates to (for $r > 2 G M$)

$$ \rho = \text{const}+r \sqrt{\frac{r-2 G M}{r}}+G M \left(\log \left(r \sqrt{\frac{r-2 G M}{r}}-G M+r\right)\right), $$

which I invite you to try and invert for $r(\rho)$. :)

In principle it can be inverted and you get the metric

$$\mathrm{d}s^2 = -\left( 1- \frac{2 G M}{r(\rho)} \right) \mathrm{d}t^2 + \mathrm{d}\rho^2 + r(\rho)^2 \mathrm{d}\Omega^2 .$$

This has absorbed the coefficient in front of the $\mathrm{d}r$ at the expense of making the rest of the metric more complicated.

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Thanks very much! –  willie.holdman Mar 4 '13 at 4:05

If your plan is to define a new time variable

$t'=\sqrt{1+\mu /r} ~t$

then when you take the derivative (to see how the metric changes) you will get

$dt=\frac{dt'}{\sqrt{1+\mu /r}}+\frac{1\mu}{2r^2}\frac{dr}{(1+\mu /r)^{3/2}}$

In other words, you will rescale both $t$ and $r$ since the function in front of t is a function of them both, rather then just a function of $t$ as $g(t)$ was. Then you will square $dt$ and get cross-terms $dt'dr$, and your metric will no longer be diagonal.

Not a problem per se, but not something we usually want to happen.

I guess if you really wanted to try and do it you would integrate

$\int dt'=\int \sqrt{1+\mu / r}dt$

but $r(t)$ in general so again, you'd have trouble.

EDIT: You are clearly following some text; check out what they say about Eddington-Finkelstein Coordinates, I think that might illustrate the problem. These coordinates come from a rescaling of the radial coordinate to get rid of the singularity at the event horizon.

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