I am looking at Wikipedia's article on deriving the Schwarzschild solution. In the section "Simplifying the components", it says,

On the hypersurfaces of constant $t$ and constant $r$, it is required that the metric be that of a 2-sphere:

$$dl^2=r^2(d\theta^2+\sin^2\theta d\phi^2)$$

My question is why does the metric have to be this particular 2-sphere with a coefficient of $r^2$? We are not necessarily dealing with Euclidean space here.

  • $\begingroup$ The Schwarzschild solution is spherically symmetric by definition. $\endgroup$ – Tom-Tom Dec 20 '13 at 17:25
  • $\begingroup$ @V.Rossetto, Yes, that is correct, but why does the coefficient have to be $r^2$? We are not necessarily dealing with Euclidean space. $\endgroup$ – Craig Feinstein Dec 20 '13 at 17:31
  • $\begingroup$ Isn't that the way the $r$ coordinate is defined? It's the circumference divided by $2\pi$. $\endgroup$ – John Rennie Dec 20 '13 at 17:48

If you want, you can go and use the ansatz:

$$ds^{2} = -A(r) dt^{2} + B(r) dr^{2} + 2C(r)\,dt\,dr + f(r)\left(d\theta^{2} + \sin^{2}\theta d\phi^{2}\right)$$

Where the functions only depend on $r$ due to the fact that $t$ generates a symmetry of the spacetime -- you are assuming a static spacetime.

Note, however, that you are free to arbitrarily rescale $r$. Well, if you choose $R = \sqrt{f(r)}$, then this is rewritten in the form

$$ds^{2} = -A'(R) dt^{2} + {\hat B}(R) dR^{2} + 2C'(R)\,dt\,dr + R^{2}\left(d\theta^{2} + \sin^{2}\theta d\phi^{2}\right)$$

Where the prime denotes that you have to feed $f^{-1}(R^{2})$ into the function, and ${\hat B} = B'(R)\left(\frac{dR}{dr}\right)^{2}$. To get rid of the $C$ term, you can make the definition:

$t = T + \int\,dR\,C'$

Which gives

$$ds^{2} = -(A'(R)-C'(R))dT^{2} + dR^{2}\left({\hat B(R)}+(1-A(R))(C'(R))^{2}\right) + R^{2}\left(d\theta^{2} + \sin^{2}\theta d\phi^{2}\right)$$

So, we can just redefine our factors, and we get the "standard" starting form of

$$ds^{2} = - \alpha(R)dT^{2} + \beta(R)dR^{2} + R^{2}\left(d\theta^{2} + \sin^{2}\theta d\phi^{2}\right)$$

without losing any generality.

| cite | improve this answer | |
  • $\begingroup$ Shouldn't it be $dR$ instead of $dr$ ? $\endgroup$ – Adam Dec 20 '13 at 17:45
  • $\begingroup$ @Adam: yes, see edit. $\endgroup$ – Jerry Schirmer Dec 20 '13 at 17:51
  • $\begingroup$ How do you know that $f$ has an inverse? $\endgroup$ – Craig Feinstein Dec 20 '13 at 18:03
  • 1
    $\begingroup$ @CraigFeinstein: If require $f(r)$ to be monotonically (strictly) increasing, which is a sensible requirement to have if you're using to to label nested $2$-spheres, then $f$ must have an inverse. $\endgroup$ – Stan Liou Dec 20 '13 at 23:29
  • $\begingroup$ But this only proves that $f$ is one-to-one. It doesn't prove that it is onto, which is required if $f$ is to have an inverse. In fact, if you look at the original Schwarzschild paper arxiv.org/pdf/physics/9905030v1.pdf you'll see that $R=(r^3+\alpha^3)^{1/3}$, where $\alpha$ is defined in their paper, so $R$ cannot be less than $\alpha$. So $f$ is not an onto function. $\endgroup$ – Craig Feinstein Dec 22 '13 at 14:55

There is no mention of Euclidean space here, because the $2$-spheres are not Euclidean. They're not going to be embedded in Euclidean space either. The Schwarzschild $r$ coordinate is defined so as to make the areas of the family of $2$-spheres $4\pi r^2$. In other words, we're simply labeling the $2$-spheres by the their areas.

Euclidean space would require a certain relationship between the area, volume, and the radius--as in, the distance from the center to the points on the sphere. But our $r$ is not the radius in that sense! It's simply a label for the $2$-spheres that sorts them by area.

Although you can obtain the metric $\mathrm{d}\Omega^2 = \mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2$ for a unit $2$-sphere by first considering it embedded in Euclidean $3$-space in spherical coordinates, once you do that there is nothing preventing you from considering its intrinsic geometry without reference to any Euclidean space. So, for example, the area element is given intrinsically $\mathrm{d}A = \sqrt{g}\,\mathrm{d}\theta\,\mathrm{d}\phi = \sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$ and the total area is $4\pi$.

Thus, it should be clear that a $2$-sphere of area $A$ can be described by the metric $$\mathrm{d}\Omega^2 = \frac{A}{4\pi}\left[\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2\right]\text{.}$$ But we are free too relabel $A/4\pi$ any way we like. There is no intrinsic difference between taking, say, $e^r\,\mathrm{d}\Omega^2$, $\tan^2 r\,\mathrm{d}\Omega^2$, and $r^2\,\mathrm{d}\Omega^2$: they will give different $r$-coordinates that have different relationships between $r$ and and the area of the $2$-sphere. That we choose the Schwarzschild $r$ to have that relationship be $A = 4\pi r^2$ is done simply for convenience.

But area is a quantity that is defined in terms of the coordinates x,y,z which describe the space. You can't just make up a quantity called area without defining it in terms of the coordinates x,y,z.

That's not quite correct. Let's say you have a metric on a $2$-manifold in orthogonal coordinates, $$ds^2 = g_{11}\,\mathrm{d}u^2 + g_{22}\,\mathrm{d}v^2\text{.}$$ That tells you that if you go an infinitesimal coordinate interval $\mathrm{d}u$ along the $u$ coordinate direction (and none along $v$), the length $\mathrm{d}s$ will be $\sqrt{g_{11}}\,\mathrm{d}u$. Similarly, going some infinitesimal coordinate interval $\mathrm{d}v$ along the $v$-direction only, the length will be $\sqrt{g_{22}}\,\mathrm{d}v$.

Since the coordinates are orthogonal, that defines an infinitesimal rectangle of area that's the product of its side lengths, i.e., $$\mathrm{d}A = \sqrt{g_{11}g_{22}}\,\mathrm{d}u\,\mathrm{d}v\text{.}$$ In general, the area element (or $n$-volume element on a manifold of dimension $n$) in arbitrary coordinates will have a factor that's the determinant of the metric, but we don't have to consider the more general case because our coordinates are orthogonal (no cross-terms $\mathrm{d}u\,\mathrm{d}v$ in the metric).

Note that we did this without mention of any Euclidean space or requiring our coordinates to be Cartesian. As an example for Euclidean $2$-space, take the polar coordinates $$\mathrm{d}s^2 = \mathrm{d}r^2 + r^2\,\mathrm{d}\theta^2\text{,}$$ which our formula immediately gives $$\mathrm{d}A = r\,\mathrm{d}r\,\mathrm{d}\theta\text{,}$$ which is the correct area element of the Euclidean $2$-space in polar coordinates. Even for the Euclidean $2$-space, we didn't need to start with Cartesian coordinates, although we could have done that too.

| cite | improve this answer | |
  • $\begingroup$ But area is a quantity that is defined in terms of the coordinates x,y,z which describe the space. You can't just make up a quantity called area without defining it in terms of the coordinates x,y,z. $\endgroup$ – Craig Feinstein Dec 20 '13 at 20:22
  • $\begingroup$ @CraigFeinstein: That's not right; I edited in some background info that hopefully will make it clearer. That the area of the unit $2$-sphere is $4\pi$ follows from $A = \int\int \mathrm{d}A$ with appropriate limits of integration. $\endgroup$ – Stan Liou Dec 20 '13 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.