0
$\begingroup$

I am learning about the Schwarzchild metric $g_{\mu\nu}(x)$ for the spacetime geometry outside a spherically symmetric source with mass $M$.

In the book by Cheng, a spherical coordinate system $(t,r,\theta,\phi)$ is used and the spherically symmetric metric tensor $g_{\mu\nu} $is defined as $$g_{\mu\nu}=diag(g_{00}, g_{rr}, r^2,r^2\sin^2\theta ),$$ where $g_{00}(r,t)$ and $g_{rr}(r,t)$ are two unknown scalar functions.

It then proceeded to say that $$ds^2=g_{00}(r,t)c^2dt^2+g_{rr}(r,t)dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2).$$ Also, the author reexpressed $g_{00}$ and $g_{rr}$ as $$g_{00}= \frac{1}{g^{00}}=-e^{v(r,t)}$$ $$g_{rr}= \frac{1}{g^{rr}}=e^{\rho(r,t)}$$ and said that the unknown metric functions are now $v(r,t)$ and $\rho(r,t)$.

I have two questions:

  1. Why did a factor of $c^2$ appear in the expression for $ds^2$? I learned that $ds^2=g_{\mu\nu}dx^\mu dx^u$. There isn't a factor of $c$ present anywhere in $g_{\mu\nu}dx^\mu dx^u$.
  2. What is the rule used to state $ \frac{1}{g^{00}}=-e^{v(r,t)} $ and $\frac{1}{g^{rr}}=e^{\rho(r,t)}$ ? Why do they have opposite signs?

References

Cheng, Relativity, Gravitation, and Cosmology

$\endgroup$
1
  • 1
    $\begingroup$ The c^2 is the constant of proportionality, it is just a dimensional correction for the units of the metric, here in the length squared definition. $\endgroup$ – Gareth Meredith Dec 16 '19 at 15:32
3
$\begingroup$
  1. The $c^2$ is included for dimensional reasons, one wants to interpret the line element, $ds^2$, as a distance squared, therefore, (speed of light $\times$ time interval) produces a distance to be added/subtracted along with spatial quantities.

  2. The second question can be answered if you realize that the metric by construction has to fulfil the equation: $$g^{\mu\beta}g_{\beta\nu} = \delta^\mu_\nu$$ so that if you have that the metric is diagonal, then the metric components with the indices raised, the inverse, corresponds to inverting the diagonal. That explains the first equal sign.

One is free to choose the form of these components, however one can add the additional physical input of the signature. Because of the causality structure coming from special relativity, one must ensure that in a certain frame your metric agrees with the Minkowski metric, $\text{diag}(-,+,+,+)$. This implies that signs are fixed in principle (I'm ignoring Black holes for now) which explains the exponential, since any real (continuous) function that does not change signs can be written as such. Following the conventions of GR, one chooses the time component to be negative and the rest are positive which are the signs that you wrote.

$\endgroup$
1
  • $\begingroup$ Thanks for the clear explanation. $\endgroup$ – TaeNyFan Dec 16 '19 at 16:20

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.