3
$\begingroup$

Often in the literature, when deriving the famous Schwarzschild solution the metric can be cast in two equivalent ways:

$$(ds)^2 = e^{2\alpha(r,t)} dt^2 - e^{2\beta(r,t)}dr^2 - r^2 d\Omega^2. $$

$$ (ds)^2 = A(r,t)dt^2 - B(r,t)dr^2 - r^2 d\Omega^2. $$

However, I don't necessarily understand why we can write it in the exponential form? There is obviously a very simple answer. Is it something to do with the unknown function never being identically zero... or at least something along these lines?

UPDATE (post comment(s))

In response to the comment by ArthurSuvorov. The exponential function is never negative for real input and will keep the manifold Lorentzian. This makes sense. However, what about if we write it in the second form? There isn't conditions on these unknown functions initially? Or is there?

$\endgroup$
2
$\begingroup$

For the Schwarzschild solution it is convenient to do so as @Arthur Suvorov pointed out. But in general the second non-exponential ansatz can also be used. Asymptotic flatness $$\lim_{r\rightarrow\infty}A(r,t)=1$$ $$\lim_{r\rightarrow\infty}B(r,t)=1$$ poses two boundary conditions for $A$ and $B$ (or $\alpha$ and $\beta$). The field equations or in general the static nature of the problem implies that $A$ and $B$ are functions of $r$ only. One can find the Schwarzschild solution using $A$ and $B$ but $\alpha$ and $\beta$ are nicer to work with.

In fact it is not uncommon to use such an exponential ansatz since they often make computations a lot simpler. Inside a spherical symmetric body the exponential form of the $g_{tt}$ becomes very important: Its radial derivative is directly related to the fluids $\log$-enthalpy gradient:

$$\frac{P'(r)}{\rho(r)+P(r)}\equiv h'(r)=-\frac{1}{2}\frac{\partial \ln(g_{tt}(r))}{\partial r}.$$

This is the general relativistic extension of the classical Bernoulli theorem (in the special case of a static, spherical symmetric ideal fluid interior). In the classical limit $\frac{1}{2}\frac{\partial \ln(g_{tt}(r))}{\partial r}$ becomes the classical Newtonian gravitational potential. So in some cases it makes more sense to use the exponential forms apart from pure mathematical convenience.

$\endgroup$
  • $\begingroup$ Quite informative. Good answer. Although I've seen the same conditions in an undergrad book ( name to come, currently away from computer ) that says in the limit the unknown temporal function should tend to $c^2$? $\endgroup$ – Rumplestillskin May 26 '17 at 9:30
  • 1
    $\begingroup$ Yes sorry I have not mentioned I was using geometrized units with $c=G=1$. The $g_{tt}$ metric potential should go to $c^2$ if the constant is not absorbed in the definition of the of the temporal coordinate. $\endgroup$ – N0va May 26 '17 at 15:07
4
$\begingroup$

The exponential function is never negative for real input, and one wants to keep the signature of the metric the same [(+,-,-,-) in your case] for all values of $t$ and $r$. Writing the coefficients as exponentials enforces this without the need to worry about domains and ranges so much.

$\endgroup$
  • $\begingroup$ I've updated the question after your helpful comment. $\endgroup$ – Rumplestillskin May 26 '17 at 2:09

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.