1
$\begingroup$

The Einstein's Field Equation is $$R_{\mu\nu}-(1/2)g_{\mu\nu}R=-8\pi T_{\mu\nu},$$ where the left hand side is the curvature term and the right hand side is the source term (see, Hartle). Now, in the case of Schwarzschild Exterior Solution (solution for empty space-time outside of a spherical body) we are taking the energy-momentum tensor $T_{\mu\nu}$ to be zero as there is no mass or energy outside the spherical body; and we are writing the field equation as $$R_{\mu\nu}=0.$$ My confusion is that what is the source of curvature here? Undoubtedly, here the source of curvature is the spherical body which have non-zero $T_{\mu\nu}$. So, does not the energy-momentum tensor refer to the source of curvature? What does it refer to then? Please clarify the meaning of Einstein's Field Equation in terms of source and its effects on space-time curvature.

$\endgroup$
  • 2
    $\begingroup$ The source of the curvature would be the spherically symmetric central body. If we are outside of the body, then there is no matter source, but that doesn't mean that there is no matter anywhere. $\endgroup$ – Jerry Schirmer Feb 27 '15 at 16:34
  • $\begingroup$ Side note, with $T_{\mu\nu}=0$, you get $R_{\mu\nu}=\frac{1}{2}Rg_{\mu\nu}$, which only equals zero iff $R=R^{\mu}_{\,\mu}=0$ $\endgroup$ – Jim Feb 27 '15 at 16:46
  • $\begingroup$ Sorry, but you did not get me. I haven't said that there is no matter anywhere; rather I said there in no matter outside the spherical body. And if the spherical body is the source of space-time curvature then why we don't take the energy-momentum tensor of that body in the field equation? I may have misconception about what is meant by $T_{\mu\nu}$ in field equation? please clarify. @Jerry $\endgroup$ – VacuuM Feb 27 '15 at 16:50
  • $\begingroup$ Sorry, but the condition $T_{\mu\nu}=0$ automatically, implies that $R=R_{\mu}^{\mu}=0$, that's why we can write, $R_{\mu\nu}=0$ @Jimdalf $\endgroup$ – VacuuM Feb 27 '15 at 16:54
  • $\begingroup$ Remember that curvature has to be continuous. Spacetime is supposed to be a manifold. So when you get outside the body, the source of curvature in a given region is the curvature on the boundaries of that region. Infinitely far away from the body, spacetime is flat an Minkowski. So each point in between has to continuously adapt the curvature from what it is inside the body to what it is at infinity $\endgroup$ – Jim Feb 27 '15 at 16:59
2
$\begingroup$

We first note that the vanishing of the Ricci tensor does not imply the vanishing of the Riemann tensor. Thus the vacuum equations, $R_{\mu\nu}=0$, do not imply that spacetime is flat. The vacuum equations tell us that certain linear combination of components of the Riemann tensor vanishes.

When solving differential equations, one usually has to worry about the boundary conditions. The EFEs are just differential equations of the metric, to bo be solved for the components $g_{\mu\nu}$. Let $D$ be a region of spacetime containing matter, i.e. $T|_D\ne 0$. General relativity seeks to solve the problem

$$ \begin{cases} G_{\mu\nu}(x) = 8 \pi T_{\mu\nu}(x) & x\in D \\ R_{\mu\nu}(x) = 0 & x\notin D \end{cases} $$ We have two solutions: $$ \begin{cases} \tilde{g}_{\mu\nu}(x) & x\in D \\ \bar{g}_{\mu\nu}(x) & x\notin D \end{cases} $$ The relevant boundary conditions are $$\tilde{g}_{\mu\nu}(p)=\bar{g}_{\mu\nu}(p),\quad\forall p\in \partial D.$$ Furthermore, we require that this be a smooth transition. The metric of spacetime is thus $$g_{\mu\nu}=\chi_D \tilde{g}_{\mu\nu}+(1-\chi_D)\bar{g}_{\mu\nu},$$ where $\chi_D$ is the characteristic function of $D$.

Let us consider a concrete example. We shall look at a spherically symmetric static star$^{1}$. By the Birkhoff theorem$^2$, we already know that the exterior solution is the Schwarzschild solution. We take the star to have a perfect fluid energy-momentum tensor in the interior. In the interior, the metric will still have spherical symmetry because of the properties of perfect fluids. Thus we must find $A$ and $B$ in $$\mathrm{d}s^2=-B(r)\mathrm{d}t^2+A(r)\mathrm{d}r^2+r^2\mathrm{d}\Omega^2.$$

The solution for $A(r)$ is $$A(r)=\left[1-\frac{2G\mathcal{M}(r)}{r}\right]^{-1},\quad\mathcal{M}(r)=\int^r \rho \,\mathrm{d}V=\int_0^r 4\pi r'^2\rho(r')\,\mathrm{d}r.$$ The solution for $B(r)$ is $$B(r)=\exp\left\{-\int_r^\infty \frac{2G}{r'^2}[\mathcal{M}(r')+4\pi r'^3 P(r')]A(r')\,\mathrm{d}r'\right\}.$$ In these equations $\rho$ is the density of the star and $P$ its pressure.

Clearly, the support$^3$ of the density and pressure is contained within the star. If the star has radius $R$, then $\mathcal{M}(R)=M$ is the total mass. Let us analyze our solution at $r=R$. For $A(r)$ we get just the standard Schwarzschild metric component. A simple exercise in calculus is to check that $$B(R)=A^{-1}(R)$$ and that $B(r)=A^{-1}(r)$ for $r\ge R$. We thus see the above general situation: we have an interior solution and an exterior vacuum solution, which are equal at the boundary of the source's support.


$^1$ See [1] section 11.1 or [2] section 6.2 for the full calculation.

$^2$ Proved in [1] on page 337.

$^3$ A quantity $f$ is said to have support in $X$ if $f(x)=0$ for all $x\notin X$.

References:

[1] S. Weinberg, Gravitation and Cosmology (1972).

[2] R. M. Wald, General Relativity (1984).

$\endgroup$
3
$\begingroup$

The fact that the energy-momentum tensor is called the source of curvature doesn't mean that there can't be any curvature where there is no energy-momentum. In fact, even if $T_{\mu\nu}=0$ across all spacetime, there are still nontrivial solutions of Einstein's equations, in the form of gravitational waves.

You should remember that $T_{\mu\nu}$ is a function of $x$, and can be nonzero inside a body but zero outside. Suppose we have a spherically symmetric star. Inside of it, the energy momentum tensor will be a very complicated object, with all sorts of pressures and flows and whatever. But outside of it, $T_{\mu\nu}$ will be simply zero. So we have divided space into two regions:

$$ \begin{cases} R_{\mu\nu}-\frac12 g_{\mu\nu} R = T_{\mu\nu} & \text{Inside} \\ R_{\mu\nu} = 0 & \text{Outside} \end{cases} $$

When we solve the outside equation assuming spherical symmetry and a static solution, we get Schwarzchild's solution, and we find that we don't need to know the metric inside the star in detail. All we need is a single number, $M$, that tells us everything we need to konw about the curvature outside of the star.

The Schwarzchild metric satisfies $R_{\mu\nu}=0$, because outside of the star (or black hole or whatever), there are no sources. But when we solve an differential equation, we need boundary conditions. And the boundary conditions at the star's surface (or at $r=0$ for a black hole) are the energy-momentum tensor's way of telling us that there is matter somewhere, even if the region where we solved the equation is empty.

$\endgroup$
  • $\begingroup$ So, is it like that both $R_{\mu\nu}-(1/2)g_{\mu\nu}R$ and $T_{\mu\nu}$ are functions of $(x,t)$ and we have to consider what is the value of $R_{\mu\nu}-(1/2)g_{\mu\nu}R$ and $T_{\mu\nu}$ at the same event and then equate them. @Javier Badia $\endgroup$ – VacuuM Feb 27 '15 at 17:39
  • $\begingroup$ @VikramadityaMondal: Yes, that's exactly it. Everything that we're considering here is a function of $x^\mu$, and Einstein's equations are partial differential equations beteen tensors that are functions of space and time. $\endgroup$ – Javier Feb 27 '15 at 17:50
0
$\begingroup$

The equations $R_{\mu\nu} = 0$ alone don't define a well-posed problem since you need to add boundary conditions. Moreover the curvature of a manifold is measured by the Riemann tensor and $R_{\mu\nu}=0$ doesn't imply ${R^{\mu}}_{\nu\sigma\rho} = 0$.

$\endgroup$
0
$\begingroup$

It's helpful to look at an analogous situation in Classical Electrodynamics where the same issues come up.

In General Relativity, the source term is the stress-energy tensor. In Classical Electrodynamics the source terms are the charge and the current. When there are no sources, one possible vacuum solution is a constant uniform nonzero electric field. Another possible vacuum solution is a constant uniform zero electric field. We can piece together vacuum solution and sew them together, and this will no longer be a vacuum solution. But for instance if you take a field uniform in the $\hat{x}$ direction $\vec{E}=E\hat{x}$ from $x=0$ to $x=1m$, and piece it together with $\vec{E}=\vec{0}$ for $x<0$ and $\vec{E}=\vec{0}$ for $x> 1m$, then this can still be a solution to Maxwell, with a source term. In this case you need an infinite sheet of surface charge in the $x=0$ plane, and an equal and opposite surface charge on an infinite sheet in the $x=1m$ plane.

So let's try this with General Relativity, you can take two Schwarzschild solutions for different mass parameters, $M$ and $m$, with $M$>$m$. As embeddings they might look like funnels. If you a surface of constant areal coordinate, you get a thin spherical shell, pick a surface with surface area $4\pi R^2$. For the solution with $M$ remove the inside interior to the shell. Repeat for the solution with $m$ find the thin shell with the exact same surface area $4\pi R^2$ (with the same numerical value of $R$) and this time remove the outside exterior to the shell. So the one looks like a funnel with the tip cut off, and the other looks like a tip to a funnel. Since they have the same surface area and are just spherical shells, we can sew them together. We now have a manifold that looks like a funnel with a kink.

Each vacuum solution was a source free solution, and we sewed them together in a surface where the geometry of the surface matched. But the result (just like for the electrodynamics case) is not a source free solution. In fact this is a solution where there is a thin shell of mass at the surface area $4\pi R^2$ spherical shell surface. It is OK for spacetime to be curved (just like it is OK to have a travelling wave in electromagnetism or a constant electric field). But only certain curvature is allowed, what a source term does is allow different kinds of curvature.

And this is not contrived in any way. If you repeated that trick with the two solutions you could cut out the inside of the $m$ solution and put a $\mu$ solution inside it and then have a solution with two places with source, one at surface $4\pi R_1^2$ and another at $4\pi R_2^2$. As you place shells at more and more radii you can approach the solution to a realistic nonrotating star or planet. All the source term does is allow you to have a continuous limit of this process, the process of sewing vacuum solutions together.

Think of $G_{\mu \nu}=0$ as a natural allowed way for spacetime to curve (just as vacuum waves or static fields are allowed electromagnetic fields) and that curvature is allowed to deviate from these natural allowed curvatures as long as there is a source term there to give the appropriate OK. Just like an electric field line isn't allowed to terminate unless there is a charge there to approve that message.

$\endgroup$

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.