# Equilibrium for a rope hanging in a Schwarzschild spacetime

Update: Trimok and MBN helped me solve most of my confusion. However, there is still an extra term $-(2/r)T$ in the final result. Brown doesn't write this term, and it seems physically wrong.

Update #2: Possible resolution of the remaining issue. See comment on MBN's answer.

Suppose we have a rope hanging statically in a Schwarzschild spacetime. It has constant mass per unit length $\mu$, and we want to find the varying tension $T$. Brown 2012 gives a slightly more general treatment of this, which I'm having trouble understanding. Recapitulating Brown's equations (3)-(5) and specializing them to this situation, I have in Schwarzschild coordinates $(t,r,\theta,\phi)$, with signature $-+++$, the metric

$$ds^2=-f^2 dt^2+f^{-2}dr^2+... \qquad , \text{ where} f=(1-2M/r)^{1/2}$$

and the stress-energy tensor

$$T^\kappa_\nu=(4\pi r^2)^{-1}\operatorname{diag}(-\mu,-T,0,0) \qquad .$$

He says the equation of equilibrium is:

$$\nabla_\kappa T^\kappa_r=0$$

He then says that if you crank the math, the equation of equilibrium becomes something that in my special case is equivalent to

$$T'+(f'/f)(T-\mu)=0 \qquad ,$$

where the primes are derivatives with respect to $r$. This makes sense because in flat spacetime, $f'=0$, and $T$ is a constant. The Newtonian limit also makes sense, because $f'$ is the gravitational field, and $T-\mu\rightarrow -\mu$.

There are at least two things I don't understand here.

First, isn't his equation of equilibrium simply a statement of conservation of energy-momentum, which would be valid regardless of whether the rope was in equilibrium?

Second, I don't understand how he gets the final differential equation for $T$. Since the upper-lower-index stress-energy tensor is diagonal, the only term in the equation of equilibrium is $\nabla_r T^r_r=0$, which means $\mu$ can't come in. Also, if I write out the covariant derivative in terms of the partial derivative and Christoffel symbols (the relevant one being $\Gamma^r_{rr}=-m/r(r-2m)$), the two Christoffel-symbol terms cancel, so I get

$$\nabla_r T^r_r = \partial _r T^r_r + \Gamma^r_{rr} T^r_r - \Gamma^r_{rr} T^r_r \qquad ,$$

which doesn't involve $f$ and is obviously wrong if I set it equal to 0.

What am I misunderstanding here?

References

Brown, "Tensile Strength and the Mining of Black Holes," http://arxiv.org/abs/1207.3342

• Brown does not use a Schwarzschild black hole, but a general static spherically-symmetric spacetime $(3)$, which has for source, a general static spherically symmetric matter distribution given by the stress-energy tensor $(4)$. So $\chi(r)$ and $f(r)$ depends on $T$ (and $\mu$), but, yes, this "equation of equilibrium" is nothing that the usual "conservation" of the stress-energy tensor. – Trimok Sep 15 '13 at 15:57
• @Trimok: He does a treatment that becomes Schwarzschild when $\chi=f$, which is the special case I present above. – Ben Crowell Sep 15 '13 at 19:31
• If it is, then the $T^\kappa_\nu$ cannot be the source of the gravitational field. – Trimok Sep 16 '13 at 8:52
• @Trimok: Right, the $T^\kappa_\nu$ being discussed here is the stress-energy tensor of the rope, not of the gravitating body. – Ben Crowell Sep 16 '13 at 19:08
• Related: physics.stackexchange.com/q/104474/2451 and links therein. – Qmechanic Apr 27 '16 at 9:01

From that you have $\nabla_r T^r_r=T'$, but there is also

$\nabla_t T^t_r=\frac{\partial}{\partial t}T^t_r+\Gamma^t_{\alpha t}T^\alpha_r-\Gamma^\alpha_{r t}T^t_\alpha=\Gamma^t_{r t}T^r_r-\Gamma^t_{r t}T^t_t=-\Gamma^t_{rt}(T-\mu)$.

So

$\nabla_k T^k_r=\nabla_rT^r_r+\nabla_tT^t_r=-T'-(f'/f)(T-\mu).$

This should be a comment, but the symbols didn't work.

My guess is that it is called equation of equilibrium because the T is the stress energy of the rope, not a stress energy that affects the space time geometry. The background is fixed and the rope lives on it.

• I think there is an other problem : we forget the $r^{-2}$ term in the definition of the stress-energy tensor, and this a problem in the derivation of $\partial _r T^r_r$ – Trimok Sep 15 '13 at 15:49
• The notations can be misleading it really is $(\nabla_i T)^j_k$. But Trimok has pointed out that I missed the $r^{-2}$. I thought it was just constants in the front. – MBN Sep 15 '13 at 20:11
• With the notations of Fouxon, diagonal $T^\kappa_\nu$ without $r^{-2}$ terms, we will have : $\nabla_k T^k_r=S'+(f'/f)(S + \rho) = 0.$ – Trimok Sep 16 '13 at 9:17
• @Trimok: Aha! I finally understand what's going on with the $-(2/r)T$ term. Not only did I leave out the $\nabla_t T^t_r$ term, but none of us realized that there were also terms $\nabla_\phi T^\phi_r$ and $\nabla_\theta T^\theta_r$. These end up canceling the $-(2/r)T$. – Ben Crowell Sep 17 '13 at 0:36
• @BenCrowell : Yes, you are right. I check that too. – Trimok Sep 17 '13 at 7:02