[Physics] Metric for infinite straight cosmic string

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A string theory question on my general relativity problem set:

Metric is given as $$\mathrm{d}s^2 = -A(r)\mathrm{d}t^2 + B(r)\mathrm{d}r^2 + r^2 \mathrm{d}\theta^2.$$

a) Solve the vacuum equations and find the metric around the massive source assuming that the Ricci curvature vanishes. Use a change of coordinates to put the metric in the form,

$$ \mathrm{d}s^2 = -\mathrm{d}t^2 + \mathrm{d}r^2 + C(r)r^2 \mathrm{d} \theta^2$$

b) Consider the metric on a spatial slice ($t$ = constant). Use the relationship between curvature and parallel transport to determine the total curvature that must be contained in the region containing the source if parallel transport around a single cosmic string rotates a vector by $2\pi/N$.

c) Write the metric at a large distance from a configuration of $k$ sources, assuming each separately gives rise to the metric described in the previous part. What is the maximum number of parallel such cosmic strings that can inhabit space-time? With this maximum number of cosmic strings, what is the total curvature over a 2D surface perpendicular to the strings?

–I attempted the first part, found all connections and Ricci tensor components. Requiring that the Ricci tensor be zero. I attempted to get what seemed to be a mess of algebra to solve for $B(r)$ and $A(r)$. But I didn't really get it to work. Regardless, I know that in the form C(r) is given by a few books to be something of the form $1-4\mu$, whose interpretation as canonical spacetime makes enough sense to me. The latter two parts are where I'm really at a loss.

Best Answer

We will work with the metric supplied,

$$\mathrm{d}s^2 = A(r)^2\mathrm{d}t^2 -B(r)^2 \mathrm{d}r^2 -r^2 \mathrm{d}\theta^2 - r^2 \sin^2\theta \, \mathrm{d}\phi^2$$

I will assume that omitting the fourth spatial coordinate (which I have added) is simply a typo in the original post. I have also redefined the arbitrary functions for convenience. We choose a basis,

$$e^{t} = A(r)\mathrm{d}t, \, \, \, e^{r}=B(r)\mathrm{d}r, \, \, \, e^{\theta} = r\mathrm{d}\theta, \, \, \, e^{\phi} = r\sin \theta \, \mathrm{d}\phi$$

We now compute the exterior derivative of all $e^{a}$, and re-express them in terms of the basis,

$$\mathrm{d}e^{t} = -\frac{A'}{AB} e^{t} \wedge e^{r}, \, \, \, \mathrm{d}e^{r} = 0, \, \, \, \mathrm{d}e^{\theta}=-\frac{1}{rB} e^{\theta} \wedge e^{r}, \, \, \, \mathrm{d}e^{\phi} =-\frac{1}{rB}e^{\phi}\wedge e^{r} - \frac{\cot \theta}{r} e^{\phi}\wedge e^{\theta}$$

The spin connection $\omega^{a}_{b}$ can be read off from Cartan's first equation,

$$\mathrm{d}e^{a} + \omega^{a}_{b}\wedge e^{b} = 0$$

As the exterior derivatives are already in terms of the basis, this is fairly straightforward:

$$\omega^t_r = \frac{A'}{AB} e^{t},\, \, \, \omega^{\theta}_{r} = \frac{1}{rB} e^{\theta}, \, \, \, \omega^{\phi}_{r} = \frac{1}{rB}e^{\phi}, \, \, \, \omega^{\phi}_{\theta} = \frac{\cot \theta}{r}e^{\phi}$$

All other connection components presumably vanish. To compute the curvature 2-form, $R^{a}_{b}$, the second Cartan equation is required,

$$R^{a}_{b} = \mathrm{d}\omega^{a}_{b} + \omega^{a}_{c}\wedge \omega^{c}_{b}$$

An example of a particular component,

$$R^{t}_{r}= -\left( \frac{A''B -A'B'}{AB^3}\right) e^{t}\wedge e^{r}$$

Often $\omega^{a}_{c}\wedge \omega^{c}_{b}$ will simply be zero because of the few non-zero spin connections. After you have found all the components of the 2-form, the Riemann tensor is given by,

$$R^{a}_{b} = \frac{1}{2}R^{a}_{bcd} e^{c}\wedge e^{d}$$

For our example, this implies,

$$R^{t}_{r t r} = -2\left( \frac{A''B -A'B'}{AB^3}\right)$$

Now recall that thus far we have been working in a particular orthonormal basis. To find the Riemann tensor in the coordinate basis, we use the relation,

$$R^{\lambda}_{\mu \nu \sigma}=(e^{-1})^{\lambda}_{a} R^{a}_{bcd} e^{b}_{\mu}e^{c}_{\nu}e^{d}_{\sigma}$$

where the l.h.s the curvature is in the coordinate basis, and $e^{a}_\mu \mathrm{d}x^\mu = e^{a}$. As you are aware, the Ricci tensor $R_{\mu \nu} = R^{\lambda}_{\mu \lambda \nu}$, which you can then use to solve $R_{\mu \nu} = 0$ which should, hopefully, be straightforward given this is a homework problem.


If you are unfamiliar with the Cartan formalism, see the gravitational physics lectures at http://perimeterscholars.org/, at approximately lecture 3-4.