Cosmic strings

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.6.

Another example of a solution to the Einstein equation is a simple model of a cosmic string. Cosmic strings are postulated objects that are left over from the big bang. They are virtually one-dimensional structures with a radius much smaller than an atomic nucleus, but with lengths of hundreds of thousands of light years. As a model of a cosmic string, suppose we have an infinite, straight string stretching along the {z} axis, and that the string is axially symmetric, that is, that its structure depends only on the radial coordinate {r} measured from the {z} axis. The metric describing the string is a generalization of the cylindrical coordinate system:

\displaystyle ds^{2}=-dt^{2}+dr^{2}+f^{2}\left(r\right)d\phi^{2}+dz^{2} \ \ \ \ \ (1)

For an ordinary cylindrical system in flat space, {f\left(r\right)=r}.

[The interpretation of the {r} coordinate is qualitatively different from the Schwarzschild metric, where we assumed spherical symmetry and used this to write down the angular components of the metric as those that apply in flat space, that is {r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}}. This choice results in the radial coordinate {r} being a circumferential coordinate, in that the circumference of a circle with radial coordinate {r} is {2\pi r}, but the distance from the origin to a point on the circle is not {r}. In the cylindrical metric here, {r} is not a circumferential coordinate because the metric component {g_{\phi\phi}=f^{2}\ne1}, so the circumference of a circle of radius {r} is {2\pi f} (as you can verify by setting {dt=dr=dz=0} and integrating over {\phi} from 0 to {2\pi} for a fixed {r}). However, because {g_{rr}=1}, the {r} coordinate here does represent the actual distance from the {z} axis to a point on a circle with coordinate {r}.]

We take the stress-energy tensor to be

\displaystyle T_{\;t}^{t}=T_{\;z}^{z}=-\sigma\left(r\right) \ \ \ \ \ (2)


[Moore doesn’t explain where these come from, but we’ll just accept this for now.] From the definition of the stress-energy tensor {T^{tt}=-T_{\;t}^{t}=\sigma} is the energy density.

The Einstein equation for a perfect fluid is

\displaystyle R_{\mu\nu}=8\pi G\left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right) \ \ \ \ \ (3)


The scalar {T} is

\displaystyle T=T_{\;\mu}^{\mu}=T_{\;t}^{t}+T_{\;z}^{z}=-2\sigma \ \ \ \ \ (4)

The non-zero components of {T_{\mu\nu}} can be found from 2 by lowering the first index:

\displaystyle T_{tt} \displaystyle = \displaystyle g_{tt}T_{\;t}^{t}=\sigma\ \ \ \ \ (5)
\displaystyle T_{zz} \displaystyle = \displaystyle g_{zz}T_{\;z}^{z}=-\sigma \ \ \ \ \ (6)

From 3 we therefore have

\displaystyle R_{tt} \displaystyle = \displaystyle 8\pi G\left(T_{tt}-\frac{1}{2}g_{tt}T\right)\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle 8\pi G\left(\sigma-\sigma\right)=0\ \ \ \ \ (8)
\displaystyle R_{rr} \displaystyle = \displaystyle 8\pi G\left(0+g_{rr}\sigma\right)\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle 8\pi G\sigma\ \ \ \ \ (10)
\displaystyle R_{\phi\phi} \displaystyle = \displaystyle 8\pi G\left(0+g_{\phi\phi}\sigma\right)\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle 8\pi Gf^{2}\sigma\ \ \ \ \ (12)
\displaystyle R_{zz} \displaystyle = \displaystyle 8\pi G\left(-\sigma+g_{zz}\sigma\right)=0 \ \ \ \ \ (13)

All off-diagonal components of {R_{\mu\nu}} are zero since both {T_{\mu\nu}} and {g_{\mu\nu}} are diagonal. Thus

\displaystyle R_{rr}=\frac{R_{\phi\phi}}{f^{2}} \ \ \ \ \ (14)


Using the Ricci tensor worksheet we can work out {R_{\mu\nu}} in terms of {g_{\mu\nu}}. The only non-zero terms are those involving a derivative of {g_{\phi\phi}} with respect to {r} on its own (that is, not multiplied by some other derivative), or in terms of the notation of the worksheet, those terms involving either {C_{1}} or {C_{11}} on their own. We have (where a subscript 1 indicates a derivative with respect to {r}):

\displaystyle C \displaystyle = \displaystyle f^{2}\ \ \ \ \ (15)
\displaystyle C_{1} \displaystyle = \displaystyle \frac{d\left(f^{2}\right)}{dr}=2ff_{1}\ \ \ \ \ (16)
\displaystyle C_{11} \displaystyle = \displaystyle \frac{d\left(2ff_{1}\right)}{dr}=2f_{1}^{2}+2ff_{11} \ \ \ \ \ (17)

The only components of {R_{\mu\nu}} involving these two derivatives on their own are {R_{rr}} and {R_{\phi\phi}}:

\displaystyle R_{rr} \displaystyle = \displaystyle -\frac{1}{2C}C_{11}+\frac{1}{4C^{2}}C_{1}^{2}\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle -\frac{f_{1}^{2}}{f^{2}}-\frac{f_{11}}{f}+\frac{4f^{2}f_{1}^{2}}{4f^{4}}\ \ \ \ \ (19)
\displaystyle \displaystyle = \displaystyle -\frac{f_{11}}{f}\ \ \ \ \ (20)
\displaystyle R_{\phi\phi} \displaystyle = \displaystyle -f_{1}^{2}-ff_{11}+\frac{4f^{2}f_{1}^{2}}{4f^{2}}\ \ \ \ \ (21)
\displaystyle \displaystyle = \displaystyle -ff_{11}\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle f^{2}R_{rr} \ \ \ \ \ (23)

Thus 14 is satisfied here as well. [Note that there are a couple of errors in Moore’s problem statement – see the errata list here.] Combining 10 and 20 gives

\displaystyle f_{11}=\frac{d^{2}f}{dr^{2}}=-8\pi Gf\left(r\right)\sigma\left(r\right) \ \ \ \ \ (24)


Moore now says that we require the metric to be non-singular at {r=0} (actually he says ‘non-singular at the origin’ although I assume he means ‘non-singular at all points on the {z} axis, since there’s nothing special about {z=0} here). It’s not entirely clear to me why we would require this since the Schwarzschild metric is singular at {r=0}. He also says that this requirement leads to the metric reducing to the flat space metric as {r\rightarrow0}. Again, this isn’t exactly obvious; there are lots of metrics that are finite at {r=0} so why choose flat space? Anyway, let’s plow onwards…

If we require {f\left(r\right)\rightarrow r} as {r\rightarrow0} to give us the flat, cylindrical metric near the {z} axis, then {f_{1}=\frac{df}{dr}\rightarrow1} as {r\rightarrow0}. We can then integrate 24 to give, for points outside the string’s radius {r_{s}}:

\displaystyle f_{1} \displaystyle = \displaystyle -4G\int_{0}^{r_{s}}2\pi f\left(r\right)\sigma\left(r\right)dr+A\ \ \ \ \ (25)
\displaystyle \displaystyle = \displaystyle -4G\mu+A \ \ \ \ \ (26)

where {A} is a constant of integration, {r_{s}} is the radius of the string and

\displaystyle \mu\left(r_{s}\right)\equiv\int_{0}^{r_{s}}2\pi f\left(r\right)\sigma\left(r\right)dr \ \ \ \ \ (27)

is the string’s energy density (energy per unit length). [Recall from above that the circumference of a circle of radius {r} is {2\pi f} and {\sigma} is the energy density (per unit volume), so the energy in a cylindrical shell of radius {r}, thickness {dr} and unit length is {2\pi f\left(r\right)\sigma\left(r\right)dr}.] In the limiting case of no string at all, {r_{s}=0} and the metric reduces to flat space where {f_{1}=1} so {A=1} and we have for {r>r_{s}}:

\displaystyle \frac{df}{dr}=1-4G\mu \ \ \ \ \ (28)

Since {\mu} is a constant for {r>r_{s}}, we can integrate this directly to get

\displaystyle f\left(r\right)=\left(1-4G\mu\right)r+K \ \ \ \ \ (29)

where {K} is a constant of integration. For very small {r} we should have {f\rightarrow r} and since {r_{s}} is very small, we’d expect {\mu} to be small, so {K} would be close to zero.

The resulting metric is

\displaystyle ds^{2}=-dt^{2}+dr^{2}+\left(1-4G\mu\right)^{2}r^{2}d\phi^{2}+dz^{2} \ \ \ \ \ (30)

We can redefine the angular coordinate {\phi} (in a way similar to the redefinition of the time coordinate used in deriving Birkhoff’s theorem) by defining

\displaystyle \tilde{\phi}\equiv\left(1-4G\mu\right)\phi \ \ \ \ \ (31)

to get what appears to be a flat space metric:

\displaystyle ds^{2}=-dt^{2}+dr^{2}+r^{2}d\tilde{\phi}^{2}+dz^{2} \ \ \ \ \ (32)

However, remember that the radial coordinate {r} is not the same as that used in flat space, since the circumference of a circle of radius {r} is given by {2\pi\left(1-4G\mu\right)r}, so is actually slightly smaller than {2\pi r}. Also, the new axial coordinate {\tilde{\phi}} covers {2\pi\left(1-4G\mu\right)<2\pi} for a complete circle.

2 thoughts on “Cosmic strings

  1. Asher Weinerman

    I’m confused by the time coordinate in this problem:
    In Schwarzschild space-time we have dtau^2 = dt^2*(1-2GM/r) so it is easy to see that proper time “tau” changes with distance from M relative to far-away time “t”. But it seems this cosmic string metric doesn’t refer to a far-away time coordinate at all – instead dtau = dt as long as the object is not moving (like SR). I guess that’s ok since the other metric components are time-independent anyway.Maybe to help shed some light on this, how would you compute the time difference as measured by stationary clocks at different “r” from the string (or one clock at r and another at infinity)?


  2. Asher Weinerman

    I think I asked the question too quickly because the answer must be that time is independent of r just like Minkowski space. I resisted that possibility because it seems so unphysical.

    So strings don’t gravitate – very unintuitive – stationary objects near a string don’t accelerate/gravitate towards it at all.. But strings can deflect MOVING objects because of conical geometry, just like global geodesics on either side of the vertex of a cone diverge and then converge again.

    So I guess I answered my own question. Sorry to bother you and thanks for all your efforts to help the rest of us understand.


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