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# Metric tensor for the stereographic projection of a sphere onto a plane

Reference: Hobson, M.P., Efstathiou, G. P. & Lasenby, A. N. (2006), General Relativity: An Introduction for Physicists; Cambridge University Press. Problem 2.4.

As another example of finding the metric tensor in a new coordinate system, consider the stereographic projection of a sphere of radius ${a}$ (problem 2.4 in Hobson’s book should begin “Consider the surface of a 2-sphere…). The projection is done by drawing lines from the south pole through every point on the sphere and extending each line until it intersects the plane tangent to the north pole. The location of the point on the tangent plane is expressed using polar coordinates, with ${\rho}$ being the distance from the north pole and ${\phi}$ being the same azimuthal angle as is used to define the original longitude on the sphere.

In order to get the line element on the tangent plane, we need the transformation equations for the coordinates (see diagram, which shows a cross-section of the sphere).

The larger right angled triangle with ${\rho}$ as its top edge is similar to the smaller right angled triangle with top edge ${a\sin\theta}$ (with ${\theta}$ being the usual spherical polar angle), so we have, by taking the ratio of the two sides opposite the hypotenuse in each triangle:

$\displaystyle \frac{\rho}{2a}=\frac{a\sin\theta}{a\left(1+\cos\theta\right)}=\frac{\sin\theta}{1+\cos\theta} \ \ \ \ \ (1)$

Solving for ${\cos\theta}$ we have

 $\displaystyle \frac{\rho}{2a}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{1-\cos^{2}\theta}}{1+\cos\theta}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\ \ \ \ \ (3)$ $\displaystyle \frac{\rho^{2}}{4a^{2}}$ $\displaystyle =$ $\displaystyle \frac{1-\cos\theta}{1+\cos\theta}\ \ \ \ \ (4)$ $\displaystyle \cos\theta$ $\displaystyle =$ $\displaystyle \frac{1-\rho^{2}/4a^{2}}{1+\rho^{2}/4a^{2}} \ \ \ \ \ (5)$

To transform the metric, we use the formula

$\displaystyle g_{cd}^{\prime}=g_{ab}\frac{\partial x^{a}}{\partial x^{\prime c}}\frac{\partial x^{b}}{\partial x^{\prime d}} \ \ \ \ \ (6)$

The spherical coordinate line element is

$\displaystyle ds^{2}=a^{2}d\theta^{2}+a^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (7)$

From 1, we can take the derivative with respect to ${\rho}$ to get

 $\displaystyle \frac{1}{2a}$ $\displaystyle =$ $\displaystyle \frac{\cos\theta\left(1+\cos\theta\right)-\sin\theta\left(-\sin\theta\right)}{\left(1+\cos\theta\right)^{2}}\frac{\partial\theta}{\partial\rho}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(1+\cos\theta\right)}{\left(1+\cos\theta\right)^{2}}\frac{\partial\theta}{\partial\rho}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1+\cos\theta}\frac{\partial\theta}{\partial\rho}\ \ \ \ \ (10)$ $\displaystyle \frac{\partial\theta}{\partial\rho}$ $\displaystyle =$ $\displaystyle \frac{1+\cos\theta}{2a}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1/a}{1+\rho^{2}/4a^{2}} \ \ \ \ \ (12)$

using 5 to get the last line. Therefore

 $\displaystyle g_{\rho\rho}^{\prime}$ $\displaystyle =$ $\displaystyle g_{\theta\theta}\left(\frac{\partial\theta}{\partial\rho}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial\rho}\right)^{2}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\left[\frac{1/a}{1+\rho^{2}/4a^{2}}\right]^{2}+0\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(1+\rho^{2}/4a^{2}\right)^{2}} \ \ \ \ \ (15)$

[Note that this is different from Hobson’s answer which is ${\frac{1}{\left(1+\rho^{2}/a^{2}\right)^{2}}}$, but I can’t see anything wrong with my derivation. Comments?

The other metric component is

 $\displaystyle g_{\phi\phi}^{\prime}$ $\displaystyle =$ $\displaystyle g_{\theta\theta}\left(\frac{\partial\theta}{\partial\phi}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial\phi}\right)^{2}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+a^{2}\sin^{2}\theta\left(1\right)^{2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\sin^{2}\theta \ \ \ \ \ (18)$

From 5, we have

 $\displaystyle \sin^{2}\theta$ $\displaystyle =$ $\displaystyle 1-\left[\frac{1-\rho^{2}/4a^{2}}{1+\rho^{2}/4a^{2}}\right]^{2}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho^{2}/a^{2}}{\left(1+\rho^{2}/4a^{2}\right)^{2}}\ \ \ \ \ (20)$ $\displaystyle g_{\phi\phi}^{\prime}$ $\displaystyle =$ $\displaystyle \frac{\rho^{2}}{\left(1+\rho^{2}/4a^{2}\right)^{2}} \ \ \ \ \ (21)$

[Again, this differs from Hobson’s answer of ${\frac{\rho^{2}}{1+\rho^{2}/a^{2}}}$, but again, I can’t see anything wrong with my derivation.]

The line element in the stereographic coordinates is

$\displaystyle ds^{2}=\frac{d\rho^{2}}{\left(1+\rho^{2}/4a^{2}\right)^{2}}+\frac{\rho^{2}d\phi^{2}}{\left(1+\rho^{2}/4a^{2}\right)^{2}} \ \ \ \ \ (22)$

If we used ordinary rectangular coordinates ${x}$ and ${y}$ in the tangent plane, then the transformation equations between them and the polar coordinates are, as usual:

 $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}}\ \ \ \ \ (23)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{y}{x} \ \ \ \ \ (24)$

The required derivatives are

 $\displaystyle \frac{\partial\rho}{\partial x}$ $\displaystyle =$ $\displaystyle \frac{x}{\sqrt{x^{2}+y^{2}}}\ \ \ \ \ (25)$ $\displaystyle \frac{\partial\rho}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{y}{\sqrt{x^{2}+y^{2}}}\ \ \ \ \ (26)$ $\displaystyle \frac{\partial\phi}{\partial x}$ $\displaystyle =$ $\displaystyle -\frac{y}{x^{2}+y^{2}}\ \ \ \ \ (27)$ $\displaystyle \frac{\partial\phi}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{x}{x^{2}+y^{2}} \ \ \ \ \ (28)$

The metric becomes

 $\displaystyle g_{xx}$ $\displaystyle =$ $\displaystyle g_{\rho\rho}^{\prime}\left(\frac{\partial\rho}{\partial x}\right)^{2}+g_{\phi\phi}^{\prime}\left(\frac{\partial\phi}{\partial x}\right)^{2}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(1+\frac{x^{2}+y^{2}}{4a^{2}}\right)^{2}}\frac{x^{2}}{x^{2}+y^{2}}+\frac{x^{2}+y^{2}}{\left(1+\frac{x^{2}+y^{2}}{4a^{2}}\right)^{2}}\frac{y^{2}}{\left(x^{2}+y^{2}\right)^{2}}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(1+\frac{x^{2}+y^{2}}{4a^{2}}\right)^{2}}\ \ \ \ \ (31)$ $\displaystyle g_{yy}$ $\displaystyle =$ $\displaystyle g_{\rho\rho}^{\prime}\left(\frac{\partial\rho}{\partial y}\right)^{2}+g_{\phi\phi}^{\prime}\left(\frac{\partial\phi}{\partial y}\right)^{2}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(1+\frac{x^{2}+y^{2}}{4a^{2}}\right)^{2}}\frac{y^{2}}{x^{2}+y^{2}}+\frac{x^{2}+y^{2}}{\left(1+\frac{x^{2}+y^{2}}{4a^{2}}\right)^{2}}\frac{x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(1+\frac{x^{2}+y^{2}}{4a^{2}}\right)^{2}}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{xx} \ \ \ \ \ (35)$

The line segment is therefore

$\displaystyle ds^{2}=\frac{1}{\left(1+\frac{x^{2}+y^{2}}{4a^{2}}\right)^{2}}\left[dx^{2}+dy^{2}\right] \ \ \ \ \ (36)$

Near the north pole, ${x\approx0}$ and ${y\approx0}$, and the line element reduces to ${ds^{2}\approx dx^{2}+dy^{2}}$, showing that flat space is a good approximation to the sphere at the point where the tangent plane touches the sphere. The farther out we go (that is, for larger ${\rho}$), the interval ${ds^{2}}$ gets smaller and smaller. This is because these points correspond to points on the sphere near the south pole, where a small change in position on the sphere results in a very large change in ${\rho}$. The same effect occurs in 22, where ${ds^{2}}$ becomes very small for large ${\rho}$. [Note that this wouldn’t be the case if we used Hobson’s ${g_{\phi\phi}=\frac{\rho^{2}}{1+\rho^{2}/a^{2}}}$, which leads me to believe that my answer is correct. Remember ${ds^{2}}$ is an invariant, so it must always get small for large ${\rho}$ in all coordinate systems.]

# M27 (Dumbbell Nebula)

M27 (also known as the Dumbbell Nebula because of its shape) is a planetary nebula in the constellation Vulpecula, just north of the bright star Altair.

My photo:

Photo location: Monifieth (near Dundee), Scotland, UK.

Date: 1 Oct 2015; 21:00 UTC.

Telescope: 11-inch Celestron SCT.

Camera: Pentax K3

Exposure: ISO 800, 40 30-second exposures stacked with DeepSkyStacker.

# Metric tensor for a non-orthogonal coordinate system

Reference: Hobson, M.P., Efstathiou, G. P. & Lasenby, A. N. (2006), General Relativity: An Introduction for Physicists; Cambridge University Press. Problem 2.3.

As another example of finding the metric tensor in a new coordinate system, suppose the unprimed coordinates are the usual 3-d rectangular coordinates and we define a new set of (primed) coordinates by

 $\displaystyle x^{1}$ $\displaystyle =$ $\displaystyle x^{\prime1}+x^{\prime2}\ \ \ \ \ (1)$ $\displaystyle x^{2}$ $\displaystyle =$ $\displaystyle x^{\prime1}-x^{\prime2}\ \ \ \ \ (2)$ $\displaystyle x^{3}$ $\displaystyle =$ $\displaystyle 2x^{\prime1}x^{\prime2}+x^{\prime3} \ \ \ \ \ (3)$

For rectangular coordinates

$\displaystyle g_{ab}=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (4)$

By applying the formula

$\displaystyle g_{cd}^{\prime}=g_{ab}\frac{\partial x^{a}}{\partial x^{\prime c}}\frac{\partial x^{b}}{\partial x^{\prime d}} \ \ \ \ \ (5)$

and calculating the derivatives we get

$\displaystyle \mathsf{G}=g_{cd}^{\prime}=\left[\begin{array}{ccc} 2+4\left(x^{\prime2}\right)^{2} & 4x^{\prime1}x^{\prime2} & 2x^{\prime2}\\ 4x^{\prime1}x^{\prime2} & 2+4\left(x^{\prime1}\right)^{2} & 2x^{\prime1}\\ 2x^{\prime2} & 2x^{\prime1} & 1 \end{array}\right] \ \ \ \ \ (6)$

Because the metric isn’t diagonal, the coordinate curves don’t intersect at right angles. [By Pythagoras, if the coordinate curves intersect at right angles, then ${ds^{2}=\sum_{i}c_{i}\left(dx^{i}\right)^{2}}$ for some coefficients ${c_{i}}$; in other words, the metric has to be diagonal.]

The volume element is most easily found using the Jacobian.

 $\displaystyle \mathsf{J}$ $\displaystyle =$ $\displaystyle \left[\frac{\partial x^{a}}{\partial x^{\prime b}}\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 1 & 0\\ 1 & -1 & 0\\ 2x^{\prime2} & 2x^{\prime1} & 1 \end{array}\right]\ \ \ \ \ (8)$ $\displaystyle J$ $\displaystyle =$ $\displaystyle \det\mathsf{J}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2 \ \ \ \ \ (10)$

The volume element is

 $\displaystyle dV$ $\displaystyle =$ $\displaystyle \left|J\right|dx^{\prime1}dx^{\prime2}dx^{\prime3}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2dx^{\prime1}dx^{\prime2}dx^{\prime3} \ \ \ \ \ (12)$

We can also find the volume element using Hobson’s equation 2.22, which requires calculating the determinant of the metric matrix 6. You can grind through this if you want to (I actually did!) and you’ll find

$\displaystyle \det\mathsf{G}=4 \ \ \ \ \ (13)$

so

$\displaystyle dV=\sqrt{\left|\det\mathsf{G}\right|}dx^{\prime1}dx^{\prime2}dx^{\prime3}=2dx^{\prime1}dx^{\prime2}dx^{\prime3} \ \ \ \ \ (14)$

# Metric tensor: conversion from spherical to cylindrical coordinates

Reference: Hobson, M.P., Efstathiou, G. P. & Lasenby, A. N. (2006), General Relativity: An Introduction for Physicists; Cambridge University Press. Problem 2.2.

Rather than defining the metric tensor as the dot products of the basis vectors as Moore does, Hobson just states in his equation 2.4 that the interval between two points has the form

$\displaystyle ds^{2}=g_{ab}\left(x\right)dx^{a}dx^{b} \ \ \ \ \ (1)$

He then derives the symmetry condition from the fact that we can split ${g_{ab}}$ into symmetric and antisymmetric parts by writing

$\displaystyle g_{ab}=\frac{1}{2}\left(g_{ab}+g_{ba}\right)+\frac{1}{2}\left(g_{ab}-g_{ba}\right) \ \ \ \ \ (2)$

Inserting this into 1 and doing the implied sums, the antisymmetric contribution vanishes so for the purposes of using ${g_{ab}}$ for defining invariant intervals, we might as well take it to be symmetric. Also from 1, using the chain rule, we get

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle g_{ab}\frac{\partial x^{a}}{\partial x^{\prime c}}\frac{\partial x^{b}}{\partial x^{\prime d}}dx^{\prime c}dx^{\prime d}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{cd}^{\prime}dx^{\prime c}dx^{\prime d} \ \ \ \ \ (4)$

which gives the usual transformation formula for the metric:

$\displaystyle g_{cd}^{\prime}=g_{ab}\frac{\partial x^{a}}{\partial x^{\prime c}}\frac{\partial x^{b}}{\partial x^{\prime d}} \ \ \ \ \ (5)$

As an example of a transformation, suppose the unprimed system consists of 3-d spherical coordinates ${\left(r,\theta,\phi\right)}$ and the primed system consists of 3-d cylindrical coordinates, for which we’ll use the coordinate names ${\left(\rho,z,\alpha\right)}$ where ${\rho}$ is the radial distance from the ${z}$ axis and ${\alpha}$ is the azimuthal angle. We can write spherical coordinates in terms of cylindrical coordinates as follows:

 $\displaystyle r$ $\displaystyle =$ $\displaystyle \sqrt{\rho^{2}+z^{2}}\ \ \ \ \ (6)$ $\displaystyle r\sin\theta$ $\displaystyle =$ $\displaystyle \rho\ \ \ \ \ (7)$ $\displaystyle r\cos\theta$ $\displaystyle =$ $\displaystyle z\ \ \ \ \ (8)$ $\displaystyle \tan\theta$ $\displaystyle =$ $\displaystyle \frac{\rho}{z}\ \ \ \ \ (9)$ $\displaystyle \theta$ $\displaystyle =$ $\displaystyle \arctan\frac{\rho}{z}\ \ \ \ \ (10)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \alpha \ \ \ \ \ (11)$

The interval in spherical coordinates is, as usual:

$\displaystyle ds^{2}=dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (12)$

so

 $\displaystyle g_{rr}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (13)$ $\displaystyle g_{\theta\theta}$ $\displaystyle =$ $\displaystyle r^{2}\ \ \ \ \ (14)$ $\displaystyle g_{\phi\phi}$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (15)$

with all other ${g_{ab}=0}$ if ${a\ne b}$. Using 5 we can verify that it gives the correct metric for cylindrical coordinates.

 $\displaystyle g_{\rho\rho}^{\prime}$ $\displaystyle =$ $\displaystyle g_{ab}\frac{\partial x^{a}}{\partial\rho}\frac{\partial x^{b}}{\partial\rho}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1\right)\frac{\rho^{2}}{\rho^{2}+z^{2}}+r^{2}\left[\frac{1/z}{1+\rho^{2}/z^{2}}\right]^{2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho^{2}}{\rho^{2}+z^{2}}+\left(\rho^{2}+z^{2}\right)\left[\frac{z}{z^{2}+\rho^{2}}\right]^{2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho^{2}+z^{2}}{\rho^{2}+z^{2}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (20)$ $\displaystyle g_{zz}^{\prime}$ $\displaystyle =$ $\displaystyle \left(1\right)\frac{z^{2}}{\rho^{2}+z^{2}}+r^{2}\left[\frac{-\rho/z^{2}}{1+\rho^{2}/z^{2}}\right]^{2}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho^{2}}{\rho^{2}+z^{2}}+\left(\rho^{2}+z^{2}\right)\left[\frac{-\rho}{z^{2}+\rho^{2}}\right]^{2}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (23)$ $\displaystyle g_{\alpha\alpha}^{\prime}$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho^{2} \ \ \ \ \ (25)$

The other components are all zero.

 $\displaystyle g_{\rho z}^{\prime}$ $\displaystyle =$ $\displaystyle g_{ab}\frac{\partial x^{a}}{\partial\rho}\frac{\partial x^{b}}{\partial z}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1\right)\frac{\rho z}{\rho^{2}+z^{2}}+r^{2}\frac{-\rho/z^{3}}{\left[1+\rho^{2}/z^{2}\right]^{2}}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho z}{\rho^{2}+z^{2}}+\left(\rho^{2}+z^{2}\right)\frac{-\rho/z^{3}}{\left[1+\rho^{2}/z^{2}\right]^{2}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho z}{\rho^{2}+z^{2}}+\left(\rho^{2}+z^{2}\right)\frac{-\rho z}{\left[z^{2}+\rho^{2}\right]^{2}}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (30)$ $\displaystyle g_{\rho\alpha}^{\prime}$ $\displaystyle =$ $\displaystyle g_{ab}\frac{\partial x^{a}}{\partial\rho}\frac{\partial x^{b}}{\partial\alpha}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\phi\phi}\frac{\partial\phi}{\partial\rho}\frac{\partial\phi}{\partial\alpha}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (33)$ $\displaystyle g_{z\alpha}^{\prime}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}\frac{\partial\phi}{\partial z}\frac{\partial\phi}{\partial\alpha}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (35)$

Thus

$\displaystyle ds^{2}=d\rho^{2}+dz^{2}+\rho^{2}d\alpha^{2} \ \ \ \ \ (36)$

which is the correct interval for cylindrical coordinates.

# Manifolds, the Jacobian matrix and coordinate transformations

Reference: Hobson, M.P., Efstathiou, G. P. & Lasenby, A. N. (2006), General Relativity: An Introduction for Physicists; Cambridge University Press. Problem 2.1.

Having worked through Moore’s General Relativity Workbook up to the point where we’ve obtained the Einstein equations and worked through a few solutions to obtain some metric tensors in several situations, I think it’s time to back up and have another look at general relativity from a slightly more mathematical viewpoint. I say ‘slightly’ as I don’t want to delve into the rigorous mathematics of differential geometry (although I’m sure that’s a rewarding pastime), but I found Moore’s derivations to rely a fair bit on hand waving to justify the results.

There are many books on general relativity that use a more mathematical approach than Moore’s (in fact, pretty well all of them do), so we’re spoiled for choice. I had a look at the immense tome Gravitation by Misner, Thorne and Wheeler but, although it’s elegantly written I found it a bit hard to follow (as well as being just toooo big) as it tends to jump around a lot between the main text and numerous ‘boxes’ and sidebars. There aren’t that many problems to work out either. Weinberg’s Gravitation and Cosmology is too advanced (and doesn’t have any problems to work through). The books by d’Inverno and Schutz that I started to work through a couple of years ago are also rather difficult to follow.

The book I’ve chosen in the one by Hobson et al referenced above. It received good reviews (well, except for the price) on Amazon and having read the first couple of chapters, it does seem to explain things clearly for the most part and strikes a good balance between a proper mathematical treatment and physical intuition. So I’ll go with it for now and see how we do.

To begin, we’ll review the concept of a manifold. What I wrote in the earlier post is still a good summary of manifolds and the curves and surfaces that can be embedded in them, so I’ll just add a few comments here.

A manifold is actually a more general mathematical object than my previous post may have implied. A completely general manifold is any set of points that can be parameterized. By ‘parameterized’, I mean that it is possible to define a finite set of parameters whose values uniquely determine each of the points in the manifold. Presumably we could pick, say, 10 points in 3-d space at random and parameterize them by assigning the values 1 through 10, in some order, to the points. If we call the parameter ${x}$ then the value of ${x}$ specifies which point we’re referring to. Since only one parameter is needed, the manifold is one-dimensional. The parameter(s) used to locate the points in the manifold are known as coordinates.

Most manifolds in physics are continuous and differentiable. These terms mean much the same thing as when they are applied to mathematical functions. A manifold is continuous if, for every point ${P}$ if by varying the coordinates that specify ${P}$ by an infinitesimal amount, we find other points that are infinitesimally distant from ${P}$. Thus our “10 random points” manifold above is not continuous. A straight line is a continuous manifold since the location on the line can be specified by one parameter, and if we vary this parameter slightly starting at a point ${P}$, we move slightly away from ${P}$ to one side or the other.

A differentiable manifold is one where it is possible to define a scalar field (that is, a scalar function of the points in the manifold) at each point ${P}$ that can be differentiated with respect to each of the coordinates.

The number ${N}$ of coordinates required to locate a point within the manifold is the dimension of the manifold. We can define a submanifold (or surface) with ${M dimensions by specifying ${N-M}$ constraints of the form

$\displaystyle f_{i}\left(x^{1},x^{2},\ldots,x^{N}\right)=0 \ \ \ \ \ (1)$

for ${i=1\ldots N-M}$. If ${M=N-1}$ then there is only one constraint and the submanifold so defined is known as a hypersurface. For example, a sphere is a hypersurface within 3-d space, a line is a hypersurface within a plane, and so on. [See the earlier post for a few examples.]

The choice of coordinates for a given manifold is completely arbitrary. As an example, suppose we’re working in 3-d Euclidean space (denoted ${R^{3}}$). We can use rectangular coordinates ${\left(x,y,z\right)}$ or spherical coordinates ${\left(r,\theta\phi\right)}$. The standard transformation equations between the two systems are

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\nonumber$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (2)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\theta\nonumber$

To transform differentials we can use the chain rule

 $\displaystyle dx$ $\displaystyle =$ $\displaystyle \sin\theta\cos\phi dr+r\cos\theta\cos\phi d\phi-r\sin\theta\sin\phi d\phi\ \ \ \ \ (3)$ $\displaystyle dy$ $\displaystyle =$ $\displaystyle \sin\theta\sin\phi dr+r\cos\theta\sin\phi d\phi+r\sin\theta\cos\phi d\phi\ \ \ \ \ (4)$ $\displaystyle dz$ $\displaystyle =$ $\displaystyle \cos\theta dr-r\sin\theta d\theta \ \ \ \ \ (5)$

In matrix form, we have the Jacobian matrix ${\mathsf{J}}$

 $\displaystyle \left[\begin{array}{c} dx\\ dy\\ dz \end{array}\right]$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi\\ \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi\\ \cos\theta & -r\sin\theta & 0 \end{array}\right]\left[\begin{array}{c} dr\\ d\theta\\ d\phi \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle \mathsf{J}$ $\displaystyle \equiv$ $\displaystyle \left[\begin{array}{ccc} \sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi\\ \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi\\ \cos\theta & -r\sin\theta & 0 \end{array}\right] \ \ \ \ \ (7)$

Hobson calls this the transformation matrix but doesn’t make clear what it transforms. It does not transform coordinates; for that we must use the equations 2. To see what ${\mathsf{J}}$ does transform, suppose we have a curve in 3-d space defined in cartesian coordinates by

$\displaystyle \mathbf{A}\left(x,y,z\right)=x\left(t\right)\mathbf{x}+y\left(t\right)\mathbf{y}+z\left(t\right)\mathbf{z} \ \ \ \ \ (8)$

where ${t}$ is a parameter which traces out the curve as it varies. The tangent vector to the curve is ${\mathbf{T}}$ defined as the vector joining two adjacent points on the curve, in the limit as the separation tends to zero, or:

 $\displaystyle \mathbf{T}$ $\displaystyle =$ $\displaystyle \lim_{\Delta t\rightarrow0}\frac{\mathbf{A}\left(x\left(t+\Delta t\right),y\left(t+\Delta t\right),z\left(t+\Delta t\right)\right)-\mathbf{A}\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)}{\Delta t}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \dot{x}\left(t\right)\mathbf{x}+\dot{y}\left(t\right)\mathbf{y}+\dot{z}\left(t\right)\mathbf{z} \ \ \ \ \ (10)$

where a dot denotes a derivative with respect to ${t}$. From 6 we can divide both sides by ${dt}$ to get

 $\displaystyle \left[\begin{array}{c} \dot{x}\\ \dot{y}\\ \dot{z} \end{array}\right]$ $\displaystyle =$ $\displaystyle \mathsf{J}\left[\begin{array}{c} \dot{r}\\ \dot{\theta}\\ \dot{\phi} \end{array}\right]\ \ \ \ \ (11)$ $\displaystyle \left[\begin{array}{c} \dot{r}\\ \dot{\theta}\\ \dot{\phi} \end{array}\right]$ $\displaystyle =$ $\displaystyle \mathsf{J}^{-1}\left[\begin{array}{c} \dot{x}\\ \dot{y}\\ \dot{z} \end{array}\right] \ \ \ \ \ (12)$

That is, provided the Jacobian matrix is invertible (non-singular), we can convert the tangent vector back and forth between the two coordinate systems. Thus the Jacobian matrix transforms derivatives of the coordinates, not the coordinates themselves.

From linear algebra, a square matrix is invertible if its determinant is non-zero, so these coordinate transformations will work at all points where ${J\equiv\det\mathsf{J}\ne0}$. This determinant is known as the Jacobian determinant or sometimes just the Jacobian. [In a previous post, we looked at the role of the Jacobian in transforming volume elements in integration.] The determinant is most easily calculated by expanding about the bottom row in 7:

 $\displaystyle J$ $\displaystyle =$ $\displaystyle \cos\theta\left[r^{2}\cos\theta\sin\theta\right]+r\sin\theta\left[r\sin^{2}\theta\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin\theta \ \ \ \ \ (14)$

This is the familiar factor multiplied into the spherical coordinate volume element.

For 2 general coordinate systems ${x^{i}}$ and ${x^{\prime i}}$ the Jacobian matrix is

$\displaystyle \mathsf{J}=\left[\begin{array}{cccc} \partial_{1}x^{\prime1} & \partial_{2}x^{\prime1} & \ldots & \partial_{n}x^{\prime1}\\ \partial_{1}x^{\prime2} & \partial_{2}x^{\prime2} & \ldots & \partial_{n}x^{\prime2}\\ \vdots & \vdots & \ddots & \vdots\\ \partial_{1}x^{\prime n} & \partial_{2}x^{\prime n} & \ldots & \partial_{n}x^{\prime n} \end{array}\right] \ \ \ \ \ (15)$

where ${\partial_{i}\equiv\frac{\partial}{\partial x^{i}}}$. The inverse transformation swaps primed and unprimed coordinates to give

$\displaystyle \mathsf{J}^{\prime}=\left[\begin{array}{cccc} \partial_{1}^{\prime}x^{1} & \partial_{2}^{\prime}x^{1} & \ldots & \partial_{n}^{\prime}x^{1}\\ \partial_{1}^{\prime}x^{2} & \partial_{2}^{\prime}x^{2} & \ldots & \partial_{n}^{\prime}x^{2}\\ \vdots & \vdots & \ddots & \vdots\\ \partial_{1}^{\prime}x^{n} & \partial_{2}^{\prime}x^{n} & \ldots & \partial_{n}^{\prime}x^{n} \end{array}\right] \ \ \ \ \ (16)$

The product (using the summation convention and the chain rule to condense line 1 to line 2) is

 $\displaystyle \left[\mathsf{J}^{\prime}\mathsf{J}\right]_{ij}$ $\displaystyle =$ $\displaystyle \partial_{k}^{\prime}x^{i}\partial_{j}x^{\prime k}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{j}^{\prime}x^{\prime i}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{j}^{i} \ \ \ \ \ (19)$

where the last line follows because ${x^{\prime i}}$ and ${x^{\prime j}}$ are independent coordinates if ${i\ne j}$ so the derivative of one with respect to the other is zero. Therefore ${\mathsf{J}^{\prime}\mathsf{J}}$ is the identity matrix and ${\mathsf{J}^{\prime}=\mathsf{J}^{-1}}$, so ${\det\mathsf{J}^{\prime}=\frac{1}{J}}$. For our example the Jacobian of the inverse transformation is therefore

$\displaystyle J^{-1}=\frac{1}{r^{2}\sin\theta} \ \ \ \ \ (20)$

The transformation is invertible except when ${r=0}$ or ${\theta=n\pi}$. When ${r=0}$, ${\theta}$ and ${\phi}$ are indeterminate (they can be anything) and similarly, when ${\theta=0}$ or ${\pi}$, we’re at the poles and ${\phi}$ is indeterminate. However, there’s nothing singular about the actual geometry of the sphere at these points; the problem is purely in our choice of coordinates.

Example As a specific example of a coordinate transformation using the Jacobian, we’ll look at the 2-d transformation between rectangular and polar coordinates. Suppose we have a curve defined parametrically in polar coordinates by

$\displaystyle \mathbf{u}\left(t\right)=\left[r,\theta\right]=\left[t,t^{2}\right] \ \ \ \ \ (21)$

Then the tangent to the curve is given by

$\displaystyle \dot{\mathbf{u}}\left(t\right)=\left[1,2t\right] \ \ \ \ \ (22)$

The curve is a spiral, as shown in blue in the plot, where the tangent at ${t=\sqrt{\pi}}$ (where ${\mathbf{u}=\left[\sqrt{\pi},\pi\right]}$ and ${\dot{\mathbf{u}}=\left[1,2\sqrt{\pi}\right]}$) is shown in red:

The tangent vector in polar coordinates is

$\displaystyle \dot{\mathbf{u}}=\hat{\mathbf{r}}+2\sqrt{\pi}\hat{\boldsymbol{\theta}} \ \ \ \ \ (23)$

since at ${\theta=\pi}$ ${\hat{\mathbf{r}}}$ points to the left and ${\hat{\boldsymbol{\theta}}}$ points downwards.

Now we can convert this tangent vector to rectangular coordinates by using the Jacobian matrix. The transformation between polar and rectangular coordinates is

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\cos\theta\ \ \ \ \ (24)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta \ \ \ \ \ (25)$

so we have

 $\displaystyle \mathsf{J}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} \partial_{r}x & \partial_{\theta}x\\ \partial_{r}y & \partial_{\theta}y \end{array}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{array}\right] \ \ \ \ \ (27)$

The transformation of the tangent vector is

 $\displaystyle \mathsf{J}\dot{\mathbf{u}}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{array}\right]\left[\begin{array}{c} 1\\ 2t \end{array}\right]\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\theta-2tr\sin\theta\\ \sin\theta+2tr\cos\theta \end{array}\right] \ \ \ \ \ (29)$

At ${t=\sqrt{\pi}}$, ${r=\sqrt{\pi}}$ and ${\theta=\pi}$ so the tangent vector in rectangular coordinates is

 $\displaystyle \dot{\mathbf{u}}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\pi-2\pi\sin\pi\\ \sin\pi+2\pi\cos\pi \end{array}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} -1\\ -2\pi \end{array}\right]\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hat{\mathbf{x}}-2\pi\hat{\mathbf{y}} \ \ \ \ \ (32)$

# NGC 7331 and four background galaxies

NGC 7331is a spiral galaxy in Pegasus that is one of the brightest galaxies not to be included in Messier’s catalogue. It’s about 40 million light years away. There are actually four other fainter galaxies visible in the same field. From top to bottom they are NGC 7336, NGC 7335, NGC 7337 and NGC 7340. These four galaxies are not associated with NGC 7331 and are about 350 million light years away.

Here’s my original photo:

The spiral structure of NGC 7331 is clearly visible in this photograph, which was a pleasant surprise as this is a stack of only 20 20-second exposures, and there was a gibbous moon visible while the photos were being taken.

Here’s another image showing where the four fainter galaxies are (they are the fuzzy patches to the left of the main galaxy above):

Photo location: Monifieth (near Dundee), Scotland, UK.

Date: 23 Sep 2015; 21:00 UTC.

Telescope: 11-inch Celestron SCT.

Camera: Pentax K3

Exposure: ISO 800, 20 20-second exposures stacked with DeepSkyStacker.

# M15 in Pegasus (23/9/15)

Messier 15(M15, or NGC 7078) is a globular cluster at the western end of the constellation Pegasus. It is one of the most compact globular clusters known. It is about 33,600 light years away, and has a visual magnitude of 6.2 so under dark skies it is almost visible to the naked eye.

Here’s my photo:

Photo location: Monifieth (near Dundee), Scotland, UK.

Date: 23 Sep 2015; 20:00 UTC.

Telescope: 11-inch Celestron SCT.

Camera: Pentax K3

Exposure: ISO 800, 40 20-second exposures stacked with DeepSkyStacker.

# 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.

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

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.

# Einstein equation solution for the interior of a spherically symmetric star

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.5.

The derivation of the Schwarzschild metric applies to the empty space outside a spherically symmetric source. We can apply a similar method to try to find a metric for the interior of a spherically symmetric, static source such as a non-rotating star. As usual, we can make a number of simplifying assumptions. Spherically symmetry implies that the metric has the general form

$\displaystyle ds^{2}=-A\left(r\right)dt^{2}+B\left(r\right)dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (1)$

where the functions ${A}$ and ${B}$ are to be determined; they depend only on ${r}$ because we’re assuming that everything is independent of time. Another simplification is the assumption that the star’s matter is a perfect fluid. We’ve looked at the Einstein equation for a perfect fluid before, which has the general form (with ${\Lambda=0}$):

$\displaystyle R^{\mu\nu}=8\pi G\left(T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T\right) \ \ \ \ \ (2)$

The stress-energy tensor is

$\displaystyle T^{\mu\nu}=\left(\rho+P\right)u^{\mu}u^{\nu}+Pg^{\mu\nu} \ \ \ \ \ (3)$

where ${u^{\mu}}$ is the four-velocity, ${\rho}$ is the energy density and ${P}$ is the pressure, both of which can be functions of ${r}$.
We have for the stress-energy scalar:

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \left(\rho+P\right)g_{\mu\nu}u^{\mu}u^{\nu}+Pg_{\mu\nu}g^{\mu\nu}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\rho+P\right)+4P\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3P-\rho \ \ \ \ \ (6)$

since ${g_{\mu\nu}g^{\mu\nu}=4}$ and ${g_{\mu\nu}u^{\mu}u^{\nu}=-1}$ in any coordinate system. This gives

$\displaystyle T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T=\left(\rho+P\right)u^{\mu}u^{\nu}+\frac{1}{2}g^{\mu\nu}\left(\rho-P\right) \ \ \ \ \ (7)$

Lowering the indices in 7, we get

$\displaystyle T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T=\left(\rho+P\right)u_{\mu}u_{\nu}+\frac{1}{2}g_{\mu\nu}\left(\rho-P\right) \ \ \ \ \ (8)$

Since the fluid is at rest, the spatial components of the four-velocity ${u_{i}}$ are all zero. From the condition ${g_{\mu\nu}u^{\mu}u^{\nu}=-1}$ we have

 $\displaystyle g_{\mu\nu}u^{\mu}u^{\nu}$ $\displaystyle =$ $\displaystyle g_{tt}u^{t}u^{t}=-A\left(u^{t}\right)^{2}=-1\ \ \ \ \ (9)$ $\displaystyle u^{t}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{A}}\ \ \ \ \ (10)$ $\displaystyle u_{t}$ $\displaystyle =$ $\displaystyle g_{t\nu}u^{\nu}=g_{tt}u^{t}=-Au^{t}=-\sqrt{A} \ \ \ \ \ (11)$

Equation 2 with indices lowered is

$\displaystyle R_{\mu\nu}=8\pi G\left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right) \ \ \ \ \ (12)$

so with the above results, we get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle 8\pi G\left(\left(\rho+P\right)u_{t}u_{t}+\frac{1}{2}g_{tt}\left(\rho-P\right)\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\pi G\left[\left(\rho+P\right)A-\frac{A}{2}\left(\rho-P\right)\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi GA\left(\rho+3P\right)\ \ \ \ \ (15)$ $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle 8\pi G\left(\left(\rho+P\right)u_{r}u_{r}+\frac{1}{2}g_{rr}\left(\rho-P\right)\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi G\left(0+B\left(\rho-P\right)\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi GB\left(\rho-P\right)\ \ \ \ \ (18)$ $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle 8\pi G\left(\left(\rho+P\right)u_{\theta}u_{\theta}+\frac{1}{2}g_{\theta\theta}\left(\rho-P\right)\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi G\left(0+r^{2}\left(\rho-P\right)\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi Gr^{2}\left(\rho-P\right) \ \ \ \ \ (21)$

We can combine these three equations to eliminate ${P}$:

$\displaystyle \frac{R_{tt}}{A}+\frac{R_{rr}}{B}+\frac{2R_{\theta\theta}}{r^{2}}=16\pi G\rho \ \ \ \ \ (22)$

When we worked out the Ricci tensor in terms of the metric, we got the equations

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (23)$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (24)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (25)$ $\displaystyle \frac{\partial_{t}B}{rB}$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (26)$

All time derivatives are zero, so these equations simplify to

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\frac{\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (27)$ $\displaystyle \frac{1}{2A}\left[-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (28)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (29)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (30)$

Applying 22 we get

 $\displaystyle \frac{\partial_{r}A}{ABr}+\frac{\partial_{r}B}{rB^{2}}-\frac{\partial_{r}A}{rAB}+\frac{\partial_{r}B}{rB^{2}}+\frac{2}{r^{2}}-\frac{2}{r^{2}B}$ $\displaystyle =$ $\displaystyle 16\pi G\rho\left(r\right)\ \ \ \ \ (31)$ $\displaystyle r\frac{\partial_{r}B}{B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho\left(r\right)\ \ \ \ \ (32)$ $\displaystyle \frac{d}{dr}\left[r\left(1-\frac{1}{B}\right)\right]$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho\left(r\right)\ \ \ \ \ (33)$ $\displaystyle r\left(1-\frac{1}{B}\right)$ $\displaystyle =$ $\displaystyle 2G\int_{0}^{r}4\pi\left(r'\right)^{2}\rho\left(r'\right)dr' \ \ \ \ \ (34)$

The integral on the last line resembles the mass of the star out to radius ${r}$, as it’s the integral of the density ${\rho}$ over a set of spherical shells of radius ${r'}$, surface area ${4\pi\left(r'\right)^{2}}$ and thickness ${dr'}$. However, the Schwarzschild radial coordinate ${r}$ isn’t equal to the actual radial distance (it’s a circumferential coordinate), so this isn’t quite accurate. With the definition

$\displaystyle m\left(r\right)\equiv\int_{0}^{r}4\pi\left(r'\right)^{2}\rho\left(r'\right)dr' \ \ \ \ \ (35)$

we therefore have

 $\displaystyle r\left(1-\frac{1}{B}\right)$ $\displaystyle =$ $\displaystyle 2Gm\left(r\right)\ \ \ \ \ (36)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \left[1-\frac{2Gm\left(r\right)}{r}\right]^{-1} \ \ \ \ \ (37)$

This gives us ${B}$, but what about ${A}$? If we use the stress-energy tensor’s conservation equation

$\displaystyle \nabla_{\nu}T^{\mu\nu}=0 \ \ \ \ \ (38)$

we can look at ${\mu=r}$ component, starting from 3. The absolute gradient is defined in terms of Christoffel symbols as

$\displaystyle \nabla_{\rho}T^{\mu\nu}=\partial_{\rho}T^{\mu\nu}+T^{\mu\alpha}\Gamma_{\alpha\rho}^{\nu}+T^{\alpha\nu}\Gamma_{\alpha\rho}^{\mu} \ \ \ \ \ (39)$

so for ${\mu=r}$ and ${\rho=\nu}$ we have

$\displaystyle \nabla_{\nu}T^{r\nu}=\partial_{\nu}T^{r\nu}+T^{\alpha\nu}\Gamma_{\alpha\nu}^{r}+T^{r\alpha}\Gamma_{\alpha\nu}^{\nu} \ \ \ \ \ (40)$

Since the covariant derivative of the metric tensor is zero and ${P}$ is a scalar (so its covariant derivative is the same as its ordinary derivative), we have

 $\displaystyle \nabla_{\nu}T^{r\nu}$ $\displaystyle =$ $\displaystyle \nabla_{\nu}\left[\left(\rho+P\right)u^{r}u^{\nu}\right]+\nabla_{\nu}\left(Pg^{r\nu}\right)\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla_{\nu}\left[\left(\rho+P\right)u^{r}u^{\nu}\right]+g^{rr}\partial_{r}P\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{\nu}\left[\left(\rho+P\right)u^{r}u^{\nu}\right]+\left(\rho+P\right)\left[u^{\alpha}u^{\nu}\Gamma_{\alpha\nu}^{r}+u^{r}u^{\alpha}\Gamma_{\alpha\nu}^{\nu}\right]\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (44)$

Because the fluid is at rest, ${u^{r}=u^{\theta}=u^{\phi}=0}$ so this reduces to

$\displaystyle \left(\rho+P\right)u^{t}u^{t}\Gamma_{tt}^{r}+g^{rr}\partial_{r}P=0 \ \ \ \ \ (45)$

For a diagonal metric ${g^{\mu\nu}=\frac{1}{g_{\mu\nu}}}$ so ${g^{rr}=\frac{1}{B}}$, from 10 ${u^{t}=\frac{1}{\sqrt{A}}}$, and from the Christoffel symbol worksheet ${\Gamma_{tt}^{r}=\Gamma_{00}^{1}=\frac{1}{2B}\partial_{r}A}$ so we have

 $\displaystyle \left(\rho+P\right)\frac{\partial_{r}A}{2AB}+\frac{\partial_{r}P}{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (46)$ $\displaystyle \frac{1}{A}\frac{dA}{dr}$ $\displaystyle =$ $\displaystyle -\frac{2}{\rho+P}\frac{dP}{dr} \ \ \ \ \ (47)$

To solve this, we need to know both ${\rho}$ and ${P}$ as functions of ${r}$. We can make some headway by using ${R_{\theta\theta}}$ in 21 and 25. We have

$\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}=4\pi Gr^{2}\left(\rho-P\right) \ \ \ \ \ (48)$

From 32 and 36

 $\displaystyle r\frac{\partial_{r}B}{B^{2}}$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho-\left(1-\frac{1}{B}\right)\ \ \ \ \ (49)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho-\frac{2Gm}{r} \ \ \ \ \ (50)$

Plugging this and 37 into 48:

 $\displaystyle -\frac{r}{2B}\left(-\frac{2}{\rho+P}\frac{dP}{dr}\right)+4\pi Gr^{2}\rho-\frac{Gm}{r}+\frac{2Gm}{r}$ $\displaystyle =$ $\displaystyle 4\pi Gr^{2}\left(\rho-P\right)\ \ \ \ \ (51)$ $\displaystyle \frac{r^{2}}{\rho+P}\left(1-\frac{2Gm}{r}\right)\frac{dP}{dr}$ $\displaystyle =$ $\displaystyle -\left(4\pi Gr^{3}P+Gm\right)\ \ \ \ \ (52)$ $\displaystyle \frac{dP}{dr}$ $\displaystyle =$ $\displaystyle -\frac{\left(\rho+P\right)}{r^{2}}\frac{\left(4\pi Gr^{3}P+Gm\right)}{1-2Gm/r} \ \ \ \ \ (53)$

This is the Oppenheimer-Volkoff equation for stellar structure. In order to solve it, we need a couple of other equations involving ${m}$ and ${\rho}$, but we’ll leave that for later. [By the way, for anyone interested in trivia, the Volkoff after whom this equation is named is George Volkoff, and he was the Dean of Science at the University of British Columbia in Vancouver when I was doing my undergraduate degree in physics and astronomy there in the 1970s. Sadly, I never had him as a professor.]

# Black hole with static charge; Reissner-Nordström solution

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.4.

The derivation of the Schwarzschild metric can be enhanced to include a source, such as a black hole, with a static electric charge ${Q}$. The resulting metric is known as the Reissner-Nordstöm solution. In order to include the effects of the charge, we have to realize that even if there is no mass outside the source, the electric field carries energy so its contribution to the stress-energy tensor must be included. We’ll begin with a quick review of the electromagnetic stress-energy tensor. The electromagnetic field tensor is

$\displaystyle F^{\mu\nu}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

where ${E_{i}}$ and ${B_{i}}$ are the spatial components of the electric and magnetic fields. We can just as well write this in spherical coordinates as

$\displaystyle F^{\mu\nu}=\left[\begin{array}{cccc} 0 & E_{r} & E_{\theta} & E_{\phi}\\ -E_{r} & 0 & B_{\phi} & -B_{\theta}\\ -E_{\theta} & -B_{\phi} & 0 & B_{r}\\ -E_{\phi} & B_{\theta} & -B_{r} & 0 \end{array}\right] \ \ \ \ \ (2)$

The electromagnetic stress-energy tensor can be written in terms of ${F^{\mu\nu}}$ (generalized to non-flat space with a metric ${g^{\mu\nu}}$):

 $\displaystyle T^{\mu\nu}$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}F_{\kappa\lambda}g^{\lambda\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}F_{\kappa}^{\;\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\alpha\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right) \ \ \ \ \ (6)$

where to get the last line, we used the anti-symmetry of ${F^{\alpha\nu}=-F^{\nu\alpha}}$. The constant ${k=1/4\pi\epsilon_{0}}$ in more conventional notation.

The Einstein equation (with ${\Lambda=0}$) is:

$\displaystyle R^{\mu\nu}=8\pi G\left(T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T\right) \ \ \ \ \ (7)$

so to work out the components of ${R^{\mu\nu}}$ we need the scalar ${T=T_{\;\mu}^{\mu}}$. For any metric ${g^{\mu\nu}}$ we have from 3

 $\displaystyle T=T_{\;\mu}^{\mu}$ $\displaystyle =$ $\displaystyle g_{\mu\nu}T^{\mu\nu}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}g_{\mu\nu}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}g_{\mu\nu}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F_{\nu\alpha}F^{\nu\alpha}-2F_{\lambda\kappa}F^{\lambda\kappa}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

The double contraction in the second line is ${g^{\mu\nu}g_{\mu\nu}=4}$ (as can be seen by working it out in a local inertial frame where ${g^{\mu\nu}=\eta^{\mu\nu}}$), and in the third line we lowered the two indices of ${F^{\mu\kappa}}$ in the first term.

To proceed, we need to make a few assumptions. First, we’ll assume that a charged black hole gives rise to a spherically symmetric field tensor with only an electric field (the magnetic field is zero). We’ll also assume that the metric obeys Birkhoff’s theorem, so that it is independent of time. In that case, the only non-zero component of ${F_{\mu\nu}}$ is the radial electric field, which is an unknown function of ${r}$ only. That is

$\displaystyle F_{\mu\nu}=\left[\begin{array}{cccc} 0 & -E\left(r\right) & 0 & 0\\ E\left(r\right) & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (12)$

Note that we’re looking at the components of ${F_{\mu\nu}}$ with both indices lowered, so that the electric field is negative for ${F_{tr}}$ and positive for ${F_{rt}}$. [I should add that the original derivation of this assumed flat space, so I’m a bit hazy on how we can make this assumption for non-flat space. I suppose, given that we’re not specifying ${E}$ at this point, we can make this assumption.]

At this stage, we’ll take the metric to be

$\displaystyle ds^{2}=-Adt^{2}+Bdr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (13)$

with ${A}$ and ${B}$ to be determined. The components of ${F^{\mu\nu}}$ with raised indices are then since the metric is diagonal,

 $\displaystyle F_{\mu\nu}$ $\displaystyle =$ $\displaystyle g_{\mu\alpha}g_{\nu\beta}F^{\alpha\beta}\ \ \ \ \ (14)$ $\displaystyle F_{tr}$ $\displaystyle =$ $\displaystyle g_{tt}g_{rr}F^{tr}\ \ \ \ \ (15)$ $\displaystyle -E$ $\displaystyle =$ $\displaystyle -ABF^{tr}\ \ \ \ \ (16)$ $\displaystyle F^{tr}$ $\displaystyle =$ $\displaystyle \frac{E}{AB}=-F^{rt} \ \ \ \ \ (17)$

Given ${F^{\mu\nu}}$ and ${F_{\mu\nu}}$ we can now work out the RHS of 7, remembering that ${T=0}$ from 11:

 $\displaystyle 8\pi GT^{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (18)$ $\displaystyle 8\pi GT_{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}\left(2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\gamma\delta}\right) \ \ \ \ \ (19)$

From 12 and 17 we have

 $\displaystyle F_{\lambda\kappa}F^{\lambda\kappa}$ $\displaystyle =$ $\displaystyle -\frac{2E^{2}}{AB}\ \ \ \ \ (20)$ $\displaystyle 8\pi GT_{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}\left(2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}+\frac{E^{2}}{AB}g^{\gamma\delta}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}+g_{\mu\nu}\frac{GE^{2}}{kAB} \ \ \ \ \ (22)$

For the individual components, we get, using ${g_{tt}=-A}$, ${g_{rr}=B}$ and ${g_{\theta\theta}=r^{2}}$

 $\displaystyle 8\pi GT_{tt}$ $\displaystyle =$ $\displaystyle \frac{G}{k}A^{2}\left(2F^{tr}BF^{tr}\right)-\frac{GE^{2}}{kB}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[2A^{2}B\frac{E^{2}}{A^{2}B^{2}}-\frac{E^{2}}{B}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GE^{2}}{kB}\ \ \ \ \ (25)$ $\displaystyle 8\pi GT_{rr}$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[2B^{2}F^{rt}\left(-A\right)F^{rt}+B\frac{E^{2}}{AB}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[-2AB^{2}\frac{E^{2}}{A^{2}B^{2}}+\frac{E^{2}}{A}\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{GE^{2}}{kA}\ \ \ \ \ (28)$ $\displaystyle 8\pi GT_{\theta\theta}$ $\displaystyle =$ $\displaystyle 0+g_{\theta\theta}\frac{GE^{2}}{kAB}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GE^{2}r^{2}}{kAB} \ \ \ \ \ (30)$

We can now plug these into 7 to get the equations that must be solved to find ${A}$ and ${B}$. We get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle \frac{GE^{2}}{kB}\ \ \ \ \ (31)$ $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle -\frac{GE^{2}}{kA}\ \ \ \ \ (32)$ $\displaystyle BR_{tt}+AR_{rr}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (33)$

When we worked out the Ricci tensor in terms of the metric, we got the equations

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (34)$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (35)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (36)$ $\displaystyle \frac{\partial_{t}B}{rB}$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (37)$

Because of Birkhoff’s theorem, all time derivatives are zero, so these equations simplify to

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\frac{\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (38)$ $\displaystyle \frac{1}{2A}\left[-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (39)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (40)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (41)$

Applying 33 we get

 $\displaystyle \frac{\partial_{r}A}{r}+\frac{A\partial_{r}B}{rB}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (42)$ $\displaystyle \frac{\partial_{r}A}{A}+\frac{\partial_{r}B}{B}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (43)$

We can replace the partial derivatives by total derivatives since ${A}$ and ${B}$ depend only on ${r}$. Multiplying through by ${AB}$ we get

$\displaystyle B\frac{dA}{dr}+A\frac{dB}{dr}=\frac{d}{dr}\left(AB\right)=0 \ \ \ \ \ (44)$

so ${AB=\mbox{constant}}$. For very large ${r}$, the metric must reduce to ${\eta_{\mu\nu}}$ so both ${A\rightarrow1}$ and ${B\rightarrow1}$. Thus the product ${AB=1}$ everywhere, which means from 17 that ${F^{tr}=-F^{rt}=E}$ and thus ${F^{\mu\nu}=-F_{\mu\nu}}$.

So far, we have established that ${A=\frac{1}{B}}$ but to get the two components separately, we need to use the fact that we’re dealing a charged black hole. Maxwell’s equations can be written in tensor form as

$\displaystyle \nabla_{\nu}F^{\mu\nu}=4\pi kJ^{\mu} \ \ \ \ \ (45)$

where the absolute gradient is defined in terms of Christoffel symbols as

$\displaystyle \nabla_{\rho}F^{\mu\nu}=\partial_{\rho}F^{\mu\nu}+F^{\mu\alpha}\Gamma_{\alpha\rho}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\rho}^{\mu} \ \ \ \ \ (46)$

Contracting ${\rho}$ with ${\nu}$ we get

$\displaystyle \nabla_{\nu}F^{\mu\nu}=\partial_{\nu}F^{\mu\nu}+F^{\mu\alpha}\Gamma_{\alpha\nu}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\nu}^{\mu} \ \ \ \ \ (47)$

In empty space, ${J^{\mu}=0}$ since there is no charge or current, so for ${\mu=t}$ we have

$\displaystyle \nabla_{\nu}F^{t\nu}=\partial_{\nu}F^{t\nu}+F^{t\alpha}\Gamma_{\alpha\nu}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\nu}^{t} \ \ \ \ \ (48)$

Unfortunately, this means working out a few Christoffel symbols, but we can use the worksheet to make things easier. The only non-zero components of ${F^{\mu\nu}}$ are ${F^{tr}=-F^{rt}}$. Because ${\Gamma_{\alpha\nu}^{t}=\Gamma_{\nu\alpha}^{t}}$, the last term is zero after the sums are done, so

$\displaystyle \nabla_{\nu}F^{t\nu}=\partial_{r}F^{tr}+F^{tr}\Gamma_{r\nu}^{\nu} \ \ \ \ \ (49)$

In the notation of the worksheet, we have, using ${B=\frac{1}{A}}$, ${C=r^{2}}$ and ${D=r^{2}\sin^{2}\theta}$ and a subscript 1 means ‘take the derivative with respect to ${r}$‘:

 $\displaystyle \Gamma_{r\nu}^{\nu}$ $\displaystyle =$ $\displaystyle \Gamma_{1\nu}^{\nu}\ \ \ \ \ (50)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}A_{1}+\frac{1}{2B}B_{1}+\frac{1}{2C}C_{1}+\frac{1}{2D}D_{1}\ \ \ \ \ (51)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}A_{1}-\frac{1}{2A}A_{1}+\frac{1}{r}+\frac{1}{r}\ \ \ \ \ (52)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{r}\ \ \ \ \ (53)$ $\displaystyle \nabla_{\nu}F^{t\nu}$ $\displaystyle =$ $\displaystyle \partial_{r}E+\frac{2E}{r}=0\ \ \ \ \ (54)$ $\displaystyle r^{2}\partial_{r}E+2rE$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (55)$ $\displaystyle \frac{d}{dr}\left(r^{2}E\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (56)$ $\displaystyle E\left(r\right)$ $\displaystyle =$ $\displaystyle \frac{b}{r^{2}} \ \ \ \ \ (57)$

where ${b}$ is a constant of integration. If this is to reduce to the Coulomb field at large ${r}$, then we require

$\displaystyle b=kQ=\frac{Q}{4\pi\epsilon_{0}} \ \ \ \ \ (58)$

From 40 and 30 with ${AB=1}$ we have

 $\displaystyle -\frac{r\partial_{r}A}{2}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle \frac{GE^{2}r^{2}}{k}\ \ \ \ \ (59)$ $\displaystyle -\frac{r\partial_{r}A}{2}+\frac{rA^{2}\partial_{r}\frac{1}{A}}{2}+1-A$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (60)$ $\displaystyle -r\frac{dA}{dr}+1-A$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (61)$ $\displaystyle -\frac{d\left(rA\right)}{dr}+1$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (62)$ $\displaystyle \frac{d\left(rA\right)}{dr}$ $\displaystyle =$ $\displaystyle 1-\frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (63)$ $\displaystyle A\left(r\right)$ $\displaystyle =$ $\displaystyle 1+\frac{GkQ^{2}}{r^{2}}+\frac{K}{r} \ \ \ \ \ (64)$

where ${K}$ is a constant of integration. In order for this to reduce to the Schwarzschild metric component ${-g_{tt}=\left(1-\frac{2GM}{r}\right)}$ when ${Q=0}$, we must have ${K=-2GM}$, so

 $\displaystyle A\left(r\right)$ $\displaystyle =$ $\displaystyle 1-\frac{2GM}{r}+\frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (65)$ $\displaystyle B\left(r\right)$ $\displaystyle =$ $\displaystyle \left[1-\frac{2GM}{r}+\frac{GkQ^{2}}{r^{2}}\right]^{-1} \ \ \ \ \ (66)$

An event horizon occurs whenever ${g_{tt}=0}$. In this case, this gives rise to a quadratic equation in ${r}$:

 $\displaystyle r^{2}-2GMr+GkQ^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (67)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[2GM\pm\sqrt{4G^{2}M^{2}-4GkQ^{2}}\right]\ \ \ \ \ (68)$ $\displaystyle$ $\displaystyle =$ $\displaystyle GM\pm\sqrt{G^{2}M^{2}-GkQ^{2}} \ \ \ \ \ (69)$

For real solutions, we must have

 $\displaystyle G^{2}M^{2}$ $\displaystyle \ge$ $\displaystyle GkQ^{2}\ \ \ \ \ (70)$ $\displaystyle GM^{2}$ $\displaystyle \ge$ $\displaystyle kQ^{2} \ \ \ \ \ (71)$

If the charge ${Q}$ is large enough to violate this condition, there are no event horizons meaning that the singularity at ${r=0}$ becomes a naked singularity.