# Map projections: equal angles and equal areas

Reference: Hobson, M.P., Efstathiou, G. P. & Lasenby, A. N. (2006), General Relativity: An Introduction for Physicists; Cambridge University Press. Problem 2.6.

The Mercator projection is one of many map projections used to represent a spherical surface on a plane. As is obvious from looking at a Mercator map, however, it does not preserve relative areas. Regions nearer to the poles are disproportionately larger than regions nearer the equator.

Suppose we have a general projection given by mapping the spherical coordinates ${\lambda}$ (the latitude, where ${\lambda=\frac{\pi}{2}-\theta}$, with ${\theta}$ the usual polar angle) and ${\phi}$ (the longitude) to a rectangular coordinate system using the general formulas ${x=x\left(\lambda,\phi\right)}$ and ${y=y\left(\lambda,\phi\right)}$. First, suppose we want the projection to preserve the angle between two vectors originating at some point ${P}$. To see how to manage this, consider an infinitesimal displacement ${ds}$ given in terms of the spherical coordinates (${a}$ is the radius of the sphere):

$\displaystyle ds^{2}=a^{2}d\lambda^{2}+a^{2}\cos^{2}\lambda d\phi^{2} \ \ \ \ \ (1)$

When mapping this to the rectangular system, we must transform the metric in the usual way (both coordinates are orthogonal systems, so only diagonal terms appear):

 $\displaystyle g_{xx}$ $\displaystyle =$ $\displaystyle g_{\lambda\lambda}\left(\frac{\partial\lambda}{\partial x}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial x}\right)^{2}\ \ \ \ \ (2)$ $\displaystyle g_{yy}$ $\displaystyle =$ $\displaystyle g_{\lambda\lambda}\left(\frac{\partial\lambda}{\partial y}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial y}\right)^{2} \ \ \ \ \ (3)$

Now suppose we have two displacements that are perpendicular to each other; one (${dx^{i}}$) has ${d\phi=0}$ and the other (${dy^{i}}$) has ${d\lambda=0}$. In spherical coordinates their scalar product is

 $\displaystyle g_{ab}dx^{a}dy^{b}$ $\displaystyle =$ $\displaystyle g_{\lambda\lambda}dx^{\lambda}dy^{\lambda}+g_{\phi\phi}dx^{\phi}dy^{\phi}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\lambda\lambda}dx^{\lambda}\left(0\right)+g_{\phi\phi}\left(0\right)dy^{\phi}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

Since this is a scalar, it’s invariant between coordinates, so is true in the ${xy}$ system as well. Thus perpendicular lines in the spherical system map into perpendicular lines in the rectangular system.

Given this, we can take a general line segment ${ds}$ in the spherical system and transform its two perpendicular components separately, giving two corresponding perpendicular segments in the rectangular system. If the ratio of the lengths of these two components is the same in the rectangular system as it was in the spherical system, then the transformed line segment must have the same angles to the ${x}$ and ${y}$ axes as the original segment had to the ${\lambda}$ and ${\phi}$ axes. If this is true, then it is true for any two different line segments, which means that these two segments must have the same angle between them in both systems.

The transformation of the component parallel to the ${\lambda}$ axis is, since ${d\phi=0}$:

 $\displaystyle g_{xx}dx_{\lambda}^{2}+g_{yy}dy_{\lambda}^{2}$ $\displaystyle =$ $\displaystyle g_{\lambda\lambda}d\lambda^{2}=a^{2}d\lambda^{2} \ \ \ \ \ (7)$

We can write the LHS as

$\displaystyle g_{xx}dx_{\lambda}^{2}+g_{yy}dy_{\lambda}^{2}=\Omega\left(x,y\right)\left(dx_{\lambda}^{2}+dy_{\lambda}^{2}\right) \ \ \ \ \ (8)$

where ${\Omega}$ is some function that is determined by the requirement 7. All we’ve done is extract the dependence on the point ${P=\left(x,y\right)}$ into the function ${\Omega}$. The quantity ${dx_{\lambda}^{2}+dy_{\lambda}^{2}}$ is ultimately just a number, as is ${g_{xx}dx_{\lambda}^{2}+g_{yy}dy_{\lambda}^{2}}$ so the function ${\Omega}$ maps one set of numbers into the other.

Now look at the transformation of the component parallel to the ${\phi}$ axis:

 $\displaystyle g_{xx}dx_{\phi}^{2}+g_{yy}dy_{\phi}^{2}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}d\phi^{2}=a^{2}\cos^{2}\lambda d\phi^{2} \ \ \ \ \ (9)$

We can write the LHS as

$\displaystyle g_{xx}dx_{\phi}^{2}+g_{yy}dy_{\phi}^{2}=\omega\left(x,y\right)\left(dx_{\phi}^{2}+dy_{\phi}^{2}\right) \ \ \ \ \ (10)$

where ${\omega}$ is some other function that maps the RHS onto the LHS. [Note that it’s possible to map the two components ${\lambda}$ and ${\phi}$ into ${x}$ and ${y}$ components that are perpendicular, but where one axis is scaled up or down relative to the other. For example, if ${\Omega=1}$ and ${\omega=2}$, we are stretching the ${\phi}$ component by a factor of 2 relative to the ${\lambda}$ component.]

Now in order for the transformed components to have the same length ratio as the original, we require

 $\displaystyle \frac{dx_{\phi}^{2}+dy_{\phi}^{2}}{dx_{\lambda}^{2}+dy_{\lambda}^{2}}$ $\displaystyle =$ $\displaystyle \frac{a^{2}\cos^{2}\lambda d\phi^{2}}{a^{2}d\lambda^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\omega\left(x,y\right)\left(dx_{\phi}^{2}+dy_{\phi}^{2}\right)}{\Omega\left(x,y\right)\left(dx_{\lambda}^{2}+dy_{\lambda}^{2}\right)} \ \ \ \ \ (12)$

Therefore

$\displaystyle \omega\left(x,y\right)=\Omega\left(x,y\right) \ \ \ \ \ (13)$

The Mercator projection has a line element of

$\displaystyle ds^{2}=\frac{4\pi^{2}a^{2}}{\cosh^{2}\left(\frac{2\pi y}{H}\right)}\left(\frac{dx^{2}}{W^{2}}+\frac{dy^{2}}{H^{2}}\right) \ \ \ \ \ (14)$

Thus if ${W=H}$ (that is, it’s a square map), the condition 13 is satisfied.

Now suppose we want a projection that preserves relative areas instead of angles. The area element in an orthogonal system ${x^{i}}$ has sides ${\sqrt{\left|g_{11}\right|}dx^{1}}$ and ${\sqrt{\left|g_{22}\right|}dx^{2}}$, so the area element is

$\displaystyle dA=\sqrt{\left|g_{11}g_{22}\right|}dx^{1}dx^{2} \ \ \ \ \ (15)$

For the original spherical system this comes out to

$\displaystyle dA=\sqrt{a^{4}\cos^{2}\lambda}d\lambda d\phi=a^{2}\cos\lambda d\lambda d\phi \ \ \ \ \ (16)$

In the rectangular system, we have from 2 and 3:

 $\displaystyle dA$ $\displaystyle =$ $\displaystyle \left[g_{\lambda\lambda}\left(\frac{\partial\lambda}{\partial x}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial x}\right)^{2}\right]^{1/2}\left[g_{\lambda\lambda}\left(\frac{\partial\lambda}{\partial y}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial y}\right)^{2}\right]^{1/2}dx\;dy\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\left[\left(\frac{\partial\lambda}{\partial x}\right)^{2}+\cos^{2}\lambda\left(\frac{\partial\phi}{\partial x}\right)^{2}\right]^{1/2}\left[\left(\frac{\partial\lambda}{\partial y}\right)^{2}+\cos^{2}\lambda\left(\frac{\partial\phi}{\partial y}\right)^{2}\right]^{1/2}dx\;dy \ \ \ \ \ (18)$

To preserve areas, the coefficient of ${dx\;dy}$ should be a constant, since there is no preferred location on the ${xy}$ plane. Thus the condition is

$\displaystyle \left[\left(\frac{\partial\lambda}{\partial x}\right)^{2}+\cos^{2}\lambda\left(\frac{\partial\phi}{\partial x}\right)^{2}\right]\left[\left(\frac{\partial\lambda}{\partial y}\right)^{2}+\cos^{2}\lambda\left(\frac{\partial\phi}{\partial y}\right)^{2}\right]=\mbox{constant} \ \ \ \ \ (19)$

One projection that satisfies this condition is the Lambert cylindrical equal-area projection, where the transformations are

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \phi-\phi_{0}\ \ \ \ \ (20)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle \sin\lambda \ \ \ \ \ (21)$

where ${\phi_{0}}$ is the central line of longitude in the map. In this projection, the area element can be calculated from

 $\displaystyle \frac{\partial\lambda}{\partial x}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (22)$ $\displaystyle \frac{\partial\phi}{\partial x}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (23)$ $\displaystyle \frac{\partial\lambda}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1-y^{2}}}=\frac{1}{\cos\lambda}\ \ \ \ \ (24)$ $\displaystyle \frac{\partial\phi}{\partial y}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (25)$

Plugging these in, we get

$\displaystyle dA=a^{2}dx\;dy \ \ \ \ \ (26)$

[Note that here ${x}$ and ${y}$ are dimensionless as they are defined in terms of angles, not distances. Thus to get a map that fits into a book, we need to reduce the radius ${a}$ of the sphere from that of the Earth to something approximating a book page.]

It doesn’t appear that it’s possible to preserve both equal areas and relative angles (if it were, I’m sure that would be the most popular projection around). To do so, we’d need ${g_{xx}=g_{yy}}$ to give equal relative angles and from 18 this would require

$\displaystyle \left(\frac{\partial\lambda}{\partial x}\right)^{2}+\cos^{2}\lambda\left(\frac{\partial\phi}{\partial x}\right)^{2}=\left(\frac{\partial\lambda}{\partial y}\right)^{2}+\cos^{2}\lambda\left(\frac{\partial\phi}{\partial y}\right)^{2}=\mbox{constant} \ \ \ \ \ (27)$

Thus we need to get rid of the ${\cos^{2}\lambda}$ while still retaining some dependence on ${\phi}$ (otherwise, we’d have no map). However, since every term in this equation is non-negative, and ${\phi}$ can contain no reference to ${\lambda}$, there’s no way of cancelling off the ${\cos^{2}\lambda}$ terms, so it would seem that we can’t satisfy this equation. [I’m sure there’s a more rigorous way of proving this, but I’m not a mathematician.]

# Metric for the Mercator projection

Reference: Hobson, M.P., Efstathiou, G. P. & Lasenby, A. N. (2006), General Relativity: An Introduction for Physicists; Cambridge University Press. Problem 2.5.

As another example of finding the metric tensor in a new coordinate system, we’ll look at the Mercator projection of the Earth’s surface onto a 2-d planar map. The Mercator projection is the one in which lines of longitude are mapped into straight vertical lines, perpendicular to the straight horizontal lines of latitude. It is also hideously inaccurate in terms of relative sizes of land masses, as it makes Greenland look larger than South America (the actual areas are ${2.166\times10^{6}\mbox{ km}^{2}}$ for Greenland and ${17.84\times10^{6}\mbox{ km}^{2}}$ for South America).

To use conventional latitude and longitude, we note that longitude is equivalent to the spherical azimuthal angle ${\phi}$, with ${\phi=0}$ corresponding to the Greenwich meridian. Latitude ${\lambda}$ is (in radians) ${\lambda=\frac{\pi}{2}-\theta}$, where ${\theta}$ is the spherical polar angle. Therefore the line element in Earth coordinates can be obtained from the spherical line element:

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle a^{2}d\theta^{2}+a^{2}\sin^{2}\theta d\phi^{2}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}d\lambda^{2}+a^{2}\sin^{2}\left(\frac{\pi}{2}-\lambda\right)d\phi^{2}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\left(d\lambda^{2}+\cos^{2}\lambda d\phi^{2}\right) \ \ \ \ \ (3)$

If ${x}$ is the horizontal coordinate on a Mercator map and ${y}$ is the vertical coordinate, the transformation is (as given in Hobson’s question):

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{W\phi}{2\pi}\ \ \ \ \ (4)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle \frac{H}{2\pi}\ln\left[\tan\left(\frac{\pi}{4}+\frac{\lambda}{2}\right)\right] \ \ \ \ \ (5)$

where ${W}$ and ${H}$ are the constant width and height of the planar map.

To get the line element in Mercator coordinates, we use the transformation of the metric

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

We need the derivatives with respect to ${x}$ and ${y}$ of ${\phi}$ and ${\lambda}$, which we can get from implicit differentiation.

 $\displaystyle \frac{\partial x}{\partial x}$ $\displaystyle =$ $\displaystyle 1=\frac{W}{2\pi}\frac{\partial\phi}{\partial x}\ \ \ \ \ (7)$ $\displaystyle \frac{\partial\phi}{\partial x}$ $\displaystyle =$ $\displaystyle \frac{2\pi}{W}\ \ \ \ \ (8)$ $\displaystyle \frac{\partial\phi}{\partial y}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle \frac{\partial y}{\partial y}$ $\displaystyle =$ $\displaystyle 1=\frac{H}{4\pi}\frac{\sec^{2}\left(\frac{\pi}{4}+\frac{\lambda}{2}\right)}{\tan\left(\frac{\pi}{4}+\frac{\lambda}{2}\right)}\frac{\partial\lambda}{\partial y}\ \ \ \ \ (10)$ $\displaystyle \frac{\partial\lambda}{\partial x}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

Using the identity ${1+\tan^{2}u=\sec^{2}u}$, we have

 $\displaystyle \frac{\partial\lambda}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{4\pi}{H}\frac{\tan\left(\frac{\pi}{4}+\frac{\lambda}{2}\right)}{1+\tan^{2}\left(\frac{\pi}{4}+\frac{\lambda}{2}\right)}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{H}\frac{e^{2\pi y/H}}{1+e^{4\pi y/H}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi}{H}\frac{2}{e^{2\pi y/H}+e^{-2\pi y/H}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi}{H\cosh\left(2\pi y/H\right)} \ \ \ \ \ (15)$

The metric becomes

 $\displaystyle g_{xx}$ $\displaystyle =$ $\displaystyle g_{\lambda\lambda}\left(\frac{\partial\lambda}{\partial x}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial x}\right)^{2}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+\frac{4\pi^{2}a^{2}}{W^{2}}\cos^{2}\lambda \ \ \ \ \ (17)$

To get ${\cos^{2}\lambda}$ in terms of ${x}$ and ${y}$ we start with 5 and define ${\alpha\equiv2\pi y/H}$. Then

 $\displaystyle e^{\alpha}$ $\displaystyle =$ $\displaystyle \tan\left(\frac{\pi}{4}+\frac{\lambda}{2}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1+\tan\frac{\lambda}{2}}{1-\tan\frac{\lambda}{2}}\ \ \ \ \ (19)$ $\displaystyle \tan\frac{\lambda}{2}$ $\displaystyle =$ $\displaystyle \frac{e^{\alpha}-1}{e^{\alpha}+1}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \tanh\frac{\alpha}{2} \ \ \ \ \ (21)$

Now we can use the trig identity ${\tan^{2}\frac{\lambda}{2}=\frac{1-\cos\lambda}{1+\cos\lambda}}$ to get

 $\displaystyle \tan^{2}\frac{\lambda}{2}$ $\displaystyle =$ $\displaystyle \frac{1-\cos\lambda}{1+\cos\lambda}\ \ \ \ \ (22)$ $\displaystyle \cos\lambda$ $\displaystyle =$ $\displaystyle \frac{1-\tan^{2}\frac{\lambda}{2}}{1+\tan^{2}\frac{\lambda}{2}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-\tanh^{2}\frac{\alpha}{2}}{1+\tanh^{2}\frac{\alpha}{2}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\cosh^{2}\frac{\alpha}{2}-\sinh^{2}\frac{\alpha}{2}}{\cosh^{2}\frac{\alpha}{2}+\sinh^{2}\frac{\alpha}{2}}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\cosh^{2}\frac{\alpha}{2}+\sinh^{2}\frac{\alpha}{2}}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\cosh^{2}\frac{\alpha}{2}-1} \ \ \ \ \ (27)$

We can expand the denominator to get

 $\displaystyle 2\cosh^{2}\frac{\alpha}{2}-1$ $\displaystyle =$ $\displaystyle 2\left(\frac{e^{\alpha/2}+e^{-\alpha/2}}{2}\right)^{2}-1\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{e^{\alpha}+e^{-\alpha}+2}{2}-1\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cosh\alpha\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cosh\frac{2\pi y}{H} \ \ \ \ \ (31)$

Therefore

 $\displaystyle \cos^{2}\lambda$ $\displaystyle =$ $\displaystyle \frac{1}{\cosh^{2}\left(\frac{2\pi y}{H}\right)}\ \ \ \ \ (32)$ $\displaystyle g_{xx}$ $\displaystyle =$ $\displaystyle \frac{4\pi^{2}a^{2}}{W^{2}\cosh^{2}\left(\frac{2\pi y}{H}\right)} \ \ \ \ \ (33)$

For ${g_{yy}}$ we have from 15

 $\displaystyle g_{yy}$ $\displaystyle =$ $\displaystyle g_{\lambda\lambda}\left(\frac{\partial\lambda}{\partial y}\right)^{2}+g_{\phi\phi}\left(\frac{\partial\phi}{\partial y}\right)^{2}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\frac{4\pi^{2}}{H^{2}\cosh^{2}\left(\frac{2\pi y}{H}\right)} \ \ \ \ \ (35)$

The line element is therefore

$\displaystyle ds^{2}=\frac{4\pi^{2}a^{2}}{\cosh^{2}\left(\frac{2\pi y}{H}\right)}\left(\frac{dx^{2}}{W^{2}}+\frac{dy^{2}}{H^{2}}\right) \ \ \ \ \ (36)$

The space is approximately Euclidean near ${y=0}$, which occurs when ${\lambda=0}$, that is, at the equator. The Mercator projection is essentially a projection onto a cylinder wrapped around the Earth, which touches the Earth at the equator, so this makes sense.

# 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.]

# 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.

# Spherically symmetric solution to the Einstein equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Box 23.1.

We’ll return now to general relativity, and build up to a derivation of the Schwarzschild metric. As a quick review, the problem is to find a solution to the Einstein equation in the form

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

where ${R^{ij}}$ is the Ricci tensor, itself a contraction of the Riemann tensor, ${T^{ij}}$ is the stress-energy tensor and ${T=g_{ij}T^{ij}}$ is the stress-energy scalar.

In a practical problem, ${T^{ij}}$ will be given, and the problem is to determine the metric ${g^{ij}}$ from the Ricci tensor. The Ricci tensor is specified in terms of Christoffel symbols, which are in turn defined in terms of the metric and its derivatives, so the Einstein equation becomes a system of coupled, non-linear partial differential equations in the components of the metric tensor.

A good starting point is to look at spacetime around a source with spherical symmetry. We can picture this spacetime as a set of nested spherical shells, on the surface of which the usual 2-d spherical metric applies:

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

This gives us several of the metric components already, in that

 $\displaystyle g_{\theta\theta}$ $\displaystyle =$ $\displaystyle r^{2}\ \ \ \ \ (3)$ $\displaystyle g_{\phi\phi}$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\ \ \ \ \ (4)$ $\displaystyle g_{\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\phi\theta}=0 \ \ \ \ \ (5)$

We can set up the coordinates on each shell such that any line with fixed values of ${\theta}$ and ${\phi}$ is perpendicular to all the surfaces. This means that the basis vectors ${\mathbf{e}_{\theta}}$ and ${\mathbf{e}_{\phi}}$ are perpendicular to the third spatial basis vector ${\mathbf{e}_{r}}$, so that ${g_{r\theta}=g_{r\phi}=0}$. With spatial symmetry, there should be no difference in the way the metric treats motions in different directions of ${\theta}$ or ${\phi}$, so we’d expect the terms ${g_{r\theta}drd\theta}$, ${g_{r\phi}drd\phi}$, ${g_{t\theta}dtd\theta}$ and ${g_{t\phi}dtd\phi}$ to all be zero, which gives us four more (well, eight, actually, since ${g_{ij}=g_{ji}}$) metric components.

We’re left with ${g_{tt}}$, ${g_{rr}}$ and ${g_{rt}=g_{tr}}$. If ${t}$ is a time coordinate, we must have ${g_{tt}<0}$ and likewise, if ${r}$ is a spatial coordinate, then ${g_{rr}>0}$. We can, in fact, eliminate ${g_{rt}}$ by making a coordinate transformation as follows:

$\displaystyle t'=t+f\left(r,t\right) \ \ \ \ \ (6)$

where ${f}$ is some function of the original ${r}$ and ${t}$ coordinates (unknown at present). We can, in principle, always determine ${f}$ so that ${g_{rt'}=0}$ and then use ${t'}$ as our new time coordinate. [Note that we can’t use the symmetry argument to claim that ${g_{rt}=0}$, since in a spherically symmetric situation, it does make a difference whether you are travelling in the plus or minus ${r}$ direction, so it isn’t necessarily so that ${g_{rt}=0}$ in all cases.]

Take the differential of this equation to get

 $\displaystyle dt'$ $\displaystyle =$ $\displaystyle dt+\partial_{r}f\; dr+\partial_{t}f\; dt\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+\partial_{t}f\right)dt+\partial_{r}f\; dr\ \ \ \ \ (8)$ $\displaystyle dt$ $\displaystyle =$ $\displaystyle \frac{dt'-\partial_{r}f\; dr}{1+\partial_{t}f}\equiv\alpha\left(dt'-\partial_{r}f\; dr\right) \ \ \ \ \ (9)$

With the deductions above, our original metric equation is

$\displaystyle ds^{2}=g_{tt}dt^{2}+2g_{rt}dr\; dt+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (10)$

Substituting 9 into the first two terms on the RHS, we get

 $\displaystyle g_{tt}dt^{2}+2g_{rt}dr\; dt$ $\displaystyle =$ $\displaystyle g_{tt}\alpha^{2}\left(dt'-\partial_{r}f\; dr\right)^{2}+2\alpha g_{rt}dr\left(dt'-\partial_{r}f\; dr\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle dr^{2}\left(g_{tt}\alpha^{2}\left(\partial_{r}f\right)^{2}-2\alpha g_{rt}\partial_{r}f\right)+\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle dr\; dt'\left(2\alpha g_{rt}-2\alpha^{2}g_{tt}\partial_{r}f\right)+\left(dt'\right)^{2}g_{tt}\alpha^{2}\nonumber$

We can now set the coefficient of ${dr\; dt'}$ to zero to get

 $\displaystyle g_{rt}$ $\displaystyle =$ $\displaystyle \alpha g_{tt}\partial_{r}f\ \ \ \ \ (13)$ $\displaystyle \frac{g_{rt}}{g_{tt}}\left(1+\partial_{t}f\right)$ $\displaystyle =$ $\displaystyle \partial_{r}f \ \ \ \ \ (14)$

Assuming this partial differential equation for ${f\left(r,t\right)}$ can be solved (which we won’t be able to do a priori, since we don’t know ${g_{rt}}$ or ${g_{tt}}$, but in principle, the equation can be solved), it is always possible to find a time coordinate ${t'}$ such that ${g_{rt'}=0}$, so we might as well use that time coordinate from the start. Relabelling this time coordinate from ${t'}$ back to ${t}$, the spherically symmetric metric is then

$\displaystyle ds^{2}=g_{tt}dt^{2}+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (15)$

We therefore have only two metric components that need to be found by solving the Einstein equation 1, which we’ll get to in the next post.

# Metric tensor: spherical coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 5; Problem 5.6.

The non-rectangular coordinate systems (semi-log and sinusoidal) we’ve looked at so far have all been flat, so it’s time to look at one in curved space. We’ll use the surface of a sphere, but rather than the usual spherical coordinates we’ll use a slight variation. We keep the azimuthal angle ${\phi}$ but use as the second coordinate the quantity ${r}$ which is the distance along the surface of the sphere measured from the north pole. If the radius of the sphere is ${R}$, then in terms of normal spherical coordinates, ${r=R\theta}$.

Curves of constant ${\phi}$ are the usual lines of longitude, while curves of constant ${r}$ are lines of latitude. The tangents to the two curves at a given point are always perpendicular, so the metric ${g_{ij}}$ will be diagonal. To find the diagonal components, consider an infinitesimal displacement ${d\mathbf{s}}$. We have

$\displaystyle d\mathbf{s}=dr\mathbf{e}_{r}+d\phi\mathbf{e}_{\phi} \ \ \ \ \ (1)$

and our job is to find the two basis vectors.

The displacement along ${\mathbf{e}_{r}}$ is just ${dr=Rd\theta}$, so ${\mathbf{e}_{r}}$ is a unit vector. A displacement along ${\mathbf{e}_{\phi}}$ depends on the radius of the constant ${r}$ curve. In spherical coordinates, this is ${R\sin\theta}$, so in our new coordinate system we get the displacement as ${R\sin\theta d\phi=R\sin\frac{r}{R}d\phi}$. Therefore the magnitude of ${\mathbf{e}_{\phi}}$ is ${R\sin\frac{r}{R}}$. The metric tensor is thus

$\displaystyle g_{ij}=\left[\begin{array}{cc} 1 & 0\\ 0 & \left(R\sin\frac{r}{R}\right)^{2} \end{array}\right] \ \ \ \ \ (2)$

# Metric tensor as a stress-energy tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 20; Problem 20.7.

A curious possibility for the stress-energy tensor is

$\displaystyle T^{ij}=-\Lambda g^{ij} \ \ \ \ \ (1)$

where ${\Lambda}$ is a positive constant and ${g^{ij}}$ is any metric tensor. In a local inertial frame (LIF), ${g^{ij}=\eta^{ij}}$ and

$\displaystyle T^{ij}=\Lambda\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (2)$

Note that this tensor satisfies the energy conservation equation ${\nabla_{i}T^{ij}=0}$ as the covariant derivative ${\nabla_{i}g^{ij}}$ is always zero.

Comparing this with the form of ${T^{ij}}$ for a perfect fluid in its LIF

$\displaystyle T^{ij}=\left[\begin{array}{cccc} \rho & 0 & 0 & 0\\ 0 & P & 0 & 0\\ 0 & 0 & P & 0\\ 0 & 0 & 0 & P \end{array}\right] \ \ \ \ \ (3)$

we see that the energy density is ${\rho=\Lambda>0}$ and the pressure is ${P=-\Lambda<0}$. Such a tensor cannot arise from a perfect fluid because for such a fluid, for example

$\displaystyle T^{xx}=\int\int\int N\left(p\right)\frac{\left(p^{x}\right)^{2}}{p^{t}V}dp^{x}dp^{y}dp^{z} \ \ \ \ \ (4)$

The integrand is instrinsically non-negative because ${N\left(p\right)}$ is the number of particles in volume ${V}$ with a magnitude of momentum ${p}$ and must be non-negative. The component ${p^{t}=\gamma m}$ for massive particles or ${p^{t}=E}$ for photons, but in either case it too is positive. The numerator ${\left(p^{x}\right)^{2}}$, being a square, is also non-negative. Thus for a perfect fluid in its LIF, ${T^{xx}\ge0}$.

Although this tensor cannot be that of a perfect fluid, it does play a role in general relativity as we’ll hopefully see a bit later.

# Covariant derivative of the metric tensor: application to a coordinate transformation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Problem P17.10.

Here’s an application of the fact that the covariant derivative of any metric tensor is always zero. Suppose we define a coordinate transformation in which:

$\displaystyle \frac{\partial x^{a}}{\partial x^{\prime m}}=\delta_{\; m}^{a}-\left[\Gamma_{\; mn}^{a}\right]_{P}\Delta x_{P}^{\prime n} \ \ \ \ \ (1)$

where ${\left[\Gamma_{\; mn}^{a}\right]_{P}}$ is the Christoffel symbol in the primed system evaluated at a particular point ${P}$ (and therefore they are constants). (In Moore’s problem P17.10, he states that this is in the unprimed system, but the problem makes no sense in that case, since we’re summing over an index ${n}$ which refers to the unprimed coordinate system in one term and the primed system in the other.) The quantity ${\Delta x_{P}^{\prime n}\equiv x^{\prime n}-x_{p}^{\prime n}}$ represents a displacement from the point ${P}$, as measured in the primed system.

Using the usual transformation equation for a tensor, we get for the metric tensor:

 $\displaystyle g_{ij}^{\prime}$ $\displaystyle =$ $\displaystyle \frac{\partial x^{a}}{\partial x^{\prime i}}\frac{\partial x^{b}}{\partial x^{\prime j}}g_{ab}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\delta_{\; i}^{a}-\left[\Gamma_{\; in}^{a}\right]_{P}\Delta x_{P}^{\prime n}\right)\left(\delta_{\; j}^{b}-\left[\Gamma_{\; jn}^{b}\right]_{P}\Delta x_{P}^{\prime n}\right)g_{ab}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ij}-g_{ib}\left[\Gamma_{\; jn}^{b}\right]_{P}\Delta x_{P}^{\prime n}-g_{aj}\left[\Gamma_{\; in}^{a}\right]_{P}\Delta x_{P}^{\prime n}+\left[\Gamma_{\; is}^{a}\right]_{P}\Delta x_{P}^{\prime s}\left[\Gamma_{\; jn}^{b}\right]_{P}\Delta x_{P}^{\prime n}g_{ab}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ij}-\Delta x_{P}^{\prime n}\left(g_{ib}\left[\Gamma_{\; jn}^{b}\right]_{P}+g_{aj}\left[\Gamma_{\; in}^{a}\right]_{P}\right)+\left[\Gamma_{\; in}^{a}\right]_{P}\left[\Gamma_{\; js}^{b}\right]_{P}\Delta x_{P}^{\prime n}\Delta x_{P}^{\prime s}g_{ab} \ \ \ \ \ (5)$

When ${x^{\prime}=x_{P}^{\prime}}$, ${\Delta x_{P}^{\prime n}=0}$, so ${g_{ij}^{\prime}=g_{ij}}$ at point ${P}$.

Now consider the second term. By renaming the dummy indices ${b\rightarrow m}$ and ${a\rightarrow m}$, we have

$\displaystyle g_{ib}\left[\Gamma_{\; jn}^{b}\right]_{P}+g_{aj}\left[\Gamma_{\; in}^{a}\right]_{P}=g_{im}\left[\Gamma_{\; jn}^{m}\right]_{P}+g_{mj}\left[\Gamma_{\; in}^{m}\right]_{P} \ \ \ \ \ (6)$

Now because the covariant derivative (with respect to the primed coordinates) of the metric is zero, we have

$\displaystyle \nabla_{n}^{\prime}g_{ij}=\partial_{n}^{\prime}g_{ij}-\Gamma_{\; in}^{m}g_{mj}-\Gamma_{\; jn}^{m}g_{im}=0 \ \ \ \ \ (7)$

Therefore, we can write

$\displaystyle g_{ij}^{\prime}=g_{ij}-\partial_{n}^{\prime}g_{ij}\Delta x_{P}^{\prime n}+\left[\Gamma_{\; in}^{a}\right]_{P}\left[\Gamma_{\; js}^{b}\right]_{P}\Delta x_{P}^{\prime n}\Delta x_{P}^{\prime s}g_{ab} \ \ \ \ \ (8)$

If we take the derivative of this with respect to a particular primed coordinate ${x^{\prime k}}$ we can use

$\displaystyle \frac{\partial\Delta x_{P}^{\prime n}}{\partial x^{\prime k}}=\delta_{\; k}^{n} \ \ \ \ \ (9)$

and then evaluate the result at ${x^{\prime}=x_{P}^{\prime}}$ so that all terms containing ${\Delta x_{P}^{\prime n}}$ vanish. We get

 $\displaystyle \frac{\partial g_{ij}^{\prime}}{\partial x^{\prime k}}$ $\displaystyle =$ $\displaystyle \frac{\partial g_{ij}}{\partial x^{\prime k}}-\frac{\partial g_{ij}}{\partial x^{\prime n}}\delta_{\; k}^{n}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

Thus all the first derivatives of ${g_{ij}^{\prime}}$ are zero in the primed coordinate system.

# Christoffel symbols in terms of the metric tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Box 17.4.

It’s time to find out how to calculate the Christoffel symbols. We start with their definition in terms of the basis vectors in some coordinate system:

$\displaystyle \boxed{\partial_{j}\mathbf{e}_{i}=\Gamma_{\;ij}^{k}\mathbf{e}_{k}} \ \ \ \ \ (1)$

We take the scalar product of this equation with another basis vector ${\mathbf{e}_{l}}$ and use the definition of the metric tensor as ${g_{ij}=\mathbf{e}_{i}\cdot\mathbf{e}_{j}}$:

 $\displaystyle \Gamma_{\;ij}^{k}\mathbf{e}_{k}\cdot\mathbf{e}_{l}$ $\displaystyle =$ $\displaystyle \left(\partial_{j}\mathbf{e}_{i}\right)\cdot\mathbf{e}_{l}\ \ \ \ \ (2)$ $\displaystyle \Gamma_{\;ij}^{k}g_{kl}$ $\displaystyle =$ $\displaystyle \partial_{j}\left(\mathbf{e}_{i}\cdot\mathbf{e}_{l}\right)-\left(\partial_{j}\mathbf{e}_{l}\right)\cdot\mathbf{e}_{i}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{j}g_{il}-\Gamma_{\;lj}^{k}\mathbf{e}_{k}\cdot\mathbf{e}_{i}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{j}g_{il}-\Gamma_{\;lj}^{k}g_{ki}\ \ \ \ \ (5)$ $\displaystyle \Gamma_{\;ij}^{k}g_{kl}+\Gamma_{\;lj}^{k}g_{ki}$ $\displaystyle =$ $\displaystyle \partial_{j}g_{il} \ \ \ \ \ (6)$

In this equation the index ${k}$ is a dummy (being summed over), so only the indices ${i}$, ${j}$ and ${l}$ are specified. We can cyclically permute these indices to generate two more equations:

 $\displaystyle \Gamma_{\;jl}^{k}g_{ki}+\Gamma_{\;il}^{k}g_{kj}$ $\displaystyle =$ $\displaystyle \partial_{l}g_{ji}\ \ \ \ \ (7)$ $\displaystyle \Gamma_{\;li}^{k}g_{kj}+\Gamma_{\;ji}^{k}g_{kl}$ $\displaystyle =$ $\displaystyle \partial_{i}g_{lj} \ \ \ \ \ (8)$

We can now use the symmetry of the Christoffel symbols to solve for ${\Gamma_{\;ij}^{k}}$ by swapping indices in 7 and 8 to get

 $\displaystyle \Gamma_{\;lj}^{k}g_{ki}+\Gamma_{\;il}^{k}g_{kj}$ $\displaystyle =$ $\displaystyle \partial_{l}g_{ji}\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\;il}^{k}g_{kj}+\Gamma_{\;ij}^{k}g_{kl}$ $\displaystyle =$ $\displaystyle \partial_{i}g_{lj} \ \ \ \ \ (10)$

We can now add 6 to 10 and subtract 9 to get

$\displaystyle 2\Gamma_{\;ij}^{k}g_{kl}=\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji} \ \ \ \ \ (11)$

Finally we can use the fact that

$\displaystyle g^{ij}g_{jk}=\delta_{\;\;k}^{i} \ \ \ \ \ (12)$

and multiply both sides of 11 by ${g^{ml}}$ to get

 $\displaystyle 2\Gamma_{\;ij}^{k}g_{kl}g^{ml}$ $\displaystyle =$ $\displaystyle g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right)\ \ \ \ \ (13)$ $\displaystyle \Gamma_{\;ij}^{k}\delta_{\;\;k}^{m}$ $\displaystyle =$ $\displaystyle \frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right)\ \ \ \ \ (14)$ $\displaystyle \Gamma_{\;ij}^{m}$ $\displaystyle =$ $\displaystyle \frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (15)$

This gives us a formula for explicitly evaluating Christoffel symbols:

$\displaystyle \boxed{\Gamma_{\;ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right)} \ \ \ \ \ (16)$

This is a bit cumbersome to use as it requires finding the inverse metric tensor ${g^{ml}}$ and has 3 sums over different derivatives.

Example As an example, we’ll work out ${\Gamma_{\;ij}^{m}}$ for 2-D polar coordinates. The metric tensor and its inverse here are:

 $\displaystyle g_{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & r^{2} \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle g^{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & r^{-2} \end{array}\right] \ \ \ \ \ (18)$

so the derivatives are

 $\displaystyle \partial_{r}g_{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 0\\ 0 & 2r \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle \partial_{\theta}g_{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (20)$

Having only one non-zero derivative helps a lot, since the only non-zero term on the RHS of 16 is ${\partial_{r}g_{\theta\theta}=2r}$. We’ll work out a couple of the symbols explicitly and then give the final result.

 $\displaystyle \Gamma_{\;r\theta}^{\theta}$ $\displaystyle =$ $\displaystyle \frac{1}{2}g^{\theta l}\left(\partial_{\theta}g_{rl}+\partial_{r}g_{l\theta}-\partial_{l}g_{\theta r}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}g^{\theta\theta}\left(\partial_{\theta}g_{r\theta}+\partial_{r}g_{\theta\theta}-\partial_{\theta}g_{\theta r}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2r^{2}}\left(2r\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Gamma_{\;\theta r}^{\theta}\ \ \ \ \ (25)$ $\displaystyle \Gamma_{\;\theta\theta}^{r}$ $\displaystyle =$ $\displaystyle \frac{1}{2}g^{rl}\left(\partial_{\theta}g_{\theta l}+\partial_{\theta}g_{l\theta}-\partial_{l}g_{\theta\theta}\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}g^{rr}\left(\partial_{\theta}g_{\theta r}+\partial_{\theta}g_{r\theta}-\partial_{r}g_{\theta\theta}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}(-2r)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -r \ \ \ \ \ (29)$

All of the other symbols are zero. The final results are

 $\displaystyle \Gamma_{\;ij}^{r}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 0\\ 0 & -r \end{array}\right]\ \ \ \ \ (30)$ $\displaystyle \Gamma_{\;ij}^{\theta}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & r^{-1}\\ r^{-1} & 0 \end{array}\right] \ \ \ \ \ (31)$

Using these in 1 gives the 4 derivatives of the basis vectors:

 $\displaystyle \partial_{r}\mathbf{e}_{r}$ $\displaystyle =$ $\displaystyle \Gamma_{\;rr}^{i}\mathbf{e}_{i}=0\ \ \ \ \ (32)$ $\displaystyle \partial_{\theta}\mathbf{e}_{r}$ $\displaystyle =$ $\displaystyle \Gamma_{\;r\theta}^{i}\mathbf{e}_{i}=\frac{1}{r}\mathbf{e}_{\theta}\ \ \ \ \ (33)$ $\displaystyle \partial_{r}\mathbf{e}_{\theta}$ $\displaystyle =$ $\displaystyle \Gamma_{\;\theta r}^{i}\mathbf{e}_{i}=\frac{1}{r}\mathbf{e}_{\theta}\ \ \ \ \ (34)$ $\displaystyle \partial_{\theta}\mathbf{e}_{\theta}$ $\displaystyle =$ $\displaystyle \Gamma_{\;\theta\theta}^{i}\mathbf{e}_{i}=-r\mathbf{e}_{r} \ \ \ \ \ (35)$