# Plane symmetric spacetime

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

A static, plane-symmetric spacetime is one in which spacetime is independent of time (static) and is composed of a set of planes, where each plane is labelled by a coordinate ${x}$. Within each plane, points are labelled by coordinates ${y}$ and ${z}$ and because the spacetime is static, the distance between two points depends only on these two coordinates:

$\displaystyle \left[ds^{2}\right]_{x}=dy^{2}+dz^{2} \ \ \ \ \ (1)$

where the subscript ${x}$ denotes the plane with coordinate ${x}$.

If the ${x}$ basis vector ${\mathbf{e}_{x}}$ is everywhere perpendicular to ${\mathbf{e}_{y}}$ and ${\mathbf{e}_{z}}$ (and ${\mathbf{e}_{y}\perp\mathbf{e}_{z}}$), then the spatial off-diagonal components of the metric are zero

 $\displaystyle g_{ij}$ $\displaystyle \equiv$ $\displaystyle \mathbf{e}_{i}\cdot\mathbf{e}_{j}\ \ \ \ \ (2)$ $\displaystyle g_{xy}$ $\displaystyle =$ $\displaystyle g_{xz}=g_{yz}=0 \ \ \ \ \ (3)$

The general metric between any two spacetime points is then

$\displaystyle ds^{2}=g_{tt}dt^{2}+2g_{tx}dt\;dx+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (4)$

Because the spacetime is static, a displacement forward in time by ${dt}$ should give the same separation as a displacement backwards by the same amount ${-dt}$. Because of this symmetry, the ${2g_{tx}dt\;dx}$ term should remain unchanged when ${dt}$ is replaced by ${-dt}$. However, since the metric is independent of time, ${g_{tx}\left(t\right)=g_{tx}\left(-t\right)}$, so the only way we can satisfy the symmetry requirement is if ${g_{tx}=0}$. Thus the plane-symmetric metric is symmetric:

$\displaystyle ds^{2}=g_{tt}dt^{2}+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (5)$

Further, ${g_{tt}}$ can depend at most on ${x}$ alone.

To work out the consequences of this metric, we need to evaluate the Christoffel symbols and Ricci tensor. The Christoffel symbol worksheet is:

 ${\Gamma_{00}^{0}=\frac{1}{2A}A_{0}}$ ${\Gamma_{10}^{0}=\Gamma_{01}^{0}=\frac{1}{2A}A_{1}}$ ${\Gamma_{20}^{0}=\Gamma_{02}^{0}=\frac{1}{2A}A_{2}}$ ${\Gamma_{30}^{0}=\Gamma_{03}^{0}=\frac{1}{2A}A_{3}}$ ${\Gamma_{11}^{0}=\frac{1}{2A}B_{0}}$ ${\Gamma_{22}^{0}=\frac{1}{2A}C_{0}}$ ${\Gamma_{33}^{0}=\frac{1}{2A}D_{0}}$ other ${\Gamma_{\mu\nu}^{0}=0}$ ${\Gamma_{01}^{1}=\Gamma_{10}^{1}=\frac{1}{2B}B_{0}}$ ${\Gamma_{11}^{1}=\frac{1}{2B}B_{1}}$ ${\Gamma_{12}^{1}=\Gamma_{21}^{1}=\frac{1}{2B}B_{2}}$ ${\Gamma_{13}^{1}=\Gamma_{31}^{1}=\frac{1}{2B}B_{3}}$ ${\Gamma_{00}^{1}=\frac{1}{2B}A_{1}}$ ${\Gamma_{22}^{1}=-\frac{1}{2B}C_{1}}$ ${\Gamma_{33}^{1}=-\frac{1}{2B}D_{1}}$ other ${\Gamma_{\mu\nu}^{1}=0}$ ${\Gamma_{02}^{2}=\Gamma_{20}^{2}=\frac{1}{2C}C_{0}}$ ${\Gamma_{12}^{2}=\Gamma_{21}^{2}=\frac{1}{2C}C_{1}}$ ${\Gamma_{22}^{2}=\frac{1}{2C}C_{2}}$ ${\Gamma_{32}^{2}=\Gamma_{23}^{2}=\frac{1}{2C}C_{3}}$ ${\Gamma_{00}^{2}=\frac{1}{2C}A_{2}}$ ${\Gamma_{11}^{2}=-\frac{1}{2C}B_{2}}$ ${\Gamma_{33}^{2}=-\frac{1}{2C}D_{2}}$ other ${\Gamma_{\mu\nu}^{2}=0}$ ${\Gamma_{03}^{3}=\Gamma_{30}^{3}=\frac{1}{2D}D_{0}}$ ${\Gamma_{13}^{3}=\Gamma_{31}^{3}=\frac{1}{2D}D_{1}}$ ${\Gamma_{23}^{3}=\Gamma_{32}^{3}=\frac{1}{2D}D_{2}}$ ${\Gamma_{33}^{3}=\frac{1}{2D}D_{3}}$ ${\Gamma_{00}^{3}=\frac{1}{2D}A_{3}}$ ${\Gamma_{11}^{3}=-\frac{1}{2D}B_{3}}$ ${\Gamma_{22}^{3}=-\frac{1}{2D}C_{3}}$ other ${\Gamma_{\mu\nu}^{3}=0}$

In this case ${\left(x^{0},x^{1},x^{2},x^{3}\right)=\left(t,x,y,z\right)}$ and

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -g_{tt}\left(x\right)\ \ \ \ \ (6)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (7)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (8)$ $\displaystyle D$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (9)$

Thus the only nonzero symbols will be those involving ${A_{1}}$, since all other derivatives are zero. These are

 $\displaystyle \Gamma_{10}^{0}$ $\displaystyle =$ $\displaystyle \Gamma_{01}^{0}=\frac{1}{2A}A_{1}=\frac{1}{2A}\frac{dA}{dx}\ \ \ \ \ (10)$ $\displaystyle \Gamma_{00}^{1}$ $\displaystyle =$ $\displaystyle \frac{1}{2B}A_{1}=\frac{1}{2}\frac{dA}{dx} \ \ \ \ \ (11)$

[We can use the total derivative rather than partial because ${A}$ depends only on ${x}$.]

From the Ricci tensor worksheet, the only nonzero components of ${R_{\mu\nu}}$ are those involving ${A_{11}}$ or ${A_{1}}$ only, so we see that

 $\displaystyle R_{00}$ $\displaystyle =$ $\displaystyle \frac{1}{2B}A_{11}-\frac{1}{4BA}A_{1}^{2}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\frac{d^{2}A}{dx^{2}}-\frac{1}{4A}\left(\frac{dA}{dx}\right)^{2}\ \ \ \ \ (13)$ $\displaystyle R_{11}$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\frac{d^{2}A}{dx^{2}}+\frac{1}{4A}\left(\frac{dA}{dx}\right)^{2} \ \ \ \ \ (14)$

with all other ${R_{\mu\nu}=0}$. In flat space, all components satisfy ${R_{\mu\nu}=0}$ so these two components both give the same condition on ${A}$:

$\displaystyle \frac{d^{2}A}{dx^{2}}=\frac{1}{2A}\left(\frac{dA}{dx}\right)^{2} \ \ \ \ \ (15)$

To examine the structure of the spacetime, we need the full Riemann tensor, which is defined in terms of the Christoffel symbols:

$\displaystyle R_{\epsilon\nu\lambda\sigma}=g_{\epsilon\mu}R_{\;\nu\lambda\sigma}^{\mu}=g_{\epsilon\mu}\left[-\partial_{\sigma}\Gamma_{\;\lambda\nu}^{\mu}+\partial_{\lambda}\Gamma_{\;\sigma\nu}^{\mu}-\Gamma_{\;\lambda\nu}^{\kappa}\Gamma_{\;\kappa\sigma}^{\mu}+\Gamma_{\;\sigma\nu}^{\kappa}\Gamma_{\;\lambda\kappa}^{\mu}\right] \ \ \ \ \ (16)$

We can work out the terms in ${R_{\;\nu\lambda\sigma}^{\mu}}$ using 10 and 11. First, we’ll expand the implied sums and label the terms:

 $\displaystyle -\Gamma_{\;\lambda\nu}^{\kappa}\Gamma_{\;\kappa\sigma}^{\mu}$ $\displaystyle =$ $\displaystyle \overbrace{-\Gamma_{\;\lambda\nu}^{0}\Gamma_{\;0\sigma}^{\mu}}^{\left[1\right]}\overbrace{-\Gamma_{\;\lambda\nu}^{1}\Gamma_{\;1\sigma}^{\mu}}^{\left[2\right]}\ \ \ \ \ (17)$ $\displaystyle \Gamma_{\;\sigma\nu}^{\kappa}\Gamma_{\;\lambda\kappa}^{\mu}$ $\displaystyle =$ $\displaystyle \overbrace{\Gamma_{\;\sigma\nu}^{0}\Gamma_{\;\lambda0}^{\mu}}^{\left[3\right]}\overbrace{+\Gamma_{\;\sigma\nu}^{1}\Gamma_{\;\lambda1}^{\mu}}^{\left[4\right]}\ \ \ \ \ (18)$ $\displaystyle -\partial_{\sigma}\Gamma_{\;\lambda\nu}^{\mu}+\partial_{\lambda}\Gamma_{\;\sigma\nu}^{\mu}$ $\displaystyle =$ $\displaystyle \overbrace{-\partial_{\sigma}\Gamma_{\;\lambda\nu}^{\mu}}^{\left[5\right]}\overbrace{+\partial_{\lambda}\Gamma_{\;\sigma\nu}^{\mu}}^{\left[6\right]} \ \ \ \ \ (19)$

Next, we’ll identify the index combinations that give (potentially) nonzero values for components of ${R_{\;\nu\lambda\sigma}^{\mu}}$ in each term, using the fact that only ${\Gamma_{10}^{0}}$ and ${\Gamma_{00}^{1}}$ are nonzero, and that only the derivative with respect to ${x}$ (index 1) is nonzero.

• Term 1:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1
• Term 2:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 0 0
• Term 3:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 1 1 1 0 0 1 0 1 1 0 1 1 0 0
• Term 4:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 0 0
• Term 5:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 1 1 0 1 0 1 1 0 0 1
• Term 6:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 1 1 0 1 1 0 1 0 1 0

From these tables, we see that there are 7 unique index combinations that can potentially give nonzero Riemann tensor components ${R_{\;\nu\lambda\sigma}^{\mu}}$. We have (remember that the Christoffel symbols are symmetric in their lower 2 indices: ${\Gamma_{\nu\lambda}^{\mu}=\Gamma_{\lambda\nu}^{\mu}}$):

 $\displaystyle R_{\;100}^{1}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{00}^{1}+\Gamma_{01}^{0}\Gamma_{00}^{1}=0\ \ \ \ \ (20)$ $\displaystyle R_{\;011}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{01}^{0}+\Gamma_{01}^{0}\Gamma_{01}^{0}-\partial_{1}\Gamma_{01}^{0}+\partial_{1}\Gamma_{01}^{0}=0\ \ \ \ \ (21)$ $\displaystyle R_{\;000}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{00}^{1}\Gamma_{10}^{0}+\Gamma_{00}^{1}\Gamma_{10}^{0}=0\ \ \ \ \ (22)$ $\displaystyle R_{\;010}^{1}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{00}^{1}+\partial_{1}\Gamma_{00}^{1}\ \ \ \ \ (23)$ $\displaystyle R_{\;001}^{1}$ $\displaystyle =$ $\displaystyle +\Gamma_{01}^{0}\Gamma_{00}^{1}-\partial_{1}\Gamma_{00}^{1}=-R_{\;010}^{1}\ \ \ \ \ (24)$ $\displaystyle R_{\;101}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{01}^{0}-\partial_{1}\Gamma_{01}^{0}\ \ \ \ \ (25)$ $\displaystyle R_{\;110}^{0}$ $\displaystyle =$ $\displaystyle \Gamma_{01}^{0}\Gamma_{01}^{0}+\partial_{1}\Gamma_{01}^{0}=-R_{\;101}^{0} \ \ \ \ \ (26)$

Thus only the last 4 can potentially be nonzero. To go further, we need the derivative terms:

 $\displaystyle \partial_{x}\Gamma_{\;10}^{0}$ $\displaystyle =$ $\displaystyle -\frac{1}{2A^{2}}\left(\frac{dA}{dx}\right)^{2}+\frac{1}{2A}\frac{d^{2}A}{dx^{2}}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2A^{2}}A_{1}^{2}+\frac{1}{2A}A_{11}\ \ \ \ \ (28)$ $\displaystyle \partial_{x}\Gamma_{\;00}^{1}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\frac{d^{2}A}{dx^{2}}=\frac{1}{2}A_{11} \ \ \ \ \ (29)$

Now we can use 10 and 11 to write these components in terms of ${A}$:

 $\displaystyle R_{\;010}^{1}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{00}^{1}+\partial_{1}\Gamma_{00}^{1}=-\frac{1}{4A}A_{1}^{2}+\frac{1}{2}A_{11}\ \ \ \ \ (30)$ $\displaystyle R_{\;001}^{1}$ $\displaystyle =$ $\displaystyle -R_{\;010}^{1}=\frac{1}{4A}A_{1}^{2}-\frac{1}{2}A_{11}\ \ \ \ \ (31)$ $\displaystyle R_{\;101}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{01}^{0}-\partial_{1}\Gamma_{01}^{0}=\frac{1}{4A^{2}}A_{1}^{2}-\frac{1}{2A}A_{11}\ \ \ \ \ (32)$ $\displaystyle R_{\;110}^{0}$ $\displaystyle =$ $\displaystyle -R_{\;101}^{0}=-\frac{1}{4A^{2}}A_{1}^{2}+\frac{1}{2A}A_{11} \ \ \ \ \ (33)$

To get the Riemann tensor with all 4 indices lowered, we multiply by the metric:

$\displaystyle R_{\epsilon\nu\lambda\sigma}=g_{\epsilon\mu}R_{\;\nu\lambda\sigma}^{\mu} \ \ \ \ \ (34)$

Here, the only two metric components we need are ${g_{00}=-A}$ and ${g_{11}=1}$ so

 $\displaystyle R_{1010}$ $\displaystyle =$ $\displaystyle g_{11}R_{\;010}^{1}=-\frac{1}{4A}A_{1}^{2}+\frac{1}{2}A_{11}\ \ \ \ \ (35)$ $\displaystyle R_{1001}$ $\displaystyle =$ $\displaystyle g_{11}R_{\;001}^{1}=\frac{1}{4A}A_{1}^{2}-\frac{1}{2}A_{11}\ \ \ \ \ (36)$ $\displaystyle R_{0101}$ $\displaystyle =$ $\displaystyle g_{00}R_{\;101}^{0}=-\frac{1}{4A}A_{1}^{2}+\frac{1}{2}A_{11}\ \ \ \ \ (37)$ $\displaystyle R_{0110}$ $\displaystyle =$ $\displaystyle g_{00}R_{\;110}^{0}=\frac{1}{4A}A_{1}^{2}-\frac{1}{2}A_{11} \ \ \ \ \ (38)$

Note that in this lowered form, the symmetries of the Riemann tensor are obeyed: ${R_{\mu\nu\lambda\sigma}=-R_{\nu\mu\lambda\sigma}=-R_{\mu\nu\sigma\lambda}}$.

Finally, if we impose the condition 15 in the form ${A_{11}=\frac{1}{2A}A_{1}^{2}}$, we find that all four of these components are zero, thus making the entire Riemann tensor zero, indicating that spacetime is completely flat. [There are a lot of indices flying about here, so I’m hoping I got them all right…]

# Schwarzschild metric: the Newtonian limit & Christoffel symbol worksheet

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

In our derivation of the Schwarzschild metric, we got as far as finding the dependence of the metric on the spacetime coordinates, giving the form

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

The final task is to find the constant ${X}$, which we can do by considering the behaviour of the metric for large ${r}$ and requiring that it reduce to the Newtonian graviational force law in that limit. [I’ve renamed the constant ${C}$ in the original post to ${X}$ here to avoid confusion with the ${C}$ that turns up in the metric tensor below.]

For an object initially at rest, the spatial comopnents of its four-velocity are all zero: ${u^{i}=0}$. However, the contraction of ${\mathbf{u}}$ with itself gives the invariant ${\mathbf{u}\cdot\mathbf{u}=-1}$, so we have

 $\displaystyle \mathbf{u}\cdot\mathbf{u}$ $\displaystyle =$ $\displaystyle g_{\mu\nu}u^{\mu}u^{\nu}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{tt}\left(u^{t}\right)^{2}=-1\ \ \ \ \ (3)$ $\displaystyle u^{t}$ $\displaystyle =$ $\displaystyle \sqrt{-\frac{1}{g_{tt}}} \ \ \ \ \ (4)$

Any object’s trajectory obeys the geodesic equation which, in terms of Christoffel symbols, is

$\displaystyle \ddot{x}^{\mu}+\Gamma_{\;\nu\sigma}^{\mu}\dot{x}^{\nu}\dot{x}^{\sigma}=0 \ \ \ \ \ (5)$

where a dot denotes a derivative with respect to proper time ${\tau}$, so that ${\dot{x}^{\nu}=u^{\nu}}$.

In our case, this reduces to

 $\displaystyle \ddot{x}^{\mu}+\Gamma_{tt}^{\mu}\left(u^{t}\right)^{2}$ $\displaystyle =$ $\displaystyle \ddot{x}^{\mu}-\frac{\Gamma_{tt}^{\mu}}{g_{tt}}=0\ \ \ \ \ (6)$ $\displaystyle \ddot{x}^{\mu}$ $\displaystyle =$ $\displaystyle \frac{\Gamma_{tt}^{\mu}}{g_{tt}}=-\frac{1}{A}\Gamma_{tt}^{\mu} \ \ \ \ \ (7)$

where ${A=-g_{tt}}$. We therefore need to calculate the Christoffel symbols ${\Gamma_{tt}^{\mu}}$, which we can do from their expression in terms of ${g_{\mu\nu}}$:

$\displaystyle \Gamma_{\;\nu\sigma}^{\mu}=\frac{1}{2}g^{\mu\lambda}\left(\partial_{\sigma}g_{\nu\lambda}+\partial_{\nu}g_{\lambda\sigma}-\partial_{\lambda}g_{\sigma\nu}\right) \ \ \ \ \ (8)$

This can get quite tedious, but Moore provides a worksheet in the Appendix which simplifies the task. The notation is the same as that used for Ricci tensor worksheet for the generic diagonal metric, written as

$\displaystyle ds^{2}=-A\left(dx^{0}\right)^{2}+B\left(dx^{1}\right)^{2}+C\left(dx^{2}\right)^{2}+D\left(dx^{3}\right)^{2} \ \ \ \ \ (9)$

where ${x^{0}}$ is the time coordinate and the other three are space coordinates. Note the minus sign in the first term: this makes explicit the fact that the metric component for time should be negative. Thus we have ${g_{00}=-A}$, ${g_{11}=B}$, ${g_{22}=C}$ and ${g_{33}=D}$.

Derivatives with respect to coordinates are written as subscripts, so that ${A_{01}=\frac{\partial^{2}A}{\partial x^{0}\partial x^{1}}}$ and so on. It’s important not to confuse this notation with tensor notation; ${A_{01}}$ is not the 01 component of a tensor. Although we don’t need all the Christoffel symbols here, I’ve produced the table for reference.

 ${\Gamma_{00}^{0}=\frac{1}{2A}A_{0}}$ ${\Gamma_{10}^{0}=\Gamma_{01}^{0}=\frac{1}{2A}A_{1}}$ ${\Gamma_{20}^{0}=\Gamma_{02}^{0}=\frac{1}{2A}A_{2}}$ ${\Gamma_{30}^{0}=\Gamma_{03}^{0}=\frac{1}{2A}A_{3}}$ ${\Gamma_{11}^{0}=\frac{1}{2A}B_{0}}$ ${\Gamma_{22}^{0}=\frac{1}{2A}C_{0}}$ ${\Gamma_{33}^{0}=\frac{1}{2A}D_{0}}$ other ${\Gamma_{\mu\nu}^{0}=0}$ ${\Gamma_{01}^{1}=\Gamma_{10}^{1}=\frac{1}{2B}B_{0}}$ ${\Gamma_{11}^{1}=\frac{1}{2B}B_{1}}$ ${\Gamma_{12}^{1}=\Gamma_{21}^{1}=\frac{1}{2B}B_{2}}$ ${\Gamma_{13}^{1}=\Gamma_{31}^{1}=\frac{1}{2B}B_{3}}$ ${\Gamma_{00}^{1}=\frac{1}{2B}A_{1}}$ ${\Gamma_{22}^{1}=-\frac{1}{2B}C_{1}}$ ${\Gamma_{33}^{1}=-\frac{1}{2B}D_{1}}$ other ${\Gamma_{\mu\nu}^{1}=0}$ ${\Gamma_{02}^{2}=\Gamma_{20}^{2}=\frac{1}{2C}C_{0}}$ ${\Gamma_{12}^{2}=\Gamma_{21}^{2}=\frac{1}{2C}C_{1}}$ ${\Gamma_{22}^{2}=\frac{1}{2C}C_{2}}$ ${\Gamma_{32}^{2}=\Gamma_{23}^{2}=\frac{1}{2C}C_{3}}$ ${\Gamma_{00}^{2}=\frac{1}{2C}A_{2}}$ ${\Gamma_{11}^{2}=-\frac{1}{2C}B_{2}}$ ${\Gamma_{33}^{2}=-\frac{1}{2C}D_{2}}$ other ${\Gamma_{\mu\nu}^{2}=0}$ ${\Gamma_{03}^{3}=\Gamma_{30}^{3}=\frac{1}{2D}D_{0}}$ ${\Gamma_{13}^{3}=\Gamma_{31}^{3}=\frac{1}{2D}D_{1}}$ ${\Gamma_{23}^{3}=\Gamma_{32}^{3}=\frac{1}{2D}D_{2}}$ ${\Gamma_{33}^{3}=\frac{1}{2D}D_{3}}$ ${\Gamma_{00}^{3}=\frac{1}{2D}A_{3}}$ ${\Gamma_{11}^{3}=-\frac{1}{2D}B_{3}}$ ${\Gamma_{22}^{3}=-\frac{1}{2D}C_{3}}$ other ${\Gamma_{\mu\nu}^{3}=0}$

In our case, we need only the ${\Gamma_{00}^{\mu}}$ terms, which occur in the first column. Since ${A=\left(1+\frac{X}{r}\right)}$ the derivatives with respect to ${t}$, ${\theta}$ and ${\phi}$ are all zero, and the only non-zero Christoffel symbol is

$\displaystyle \Gamma_{tt}^{r}=\Gamma_{00}^{1}=\frac{1}{2B}A_{1}=\frac{1}{2}\left(1+\frac{X}{r}\right)\frac{\partial A}{\partial r}=-\frac{X}{2r^{2}}\left(1+\frac{X}{r}\right) \ \ \ \ \ (10)$

Therefore from 7 we have

$\displaystyle \ddot{x}^{r}=\frac{d^{2}r}{d\tau^{2}}=-\frac{1}{A}\left(-\frac{X}{2r^{2}}\left(1+\frac{X}{r}\right)\right)=\frac{X}{2r^{2}} \ \ \ \ \ (11)$

For large ${r}$, the Schwarzschild metric reduces to flat space, so the radial coordinate becomes the Newtonian radial coordinate and the proper time ${\tau}$ becomes the Newtonian time ${t}$, so

$\displaystyle \frac{d^{2}r}{d\tau^{2}}\rightarrow\frac{d^{2}r}{dt^{2}}=\frac{X}{2r^{2}} \ \ \ \ \ (12)$

This is equivalent to Newton’s law of gravity for a mass a distance ${r}$ from a mass ${M}$ if

$\displaystyle X=-2GM \ \ \ \ \ (13)$

[The minus sign indicates that the test mass accelerates towards ${M}$, that is, in the direction of decreasing ${r}$.]

Making this substitution in 1 we get the final form of the Schwarzschild metric

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (14)$

# Christoffel symbols for a general diagonal metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Appendix.

In the appendix to Moore’s book, he gives some worksheets which can be used to calculate Christoffel symbols and the Ricci tensor for a general diagonal metric. I thought it would be useful to show where these formulas come from.

The key to calculating Christoffel symbols is to compare two forms of the geodesic equation. Suppose our diagonal metric is

$\displaystyle ds^{2}=-A\left(dx^{0}\right)^{2}+B\left(dx^{1}\right)^{2}+C\left(dx^{2}\right)^{2}+D\left(dx^{3}\right)^{2} \ \ \ \ \ (1)$

where ${A,B,C}$ and ${D}$ are positive functions of the coordinates. Note the minus sign in the first term; this ensures that ${x^{0}}$ is a time coordinate.

The two forms of the geodesic equation are

$\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (2)$

and in terms of the Christoffel symbols:

$\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (3)$

For a diagonal metric, 2 becomes (no sum over ${a}$):

$\displaystyle g_{aa}\ddot{x}^{a}+\sum_{i}\left[\partial_{i}g_{aa}\dot{x}^{a}\dot{x}^{i}-\frac{1}{2}\partial_{a}g_{ii}\left(\dot{x}^{i}\right)^{2}\right]=0 \ \ \ \ \ (4)$

Therefore, for ${a=0}$ we have, using the notation ${A_{i}\equiv\partial_{i}A}$:

$\displaystyle -A\ddot{x}^{0}-\sum_{i}A_{i}\dot{x}^{a}\dot{x}^{i}-\frac{1}{2}\left[-A_{0}\left(\dot{x}^{0}\right)^{2}+B_{0}\left(\dot{x}^{1}\right)^{2}+C_{0}\left(\dot{x}^{2}\right)^{2}+D_{0}\left(\dot{x}^{3}\right)^{2}\right]=0 \ \ \ \ \ (5)$

Dividing through by ${-A}$ we get

$\displaystyle \ddot{x}^{0}+\frac{1}{A}\sum_{i}A_{i}\dot{x}^{0}\dot{x}^{i}+\frac{1}{2A}\left[-A_{0}\left(\dot{x}^{0}\right)^{2}+B_{0}\left(\dot{x}^{1}\right)^{2}+C_{0}\left(\dot{x}^{2}\right)^{2}+D_{0}\left(\dot{x}^{3}\right)^{2}\right]=0 \ \ \ \ \ (6)$

We can now compare this term by term with 3 to read off the Christoffel symbols ${\Gamma_{\; ij}^{0}}$:

 $\displaystyle \Gamma_{\;00}^{0}$ $\displaystyle =$ $\displaystyle \frac{A_{0}}{A}-\frac{A_{0}}{2A}=\frac{A_{0}}{2A}\ \ \ \ \ (7)$ $\displaystyle \Gamma_{\;01}^{0}$ $\displaystyle =$ $\displaystyle \Gamma_{\;10}^{0}=\frac{A_{0}}{2A}\ \ \ \ \ (8)$ $\displaystyle \Gamma_{\;11}^{0}$ $\displaystyle =$ $\displaystyle \frac{B_{0}}{2A} \ \ \ \ \ (9)$

and so on. The off-diagonal symbols such as ${\Gamma_{\;01}^{0}=\Gamma_{\;10}^{0}}$ are obtained from the ${\frac{1}{A}\sum_{i}A_{i}\dot{x}^{0}\dot{x}^{i}}$ term, remembering that each term in the sum contributes to two Christoffel symbols. For example

$\displaystyle \frac{1}{A}A_{1}\dot{x}^{0}\dot{x}^{1}=\left(\Gamma_{\;01}^{0}+\Gamma_{\;10}^{0}\right)\dot{x}^{0}\dot{x}^{1} \ \ \ \ \ (10)$

The complete worksheet for the Christoffel symbols and Ricci tensor is available as a PDF from Moore’s website.

# Riemann and Ricci tensors in the weak field limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Boxes 22.2-22.4.

The Einstein equation is

$\displaystyle G^{ij}+\Lambda g^{ij}=8\pi GT^{ij} \ \ \ \ \ (1)$

where ${\Lambda}$ is the cosmological constant and ${T^{ij}}$ is the stress-energy tensor. The Einstein tensor ${G^{ij}}$ is defined in terms of the Ricci tensor and curvature scalar as

$\displaystyle G^{ij}\equiv R^{ij}-\frac{1}{2}g^{ij}R \ \ \ \ \ (2)$

The Einstein equation can be written in the alternative form

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (3)$

where

$\displaystyle T\equiv g_{ij}T^{ij} \ \ \ \ \ (4)$

In preparation for examining the behaviour of this equation in the weak field limit, we need to see how the Ricci tensor behaves for metrics that are nearly flat. Since the Ricci tensor is a contraction of the Riemann tensor, we need first to see how the Riemann tensor behaves in this limit.

The metric for a weak field can be written as

$\displaystyle g_{ij}=\eta_{ij}+h_{ij} \ \ \ \ \ (5)$

where ${\eta_{ij}}$ is the flat space metric and ${h_{ij}}$ is a small perturbation, where small means ${\left|h_{ij}\right|\ll1}$. Note that since ${g_{ij}}$ and ${\eta_{ij}}$ are both symmetric, we must have

$\displaystyle h_{ij}=h_{ji} \ \ \ \ \ (6)$

As an approximation, we’ll discard all terms of order ${\left|h_{ij}\right|^{2}}$ or higher. First, we’ll get an approximation for the inverse metric ${g^{ij}}$. Since ${\eta_{ij}}$ is diagonal with elements ${\left[-1,1,1,1\right]}$ it is its own inverse so

$\displaystyle \eta^{ij}=\eta_{ij} \ \ \ \ \ (7)$

Therefore it’s reasonable to assume that

$\displaystyle g^{ij}=\eta^{ij}+b^{ij} \ \ \ \ \ (8)$

where ${\left|b^{ij}\right|\ll1}$. We can find ${b^{ij}}$ from the condition that

$\displaystyle g^{ij}g_{jk}=\delta_{k}^{i} \ \ \ \ \ (9)$

Therefore

 $\displaystyle g^{ij}g_{jk}$ $\displaystyle =$ $\displaystyle \eta^{ij}\eta_{jk}+\eta^{ij}h_{jk}+b^{ij}\eta_{jk}+b^{ij}h_{jk}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \delta_{k}^{i}+\eta^{ij}h_{jk}+b^{ij}\eta_{jk} \ \ \ \ \ (11)$

to first order (since we’re assuming both ${b^{ij}}$ and ${h_{jk}}$ are very small). Therefore

 $\displaystyle b^{ij}\eta_{jk}$ $\displaystyle =$ $\displaystyle -\eta^{ij}h_{jk}\ \ \ \ \ (12)$ $\displaystyle b^{ij}\eta^{kl}\eta_{jk}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (13)$ $\displaystyle b^{ij}\delta_{j}^{l}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (14)$ $\displaystyle b^{il}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -h^{il} \ \ \ \ \ (16)$

where

$\displaystyle h^{il}\equiv\eta^{kl}\eta^{ij}h_{jk} \ \ \ \ \ (17)$

Thus the perturbation ${h^{ij}}$ in the inverse metric is of the same order as the perturbation ${h_{jk}}$ in the original metric which means that, to first order, we can raise an index of a tensor whose components are all of order ${h_{jk}}$ by simply multiplying by the flat space metric ${\eta^{ij}}$ rather than the full metric ${g^{ij}}$. That is, for some tensor ${A_{kl}}$ where ${\left|A_{kl}\right|\ll1}$ we have, to first order

$\displaystyle A_{\; l}^{j}=g^{jk}A_{kl}=\left(\eta^{jk}-h^{jk}\right)A_{kl}=\eta^{jk}A_{kl}-\mathcal{O}\left(\left|h\right|^{2}\right)\approx\eta^{jk}A_{kl} \ \ \ \ \ (18)$

From 3 with ${\Lambda=0}$ and indices lowered, the Einstein equation is

$\displaystyle R_{ij}=8\pi G\left(T_{ij}-\frac{1}{2}g_{ij}T\right) \ \ \ \ \ (19)$

so we need to see what ${R_{ij}}$ looks like in the weak field limit. We’ll start with the Christoffel symbols

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (20)$

In the weak field limit

$\displaystyle \partial_{j}g_{il}=\partial_{j}\left(\eta_{il}+h_{il}\right)=\partial_{j}h_{il} \ \ \ \ \ (21)$

since ${\partial_{j}\eta_{il}=0}$. Therefore,

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}h_{il}+\partial_{i}h_{lj}-\partial_{l}h_{ji}\right) \ \ \ \ \ (22)$

This result is exact so far. Now if we want only first order terms, we can replace ${g^{ml}=\eta^{ml}-h^{ml}}$ by just ${\eta^{ml}}$ since the ${h^{ml}}$ multiplied into the derivatives gives only second order terms. Therefore, to first order

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}\eta^{ml}\left(\partial_{j}h_{il}+\partial_{i}h_{lj}-\partial_{l}h_{ji}\right) \ \ \ \ \ (23)$

Now the Riemann tensor is defined as

$\displaystyle R_{\; jlm}^{i}\equiv-\left[\partial_{m}\Gamma_{\; lj}^{i}-\partial_{l}\Gamma_{\; mj}^{i}+\Gamma_{\; lj}^{k}\Gamma_{\; km}^{i}-\Gamma_{\; mj}^{k}\Gamma_{\; lk}^{i}\right] \ \ \ \ \ (24)$

Since the Christoffel symbols are all of order ${h_{ij}}$ the last two terms in the Riemann tensor are of order ${\left|h_{ij}\right|^{2}}$ so we can drop them. With the first index lowered, we get

 $\displaystyle R_{njlm}$ $\displaystyle =$ $\displaystyle -g_{ni}\left[\partial_{m}\Gamma_{\; lj}^{i}-\partial_{l}\Gamma_{\; mj}^{i}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\eta_{ni}\left[\eta^{ik}\partial_{m}\left(\partial_{j}h_{lk}+\partial_{l}h_{kj}-\partial_{k}h_{jl}\right)-\eta^{ik}\partial_{l}\left(\partial_{j}h_{mk}+\partial_{m}h_{kj}-\partial_{k}h_{jm}\right)\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left[\partial_{m}\left(\partial_{j}h_{ln}+\partial_{l}h_{nj}-\partial_{n}h_{jl}\right)-\partial_{l}\left(\partial_{j}h_{mn}+\partial_{m}h_{nj}-\partial_{n}h_{jm}\right)\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\partial_{j}\partial_{l}h_{nm}+\partial_{n}\partial_{m}h_{jl}-\partial_{n}\partial_{l}h_{jm}-\partial_{j}\partial_{m}h_{nl}\right] \ \ \ \ \ (28)$

where we used 6 to get the last line.

Now to get the Ricci tensor in the weak field limit. We have, to first order

 $\displaystyle R_{jm}$ $\displaystyle =$ $\displaystyle g^{nl}R_{njlm}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\eta^{nl}\left[\partial_{j}\partial_{l}h_{nm}+\partial_{n}\partial_{m}h_{jl}-\partial_{n}\partial_{l}h_{jm}-\partial_{j}\partial_{m}h_{nl}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\left(\partial_{j}\partial_{l}h_{nm}-\frac{1}{2}\partial_{j}\partial_{m}h_{nl}\right)+\eta^{nl}\left(\partial_{n}\partial_{m}h_{jl}-\frac{1}{2}\partial_{j}\partial_{m}h_{nl}\right)\right]-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\eta^{nl}\partial_{n}\partial_{l}h_{jm}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\partial_{j}\left(\partial_{l}h_{nm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{jl}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{ln}\partial_{j}\left(\partial_{l}h_{nm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{lj}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\partial_{j}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{lj}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (35)$

where

$\displaystyle H_{m}\equiv\eta^{nl}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right) \ \ \ \ \ (36)$

The index magic we performed above includes using ${\eta^{nl}=\eta^{ln}}$ and ${h_{nl}=h_{ln}}$ in 33 and then swapping the bound indices ${l}$ and ${n}$ and then using ${h_{ln}=h_{nl}}$ again in the first term in 34. The Ricci tensor in the weak field limit is thus

$\displaystyle R_{jm}=\frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (37)$

# Ricci tensor and curvature scalar for a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.6.

As an example of calculating the Ricci tensor and curvature scalar we’ll find them for the 2-d surface of a sphere. The Ricci tensor is calculated from the Riemann tensor, and that in turn depends on the Christoffel symbols, so we’ll need them first. It’s easiest to find them from the geodesic equation

$\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (1)$

which is formally equivalent to

$\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (2)$

The metric for a sphere is

$\displaystyle ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)$

where ${r}$ is the constant radius of the sphere, so

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

We have two equations arising from 1. For ${a=\theta}$

$\displaystyle r^{2}\ddot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (6)$

Comparing with 2 we get, after dividing out the ${r^{2}}$:

$\displaystyle \Gamma_{\;\phi\phi}^{\theta}=-\sin\theta\cos\theta=-\frac{1}{2}\sin2\theta \ \ \ \ \ (7)$

For ${a=\phi}$:

$\displaystyle r^{2}\sin^{2}\theta\ddot{\phi}^{2}+2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}=0 \ \ \ \ \ (8)$

Dividing through by ${r^{2}\sin^{2}\theta}$ and comparing with 2 we get (remember that the second term is ${\left(\Gamma_{\;\theta\phi}^{\phi}+\Gamma_{\;\phi\theta}^{\phi}\right)\dot{\theta}\dot{\phi}}$ and that ${\Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}}$):

$\displaystyle \Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}=\cot\theta \ \ \ \ \ (9)$

All other Christoffel symbols are zero.

In 2D, the Riemann tensor has only one independent component, which we can take to be ${R_{\theta\phi\theta\phi}}$, which can be calculated from

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (10)$

Lowering the first index, we have

 $\displaystyle R_{\theta\phi\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\theta i}R_{\;\phi\theta\phi}^{i}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(-\cos2\theta-0+0-\Gamma_{\;\theta\phi}^{\phi}\Gamma_{\;\phi\phi}^{\theta}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\sin^{2}\theta-\cos^{2}\theta+\cos^{2}\theta\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (16)$

We can now find the Ricci tensor.

$\displaystyle R_{ij}=g^{ab}R_{aibj} \ \ \ \ \ (17)$

Since the metric is diagonal and ${g^{ab}}$ is the inverse of ${g_{ab}}$, we have

 $\displaystyle g^{\theta\theta}$ $\displaystyle =$ $\displaystyle \frac{1}{g^{\theta\theta}}=\frac{1}{r^{2}}\ \ \ \ \ (18)$ $\displaystyle g^{\phi\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{g^{\phi\phi}}=\frac{1}{r^{2}\sin^{2}\theta} \ \ \ \ \ (19)$

so, using the symmetries of the Riemann tensor,

 $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle g^{ab}R_{a\theta b\theta}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\phi\phi}R_{\phi\theta\phi\theta}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}\sin^{2}\theta}r^{2}\sin^{2}\theta\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (23)$ $\displaystyle R_{\phi\phi}$ $\displaystyle =$ $\displaystyle g^{ab}R_{a\phi b\phi}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\theta\theta}R_{\theta\phi\theta\phi}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sin^{2}\theta\ \ \ \ \ (26)$ $\displaystyle R_{\theta\phi}$ $\displaystyle =$ $\displaystyle R_{\phi\theta}=0 \ \ \ \ \ (27)$

We can get the upstairs version of the Ricci tensor as well:

 $\displaystyle R^{\theta\theta}$ $\displaystyle =$ $\displaystyle g^{\theta i}g^{\theta j}R_{ij}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\theta\theta}g^{\theta\theta}R_{\theta\theta}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}}\ \ \ \ \ (30)$ $\displaystyle R^{\phi\phi}$ $\displaystyle =$ $\displaystyle g^{\phi\phi}g^{\phi\phi}R_{\phi\phi}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}\sin^{2}\theta} \ \ \ \ \ (32)$

The curvature scalar is

 $\displaystyle R$ $\displaystyle =$ $\displaystyle g_{ij}R^{ij}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\frac{1}{r^{4}}+r^{2}\sin^{2}\theta\frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{r^{2}} \ \ \ \ \ (35)$

As we would expect, the curvature of a sphere decreases as its radius gets larger.

# The Bianchi identity for the Riemann tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; d.

Another relation of the Riemann tensor involves the covariant derivative of the tensor, and is known as the Bianchi identity (actually the second Bianchi identity; the first
identity is the symmetry
relation
${R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}=0}$ that we saw earlier). The identity is easiest to derive at the origin of a locally inertial frame (LIF), where the first derivatives of the metric tensor, and thus the Christoffel symbols, are all zero. At this point, we have

$\displaystyle R_{nj\ell m}=\frac{1}{2}\left(\partial_{\ell}\partial_{j}g_{mn}+\partial_{m}\partial_{n}g_{j\ell}-\partial_{\ell}\partial_{n}g_{jm}-\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (1)$

If the Christoffel symbols are all zero, then the covariant derivative becomes the ordinary derivative

$\displaystyle \nabla_{j}A^{k}\equiv\partial_{j}A^{k}+A^{i}\Gamma_{\; ij}^{k}=\partial_{j}A^{k} \ \ \ \ \ (2)$

Therefore, we get, at the origin of a LIF:

 $\displaystyle \nabla_{k}R_{nj\ell m}$ $\displaystyle =$ $\displaystyle \partial_{k}R_{nj\ell m}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{k}\partial_{\ell}\partial_{j}g_{mn}+\partial_{k}\partial_{m}\partial_{n}g_{j\ell}-\partial_{k}\partial_{\ell}\partial_{n}g_{jm}-\partial_{k}\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (4)$

By cyclically permuting the index of the derivative with the last two indices of the tensor, we get

 $\displaystyle \nabla_{\ell}R_{njmk}$ $\displaystyle =$ $\displaystyle \partial_{\ell}R_{njmk}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{\ell}\partial_{m}\partial_{j}g_{kn}+\partial_{\ell}\partial_{k}\partial_{n}g_{jm}-\partial_{\ell}\partial_{m}\partial_{n}g_{jk}-\partial_{\ell}\partial_{k}\partial_{j}g_{mn}\right)\ \ \ \ \ (6)$ $\displaystyle \nabla_{m}R_{njk\ell}$ $\displaystyle =$ $\displaystyle \partial_{m}R_{njk\ell}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{m}\partial_{k}\partial_{j}g_{\ell n}+\partial_{m}\partial_{\ell}\partial_{n}g_{jk}-\partial_{m}\partial_{k}\partial_{n}g_{j\ell}-\partial_{m}\partial_{\ell}\partial_{j}g_{kn}\right) \ \ \ \ \ (8)$

By adding up 4, 6 and 8 and using the commutativity of partial derivatives, we see that the terms cancel in pairs, so we get

$\displaystyle \boxed{\nabla_{k}R_{nj\ell m}+\nabla_{\ell}R_{njmk}+\nabla_{m}R_{njk\ell}=0} \ \ \ \ \ (9)$

As usual we can use the argument that since we can set up a LIF with its origin at any non-singular point in spacetime, this equation is true everywhere and since the covariant derivative is a tensor, this is a tensor equation and is thus valid in all coordinate systems. This is the Bianchi identity.

# Riemann tensor for surface of a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem 19.4.

As an example of the Riemann tensor in 2-d curved space we can use our old standby of the surface of a sphere. As usual, we need the Christoffel symbols and we get them by comparing the two forms of the geodesic equation.

 $\displaystyle \frac{d}{d\tau}\left(g_{aj}\dot{x}^{j}\right)-\frac{1}{2}\partial_{a}g_{ij}\dot{x}^{i}\dot{x}^{j}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (2)$

where as usual a dot denotes a derivative with respect to proper time ${\tau}$.

For a sphere, the interval is

$\displaystyle ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)$

Note that ${r}$ (the radius of the sphere) is a constant here.

From 1 we get, with ${a=\theta}$:

$\displaystyle r^{2}\ddot{x}^{j}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (4)$

Dividing through by ${r^{2}}$ and comparing with 2 we get

 $\displaystyle \Gamma_{\phi\phi}^{\theta}$ $\displaystyle =$ $\displaystyle -\sin\theta\cos\theta\ \ \ \ \ (5)$ $\displaystyle \Gamma_{\theta\phi}^{\theta}$ $\displaystyle =$ $\displaystyle \Gamma_{\phi\theta}^{\theta}=\Gamma_{\theta\theta}^{\theta}=0 \ \ \ \ \ (6)$

With ${a=\phi}$ we have

 $\displaystyle 2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}+r^{2}\sin^{2}\theta\ddot{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle 2\cot\theta\dot{\theta}\dot{\phi}+\ddot{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \Gamma_{\theta\phi}^{\phi}=\Gamma_{\phi\theta}^{\phi}$ $\displaystyle =$ $\displaystyle \cot\theta\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\theta\theta}^{\phi}=\Gamma_{\phi\phi}^{\phi}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (10)$

We can use these results to get the Riemann tensor. Unfortunately, in the form ${R_{\; bcd}^{a}}$, the Riemann tensor doesn’t have all the symmetries of the form ${R_{abcd}}$, so if we want the latter form, we need to work out the former form first and then use

 $\displaystyle R_{abcd}$ $\displaystyle =$ $\displaystyle g_{af}R_{\; bcd}^{f}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{af}\left(\partial_{c}\Gamma_{\; db}^{f}-\partial_{d}\Gamma_{\; cb}^{f}+\Gamma_{\; db}^{k}\Gamma_{\; ck}^{f}-\Gamma_{\; cb}^{k}\Gamma_{\; kd}^{f}\right) \ \ \ \ \ (12)$

Although we know there is only one independent component in 2-d, we can work out all four non-zero components to see how the calculations go.

 $\displaystyle R_{\theta\phi\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\theta f}R_{\;\phi\theta\phi}^{f}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\sin^{2}\theta-\cos^{2}\theta-0+0+\cos^{2}\theta\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\ \ \ \ \ (17)$ $\displaystyle R_{\theta\phi\phi\theta}$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\phi\theta}^{\theta}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}-\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}+\Gamma_{\;\theta\phi}^{k}\Gamma_{\;\phi k}^{\theta}-\Gamma_{\;\phi\phi}^{k}\Gamma_{\; k\theta}^{\theta}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{\theta\phi\theta\phi}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -r^{2}\sin^{2}\theta\ \ \ \ \ (21)$ $\displaystyle R_{\phi\theta\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}R_{\;\theta\theta\phi}^{\phi}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\left(\partial_{\theta}\Gamma_{\;\phi\theta}^{\phi}-\partial_{\phi}\Gamma_{\;\theta\theta}^{\phi}+\Gamma_{\;\phi\theta}^{k}\Gamma_{\;\theta k}^{\phi}-\Gamma_{\;\theta\theta}^{k}\Gamma_{\; k\phi}^{\phi}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\left(-\frac{1}{\sin^{2}\theta}-0+0+\frac{\cos^{2}\theta}{\sin^{2}\theta}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -r^{2}\sin^{2}\theta\ \ \ \ \ (25)$ $\displaystyle R_{\phi\theta\phi\theta}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}R_{\;\theta\phi\theta}^{\phi}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\left(\partial_{\phi}\Gamma_{\;\theta\theta}^{\phi}-\partial_{\theta}\Gamma_{\;\phi\theta}^{\phi}+\Gamma_{\;\theta\theta}^{k}\Gamma_{\; k\phi}^{\phi}-\Gamma_{\;\phi\theta}^{k}\Gamma_{\;\theta k}^{\phi}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{\phi\theta\theta\phi}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (29)$

Finally, we can calculate one of the other components to verify that it’s zero.

 $\displaystyle R_{\theta\theta\theta\theta}$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\theta\theta\theta}^{\theta}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\theta\theta}^{\theta}-\partial_{\theta}\Gamma_{\;\theta\theta}^{\theta}+\Gamma_{\;\theta\theta}^{k}\Gamma_{\; k\theta}^{\theta}-\Gamma_{\;\theta\theta}^{k}\Gamma_{\;\theta k}^{\theta}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (32)$

# Riemann tensor for an infinite plane of mass

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 18; 3.

We’ve seen that an infinite plane of mass causes no tidal effects in Newtonian physics. If this were also true in general relativity, that is, if there were no geodesic deviation, then according to GR, space is flat.

If we take the ${x}$ axis perpendicular to the plane, then the system is completely symmetric in the ${t}$, ${y}$ and ${z}$ dimensions, so any metric tensor describing spacetime in this system can depend only on ${x}$. If take the metric tensor to be diagonal, then we get

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle -dt^{2}+f\left(x\right)dx^{2}+dy^{2}+dz^{2}\ \ \ \ \ (1)$ $\displaystyle g_{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & f\left(x\right) & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (2)$

where ${f\left(x\right)}$ is an arbitrary function of ${x}$ only. If the Riemann tensor for this metric is identically zero, then we can say that this metric describes a flat spacetime. The Riemann tensor is

$\displaystyle R_{\; j\ell m}^{i}\equiv\pm\left[\partial_{m}\Gamma_{\;\ell j}^{i}-\partial_{\ell}\Gamma_{\; mj}^{i}+\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i}-\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}\right] \ \ \ \ \ (3)$

so we need the Christoffel symbols, which we can get by comparing the two forms of the geodesic equation. These equations are

 $\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (4)$ $\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (5)$

Using the first equation, we see that if ${a=t,\; y}$ or ${z}$, the geodesic equation is

$\displaystyle g_{aa}\ddot{x}^{a}=0\mbox{ (no sum)} \ \ \ \ \ (6)$

For ${a=x}$ we get

$\displaystyle f\left(x\right)\ddot{x}+\frac{1}{2}f'\left(x\right)\dot{x}^{2}=0 \ \ \ \ \ (7)$

Dividing through by ${f\left(x\right)}$ and comparing with 5, we see that the only non-zero Christoffel symbol is

$\displaystyle \Gamma_{\; xx}^{x}=\frac{f'\left(x\right)}{2f\left(x\right)} \ \ \ \ \ (8)$

From 3, the only possibly non-zero component of the Riemann tensor is ${R_{\; xxx}^{x}}$, since the only non-zero Christoffel symbol is ${\Gamma_{\; xx}^{x}}$ and the only non-zero derivative of the Christoffel symbols is with respect to ${x}$. However, in this case the first two terms in 3 cancel, as do the last two, so ${R_{\;\ell mj}^{i}=0}$ for all components. Thus spacetime is indeed flat for this metric.

# Covariant derivative of the metric tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; 9.

One interesting and useful theorem is that the covariant derivative of any metric tensor is always zero. We can show this by using the expression for the covariant derivative of a general tensor to say:

$\displaystyle \nabla_{j}g_{kl}=\partial_{j}g_{kl}-\Gamma_{\; jk}^{m}g_{ml}-\Gamma_{\; jl}^{m}g_{km} \ \ \ \ \ (1)$

We can combine this with the explicit expression for the Christoffel symbols:

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

Substituting, we get

$\displaystyle \nabla_{j}g_{kl}=\partial_{j}g_{kl}-\frac{1}{2}g^{mn}\left(\partial_{k}g_{jn}+\partial_{j}g_{nk}-\partial_{n}g_{kj}\right)g_{ml}-\frac{1}{2}g^{mn}\left(\partial_{l}g_{jln}+\partial_{j}g_{nl}-\partial_{n}g_{lj}\right)g_{km} \ \ \ \ \ (3)$

Since ${g^{mn}g_{ml}=\delta_{\; l}^{n}}$, we get

$\displaystyle \nabla_{j}g_{kl}=\partial_{j}g_{kl}-\frac{1}{2}\left(\partial_{k}g_{jl}+\partial_{j}g_{lk}-\partial_{l}g_{kj}\right)-\frac{1}{2}\left(\partial_{l}g_{jk}+\partial_{j}g_{kl}-\partial_{k}g_{lj}\right) \ \ \ \ \ (4)$

Now we use the symmetry of the metric tensor: ${g_{kl}=g_{lk}}$:

 $\displaystyle \nabla_{j}g_{kl}$ $\displaystyle =$ $\displaystyle \partial_{j}g_{lk}-\frac{1}{2}\left(\partial_{k}g_{jl}+\partial_{j}g_{lk}-\partial_{l}g_{kj}\right)-\frac{1}{2}\left(\partial_{l}g_{kj}+\partial_{j}g_{lk}-\partial_{k}g_{jl}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

# Covariant derivative of a vector in the Schwarzschild metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; 8.

Here’s another example of calculating the covariant derivative in the Schwarzschild (S) metric. We’re given a vector with coordinates in the S metric of:

$\displaystyle \mathbf{v}=\left[1-\frac{2GM}{r},0,0,0\right] \ \ \ \ \ (1)$

The covariant derivative is given by

$\displaystyle \nabla_{j}v^{k}\equiv\frac{\partial v^{k}}{\partial x^{j}}+v^{i}\Gamma_{\; ij}^{k} \ \ \ \ \ (2)$

Since the only non-zero component of ${\mathbf{v}}$ is ${v^{t}}$ and it depends only on ${r}$, most of the terms are zero.

$\displaystyle \Gamma_{\; ij}^{t}=\left[\begin{array}{cccc} 0 & \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (3)$

$\displaystyle \Gamma_{\; ij}^{r}=\left[\begin{array}{cccc} \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right) & 0 & 0 & 0\\ 0 & -\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ 0 & 0 & -r\left(1-\frac{2GM}{r}\right) & 0\\ 0 & 0 & 0 & -r\sin^{2}\theta\left(1-\frac{2GM}{r}\right) \end{array}\right] \ \ \ \ \ (4)$

$\displaystyle \Gamma_{\; ij}^{\theta}=\left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & \frac{1}{r} & 0\\ 0 & \frac{1}{r} & 0 & 0\\ 0 & 0 & 0 & -\sin\theta\cos\theta \end{array}\right] \ \ \ \ \ (5)$

$\displaystyle \Gamma_{\; ij}^{\phi}=\left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{r}\\ 0 & 0 & 0 & \cot\theta\\ 0 & \frac{1}{r} & \cot\theta & 0 \end{array}\right] \ \ \ \ \ (6)$

The one non-zero derivative is

$\displaystyle \frac{\partial v^{t}}{\partial r}=\frac{2GM}{r^{2}} \ \ \ \ \ (7)$

and the values of the second term in 2 are

 $\displaystyle v^{i}\Gamma_{\; ir}^{t}$ $\displaystyle =$ $\displaystyle \frac{GM}{r^{2}}\ \ \ \ \ (8)$ $\displaystyle v^{i}\Gamma_{\; it}^{r}$ $\displaystyle =$ $\displaystyle \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{2} \ \ \ \ \ (9)$

with all other terms being zero.

The covariant derivative is then (with ${j}$ the row index and ${k}$ the column index):

$\displaystyle \nabla_{j}v^{k}=\left[\begin{array}{cccc} 0 & \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{2} & 0 & 0\\ \frac{3GM}{r^{2}} & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (10)$