## The Bianchi identity for the Riemann tensor

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

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.

## A million visitors

At around 7 AM UK time today physicspages got its one millionth visitor. I don’t know who this was so I can’t give them any prizes, but I’d like to express my appreciation for all the visitors I’ve had over the past 2 or 3 years. I’ve learned a lot writing all the posts and it’s gratifying that others find the information useful as well.

On to 2 million now…

## 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 ${a=\theta}$:

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

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: counting components in general

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problems 19.2, 19.3.

We can generalize the method for counting the number of independent components in the Riemann tensor to ${n}$-dimensional spacetime. As before, we know that the first pair and last pair of indices must both consist of different values in order for the component to be (possibly) non-zero. With ${n}$ components to choose from, this gives us ${\binom{n}{2}^{2}=\left[\frac{1}{2}n\left(n-1\right)\right]^{2}}$ components. If we arrange these components in a ${\frac{1}{2}n\left(n-1\right)\times\frac{1}{2}n\left(n-1\right)}$ matrix with the rows and columns labelled by the first and second pairs of indices, respectively, then due to the condition

$\displaystyle R_{nj\ell m}=R_{\ell mnj} \ \ \ \ \ (1)$

the lower triangle of this matrix is a mirror of the upper triangle, so the possible number of independent components is reduced to at most ${\sum_{i=1}^{n\left(n-1\right)/2}=\frac{1}{2}\left(\frac{1}{2}n\left(n-1\right)\right)\left(\frac{1}{2}n\left(n-1\right)+1\right)}$. This gives

$\displaystyle \frac{1}{2}\left(\frac{1}{2}n\left(n-1\right)\right)\left(\frac{1}{2}n\left(n-1\right)+1\right)=\frac{1}{8}n\left(n-1\right)\left[n\left(n-1\right)+2\right] \ \ \ \ \ (2)$

We now need to apply the final symmetry condition, which is

$\displaystyle R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}=0 \ \ \ \ \ (3)$

As we saw in the last post, this equation gives new constraints only if all four indices are different, and the order in which these indices appear in the first term doesn’t matter. Therefore, this equation provides a total of ${\binom{n}{4}=\frac{1}{24}n\left(n-1\right)\left(n-2\right)\left(n-3\right)}$ constraints, so the total number of independent components is

 $\displaystyle N\left(n\right)$ $\displaystyle =$ $\displaystyle \frac{1}{8}n\left(n-1\right)\left[n\left(n-1\right)+2\right]-\frac{1}{24}n\left(n-1\right)\left(n-2\right)\left(n-3\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{24}\left(n^{2}-n\right)\left(n^{2}-n+2\right)-\frac{1}{24}\left(n^{2}-n\right)\left(n^{2}-5n+6\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{24}\left(n^{2}-n\right)\left(2n^{2}+2n\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{12}\left(n^{2}-n\right)\left(n^{2}+n\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{12}\left(n^{4}-n^{2}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{12}n^{2}\left(n^{2}-1\right) \ \ \ \ \ (9)$

This formula works even if ${n<4}$, since the second term in 4 is zero in this case.

The numbers of independent components for the first few dimensions are

 ${n}$ ${N\left(n\right)}$ 2 1 3 6 4 20 5 50 6 105

\

## Riemann tensor: counting independent components

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.3.

The symmetries of the Riemann tensor mean that only some of its components are independent. The two conditions

 $\displaystyle R_{jn\ell m}$ $\displaystyle =$ $\displaystyle -R_{nj\ell m}\ \ \ \ \ (1)$ $\displaystyle R_{njm\ell}$ $\displaystyle =$ $\displaystyle -R_{nj\ell m} \ \ \ \ \ (2)$

show that all components where either the first and second indices, or the third and fourth indices are equal must be zero. In four dimensional spacetime, this means that at most ${\binom{4}{2}^{2}=36}$ components can be non-zero, since we can choose two distinct values for both the first and last pairs of indices.

The condition

$\displaystyle R_{nj\ell m}=R_{\ell mnj} \ \ \ \ \ (3)$

means that ${R_{nl\ell m}}$ is symmetric with respect to its two pairs of indices. If we arrange the 36 non-zero components in a ${6\times6}$ matrix where the rows and columns are labelled by the distinct pairs of values 01, 02, 03, 12, 13, 23, then the lower triangle of this matrix is the mirror image of the upper triangle, meaning we can eliminate ${\sum_{i=1}^{5}i=15}$ more components, leaving ${36-15=21}$ possibly independent components.

There is one final symmetry condition for the Riemann tensor, and it is the trickiest to handle.

$\displaystyle R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}=0 \ \ \ \ \ (4)$

The first thing we need to show about this condition is that if any two indices are equal, then 4 follows from the other three conditions and tells us nothing new. To see this, consider the various ways in which two indices can be equal.

First, suppose ${n=j}$. Then ${R_{nn\ell m}=0}$ from 1, so we’re left with

 $\displaystyle R_{n\ell mn}+R_{nmn\ell}$ $\displaystyle =$ $\displaystyle R_{n\ell mn}+R_{n\ell nm}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{n\ell mn}-R_{n\ell mn}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

The first line uses 3 and the second line uses 2. Because 4 cyclically permutes the last three indices, the same argument also applies to the cases ${n=\ell}$ and ${n=m}$.

Now suppose ${j=\ell}$. Then the third term in 4 is ${R_{nmjj}=0}$ using 2, so we’re left with

$\displaystyle R_{njjm}+R_{njmj}=R_{njjm}-R_{njjm}=0 \ \ \ \ \ (8)$

using 2.

For ${j=m}$ the second term in 4 is ${R_{n\ell jj}=0}$ so

$\displaystyle R_{nj\ell j}+R_{njj\ell}=R_{nj\ell j}-R_{nj\ell j}=0 \ \ \ \ \ (9)$

Finally, if ${\ell=m}$, the first term in 4 is ${R_{njmm}=0}$ and

$\displaystyle R_{nmmj}+R_{nmjm}=R_{nmmj}-R_{nmmj}=0 \ \ \ \ \ (10)$

Now that we know that 4 gives us new information only if all the indices are different, how many ways can we choose these indices? Given that we have four indices, we might think that there are ${4!=24}$ possibilities, depending on the ordering of the indices. In fact, for any set of four indices, the condition gives us only one independent constraint, as the four different indices can be placed in any order in the first term. Starting with 4 with all indices different, suppose we put ${j}$ first instead of ${n}$. We can do this by swapping ${n}$ and ${j}$ to get

 $\displaystyle R_{jn\ell m}+R_{j\ell mn}+R_{jmn\ell}$ $\displaystyle =$ $\displaystyle -R_{nj\ell m}+R_{mnj\ell}+R_{n\ell jm}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{nj\ell m}-R_{nmj\ell}-R_{n\ell mj}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (13)$

We used the first three symmetries in the first and second lines and then 4 to get the zero in the third line.

If we swap ${n}$ with ${\ell}$ we get

 $\displaystyle R_{\ell jnm}+R_{\ell nmj}+R_{\ell mjn}$ $\displaystyle =$ $\displaystyle R_{nm\ell j}-R_{n\ell mj}+R_{jn\ell m}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{nmj\ell}-R_{n\ell mj}-R_{nj\ell m}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (16)$

If we swap ${n}$ and ${m}$ we get

 $\displaystyle R_{mj\ell n}+R_{m\ell nj}+R_{mnj\ell}$ $\displaystyle =$ $\displaystyle R_{\ell nmj}+R_{njm\ell}-R_{nmj\ell}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{n\ell mj}-R_{nj\ell m}-R_{nmj\ell}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (19)$

Swapping any two of the last three indices gives the same result, since these three indices are present in a cyclic permutation, so we need to consider only one such case, say swapping ${j}$ with ${m}$:

 $\displaystyle R_{nm\ell j}+R_{n\ell jm}+R_{njm\ell}$ $\displaystyle =$ $\displaystyle -R_{nmj\ell}-R_{n\ell mj}-R_{nj\ell m}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (21)$

Therefore, the condition 4 can give us only ${\binom{4}{4}=1}$ extra constraint. The total number of independent components in four-dimensional spacetime is therefore ${21-1=20}$.

Example As another example, we can apply this reasoning to find the number of independent components in two dimensions. First, the number of possible pairs with distinct values is ${\binom{2}{2}=1}$, so the matrix referred to above is only ${1\times1}$. To verify that the condition 4 doesn’t reduce the number of independent components any further, note that with only 2 possible values for indices, we must repeat at least one of them when choosing the indices in ${R_{nj\ell m}}$, so this condition doesn’t in fact tell us anything new. Thus there is only one independent component in two dimensions.

## Riemann tensor: symmetries

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Boxes 19.1, 19.2.

We can derive a few useful symmetries of the Riemann tensor by looking at its form in a locally inertial frame (LIF). At the origin of such a frame, all first derivatives of ${g_{ij}}$ are zero, which means the Christoffel symbols are all zero there. However, the second derivatives of ${g_{ij}}$ are not, in general, zero, so the derivatives of the Christoffel symbols will not, in general, be zero either.

Using the definition of the Riemann tensor:

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (1)$

we can write it at the origin of a LIF:

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i} \ \ \ \ \ (2)$

The Christoffel symbols are

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (3)$

The symmetries of the Riemann tensor are easiest to write if we look at its form with all indices lowered, that is:

 $\displaystyle R_{nj\ell m}$ $\displaystyle =$ $\displaystyle g_{nk}R_{\; j\ell m}^{k}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{nk}\left(\partial_{\ell}\Gamma_{\; mj}^{k}-\partial_{m}\Gamma_{\;\ell j}^{k}\right) \ \ \ \ \ (5)$

First, we calculate the derivative:

$\displaystyle \partial_{\ell}\Gamma_{\; mj}^{k}=\frac{1}{2}\partial_{\ell}g^{ki}\left(\partial_{j}g_{mi}+\partial_{m}g_{ij}-\partial_{i}g_{jm}\right)+\frac{1}{2}g^{ki}\left(\partial_{\ell}\partial_{j}g_{mi}+\partial_{\ell}\partial_{m}g_{ij}-\partial_{\ell}\partial_{i}g_{jm}\right) \ \ \ \ \ (6)$

At the origin of a LIF, the first term is zero since all first derivatives of ${g_{ij}}$ are zero, so we’re left with

$\displaystyle \partial_{\ell}\Gamma_{\; mj}^{k}=\frac{1}{2}g^{ki}\left(\partial_{\ell}\partial_{j}g_{mi}+\partial_{\ell}\partial_{m}g_{ij}-\partial_{\ell}\partial_{i}g_{jm}\right) \ \ \ \ \ (7)$

Multiplying this by ${g_{kn}}$ and using ${g_{kn}g^{ik}=\delta_{n}^{i}}$, we have

 $\displaystyle g_{kn}\partial_{\ell}\Gamma_{\; mj}^{k}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\delta_{n}^{i}\left(\partial_{\ell}\partial_{j}g_{mi}+\partial_{\ell}\partial_{m}g_{ij}-\partial_{\ell}\partial_{i}g_{jm}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{\ell}\partial_{j}g_{mn}+\partial_{\ell}\partial_{m}g_{nj}-\partial_{\ell}\partial_{n}g_{jm}\right) \ \ \ \ \ (9)$

By substituting indices, we can get the second term in 5:

$\displaystyle g_{nk}\partial_{m}\Gamma_{\;\ell j}^{k}=\frac{1}{2}\left(\partial_{m}\partial_{j}g_{\ell n}+\partial_{m}\partial_{\ell}g_{nj}-\partial_{m}\partial_{n}g_{j\ell}\right) \ \ \ \ \ (10)$

Subtracting 10 from 9 we see that the middle terms cancel, so we’re left with

$\displaystyle \boxed{R_{nj\ell m}=\frac{1}{2}\left(\partial_{\ell}\partial_{j}g_{mn}+\partial_{m}\partial_{n}g_{j\ell}-\partial_{\ell}\partial_{n}g_{jm}-\partial_{m}\partial_{j}g_{\ell n}\right)} \ \ \ \ \ (11)$

This equation is valid only at the origin on a LIF.

From this we can get some symmetry properties. First, if we interchange the first two indices ${n}$ and ${j}$ we see that the first and fourth terms in 11 swap, as do the second and third, so we end up with the negative of what we started with. That is

$\displaystyle \boxed{R_{jn\ell m}=-R_{nj\ell m}} \ \ \ \ \ (12)$

If we interchange the last two indices ${\ell}$ and ${m}$, again the first term swaps with the fourth, and the second with the third, so we get the same result:

$\displaystyle \boxed{R_{njm\ell}=-R_{nj\ell m}} \ \ \ \ \ (13)$

A third symmetry property is a bit more subtle. If we cyclically permute the last 3 indices ${j}$, ${\ell}$ and ${m}$ and add up the 3 terms, we get

 $\displaystyle R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{\ell}\partial_{j}g_{mn}+\partial_{m}\partial_{n}g_{j\ell}-\partial_{\ell}\partial_{n}g_{jm}-\partial_{m}\partial_{j}g_{\ell n}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\left(\partial_{m}\partial_{\ell}g_{jn}+\partial_{j}\partial_{n}g_{\ell m}-\partial_{m}\partial_{n}g_{\ell j}-\partial_{j}\partial_{\ell}g_{mn}\right)+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\left(\partial_{j}\partial_{m}g_{\ell n}+\partial_{\ell}\partial_{n}g_{mj}-\partial_{j}\partial_{n}g_{m\ell}-\partial_{\ell}\partial_{m}g_{jn}\right)\nonumber$

Using the symmetry of ${g_{ij}=g_{ji}}$ and the fact that partial derivatives commute, we find that the first two terms in the first line cancel with the last two terms in the second line, the first two in the second line cancel with the last two in the third line, and the first two in the third line cancel with the last two in the first line, giving the result:

$\displaystyle \boxed{R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}=0} \ \ \ \ \ (15)$

We’ve derived these results for the special case at the origin of a LIF. However, the origin of a LIF defines one particular event in spacetime and since all these symmetries are tensor equations, they must be true for that particular event, regardless of which coordinate system we’re using. Further, in our discussion of LIFs, we showed that we could define a LIF with its origin at any point in spacetime, provided that point is locally flat (that is, that there is no singularity at that point). So the argument shows that these symmetries are true for all non-singular points in spacetime.

Incidentally, it might be confusing that we can say that these symmetries are universally valid at all points in all coordinate systems just because they are tensor equations, while we say that 11 is valid only at the origin of a LIF. The difference is that 11 is written explicitly in terms of a particular metric ${g_{ij}}$ and that metric is defined precisely so that all its first derivatives are zero at the origin of the LIF. If we wanted an equation for ${R_{nj\ell m}}$ at some other point in spacetime, we could write it in the same form, but we’d need to find a different metric ${g_{ij}}$ whose first derivatives are zero at this other point. If we wanted to use the original metric, then since this other point is not at the origin of the original LIF, the ${\Gamma_{\; k\ell}^{j}}$ would not be zero at this point since the derivatives of ${g_{ij}}$ wouldn’t be zero there, and the expression for ${R_{nj\ell m}}$ would be more complicated in terms of the original metric.

## Covariant derivative: commutativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 18; Problem P18.8.

The second absolute gradient (or covariant derivative) of a four-vector is not commutative, as we can show by a direct derivation. Starting with the formula for the absolute gradient of a four-vector:

$\displaystyle \nabla_{j}A^{k}\equiv\frac{\partial A^{k}}{\partial x^{j}}+A^{i}\Gamma_{\; ij}^{k} \ \ \ \ \ (1)$

and the formula for the absolute gradient of a mixed tensor:

$\displaystyle \nabla_{l}C_{j}^{i}=\partial_{l}C_{j}^{i}+\Gamma_{lm}^{i}C_{j}^{m}-\Gamma_{lj}^{m}C_{m}^{i} \ \ \ \ \ (2)$

we can write out the second absolute gradient of a four-vector:

$\displaystyle \nabla_{i}\left(\nabla_{j}A^{k}\right)=\partial_{i}\partial_{j}A^{k}+\Gamma_{j\ell}^{k}\partial_{i}A^{\ell}+A^{\ell}\partial_{i}\Gamma_{j\ell}^{k}-\Gamma_{ji}^{m}\left(\partial_{m}A^{k}+A^{\ell}\Gamma_{m\ell}^{k}\right)+\Gamma_{im}^{k}\left(\partial_{j}A^{m}+A^{\ell}\Gamma_{j\ell}^{m}\right) \ \ \ \ \ (3)$

If we now swap ${i}$ and ${j}$, we get, using the commutativity of ordinary derivatives and the symmetry of ${\Gamma_{ji}^{m}}$:

$\displaystyle \nabla_{j}\left(\nabla_{i}A^{k}\right)=\partial_{i}\partial_{j}A^{k}+\Gamma_{i\ell}^{k}\partial_{j}A^{\ell}+A^{\ell}\partial_{j}\Gamma_{i\ell}^{k}-\Gamma_{ji}^{m}\left(\partial_{m}A^{k}+A^{\ell}\Gamma_{m\ell}^{k}\right)+\Gamma_{jm}^{k}\left(\partial_{i}A^{m}+A^{\ell}\Gamma_{i\ell}^{m}\right) \ \ \ \ \ (4)$

Subtracting these two equations gives

$\displaystyle \left(\nabla_{i}\nabla_{j}-\nabla_{j}\nabla_{i}\right)A^{k}=\left(\partial_{i}\Gamma_{j\ell}^{k}-\partial_{j}\Gamma_{i\ell}^{k}+\Gamma_{im}^{k}\Gamma_{j\ell}^{m}-\Gamma_{jm}^{k}\Gamma_{i\ell}^{m}\right)A^{\ell} \ \ \ \ \ (5)$

Using the definition of the Riemann tensor:

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (6)$

we have

$\displaystyle \left(\nabla_{i}\nabla_{j}-\nabla_{j}\nabla_{i}\right)A^{k}=R_{\;\ell ij}^{k}A^{\ell} \ \ \ \ \ (7)$

Thus the covariant derivative commutes only if the Riemann tensor is zero, which occurs only in flat spacetime.

## Geodesic deviation in a locally inertial frame

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 18; Problem P18.7.

Consider a coordinate frame attached to an object that is freely falling (that is, it’s following a geodesic). We can choose the coordinates such that they satisfy the conditions for a locally inertial frame (LIF), and they remain so as long as the object remains in free fall. This means that the Christoffel symbols are zero, since they are defined in terms of the metric tensor by

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

and in a LIF all first derivatives of ${g_{ij}}$ are zero.

An object at rest in a LIF has a four velocity of ${u^{i}=\left[1,0,0,0\right]}$ in that frame, so the equation of geodesic deviation becomes

 $\displaystyle \ddot{\mathbf{n}}^{i}$ $\displaystyle =$ $\displaystyle -R_{\; j\ell m}^{i}u^{m}u^{j}n^{\ell}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{\; t\ell t}^{i}n^{\ell} \ \ \ \ \ (3)$

To an object at rest in the LIF, its coordinate time ${t}$ is the same as its proper time ${\tau}$ so we can write this as

$\displaystyle \frac{d^{2}n^{i}}{dt^{2}}=-R_{\; t\ell t}^{i}n^{\ell} \ \ \ \ \ (4)$

Note that although the Riemann tensor is defined in terms of the Christoffel symbols, the fact that all these symbols are zero in a LIF doesn’t necessarily mean the Riemann tensor is also zero. The Riemann tensor is

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (5)$

Since it involves the derivatives of the Christoffel symbols, this in turn means we must have the second derivatives of ${g_{ij}}$ and we’ve seen that these are not zero for curved spacetime.

Incidentally, in Moore’s statement of this problem, he gives a hint that the time derivatives of the Christoffel symbols are zero. Although this may be true (since the symbols at the origin of the LIF remain zero as the object continues to free-fall), I don’t see how this figures into the derivation above. Any thoughts anyone?

## Locally inertial frames (LIFs)

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Box 17.7.

We’ve seen that for a smooth metric, we can define a locally flat coordinate system at a given event in which the metric is the flat spacetime metric ${\eta_{ij}}$. Although we specify the value of the metric tensor at that point, we haven’t imposed any conditions on its derivatives with respect to the various coordinates. If we impose the addition constraint that all the derivatives of the metric are zero, then we have a locally inertial frame or LIF.

It’s not immediately obvious that we can impose this restriction, so let’s see how we can demonstrate this. We start with some arbitrary metric ${g_{ij}}$ which doesn’t satisfy the LIF conditions and consider the transformation to another metric ${g'_{ij}}$ on which we try to impose these conditions. As usual, the transformation is

$\displaystyle g'_{ij}=\partial_{i}^{'}x^{a}\partial_{j}^{'}x^{b}g_{ab} \ \ \ \ \ (1)$

Each of the partial derivatives is a function of the primed coordinates so, for a region close to the event point ${P}$, we can expand these derivatives in Taylor series:

$\displaystyle \partial_{i}^{'}x^{a}\left(x'_{P}+\Delta x'\right)=a_{i}^{a}+b_{ij}^{a}\Delta x'^{j}+c_{ijk}^{a}\Delta x'^{j}\Delta x'^{k}+... \ \ \ \ \ (2)$

where the coefficients are defined in terms of ${\partial'_{i}x^{a}}$ and its derivatives, all evaluated at ${P}$:

 $\displaystyle a_{i}^{a}$ $\displaystyle =$ $\displaystyle \partial_{i}^{'}x^{a}\left(x'_{P}\right)\ \ \ \ \ (3)$ $\displaystyle b_{ij}^{a}$ $\displaystyle =$ $\displaystyle \partial'_{j}\partial_{i}^{'}x^{a}\left(x'_{P}\right)\ \ \ \ \ (4)$ $\displaystyle c_{ijk}^{a}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\partial'_{k}\partial'_{j}\partial_{i}^{'}x^{a}\left(x'_{P}\right) \ \ \ \ \ (5)$

We can also think of the unprimed metric as a function of the primed coordinates, and expand it in a Taylor series as well:

$\displaystyle g_{ab}\left(x'_{P}+\Delta x'\right)=g_{ab}\left(x'_{P}\right)+\Delta x'^{c}\partial'_{c}g_{ab}\left(x'_{P}\right)+\frac{1}{2}\Delta x'^{d}\Delta x'^{c}\partial'_{d}\partial'_{c}g_{ab}\left(x'_{P}\right)+... \ \ \ \ \ (6)$

We can now substitute these two series into 1 and collect terms. Due to the large number of indices floating about, it’s easier to use a condensed notation for this step. We can temporarily drop the indices to get

 $\displaystyle \partial'x$ $\displaystyle =$ $\displaystyle a+b\Delta x'+c\left(\Delta x'\right)^{2}\ \ \ \ \ (7)$ $\displaystyle g$ $\displaystyle =$ $\displaystyle g_{P}+\Delta x'\partial'g_{P}+\frac{1}{2}\left(\Delta x'\right)^{2}\partial'^{2}g_{P} \ \ \ \ \ (8)$

The transformation now becomes, up to second order terms:

 $\displaystyle g'\left(x'_{P}+\Delta x'\right)$ $\displaystyle =$ $\displaystyle \left(\partial'x\right)^{2}g\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(a+b\Delta x'+c\left(\Delta x'\right)^{2}\right)\left(a+b\Delta x'+c\left(\Delta x'\right)^{2}\right)\left(g_{P}+\Delta x'\partial'g_{P}+\frac{1}{2}\left(\Delta x'\right)^{2}\partial'^{2}g_{P}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}g_{P}+\Delta x'\left[abg_{P}+bag_{P}+a^{2}\partial'g_{P}\right]+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(\Delta x'\right)^{2}\left[\frac{1}{2}a^{2}\partial'^{2}g_{P}+ab\partial'g_{P}+acg_{P}+ba\partial'g_{P}+b^{2}g_{P}+cag_{P}\right] \ \ \ \ \ (11)$

The last equation is the Taylor expansion of ${g'\left(x'_{P}+\Delta x'\right)}$ where every factor is now a function of ${x'}$. That is

$\displaystyle g'_{ij}\left(x'_{P}+\Delta x'\right)=g'_{ij}\left(x'_{P}\right)+\Delta x'^{k}\partial'_{k}g'_{ij}\left(x'_{P}\right)+\Delta x'^{\ell}\Delta x'^{k}\partial'_{\ell}\partial'_{k}g'_{ij}\left(x'_{P}\right)$

We therefore have, taking each term separately and restoring the indices:

 $\displaystyle g'_{ij}\left(x'_{P}\right)$ $\displaystyle =$ $\displaystyle a^{2}g_{P}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{i}^{a}a_{j}^{b}\left(g_{ab}\right)_{P}\ \ \ \ \ (13)$ $\displaystyle \partial'_{k}g'_{ij}\left(x'_{P}\right)$ $\displaystyle =$ $\displaystyle abg_{P}+bag_{P}+a^{2}\partial'g_{P}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[a_{i}^{a}b_{jk}^{b}+b_{ik}^{a}a_{j}^{b}\right]\left(g_{ab}\right)_{P}+a_{i}^{a}a_{j}^{b}\partial'_{k}\left(g_{ab}\right)_{P}\ \ \ \ \ (15)$ $\displaystyle \partial'_{\ell}\partial'_{k}g'_{ij}\left(x'_{P}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}a^{2}\partial'^{2}g_{P}+ab\partial'g_{P}+acg_{P}+ba\partial'g_{P}+b^{2}g_{P}+cag_{P}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}a_{i}^{a}a_{j}^{b}\partial'_{k}\partial'_{\ell}\left(g_{ab}\right)_{P}+a_{i}^{a}b_{jk}^{b}\partial'_{\ell}\left(g_{ab}\right)_{P}+a_{i}^{a}c_{jk\ell}^{b}\left(g_{ab}\right)_{P}+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle b_{ik}^{a}a_{j}^{b}\partial'_{\ell}\left(g_{ab}\right)_{P}+b_{ik}^{a}b_{j\ell}^{b}\left(g_{ab}\right)_{P}+c_{ik\ell}^{a}a_{j}^{b}\left(g_{ab}\right)_{P} \ \ \ \ \ (17)$

We won’t actually try to solve these equations, but what we need to do is see if we have enough freedom to impose the conditions above, that is, that ${g'_{ij}\left(x'_{P}\right)=\eta_{ij}}$ and ${\partial'_{k}g'_{ij}\left(x'_{P}\right)=0}$. To do this, we need to see how many independent equations each of these equations provides.

Consider 13 first. Each index ${i}$ and ${j}$ has 4 possible values, but because ${g'_{ij}=g'_{ji}}$ we have to exclude permutations of ${i}$ and ${j}$. We can get these using binomial coefficients:

$\displaystyle N=\binom{4}{1}+\binom{4}{2}=4+6=10$

For 15, the symmetry of ${g'}$ means we have ${4\times10=40}$ (10 choices for ${i}$ and ${j}$, times the 4 different values of ${k}$) independent equations. Finally, for 17, since the order of partial derivatives doesn’t matter and the metric is symmetric, we have ${10\times10=100}$ independent equations.

The original metric ${g_{ij}}$ is assumed to be given, so we can’t change that, so how many variables can we solve for in these equations? The transformation is effectively determined (up to second order) by the coefficients ${a_{i}^{a}}$, ${b_{ij}^{a}}$ and ${c_{ijk}^{a}}$, so we need to figure out how many of these are independent of each other.

For ${a_{i}^{a}}$ every choice of the two indices gives an independent quantity, so there are 16 different variables. Thus we have 6 more degrees of freedom than we need to solve 13 and set ${g'_{ij}=\eta_{ij}}$.

For ${b_{ij}^{a}}$, the order of the partial derivatives doesn’t matter so there are 10 possible choices of ${i}$ and ${j}$ for each value of ${a}$, giving ${10\times4=40}$. Having chosen the ${a_{i}^{a}}$ in solving 13, we can plug these into the 40 equations specified by 15 and solve for the 40 ${b_{ij}^{a}}$ coefficients. Therefore, in principle, we can set ${\partial'_{k}g'_{ij}\left(x'_{P}\right)=0}$.

When it comes to ${c_{ijk}^{a}}$, again the order of the 3 partial derivatives doesn’t matter, so for each value of ${a}$, we can have ${i=j=k}$, or two indices equal with the third different, or all three indices different. The total number of choices for each value of ${a}$ is then

$\displaystyle N_{c}=\binom{4}{1}+2\binom{4}{2}+\binom{4}{3}=20$

To get the middle term, note that ${\binom{4}{2}}$ is the number of ways of choosing 2 numbers out of 4, but it assumes there are only 2 slots into which these numbers can be placed. In our case, one of the numbers is used twice, so this term would imply that, for example, 1-1-2 and 2-2-1 are just permutations of the same combination, whereas we want them to be distinct, so we have to multiply by 2. The total number of independent ${c_{ijk}^{a}}$ is therefore ${4\times20=80}$. The fact that there are 100 independent equations specified by 17 means that we can’t impose values on all of the ${\partial'_{\ell}\partial'_{k}g'_{ij}\left(x'_{P}\right)}$ so in general even if we require 80 of them to be zero, the other 20 may have non-zero values. It turns out that this is a result of the inherent curvature of the spacetime, so it’s not surprising that we can’t get rid of that.

The LIF imposes stricter conditions on the local metric than the locally flat frame we mentioned at the start. The latter just specifies the metric at a given point, while the LIF specifies both the metric and its derivatives.

## Riemann tensor in the Schwarzschild metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 18; Problem P18.6.

We’ll calculate one component of the Riemann tensor for the Schwarzschild metric. The tensor is

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (1)$

As usual, we need the Christoffel symbols, but we’ve already worked these out.

 $\displaystyle \Gamma_{\; ij}^{t}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \Gamma_{\; ij}^{r}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right) & 0 & 0 & 0\\ 0 & -\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ 0 & 0 & -r\left(1-\frac{2GM}{r}\right) & 0\\ 0 & 0 & 0 & -r\sin^{2}\theta\left(1-\frac{2GM}{r}\right) \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle \Gamma_{\; ij}^{\theta}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & \frac{1}{r} & 0\\ 0 & \frac{1}{r} & 0 & 0\\ 0 & 0 & 0 & -\sin\theta\cos\theta \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \Gamma_{\; ij}^{\phi}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{r}\\ 0 & 0 & 0 & \cot\theta\\ 0 & \frac{1}{r} & \cot\theta & 0 \end{array}\right] \ \ \ \ \ (5)$

We can plug these into the formula above to get ${R_{\; rtr}^{t}}$. We have

$\displaystyle R_{\; rtr}^{t}=\partial_{t}\Gamma_{\; rr}^{t}-\partial_{r}\Gamma_{\; tr}^{t}+\Gamma_{\; rr}^{k}\Gamma_{\; kt}^{t}-\Gamma_{\; tr}^{k}\Gamma_{\; rk}^{t} \ \ \ \ \ (6)$

We can work out these terms one at a time (only the index ${k}$ is summmed):

 $\displaystyle \partial_{t}\Gamma_{\; rr}^{t}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle -\partial_{r}\Gamma_{\; tr}^{t}$ $\displaystyle =$ $\displaystyle \frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1}+\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-2}\left(\frac{2GM}{r^{2}}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1}+\frac{2G^{2}M^{2}}{r^{4}}\left(1-\frac{2GM}{r}\right)^{-2}\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\; rr}^{k}\Gamma_{\; kt}^{t}$ $\displaystyle =$ $\displaystyle \Gamma_{\; rr}^{r}\Gamma_{\; rt}^{t}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{G^{2}M^{2}}{r^{4}}\left(1-\frac{2GM}{r}\right)^{-2}\ \ \ \ \ (11)$ $\displaystyle -\Gamma_{\; tr}^{k}\Gamma_{\; rk}^{t}$ $\displaystyle =$ $\displaystyle -\left(\Gamma_{\; rt}^{t}\right)^{2}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{G^{2}M^{2}}{r^{4}}\left(1-\frac{2GM}{r}\right)^{-2} \ \ \ \ \ (13)$

$\displaystyle R_{\; rtr}^{t}=\frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1} \ \ \ \ \ (14)$
Since this is never zero, Schwarzschild spacetime is curved everywhere, but as ${r\rightarrow\infty}$, ${R_{\; rtr}^{t}\rightarrow0}$ so the further we get from the mass, the less curved the spacetime becomes.