Author Archives: growescience

Riemann tensor in a 2-d curved space

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.6.

Here’s another example of the Riemann tensor in a 2-d coordinate system. 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, 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\ \ \ \ \ (2)
\displaystyle  \ddot{x}^{a}+\Gamma_{\; ij}^{a}\dot{x}^{j}\dot{x}^{i} \displaystyle  = \displaystyle  0 \ \ \ \ \ (3)

The metric is

\displaystyle  ds^{2}=dp^{2}+e^{2p/p_{0}}dq^{2} \ \ \ \ \ (4)

so {g_{pp}=1} and {g_{qq}=e^{2p/p_{0}}}. For the two coordinates, 2 gives us

\displaystyle   \ddot{p}-\frac{1}{p_{0}}e^{2p/p_{0}}\dot{q}^{2} \displaystyle  = \displaystyle  0\ \ \ \ \ (5)
\displaystyle  e^{2p/p_{0}}\ddot{q}+\frac{2}{p_{0}}e^{2p/p_{0}}\dot{p}\dot{q} \displaystyle  = \displaystyle  0 \ \ \ \ \ (6)

Dividing through by the coefficient of the second derivative in the second equation case gives:

\displaystyle   \ddot{p}-\frac{1}{p_{0}}e^{2p/p_{0}}\dot{q}^{2} \displaystyle  = \displaystyle  0\ \ \ \ \ (7)
\displaystyle  \ddot{q}+\frac{2}{p_{0}}\dot{p}\dot{q} \displaystyle  = \displaystyle  0 \ \ \ \ \ (8)

Comparing with 3 we get

\displaystyle   \Gamma_{\; qq}^{p} \displaystyle  = \displaystyle  -\frac{1}{p_{0}}e^{2p/p_{0}}\ \ \ \ \ (9)
\displaystyle  \Gamma_{\; pq}^{q} \displaystyle  = \displaystyle  \Gamma_{\; qp}^{q}=\frac{1}{p_{0}} \ \ \ \ \ (10)

with all other Christoffel symbols equal to zero.

The only independent Riemann tensor component in 2-d is {R_{\; qpq}^{p}} :

\displaystyle   R_{\; qpq}^{p} \displaystyle  = \displaystyle  \partial_{p}\Gamma_{\; qq}^{p}-\partial_{q}\Gamma_{\; pq}^{p}+\Gamma_{\; kp}^{p}\Gamma_{\; qq}^{k}-\Gamma_{\; qk}^{p}\Gamma_{\; pq}^{k}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \partial_{p}\Gamma_{\; qq}^{p}-0+0-\Gamma_{\; qq}^{p}\Gamma_{\; pq}^{q}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  -\frac{2}{p_{0}^{2}}e^{2p/p_{0}}+\frac{1}{p_{0}^{2}}e^{2p/p_{0}}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{p_{0}^{2}}e^{2p/p_{0}} \ \ \ \ \ (14)

Any non-zero component indicates that the space is curved, so this metric represents a curved space.

Riemann tensor in 2-d flat space

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.5.

Another example of the Riemann tensor in a 2-d space. The metric is

\displaystyle ds^{2}=dp^{2}+\frac{dq^{2}}{b^{2}q^{2}} \ \ \ \ \ (1)

where {b} is a constant. The metric tensor is therefore {g_{pp}=1}, {g_{qq}=1/b^{2}q^{2}}. By comparing the two forms of the geodesic equation, we can calculate the Christoffel symbols.

\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\ \ \ \ \ (2)
\displaystyle \ddot{x}^{a}+\Gamma_{\; ij}^{a}\dot{x}^{j}\dot{x}^{i} \displaystyle = \displaystyle 0 \ \ \ \ \ (3)

With {a=p}, we get from 2

\displaystyle \ddot{p}=0 \ \ \ \ \ (4)

From 3, we see that {\Gamma_{\; ij}^{p}=0} for all {i} and {j}.

With {a=q}, we have

\displaystyle \frac{1}{b^{2}q^{2}}\ddot{q}-\frac{2}{b^{2}q^{3}}\dot{q}^{2}+\frac{1}{b^{2}q^{3}}\dot{q}^{2} \displaystyle = \displaystyle 0\ \ \ \ \ (5)
\displaystyle \ddot{q}-\frac{1}{q}\dot{q}^{2} \displaystyle = \displaystyle 0 \ \ \ \ \ (6)

Comparing with 3 we find

\displaystyle \Gamma_{qq}^{q} \displaystyle = \displaystyle -\frac{1}{q}\ \ \ \ \ (7)
\displaystyle \Gamma_{pq}^{q} \displaystyle = \displaystyle \Gamma_{qp}^{q}=\Gamma_{pp}^{q}=0 \ \ \ \ \ (8)

The only independent component of the Riemann tensor in 2-d is {R_{\; qpq}^{p}} :

\displaystyle R_{\; qpq}^{p} \displaystyle = \displaystyle \partial_{p}\Gamma_{\; qq}^{p}-\partial_{q}\Gamma_{\; pq}^{p}+\Gamma_{\; kp}^{p}\Gamma_{\; qq}^{k}-\Gamma_{\; qk}^{p}\Gamma_{\; pq}^{k}\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle 0 \ \ \ \ \ (10)

since all terms involve components of the form {\Gamma_{ij}^{p}}. Therefore, the Ricci tensor is also zero: {R_{ij}=0} as is the curvature scalar {R=0}. The space is flat.

Ricci tensor and curvature scalar

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.5, Problem P19.1.

We can form contractions over the indices of the Riemann tensor to get some other useful quantities.

First, we can contract the first and third indices to get the Ricci tensor {R_{bc}}:

\displaystyle   R_{\; bac}^{a} \displaystyle  = \displaystyle  g^{ad}R_{dbac}\ \ \ \ \ (1)
\displaystyle  \displaystyle  \equiv \displaystyle  R_{bc} \ \ \ \ \ (2)

Using the symmetry relation {R_{dbac}=R_{acdb}} and the symmetry of the metric, we have

\displaystyle  R_{bc}=g^{ad}R_{dbac}=g^{da}R_{acdb}=R_{cb} \ \ \ \ \ (3)

so the Ricci tensor is symmetric.

We can contract the Ricci tensor in turn to get the curvature scalar {R}:

\displaystyle   R_{\; c}^{b} \displaystyle  = \displaystyle  g^{ab}R_{ac}\ \ \ \ \ (4)
\displaystyle  R_{\; b}^{b} \displaystyle  = \displaystyle  g^{ab}R_{ab}\ \ \ \ \ (5)
\displaystyle  \displaystyle  \equiv \displaystyle  R \ \ \ \ \ (6)

Since the Riemann tensor is identically zero in flat spacetime, the Ricci tensor and curvature scalar are also both zero there. However, although the Riemann tensor always has at least one non-zero component in curved spacetime, the Ricci tensor and curvature scalar can both be zero in curved spacetime. Thus we can say that if the Ricci tensor or the curvature tensor are non-zero, the spacetime is curved, but we can’t draw any conclusions if they are zero; we then need to work out the full Riemann tensor.

One other contraction of the Riemann tensor is over its first and second indices:

\displaystyle  R_{\; abc}^{a}=g^{ad}R_{dabc}=-g^{da}R_{adbc}=0 \ \ \ \ \ (7)

In the second equation we’ve used the antisymmetry relation {R_{dabc}=-R_{adbc}} and the symmetry of the metric. Thus this contraction doesn’t tell us anything useful.

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.

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