Non-zero components of the Riemann tensor for the Schwarzschild metric

Question

Can anyone tell me which are the non-zero components of the Riemann tensor for the Schwarzschild metric? I've been searching for these components for about 2 weeks, and I've found a few sites, but the problem is that each one of them shows different components, in number and form. I´ve calculated a few components but I don't know if they are correct. I'm using the form of the metric:

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

Answer 1

According to Mathematica, and assuming I haven't made any silly errors typing in the metric, I get the non-zero components of $R^\mu{}_{\nu\alpha\beta}$ to be:

{1, 2, 1, 2} -> (2 G M)/(r^2 (-2 G M + c^2 r)),
{1, 2, 2, 1} -> -((2 G M)/(r^2 (-2 G M + c^2 r))),
{1, 3, 1, 3} -> -((G M)/(c^2 r)),
{1, 3, 3, 1} -> (G M)/(c^2 r),
{1, 4, 1, 4} -> -((G M Sin[\[Theta]]^2)/(c^2 r)),
{1, 4, 4, 1} -> (G M Sin[\[Theta]]^2)/(c^2 r),
{2, 1, 1, 2} -> (2 G M (-2 G M + c^2 r))/(c^4 r^4),
{2, 1, 2, 1} -> -((2 G M (-2 G M + c^2 r))/(c^4 r^4)),
{2, 3, 2, 3} -> -((G M)/(c^2 r)),
{2, 3, 3, 2} -> (G M)/(c^2 r),
{2, 4, 2, 4} -> -((G M Sin[\[Theta]]^2)/(c^2 r)),
{2, 4, 4, 2} -> (G M Sin[\[Theta]]^2)/(c^2 r),
{3, 1, 1, 3} -> (G M (2 G M - c^2 r))/(c^4 r^4),
{3, 1, 3, 1} -> (G M (-2 G M + c^2 r))/(c^4 r^4),
{3, 2, 2, 3} -> (G M)/(r^2 (-2 G M + c^2 r)),
{3, 2, 3, 2} -> (G M)/(r^2 (2 G M - c^2 r)),
{3, 4, 3, 4} -> (2 G M Sin[\[Theta]]^2)/(c^2 r),
{3, 4, 4, 3} -> -((2 G M Sin[\[Theta]]^2)/(c^2 r)),
{4, 1, 1, 4} -> (G M (2 G M - c^2 r))/(c^4 r^4),
{4, 1, 4, 1} -> (G M (-2 G M + c^2 r))/(c^4 r^4),
{4, 2, 2, 4} -> (G M)/(r^2 (-2 G M + c^2 r)),
{4, 2, 4, 2} -> (G M)/(r^2 (2 G M - c^2 r)),
{4, 3, 3, 4} -> -((2 G M)/(c^2 r)),
{4, 3, 4, 3} -> (2 G M)/(c^2 r),

Answer 2

The answer given by @John Rennie is correct. But maybe a note on how one can actually compute the Riemann most efficiently (by hand or with computer algebra). To compute it fast it is convenient to first compute $R_{abcd}$ because it has the most symmetries:

• Skrew symmetry $R_{abcd}=-R_{bacd}=-R_{abdc}=R_{badc}$
• Interchange symmetry $R_{abcd}=R_{cdab}$

This means that in four dimensions one has only 21 independent components to compute: Those can be written in a $6\times6$ symmetric matrix in respect to tuples of antisymmetric index pairs $\left\{(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)\right\}$. For the Schwarzschild metric this matrix looks like:

So there are only 6 nonvanishing, independent components of $R_{abcd}$ for the Schwarzschild metric. From those one can construct the remaining dependent ones.

In the special case of the Schwarzschild metric, the interchange symmetry gives no new nonvanishing components, since the matrix of tuples is diagonal. This leaves skrew symmetry for the six diagonal components, which leads to three new nonvanishing components per diagonal component. So in total $6\times4=24$ nonvanishing components of $R^{a}{}_{bcd}$.

To get to $R^{a}{}_{bcd}$ one needs to raise the first index with the inverse metric, which in the present case is just multiplying $R_{abcd}$ with $g^{aa}=1/g_{aa}$, since $g_{ab}$ is symmetric.

John Rennie gave those 24 nonvanishing component of $R^{a}{}_{bcd}$.

A last comment on those 21 independent components in four dimensions: if one considers the first Bianchi identity one gets down to 20 independent components in four dimensions for $R_{abcd}$. So still some computing to do but 21 or 20 is better than 256.

Answer 3

The non-zero components of the Riemann tensor of the Schwarzschild metric are:

\begin{array}{lcl} \mathrm{R}_{ \phantom{\, t} \, r \, t \, r }^{ \, t \phantom{\, r} \phantom{\, t} \phantom{\, r} } & = & \frac{2 \, {G} m}{c^{2} r^{3} - 2 \, {G} m r^{2}} \\ \mathrm{R}_{ \phantom{\, t} \, r \, r \, t }^{ \, t \phantom{\, r} \phantom{\, r} \phantom{\, t} } & = & -\frac{2 \, {G} m}{c^{2} r^{3} - 2 \, {G} m r^{2}} \\ \mathrm{R}_{ \phantom{\, t} \, {\theta} \, t \, {\theta} }^{ \, t \phantom{\, {\theta}} \phantom{\, t} \phantom{\, {\theta}} } & = & -\frac{{G} m}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, t} \, {\theta} \, {\theta} \, t }^{ \, t \phantom{\, {\theta}} \phantom{\, {\theta}} \phantom{\, t} } & = & \frac{{G} m}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, t} \, {\phi} \, t \, {\phi} }^{ \, t \phantom{\, {\phi}} \phantom{\, t} \phantom{\, {\phi}} } & = & -\frac{{G} m \sin\left({\theta}\right)^{2}}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, t} \, {\phi} \, {\phi} \, t }^{ \, t \phantom{\, {\phi}} \phantom{\, {\phi}} \phantom{\, t} } & = & \frac{{G} m \sin\left({\theta}\right)^{2}}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, r} \, t \, t \, r }^{ \, r \phantom{\, t} \phantom{\, t} \phantom{\, r} } & = & \frac{2 \, {\left({G} c^{2} m r - 2 \, {G}^{2} m^{2}\right)}}{c^{4} r^{4}} \\ \mathrm{R}_{ \phantom{\, r} \, t \, r \, t }^{ \, r \phantom{\, t} \phantom{\, r} \phantom{\, t} } & = & -\frac{2 \, {\left({G} c^{2} m r - 2 \, {G}^{2} m^{2}\right)}}{c^{4} r^{4}} \\ \mathrm{R}_{ \phantom{\, r} \, {\theta} \, r \, {\theta} }^{ \, r \phantom{\, {\theta}} \phantom{\, r} \phantom{\, {\theta}} } & = & -\frac{{G} m}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, r} \, {\theta} \, {\theta} \, r }^{ \, r \phantom{\, {\theta}} \phantom{\, {\theta}} \phantom{\, r} } & = & \frac{{G} m}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, r} \, {\phi} \, r \, {\phi} }^{ \, r \phantom{\, {\phi}} \phantom{\, r} \phantom{\, {\phi}} } & = & -\frac{{G} m \sin\left({\theta}\right)^{2}}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, r} \, {\phi} \, {\phi} \, r }^{ \, r \phantom{\, {\phi}} \phantom{\, {\phi}} \phantom{\, r} } & = & \frac{{G} m \sin\left({\theta}\right)^{2}}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, {\theta}} \, t \, t \, {\theta} }^{ \, {\theta} \phantom{\, t} \phantom{\, t} \phantom{\, {\theta}} } & = & -\frac{{G} c^{2} m r - 2 \, {G}^{2} m^{2}}{c^{4} r^{4}} \\ \mathrm{R}_{ \phantom{\, {\theta}} \, t \, {\theta} \, t }^{ \, {\theta} \phantom{\, t} \phantom{\, {\theta}} \phantom{\, t} } & = & \frac{{G} c^{2} m r - 2 \, {G}^{2} m^{2}}{c^{4} r^{4}} \\ \mathrm{R}_{ \phantom{\, {\theta}} \, r \, r \, {\theta} }^{ \, {\theta} \phantom{\, r} \phantom{\, r} \phantom{\, {\theta}} } & = & \frac{{G} m}{c^{2} r^{3} - 2 \, {G} m r^{2}} \\ \mathrm{R}_{ \phantom{\, {\theta}} \, r \, {\theta} \, r }^{ \, {\theta} \phantom{\, r} \phantom{\, {\theta}} \phantom{\, r} } & = & -\frac{{G} m}{c^{2} r^{3} - 2 \, {G} m r^{2}} \\ \mathrm{R}_{ \phantom{\, {\theta}} \, {\phi} \, {\theta} \, {\phi} }^{ \, {\theta} \phantom{\, {\phi}} \phantom{\, {\theta}} \phantom{\, {\phi}} } & = & \frac{2 \, {G} m \sin\left({\theta}\right)^{2}}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, {\theta}} \, {\phi} \, {\phi} \, {\theta} }^{ \, {\theta} \phantom{\, {\phi}} \phantom{\, {\phi}} \phantom{\, {\theta}} } & = & -\frac{2 \, {G} m \sin\left({\theta}\right)^{2}}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, {\phi}} \, t \, t \, {\phi} }^{ \, {\phi} \phantom{\, t} \phantom{\, t} \phantom{\, {\phi}} } & = & -\frac{{G} c^{2} m r - 2 \, {G}^{2} m^{2}}{c^{4} r^{4}} \\ \mathrm{R}_{ \phantom{\, {\phi}} \, t \, {\phi} \, t }^{ \, {\phi} \phantom{\, t} \phantom{\, {\phi}} \phantom{\, t} } & = & \frac{{G} c^{2} m r - 2 \, {G}^{2} m^{2}}{c^{4} r^{4}} \\ \mathrm{R}_{ \phantom{\, {\phi}} \, r \, r \, {\phi} }^{ \, {\phi} \phantom{\, r} \phantom{\, r} \phantom{\, {\phi}} } & = & \frac{{G} m}{c^{2} r^{3} - 2 \, {G} m r^{2}} \\ \mathrm{R}_{ \phantom{\, {\phi}} \, r \, {\phi} \, r }^{ \, {\phi} \phantom{\, r} \phantom{\, {\phi}} \phantom{\, r} } & = & -\frac{{G} m}{c^{2} r^{3} - 2 \, {G} m r^{2}} \\ \mathrm{R}_{ \phantom{\, {\phi}} \, {\theta} \, {\theta} \, {\phi} }^{ \, {\phi} \phantom{\, {\theta}} \phantom{\, {\theta}} \phantom{\, {\phi}} } & = & -\frac{2 \, {G} m}{c^{2} r} \\ \mathrm{R}_{ \phantom{\, {\phi}} \, {\theta} \, {\phi} \, {\theta} }^{ \, {\phi} \phantom{\, {\theta}} \phantom{\, {\phi}} \phantom{\, {\theta}} } & = & \frac{2 \, {G} m}{c^{2} r} \end{array}

Answer 4

In case some student is copying from this page, it is important to note that the original post has a typo/error, one or three of the metric terms of course must have a negative sign.