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

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:

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

• "each one of them show differents components" - The components of $R^\mu_{\nu\alpha\beta}$ aren't the same as the components of $R_{\mu\nu\alpha\beta}$ Commented Nov 30, 2016 at 18:08
• Are you calculating the coeficients using that metric? The signs can not be all positive. Commented Aug 4, 2020 at 0:54

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),

• Would you be able to share your code? :) Commented Mar 13, 2021 at 10:30
• @zabop ping me in the Physics chat room and I'll pop the notebook on a server for you to download. Commented Mar 13, 2021 at 17:11
• Alright, thank you! Commented Mar 13, 2021 at 17:17

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.

• Yes sorry; I use geometrized units with $c=G=1$, like OP. I just changed $m$ of OP to $M$.
– N0va
Commented Dec 1, 2016 at 16:00
• @AccidentalFourierTransform: very, very close to 100% of relativists will be working in $c = G = 1$ Commented Dec 1, 2016 at 16:32
• How do you define the indices of the coordinates in your answer? Does the time index correspond to "1" in your result or "4"? (The usual convention is t=0, r=1, theta=2, phi=3, but your index range is between 1 and 4.) Commented Oct 4, 2018 at 12:06

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}$$

• The question was about the non-zero components of the Riemann tensor of the Schwarzschild metric
– Okba
Commented Aug 4, 2020 at 21:42

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.