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Using the usual coordinates on a 2-sphere of radius $r$, I get the metric tensor $g_{\mu\nu}=\text{diag}(r^2, r^2\sin^2\theta)$ and so $g^{\mu\nu}=\text{diag}(1/r^2,1/r^2\sin^2\theta)$.

Hence the only nonzero Christoffel symbols are $\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$ and $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta$.

I'm fairly sure about these because I've been able to corroborate the same from various online sources.

Now using the formula for the Ricci tensor (e.g., Dirac's "General Theory of Relativity", Eq. (14.4)):

$$R_{\mu\nu} = \Gamma^\alpha_{\mu\alpha,\nu} - \Gamma^\alpha_{\mu\nu,\alpha} - \Gamma^\alpha_{\mu\nu}\Gamma^\beta_{\alpha\beta} + \Gamma^\alpha_{\mu\beta}\Gamma^\beta_{\nu\alpha}.\tag{14.4}$$

I get $R_{\mu\nu} = \text{diag}(-1, -\sin^2\theta)$, whence $R=g^{\mu\nu}R_{\mu\nu}=-2/r^2$.

Obviously the minus sign is wrong, but I've checked my work on $R_{\mu\nu}$ twice and I can't seem to find the error. Can anyone please spot it and advise?

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    $\begingroup$ Your Christoffel symbols are correct. The Ricci tensor should be $\mathrm{diag}(1,\sin^2\theta)$, so that's where the mistake is. Checking the indexes in $(14.4)$ it seems that is minus the Ricci tensor. $\endgroup$ Commented Dec 30, 2022 at 12:05
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    $\begingroup$ There is a sign convention in the definition of the Ricci tensor. Some authors choose to work in a convention where the Ricci scalar for a sphere is negative. It might happen that Dirac chooses this convention. Try using the expressions from other books $\endgroup$ Commented Dec 30, 2022 at 12:26
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    $\begingroup$ I've just found out the existence of that sign convention, which arises from the choice of the lower index that is contracted with the upper index of the Riemann tensor to derive the Ricci tensor (and the minus sign is a consequence of the antisymmetry of the second pair of indexes). As @NíckolasAlves suggests Dirac is using the convention opposite to mine in the above comment, which differs by a minus sign in the definition of the Ricci tensor. $\endgroup$ Commented Dec 30, 2022 at 12:30
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    $\begingroup$ The famous book by Misner, Thorne and Wheeler has a discussion about the sign conventions in Relativity right at the beginning of the book. It might be worth a look, and perhaps it even lists Dirac's convention in there (it has a list of conventions in famous textbooks) $\endgroup$ Commented Dec 30, 2022 at 12:53
  • $\begingroup$ What's bothering me is that Dirac writes (p. 24) "$R$ ... is the scalar curvature ... is defined in such a way that it is positive for the surface of a sphere in three dimensions." So I should be getting $R=2/r^2$ using Dirac's definition of the Ricci tensor. .......................... EDIT: see comments below. Dirac's definiton of $R_{\mu\nu}$ is the more common one, and I am now sure that $R=-2/r^2$; in other words, he was just mistaken and should have said that the scalar curvature of $S^2$ is negative. $\endgroup$
    – Khun Chang
    Commented Dec 31, 2022 at 12:43

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As per the comments, it appears that your equation (14.4) is the negative of what other authors define as the Ricci tensor: see e.g. this math SE post.

We can explicitly calculate e.g. $R_{\phi\phi}$ using equation (14.4), noting that most of the terms vanish:

$R_{\phi\phi}= -\partial_{\theta}{\Gamma^{\theta}}_{\phi\phi} - {\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\phi}}_{\theta\phi} + {\Gamma^{\phi}}_{\phi\theta}{\Gamma^{\theta}}_{\phi\phi} + {\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\phi}}_{\phi\theta} \\ = -(\sin^2\theta- \cos^2\theta) -(-\sin\theta\cos\theta\cot\theta)+(-\sin\theta\cos\theta\cot\theta)+(-\sin\theta\cos\theta\cot\theta) \\ = -\sin^2\theta$,

and similarly for $R_{\theta\theta}$.

I also note that the claim in Dirac's book that the scalar curvature $R$ is defined in a way to be positive comes before the definition of eq. (14.4). I have not performed the calculation, but it is possible that the contraction of the Riemann tensor as defined in the book will yield a positive scalar curvature. Alternatively it is possible that Dirac simply made an (odd number of) sign errors.

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  • $\begingroup$ Dirac's formula for the Ricci tensor $R_{\mu\nu}$ (Eq. (14.4)) is consistent with his definition of the curvature tensor $R^\beta_{\nu\rho\sigma}$ (Eq. (11.3)), being the tensor which appears in the "commutator" of second covariant derivatives: $A_{\nu:\rho:\sigma} - A_{\nu:\sigma:\rho} = A_\beta R^\beta_{\nu\rho\sigma}$ (Eq. (11.2)). Weinberg's curvature tensor (Eqn. (6.1.5), p. 133) is the OPPOSITE of Dirac's. Ohanian-Ruffini's definition (Eq. (6.87), p. 244) is the same as Dirac's. So clearly definitions vary. What puzzles me is why Dirac says the sphere has positive $R$. $\endgroup$
    – Khun Chang
    Commented Dec 31, 2022 at 11:18
  • $\begingroup$ As noted above, I get $R_{\mu\nu} = \text{diag}(-1, -\sin^2\theta)$ -- the same as you for $R_{\phi\phi}$ -- hence, effecting the contraction: $R=g^{\mu\nu}R_{\mu\nu}=-2/r^2$. $\endgroup$
    – Khun Chang
    Commented Dec 31, 2022 at 11:24
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    $\begingroup$ A final check: Landau & Lifshitz and Zee both define the curvature tensor the same way as Dirac. So, having checked 5 sources, it's 4-1 in favor of Dirac's definition (Weinberg is the lone dissenter). I must conclude that Dirac erred in saying that $R$ is positive for the surface of a sphere in 3 dimensions. $\endgroup$
    – Khun Chang
    Commented Dec 31, 2022 at 12:06

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