0

In Sean Carroll's GR book, page 101, he says that

It is straightforward to show that the Christoffel connection satisfies $$ \Gamma^{\mu}_{\mu\lambda}=\frac{1}{\sqrt{\left | g \right |}}\partial_{\lambda}\sqrt{\left | g \right |}. $$

I don't understand how this is straightforward. Can someone give a clue?

Qmechanic
  • 220,844
Lagrange629
  • 79

2 Answers2

2

The christoffel symbols are defined by \begin{align*} \Gamma^\rho_{\mu\nu} = \frac{1}{2}g^{\rho\lambda}\Big(\partial_\mu g_{\nu\lambda} + \partial_\nu g_{\mu\lambda} - \partial_\lambda g_{\mu\nu}\Big) \end{align*} In the sought after expression we sum over $\mu$ i.e. \begin{align*} \Gamma^\mu_{\mu\nu} &= \frac{1}{2}g^{\mu\lambda}\Big(\partial_\mu g_{\nu\lambda} + \partial_{\nu} g_{\mu\lambda} - \partial_\lambda g_{\mu\nu}\Big) \\ &= \frac{1}{2}\Big(\partial^\lambda g_{\lambda\nu} + g^{\mu\lambda} \partial_\nu g_{\mu\lambda} - \partial^{\mu} g_{\mu\nu}\Big) \\ &= \frac{1}{2} g^{\mu\lambda}\,\partial_\nu\, g_{\mu\lambda} \end{align*} Now note that the derivative of a determinant is given by $\delta g = g\, g^{\mu\nu}\delta g_{\mu\nu}$ (see 4.68 p 163 in Carroll) Where we choose $x^\mu \to x^\mu + \varepsilon^\mu$ st. $g_{\mu\nu}(x) \to g_{\mu\nu}(x + \varepsilon) = g_{\mu\nu}(x) + \varepsilon^{\lambda}\partial_\lambda g_{\mu\nu}(x) + \mathcal{O}(\varepsilon^2)$ so we see that $\delta g_{\mu\nu} = \partial_\lambda g_{\mu\nu}$ (and similarly for the determinant). So we can express the derivative of the metric as $\partial_\lambda g = g\, g^{\mu\nu}\partial_\lambda g_{\mu\nu}$, which we then plug into the formula above \begin{align*} \Gamma^\mu_{\mu\nu} = \frac{1}{2g} \partial_\lambda g = \frac{1}{\sqrt{g}}\partial_\lambda \sqrt{g} \end{align*}

Wihtedeka
  • 2,078
2

It's quite immediate to see that when you write explicitly the definition for the Christoffel symbols, when contracting the first indices, you will obtain the following form: $$ \Gamma^\mu_{\mu\nu}=\dfrac{1}{2}g^{\mu\sigma}\partial_\nu g_{\mu\sigma}=\dfrac{1}{2}\text{tr}(g^{-1}\partial_\nu g) $$

Now, the following matrix identity can be exploited: $$ \text{tr}(\ln A)=\ln(\text{det}A) $$

Furthermore, if A turns out to be differentiable, then $$ \dfrac{d}{d\lambda}\ln(\text{det}A)=\text{tr}\left(A^{-1}\dfrac{d}{d\lambda}A\right) $$

so that at the end: $$ \Gamma^\mu_{\mu\nu}=\dfrac{1}{2}\partial_\nu \ln(\text{det} \,g)=\partial_\nu \ln\sqrt{|{g|}} $$

Gianluca
  • 629