6
$\begingroup$

The vielbein postulate says that

$$\nabla_\mu e_v^{\,a}=\partial_{\mu}e_\nu^{\,a}+\omega_{\mu\,\, b}^{\,\,a}\,e^b_\nu-\Gamma^\sigma_{\mu\nu}\,e^{\,a}_\sigma=0.$$

$\nabla$ is the coordinate covariant derivative, and $\Gamma^\sigma_{\mu\nu}$ is connection defined as $$\Gamma^\sigma_{\mu\nu}=\tilde{\Gamma}^\sigma_{\mu\nu}+K^\sigma_{\mu\nu},$$

where $\tilde{\Gamma}^\sigma_{\mu\nu}$ is the Christoffel symbol and $K^\sigma_{\mu\nu}$ is the contorsion.

The Lorentz covariant derivative however is defined as $$D_\mu e_v^{\,a}=\partial_{\mu}e_\nu^{\,a}+\omega_{\mu\,\, b}^{\,\,a}\,e^b_\nu,$$ and the vielbein postulate then implies that $$D_\mu e_v^{\,a}=\Gamma^\sigma_{\mu\nu}\,e^{\,a}_\sigma.$$

Now, I want to take the Lorentz covariant deriavtive of the vielbein determinant, namely $$D_\mu\left(\sqrt{|g|}\right),$$ which we can take in the local coordinates to be $$D_\mu\left(\sqrt{-\text{det}(e_\alpha^{\,\,a}e_\beta^{\,\,b}\eta_{ab})}\right)=D_\mu\sqrt{e^2}=D_\mu|e|.$$

I am stuck here though as I'm unsure how one exactly takes this derivative. I would assume it should become a partial derivative as this quantity is just a scalar. Also, if one goes back a few steps and considers the relationship between the Christoffel symbols and the metric, we could just write this as

$$D_\mu\left(\sqrt{|g|}\right)=\partial_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\,\tilde{\Gamma}^\sigma_{\sigma\mu}.$$

Is this correct, or is there a better way to express this derivative?

EDIT:

Just in response to the answer given by @Immanuel (to avoid a long comment section), I will try to make clear my confusion with their answer.

The Lorentz covariant derivative is $$D_\mu V^a=\partial_\mu V^a+\omega_{\mu\,\,b}^{\,\,a} V^b,$$ where the spin connection $\omega$ looks like $$\omega_{\mu\,\,b}^{\,\,a}=\tilde{\omega}_{\mu\,\,b}^{\,\,a}+K_{\mu\,\,b}^{\,\,a}.$$

So I suppose my first point of confusion is why you have a $\Gamma$ instead of an $\omega$ in your definition of the Lorentz covariant derivative of a density.

If my definition is correct, shouldn't the Lorentz covariant derivative of a density be

$$D_\mu\left(\sqrt{|g|}\right)=\partial_\mu\left(\sqrt{|g|}\right)+\left(\tilde{\omega}_{\mu\,\,a}^{\,\,a}+K_{\mu\,\,a}^{\,\,a}\right)\sqrt{|g|},$$ which becomes $$D_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\tilde{\Gamma}_{\mu\lambda}^{\lambda}+\left(\tilde{\omega}_{\mu\,\,a}^{\,\,a}+K_{\mu\,\,a}^{\,\,a}\right)\sqrt{|g|}.$$

If $\omega_{\mu\,\,a}^{\,\,a}=0$, then we would have $$D_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\tilde{\Gamma}_{\mu\lambda}^\lambda,$$

so I hope that clears up some of my confusion.

$\endgroup$
4
  • $\begingroup$ I think all you have written is correct $\endgroup$ Commented Mar 2, 2022 at 23:16
  • $\begingroup$ What is $K_\mu^a{}_b$? $\endgroup$
    – Immanuel
    Commented Mar 3, 2022 at 17:28
  • $\begingroup$ In my notes the spin connection is defined in curved indices as $$\omega_{\mu\nu\rho}=\tilde{\omega}_{\mu\nu\rho}+K_{\mu\nu\rho},$$ where $K_{\mu\nu\rho}$ is the contorsion. So I was just writing it in terms of the lorentz indices, which maybe I am doing incorrectly. $\endgroup$ Commented Mar 3, 2022 at 17:31
  • $\begingroup$ OK. I see. I was confused. See my edited answer in some minutes. $\endgroup$
    – Immanuel
    Commented Mar 3, 2022 at 17:41

1 Answer 1

0
$\begingroup$

$\sqrt{|g|}$ is not a scalar. It is a density. Therefore, the covaraint derivative acts differently on it, i.e., you will get an extra term. But it turns out that w.r.t. Lorentz covariant derivative one has only one term.

Let $\rho$ be a density. Then, we know that $$ D_\mu \rho = (\partial_\mu - \omega^a_{\mu a}) \rho. $$ For the similar case of coordinate covariant derivative it is easy to show this ssing the definition of parallel transport as definition of covariant derivative.

However, we know that metricity implies $\omega^a_{\mu a} = 0$. Hence, we have $ D_\mu \rho = \partial_\mu \rho$.

Therefore, in this frame ("Lorentz frame") we find \begin{align} D_\mu \left( \sqrt{|g|} \right) = \partial_\mu \sqrt{|g|} = \sqrt{|g|} \tilde{\Gamma}^\nu_{\mu \nu}. \end{align} However, this is not true in natural coordinates. In that case we have $\nabla_\mu \sqrt{|g|} = 0$.

$\endgroup$
6
  • $\begingroup$ But with torsion the connection coefficients of $\nabla$ are not just the Christoffel symbols, they are the Christoffel symbols plus the contorsion. $D_\mu$ has the spin connection $\omega$ plus the contorsion as it's connection coefficient. Right? $\endgroup$ Commented Mar 3, 2022 at 15:03
  • $\begingroup$ Well, is the contraction of $\omega$ zero? I.e, $\omega^a_{\mu a} = 0$? I think so, right? $\endgroup$
    – Immanuel
    Commented Mar 3, 2022 at 15:06
  • $\begingroup$ What I'm saying is that if there is no torsion, then $D_\mu $ and $\nabla_\mu$ are equivalent if we impose the vielbein postulate. And this is the whole point of that postulate, I think. In your case the torsion doesn't vanish. Therefore, things are slightly different. If the torsion vanishes then the covariant derivative of the determinant of metric vanishes and two covariant derivatives are equivalent under the assumption of validity of vielbein postulate. $\endgroup$
    – Immanuel
    Commented Mar 3, 2022 at 15:19
  • $\begingroup$ The spin covariant derivative is defined like $$D_\mu V^a=\partial_\mu V^a+\omega_{\mu\,\,b}^{\,\,a} V^b ,$$ where $\omega_{\mu\,\,b}^{\,\,a} $ contains a a term like the Christoffel symbol (which as you say is maybe zero when we take the covariant derivative of the density) plus a contorsion term, so if we gave the covariant derivative of the density won't the contorsion term be left over? I believe you but I am just not seeing it, maybe my definition are incorrect? $\endgroup$ Commented Mar 3, 2022 at 15:25
  • $\begingroup$ No don't believe me! We have to derive things rather than believing someone. So you are saying exactly what I'm saying. I said that $\Gamma$ which is sum of contorsion and Christoffel symbol are the connection coefficients of $D$. And based on the fact that (coordinate) covariant derivative of the density would look like my first equation, we simply could replace $\nabla$ by $D$ and $\tilde{\Gamma}$ by $\Gamma$. $\endgroup$
    – Immanuel
    Commented Mar 3, 2022 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.