Bei's Study Notes

General Relativity 08
Last updated: 2016-08-21 22:20:50 PDT.

Reference: UCI OpenCourseWare - GR 09 (Playlist)

Covariant differentiation

We are attempting to find a new definition of derivative that is covariant to coordinate change. Consider a vector:

$$\begin{align*} V'^\mu &= \frac{\partial x'^\mu}{\partial x^\nu}\, V^\nu\\ \end{align*}$$

Its partial derivative:

$$\begin{align*} \frac{\partial V'^\mu}{\partial x'^\lambda} &= \frac{\partial}{\partial x'^\lambda}\left(\frac{\partial x'^\mu}{\partial x^\nu}\, V^\nu\right)\\ &= \frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial V^\nu}{\partial x'^\lambda} + \frac{\partial^2 x'^\mu}{\partial x'^\lambda\partial x^\nu}\, V^\nu\\ &= \frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial x^\rho}{\partial x'^\lambda}\, \frac{\partial V^\nu}{\partial x^\rho} + \frac{\partial^2 x'^\mu}{\partial x^\nu\partial x^\rho} \,\frac{\partial x^\rho}{\partial x'^\lambda} \, V^\nu\\ \end{align*}$$

The appearance of the second derivative indicates that this differentiation is not covariant. Goes back to affine connection:

$$\begin{align*} \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa &=\frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial x^\rho}{\partial x'^\lambda}\, \frac{\partial x^\sigma}{\partial x'^\kappa}\, \Gamma^\nu_{\rho\sigma}\, V'^\kappa + \frac{\partial x'^\mu}{\partial x^\rho} \, \frac{\partial^2 x^\rho}{\partial x'^\lambda \partial x'^\kappa}\, V'^\kappa \\ &=\frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial x^\rho}{\partial x'^\lambda}\, \Gamma^\nu_{\rho\sigma}\, V^\sigma + \frac{\partial x'^\mu}{\partial x^\rho}\, \frac{\partial}{\partial x'^\kappa} \left(\frac{\partial x^\rho}{\partial x'^\lambda} \right)\, V'^\kappa \\ \end{align*}$$

Add them together:

$$\begin{align*} \frac{\partial V'^\mu}{\partial x'^\lambda} + \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa &= \frac{\partial x'^\mu}{\partial x^\nu}\,\frac{\partial x^\rho}{\partial x'^\lambda}\, \left(\frac{\partial V'^\mu}{\partial x'^\lambda} + \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa\right) + V^\nu\cdot\frac{\partial}{\partial x^\nu} \left( \frac{\partial x'^\mu}{\partial x^\rho}\, \frac{\partial x^\rho}{\partial x'^\lambda} \right)\\ &= \frac{\partial x'^\mu}{\partial x^\nu}\,\frac{\partial x^\rho}{\partial x'^\lambda}\, \left(\frac{\partial V'^\mu}{\partial x'^\lambda} + \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa\right) + \cancel{\frac{\partial \delta^\mu{}_\lambda}{\partial x^\nu} V^\nu} \end{align*}$$

An alternative derivation can be found here: Wikipedia.

So we can define the covariant derivative as:

$$ V^\mu{}_{;\lambda} \stackrel{def} = \frac{\partial V^\mu}{\partial x^\lambda} + \Gamma^\mu_{\lambda\kappa}V^\kappa $$


$$ \nabla_\lambda V^\mu{} \stackrel{def} = \partial_\lambda V^\mu + \Gamma^\mu_{\lambda\kappa}V^\kappa $$

It is a mixed tensor.

Similarly for covariant vector:

$$ V_{\mu;\lambda} \stackrel{def}= \frac{\partial V_\mu}{\partial x^\lambda} - \Gamma^\kappa_{\mu\lambda}V_\kappa $$

More generally:

$$\begin{align*} T^{\alpha_1\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\beta_s;\gamma} \stackrel{def}= T^{\alpha_1\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\beta_s,\gamma} &+\Gamma^{\alpha_1}_{\delta\gamma}T^{\delta\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\beta_s,\gamma} + ... + \Gamma^{\alpha_r}_{\delta\gamma}T^{\alpha_1\alpha_2...\delta}{}_{\beta_1\beta_2...\beta_s,\gamma}\\ &-\Gamma^{\delta}_{\beta_1\gamma}T^{\alpha_1\alpha_2...\alpha_r}{}_{\delta\beta_2...\beta_s,\gamma} -...-\Gamma^{\delta}_{\beta_s\gamma}T^{\alpha_1\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\delta,\gamma} \end{align*}$$

Apply this to metric:

$$\begin{align*} g_{\mu\nu;\lambda}&=\frac{\partial g_{\mu\nu}}{\partial x_\lambda} -\Gamma^\rho_{\lambda\mu}g_{\rho\nu} -\Gamma^\rho_{\lambda\nu}g_{\mu\rho}\\ &=0 \end{align*}$$

Since $g_{\mu\nu}$ is constant in freely falling frame, the covariant derivative has to be $0$ everywhere.

Taking covariant derivative commutes with raising/lowering indices:

$$ (g^{\mu\nu}V_\nu)_{;\lambda}=g^{\mu\nu}V_{\nu;\lambda} $$

Recipes for equations in GR

  1. Take an equations valid in SR;
  2. $\eta_{\alpha\beta} \to g_{\alpha\beta}$ ;
  3. $\partial_\mu \to {}_{;\mu}(\text{or }\nabla_\lambda)$ .

"Minimal substitution"

$$ p_\mu \to p_\mu - \frac{e}{c}A_\mu $$

in $H=\frac{1}{2m}(\vec p - \frac{e}{c} \vec A)^2$ and

$$ \partial_\mu \to \partial_\mu - \frac{ie}{\hbar c}A_\mu $$

This substitution implies the Gauge invariance.

More properties of Covariant derivative

Covariant divergence:

$$\begin{align*} V^\mu{}_{;\mu} &= V^\mu{}_{,\mu}+\Gamma^\mu_{\mu\lambda}V^\lambda\\ &=\partial_\mu V^\mu + \frac{1}{2}g^{\mu\rho}\frac{\partial g_{\rho\mu}}{\partial x^\lambda} \\ &=\frac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}V^\mu)\\ \end{align*}$$

useful for conservation laws:

$$\begin{align*} \int\,d^4 \sqrt{g}\,V^\mu{}_{;\mu} &= \int\,d^4\,\partial_\mu(\sqrt{g}V^\mu) \\ &= \text{boundary term} \end{align*}$$

Covariant differentiation along a curve

Transporting an vector along a curve $x(\tau)$ , we get a family of vectors:

$$ A'^\mu(\tau) = \left.\frac{\partial x'^\mu}{\partial x^\nu}\right|_{x=x(\tau)}\,A^\nu(\tau) $$

differentiation with respect to $\tau$ .

$$\begin{align*} \frac{dA'^\mu}{d\tau}&=\frac{d}{d\tau}\left(\frac{\partial x'^\mu}{\partial x^\nu}A^\nu\right) \\ &= \frac{\partial x'^\mu}{\partial x^\nu}\frac{dA^\nu}{d\tau}+ \frac{\partial^2 x'^\mu}{\partial x^\nu\partial x^\lambda} \frac{dx^\lambda}{d\tau} A^\nu \end{align*}$$

$$ \frac{DA'^\mu}{D\tau} \stackrel{def}= \frac{dA^\mu}{d\tau} + \Gamma^\mu_{\nu\lambda}\frac{dx^\lambda}{d\tau}A^\nu $$

This leads to the notion of parallel transport.

Parallel transport

A path that $A^\mu$ does not change in the frame attached to particle. Which is supposed to be local inertial frame.

In that local inertial frame:

  1. $\Gamma = 0$ (F.F frame). (This has not been mentioned before, but can be found easily).
  2. $\frac{dA^\mu}{d\tau} = 0$ if $A^\mu$ does not change direction.
  3. $\frac{DA^\mu}{D\tau} = 0$ in any frame. This means:

    $$ \frac{dA^\mu}{d\tau}=-\Gamma^{\mu}_{\nu\lambda}\frac{dx^\lambda}{d\tau}A^\nu $$

The geodesic equation is of this type:

$$\begin{align*} \frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\lambda\sigma}\frac{dx^\lambda}{d\tau}\frac{dx^\sigma}{d\tau} = 0\\ \frac{D}{D\tau}\left(\frac{dx^\mu}{d\tau}\right)=0 \end{align*}$$

Geodesic is not only the shortest path, but also the straightest path.

Parallel transporting a vector along a closed curve, the result vector may not coincide with the original one.

Integrating equation of parallel transport

$$ A^\mu \to S_\mu $$

Spin is a covariant vector, therefore

$$\begin{align*} \frac{dS_\mu}{d\tau} = \Gamma^\lambda_{\mu\nu}\,\frac{dx^\nu}{d\tau}\,S_\lambda \\ dS_\mu = \Gamma^\lambda_{\mu\nu}\,S_\lambda\,dx^\nu \\ S' = S + dS = (1+\Gamma^\lambda_{\mu\nu}\,dx^\nu)S \\ \end{align*}$$

This can be integrated in terms of a rotation matrix

$$\begin{align*} S'^\mu &= R^\mu{}_\nu S^\nu\\ R^\mu{}_\nu=[\mathcal P \exp\left(-\oint dx'^\lambda\, \Gamma^\mu_{\lambda\nu} \right)] \end{align*}$$

where $\mathcal P$ is the path-ordering meta operator.

Parallel transport can be used to detect curvature.

Pursue analogy with EM

  1. Local gauge invariance in E.M. with EM field $A_\mu(x)$ , and matter field $\psi(x)$ . Equations are preserved if:

    $$ \left\{\begin{align*} A_\mu(x) &\to A_\mu(x) + \frac{\partial}{\partial^\mu}\theta(x)\\ \psi(x) &\to e^{ie\theta(x)}\cdot \psi(x) \end{align*}\right. $$

    where $\theta(x)$ is called the gauge function. A useful tool to deal with gauge invariance is to introduce a "gauge invariant derivative":

    $$\begin{align*} D_\alpha \psi(x) &\stackrel{def} = \left(\frac{\partial}{\partial x^\alpha}-ie A_\alpha(x)\right)\psi(x)\\ D_\alpha (e^{ie\theta(x)}\psi(x)) &= e^{ie\theta(x)}\left( \frac{\partial}{\partial x^\alpha}\psi(x) - ie A_\alpha(x)\psi(x) \right)\\ &= e^{ie\theta(x)}D_\alpha(\psi(x))\\ \end{align*}$$

So when we construct the Lagrangian density $\mathcal L$ , we can use this derivative as the first derivative and the equation will be invariant modular a rotation factor:

$$ \mathcal L \propto |D_\mu\psi(x)|^2 \quad (\text{Higgs field}) \\ \mathcal L \propto \bar\psi(x)\gamma^\mu D_\mu\psi(x) \quad (\text{Dirac field}) \\ $$

It is alway the covariant derivative that makes sure the equations is gauge invariant. (Noether's theorem)

The local gauge invariance leads to conserved EM current:

$$\begin{align*} J_\alpha(x) &= -ie\left[\psi^+D_\alpha \psi - \psi(D_\alpha \psi)^+\right]\\ \frac{\partial}{\partial x^\alpha} J^\alpha(x) &= 0\\ \end{align*}$$

Similarly, the local invariance of GR give raise to a local conserved current, that is the energy-momentum tensor.

$$ \begin{array}{|c|c|c|} \hline \text{Theory}&\text{EM}&\text{GR}\\\hline \text{Gauge field} & A^\mu & \Gamma^\lambda_{\mu\nu}\\\hline \text{Fundamental field} & A^\mu & g_{\mu\nu}\\\hline \text{Field strength} & F_{\mu\nu} & R_{\mu\nu}\\\hline \text{Invariance}&\text{Phase}&\text{Coordinate}\\\hline \text{Conserved quantity}&\text{charge}&\text{energy-momentum tensor}\\\hline \end{array} $$

The local invariance can not be consistent unless the boson is massless.