Bei's Study Notes

General Relativity 07
Last updated: 2016-08-21 22:20:50 PDT.

Reference: UCI OpenCourseWare - GR 08 (Playlist)


Principle of General Covariance

  1. Starting from an equation that is valid in an inertial frame.
  2. Generalize them using symmetry (invariance).

We used Lorentz invariant equation to describe SR previously. In GR, we also started with a equation in an inertial frame and generalized it to an arbitrary lab frame. The difference, is that the coordinate transformation in SR is global and in GR is local. This is similar to QED. Schroedinger Equation and Dirac Equation are invariant under phase rotation. When we promote that to local symmetry in QED, that requires us to have a electromagnetic field. That's the reason when we couple the particle into the EM field, we have to change the derivative to:

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

(This is a part that I don't understand.)

Similar to the EM field in QED, we have two field $g_{\mu\nu}(x)$ and $\Gamma^\lambda{}_{\mu\nu}(x)$ , and they are gauge fields. They ensure that the theory has the local invariance.

We are going to use the principle of general covariance to construct equations that obeys GR local invariance, similar to what we do in QED.

Noether's theorem states that each continuous symmetry corresponds to a conserved quantity. This is also true to general relativity. The local invariance corresponds to energy momentum conservation.

From the previous observation, we can describe the motion of a free particle basing on the metric tensor $g_{\mu\nu}$ and affine connection $\Gamma^\lambda_{\mu\nu}$ . The classical limit of metric tensor $g_{00}=-(1+2\phi)$ , also we have Gaussian's law of gravity $-\nabla\cdot\phi=4\pi G\rho$ where $\rho$ is the density. This is, of course, not Lorentz invariant. We can guess the form of the equation that relates the metric tensor and energy-momentum tensor:

$$ \cancel{\square g_{\mu\nu} \stackrel{?}= 8 \pi G T_{\mu\nu}}\,, $$

which is similar to EM but this doesn't work because $\square g_{\mu\nu}=0$ . It turns out that eventually it is the affine connection and Riemann tensor that show up in the equation, which is also non-linear.

(Feynmann) Einstein's equation is the only one that can be written down if massless spin-2 particle is allowed.


Vectors and Tensors

Consider general coordinate transformation:

$$ x^\mu = x'^\mu $$

$$ \begin{array}{|r|c|c|} \hline \text{Scalars} & \pi, \tau, ... & \text{(unchanged)} \\ \hline \text{Contravariant vector} & V^\mu(x) & V'_\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu \\ \hline \text{Covariant vector} & U_\mu(x) & U'_\mu = \frac{\partial x^\nu}{\partial x'^\mu} U_\nu \\ \hline \text{Derivative of a scala field} & \frac{\partial\phi}{\partial x'^\mu} & \frac{\partial\phi}{\partial x'^\mu} = \frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial\phi}{\partial x^\nu} \\ \hline \text{Tensor}&T^\mu{}_\nu{}^\lambda& T^\mu{}_\nu{}^\lambda= \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial x^\nu}{\partial x^{\nu'}} \frac{\partial x^{\lambda'}}{\partial x^{\lambda}} T'^{\mu'}{}_{\nu'}{}^{\lambda'} \\ \hline \end{array} $$


Metric tensor

$$ g_{\mu\nu}(x) \stackrel{def} = \eta_{\alpha\beta}\frac{\partial \xi^\alpha}{\partial x^\mu}\frac{\partial \xi^\beta}{\partial x^\nu}\\ g'_{\mu\nu}(x) = g_{\rho\sigma}\frac{\partial x^\rho}{\partial x^\mu}\frac{\partial x^\sigma}{\partial x^\nu}\\ $$

Inverse of metric tensor

$$ g^{\mu\lambda}g_{\nu\mu} = \delta^\lambda{}_\nu\\ g'^{\mu\lambda} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x'^\lambda}{\partial x^\sigma} g^{\rho\sigma} $$

Kronecker symbol

$\delta^\mu{}_\nu$ is mixed.

$$\begin{align*} \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\sigma}{\partial x'^\nu} \delta^\rho{}_\sigma &= \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\rho}{\partial x'^\nu} \\ &= \frac{\partial x'^\mu}{\partial x'^\nu}\\ &= \delta^\mu{}_\nu \end{align*}$$

Write down the covariant equation

Covariant equation only contains:

  • Scalar, Vectors, Tensors
  • Linear Combination
  • Direct product
  • Contraction

For example:

$$ R^{\nu\rho}{}_\sigma = g^{\mu\nu} S_\mu{^\rho}_{\sigma} $$

If certain equation can be written as equality of tensors, it will look the same in all kinds of coordinate system.

The metric tensor is useful for lowering and raising indices:

$$ g^{\rho\sigma}g_{\sigma\mu}T^{\mu\nu} = T^{\rho\nu} \\ g_{\rho\nu}g_{\sigma\mu}T^{\mu\nu} = T_{\sigma\rho} $$

Quantities that are almost tensors (Tensor Densities)

$$ g(x) = - \det g_{\mu\nu} $$

(Convention is to put minus before metric tensor.) $g$ is not a scalar:

$$\begin{align*} g'_{\rho\sigma} &= \frac{\partial x'_\rho}{\partial x_\mu} \frac{\partial x'_\sigma}{\partial x_\nu} g_{\mu\nu}\\ \det g'_{\mu\nu} &= \left|\frac{\partial x_\rho}{\partial x'_\mu}\right| \cdot \left|\frac{\partial x_\sigma}{\partial x'_\nu}\right| \cdot g \\ &=\left|\frac {\partial x}{\partial x'}\right|^2\cdot g \end{align*}$$

The determine of the metric is a scalar density.

Define "tensor density" of weight $w$ :

$$ \mathcal T'^{\mu}{}_{\nu} = \left|\frac{\partial x'}{\partial x}\right|^w \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\sigma}{\partial x'^\nu} \mathcal T^\rho{}_\sigma $$

weight of metric density $w=-2$ .

(!!) Volume element in 4d space time: $d^4x$ under $x^\mu \to x'^\mu$ :

$$ d^4 x' = \left|{\frac{\partial x'}{\partial x}}\right|d^4 x $$

$w=1$ Therefore it is not invariant. But if we multiply it with a tensor of weight $-1$ , then it will become invariant, for example:

$$ \sqrt{-g}\,d^4x \stackrel{def} = dV $$


Tensors so far: $g_{\mu\nu},g^{\mu\nu},\delta^\mu_\nu$ , $\frac{1}{\sqrt{g}}\epsilon_{\mu\nu\lambda\sigma}(\text{Levi-Civita density})$ .

Affine Connection:

$$\begin{align*} \Gamma^\lambda_{\mu\nu} &=\frac{\partial x^\lambda}{\partial \xi^\nu}\cdot \frac{\partial^2 \xi^\nu}{\partial x^\mu\partial x^\nu}\\ \frac{\partial}{\partial x}&=\frac{\partial x'}{\partial x}\frac{\partial }{\partial x'}\\ \frac{\partial}{\partial x'}&=\frac{\partial x}{\partial x'}\frac{\partial }{\partial x}\\ \\ \Gamma'^\lambda_{\mu\nu} &= \frac{\partial x'^\lambda}{\partial \xi^\nu} \cdot \frac{\partial^2 \xi^\nu}{\partial x'^\mu\partial x'^\nu}\\ &= \frac{\partial x^\rho}{\partial \xi^\nu}\frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \frac{\partial }{\partial x^\tau} \left( \frac{\partial x^\sigma}{\partial x'^\nu} \,\frac{\partial}{\partial x^\sigma} \xi^\nu \right) \\ &= \frac{\partial x^\rho}{\partial \xi^\nu} \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \, \left( \frac{\partial^2 x^\sigma}{\partial x^\tau\partial x'^\nu}\, \frac{\partial}{\partial x^\sigma}\xi^\nu + \frac{\partial x^\sigma}{\partial x'^\nu} \frac{\partial^2}{\partial x^\sigma\partial x^\tau}\xi^\nu \right)\\ &= \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \, \left( \frac{\partial x^\rho}{\partial \xi^\nu} \frac{\partial^2 x^\sigma}{\partial x^\tau\partial x'^\nu}\, \frac{\partial \xi^\nu}{\partial x^\sigma} + \frac{\partial x^\sigma}{\partial x'^\nu} \frac{\partial x^\rho}{\partial \xi^\nu} \frac{\partial^2 \xi^\nu}{\partial x^\sigma\partial x^\tau} \right)\\ &= \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \, \left( \frac{\partial^2 x^\rho}{\partial x^\tau\partial x'^\nu}\, + \frac{\partial x^\sigma}{\partial x'^\nu} \Gamma^\rho_{\sigma\tau} \right)\\ &= \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \frac{\partial x^\sigma}{\partial x'^\nu}\, \Gamma^\rho_{\sigma\tau} + \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial^2 x^\rho}{\partial x'^\mu\partial x'^\nu}\,\\ \end{align*}$$

This is means affine connection is not a tensor.

This raises the concern that the equation of free fall:

$$ \frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\lambda\sigma}\cdot\frac{dx^\mu}{d\tau}\cdot\frac{dx^\mu}{d\tau}=0\,, $$

may not be invariant. Fortunately, this equation is indeed invariant. Also the derivative of $x$ is tensor, the second derivative of $x$ is not a tensor. This miraculously reconcile the issue of affine connection.