Bei's Study Notes

General Relativity 09
Last updated: 2017-04-19 21:55:03 PDT.

Reference: UCI OpenCourseWare - GR 10 (Playlist)

Some exercises

Covariant derivative:

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

is a tensor.

With this, we can rewrite the equations in special relativity with this procedure:

  1. $\eta_{\alpha\beta} \to g_{\alpha\beta}(x) $
  2. $\partial_\alpha \to \nabla_\alpha$

Promote equation of motion for free particles

In SR, a free particle with velocity $u^\alpha$ and spin $S^\alpha$ :

$$ \frac{d u^\alpha}{d\tau}=0\\ \frac{d S^\alpha}{d\tau}=0\\ u_\alpha S^\alpha = 0 $$

Promoted to:

  1. $$ \frac{D u^\mu}{D\tau}=\frac{d u^\mu}{d\tau}+\Gamma^\mu_{\lambda\sigma}u^\mu u^\sigma = 0\\ $$

  2. $$ \frac{D S_\mu}{D\tau}=\frac{d S_\mu}{d\tau}-\Gamma^\lambda_{\mu\sigma}S_\lambda u^\sigma = 0\\ $$

  3. $$ u^\mu S_\mu = 0 $$

Other force $f^\alpha$ :

$$ \frac{Du^\mu}{D\tau}=\frac{f^\alpha}{m} \\ $$

For Lorentz force:

$$ f^\alpha = e F^\mu{}_\nu u^\nu $$

Or

$$ m\frac{d^2 x^\mu}{d\tau^2}=f^\mu-m\Gamma^\mu_{\lambda\sigma}u^\lambda u^\sigma $$

For spin $S_\mu$ :

$$ \frac{dS^\alpha}{d\tau} = \left(\frac{u^\alpha f^\beta}{m}\right)S_\beta $$

in GR (Fermi-Walker Transport):

$$ \frac{DS^\alpha}{D\tau} = \left(\frac{u^\alpha f^\beta}{m}\right)S_\beta $$


Promote E.M. to G.C. form

E.M.:

$$\begin{align*} \frac{\partial}{\partial x^\alpha} F^{\alpha\beta} &= - J^\beta\\ \partial_\alpha F_{\beta\gamma} + \partial_\gamma F_{\alpha\beta} + \partial_\beta F_{\gamma\alpha} &= 0 \end{align*}$$

Promotes to:

$$\begin{align*} F^{\mu\nu}{}_{;\nu} &= - J^\mu\\ F_{\mu\nu}{}_{;\lambda}+F_{\lambda\mu}{}_{;\nu}+F_{\nu\lambda}{}_{;\mu}&=0 \end{align*}$$

The covariance divergence can be written as:

$$\begin{align*} \frac{\partial}{\partial x^\mu}\sqrt{g}\,F^{\mu\nu} &= - J^\nu\\ F_{\mu\nu}{}_{,\lambda}+F_{\lambda\mu}{}_{,\nu}+F_{\nu\lambda}{}_{,\mu}&=0 \end{align*}$$

Also, current conservation:

$$ \frac{\partial}{\partial x^\mu}(\sqrt{g} J^\mu) = 0 $$

the change is very minimal (!!).


Promote Energy Momentum Tensor

$$ \frac{\partial T^{\alpha\beta}}{\partial x^\alpha} = G^\beta\quad (\text{external force}) $$

Promotes to:

$$ T^{\mu\nu}{}_{;\nu} = G^\mu\quad (\text{external force}) $$

simplifies to:

$$ \frac{1}{\sqrt{g}}\frac{\partial}{\partial x^\mu}(\sqrt{g}T^{\mu\nu}) = G^\nu - \Gamma^\nu_{\lambda\sigma} T^{\lambda\sigma} $$


Hydrodynamics

Perfect fluid

$$ T^{\alpha\beta} = p \eta^{\alpha\beta} + (p + \rho)u^\alpha u^\beta $$

Promotes to:

$$ T^{\mu\nu} = p g^{\mu\nu} + (p + \rho)u^\mu u^\nu $$

Divergence of above:

$$ T^{\mu\nu}{}_{;\mu}= \frac{\partial p}{\partial x^\nu} g^{\mu\nu} + \frac{1}{\sqrt{g}}\frac{\partial}{\partial x^\nu} \left(\sqrt{g} (p + \rho)u^\mu u^\nu\right) + \Gamma^\mu_{\nu\lambda}(p + \rho)u^\mu u^\nu = 0 $$

My note: it looks like that $\Gamma$ tends show up in non-linear parts of the equation.


Curvature

In ME, curvature is described as:

$$ F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu\quad. $$

The derivative of the field gives the curvature of the space. The problem in GR is that

$$ g_{\mu\nu;\lambda}=0\quad. $$

This problem is solved by Riemann and Christoffel by taking derivative of the affine connection. But since affine connection is not a tensor, the form is rather complicated. Einstein started to notice this in 1912, when examine the Ehrenfest paradox and realized that the spacetime has to be curved. Grossman pointed this out to Einstein. At around the same time, Einstein realized that the gravitational field should not be described by a single field $\phi(x)$ , but a field of the metric tensors $g_{\mu\nu}(x)$ , which turned out to be the most fundamental field in GR.

The point of it, is we write down:

$$ \Gamma^\lambda{}_{\mu\nu} (\text{not a tensor})\, $$

so if we take the derivative of it, say:

$$ \frac{\partial}{\partial x^\sigma}\Gamma^\lambda{}_{\mu\nu}\quad,$$

it will not be a tensor either. The hope was that by twiddling the indices, we can cancel out the non-tensor part of them and obtain a tensor. This is what Riemann discovered:

$$ R^\lambda{}_{\mu\nu\kappa}\stackrel{def}= \Gamma^\lambda_{\mu\nu,\kappa} +\Gamma^\eta_{\mu\nu}\Gamma^\lambda_{\kappa\eta} - (\nu \leftrightarrow \kappa) \quad(\text{Riemann tensor}) $$

Useful contractions of $R^\lambda{}_{\mu\nu\sigma}$

$$\begin{align} R_{\mu\kappa}&\stackrel{def}=R^\lambda{}_{\mu\lambda\kappa}\quad&(\text{Ricci tensor})\\ R&\stackrel{def}=g^{\mu\nu} R_{\mu\nu}\quad&(\text{Ricci Scalar, or Scalar Curvature}) \end{align}$$

Parallel transport around a (small) closed loop

Use the analog of Stoke's theorem:

$$ \oint_C\,d\vec e \cdot \vec A = \int_\Sigma (\vec \nabla \times \vec A)\cdot \vec n\,d\sigma $$

the result is:

$$ \Delta S_\mu = \frac{1}{2} R^\sigma{}_{\mu\nu\rho}\cdot S_\sigma\cdot u^{\rho\nu} $$

where $u^{\rho\nu}$ is call the "area bivector":

$$ u^{\mu\nu}\stackrel{def}=\oint \,x^\mu\,dx^\nu $$

e.g. around a parallelogram of sides $\delta a$ and $\delta b$ :

$$ u^{\mu\nu} = \delta a^\mu \delta b^\nu - (\mu \leftrightarrow \nu)\\ u_{\mu\nu}u^{\mu\nu} = -2 A $$


Flat spacetime vs. Riemann tensor

N.S condition for spacetime to be flat ($g_{\mu\nu} \to \eta_{\alpha\beta}$ ), is:

  1. $R^\lambda_{\mu\nu\sigma}(x) = 0$
  2. $g_{\mu\nu}$ has the same signature as $\eta_{\alpha\beta}$ .

Some identities of Riemann tensor

Recall that $ R^\lambda{}_{\mu\nu\kappa}=\left(\partial_\kappa \Gamma^\lambda_{\mu\nu} + \Gamma^\eta_{\mu\nu}\Gamma^\lambda_{\kappa\eta}\right)_{[\nu,\kappa]} $ and $\Gamma^\lambda_{\mu\nu}=-\frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma_mu} - \partial_\sigma g_{\mu\nu})$ . Then:

  1. (Antisymmetric on both pairs, Skew symmetry)

    $$ R_{\lambda\mu\nu\kappa}=-R_{\mu\lambda\nu\kappa}=-R_{\lambda\mu\kappa\nu}\,. $$

  2. (Swap two pairs, Interchange symmetry)

    $$ R_{\lambda\mu\nu\kappa}=R_{\nu\kappa\lambda\mu}. $$

  3. (First Bianchi identity)

    $$ R_{\lambda\mu\nu\kappa}+R_{\lambda\nu\kappa\mu}+R_{\lambda\kappa\mu\nu}=0 $$

    or

    $$ R_{\lambda[\mu\nu\kappa]}=0 $$

  4. (Bianchi identity)

    $$ (R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R)_{;\mu}=0 $$