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General Relativity - UCI OCW

General Relativity - UCI OCW

General Relativity 10
Last updated: 2016-08-25 18:41:25 PDT.

Reference: UCI OpenCourseWare - GR 11 (Playlist)

Einstein's field equation

$$ g_{00} = -(1+2\phi(x)) \quad(\text{Weak fields})\\ $$

$$ \Delta\phi = 4\pi G\,\rho\\ \Delta g_{00} = -8\pi G\,\rho\\ $$

$$ T_{00}=\rho $$


$$ \Delta g_{00} = -8\pi G T_{00} $$

An extension might be:

$$ G_{\alpha\beta}=-8\pi G T_{\alpha\beta} $$

where $G_{\alpha\beta}$ is an unknown object that relates to gravitational field.

Promotes to G.C.:

$$ G_{\mu\nu}=-8\pi G T_{\mu\nu} $$

To determine $G_{\mu\nu}$ , we write down the properties of $G_{\mu\nu}$ :

  1. $G_{\mu\nu}$ is a tensor.
  2. $G_{\mu\nu}$ is made out of at most second derivatives of the metric tensor, e.g. $\left(\frac{\partial g}{\partial x}\right)^2$ , or $\frac{\partial^2 g}{\partial x^2}$ .
  3. $G_{\mu\nu}=G_{\nu\mu}$ .
  4. $G^{\mu\nu}{}_{;\nu} = 0$
  5. On non-relativistic limit, $G_{00}\to\Delta g_{00}$ .

These requirements uniquely determines the form of $G_{\mu\nu}$ .

$$ G_{\mu\nu} = C_1 R_{\mu\nu}+ C_2 g_{\mu\nu} R $$

$$ G^{\mu}{}_{\nu;\mu}=\left(\frac{1}{2}C_1+C_2\right)R_{;\nu} $$

therefore, $G_{\mu\nu}$ has the form $C_1(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R)$ . Take the non-relativistic limit, that is, $u^\mu \approx 0$ and the field is weak ($g_{\mu\nu} = \eta_{\mu\nu}+\delta h_{\mu\nu}$ ), and so is the affine connection and the Riemann tensor. Also, $ R_{ij} \approx \frac{1}{2}g_{ij}R$ .

$$ R\equiv g^{\mu\nu} R_{\mu\nu}\approx R_{kk}-R_{00}=\frac{3}{2}R - R_{00}\\ R \approx 2R_{00} $$

$$ G_{00} = C_1\left(R_{00}-\frac{1}{2}\eta_{00}R\right)=2C_1R_{00}\\ R_{\mu\nu}\approx \eta^{\lambda\sigma}R_{\lambda\mu\sigma\nu}\\ R_{00}\approx \eta^{\lambda\sigma}R_{\lambda 0\sigma 0}\\ R_{\lambda\mu\nu\kappa} = \frac{1}{2}\left[ \left(\frac{\partial^2g_{\lambda\nu}}{\partial x^\kappa\partial x^\nu}\right) _{[\lambda,\mu]} \right]_{[\nu,\kappa]} + o(\Gamma^2)\\ R_{0000} \approx 0\\ R_{i0j0} \approx \frac{1}{2}\frac{\partial^2g_{00}}{\partial x^i\partial x^j}\\ G_{00} \approx 2C_1(\frac{1}{2}g_{00}) = C_1 \Delta g_{00} \approx -8\pi G \rho\\ C_1 = 1 $$

Conclusion (Einstein's field equation):

$$ R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=-8\pi G T_{\mu\nu} $$

Equivalent form:

$$\begin{align*} g^{\mu\nu}R_{\mu\nu}-\frac{1}{2}g^{\mu\nu}g_{\mu\nu}R&=-8\pi G T_{\mu\nu} g^{\mu\nu}\\ R-2R&=-8\pi G T^{\nu}{}_{\nu}\\ R&=8\pi G T^{\nu}{}_{\nu}\\ R_{\mu\nu}&=-8\pi G \left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T^\lambda{}_\lambda\right)\\ \end{align*}$$

In vacuum, $T=0$ , therefore $R_{\mu\nu}=0$ , this is how gravitational waves are calculated, which does not imply Riemann tensor is $0$ . In fact, even when $R_{\mu\nu}$ is $0$ , $g_{\mu\nu}$ can still oscillate through spacetime.

(then a discussion about the cosmological constant).

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:

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


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

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


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

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:

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


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.


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)

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


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

  4. (Bianchi identity)

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

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.

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.

General Relativity 06
Last updated: 2016-08-14 19:21:06 PDT.

Reference: UCI OpenCourseWare - GR 07 (Playlist)

Some remarks (Mach's Principle)

Starting from 39:00.

My Note: I've never been convinced of Mach's Principle. In fact, there has been many null results in testing Mach's principle. I don't know this has to be a thing in the first place.

Non-Cartitian coordinate system

Starting from 1:07:55.

Starting from an example:

$$ \xi^\alpha = \begin{cases} x &=& r \sin \theta \cos \phi \\ y &=& r \sin \theta \sin \phi \\ z &=& r \cos \theta \end{cases} \\ x^\alpha = \begin{cases} r &=& \sqrt{x^2 + y^2 + z^2} \\ \theta &=& \cos^{-1}(\frac{z}{r})\\ \phi &=& \tan^{-1}(\frac{y}{x}) \end{cases} \\ \eta_{\alpha\beta} =\delta_{\alpha\beta} $$

We will try to calculate the metric in this case. The Jacobian:

$$\begin{align*} \frac{\partial(x,y,z)}{\partial(r,\theta,\phi)} &= \left(\begin{array}{lll} \frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}&\frac{\partial x}{\partial \phi}\\ \frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}&\frac{\partial y}{\partial \phi}\\ \frac{\partial z}{\partial r}&\frac{\partial z}{\partial \theta}&\frac{\partial z}{\partial \phi}\\ \end{array}\right)\\ &=\left(\begin{array}{lll} \sin \theta \cos \phi & r \cos \theta \cos\phi & -r \sin \theta \sin \phi \\ \sin \theta \sin \phi & r \cos \theta \sin\phi & r \sin \theta \cos \phi \\ \cos \theta & -r \sin \theta \cos\phi & 0 \\ \end{array}\right)\\ g_{\mu\nu} &= \eta_{\alpha\beta} \frac{\partial \xi^\alpha}{\partial x^\mu} \frac{\partial \xi^\beta}{\partial x^\nu}\\ &= \frac{\partial x}{\partial x^\mu}\frac{\partial x}{\partial x^\nu}+ \frac{\partial y}{\partial x^\mu}\frac{\partial y}{\partial x^\nu} + \frac{\partial z}{\partial x^\mu}\frac{\partial z}{\partial x^\nu} \\ &= \left( \begin{array}{lll} 1 & 0 & 0\\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2\theta \\ \end{array} \right)\\ \sqrt{\det g_{\mu\nu}} &= r^2 sin^\theta\\ ds^2 &= g_{\mu\nu}dx^\mu dx^\nu = dr^2 + r^2(d\theta^2 + sin^2\theta d\phi^2) \end{align*}$$

The inverse is simple:

$$\begin{align*} g^{\mu\nu} &= \left( \begin{array}{lll} 1 & 0 & 0\\ 0 & \frac{1}{r^2} & 0 \\ 0 & 0 & \frac{1}{r^2 \sin^2\theta} \\ \end{array} \right)\\ \Gamma^\lambda{}_{\mu\nu} &= -\frac{1}{2}g^{\mu\nu}(\partial g+\partial g-\partial g)\\ &= ... \end{align*}$$

The important takeaway is that $\Gamma$ is in fact not a tensor, for that it can be $0$ in some coordinate systems but non-zero in others, even in the same planar space.

General Relativity 05
Last updated: 2016-08-14 19:21:06 PDT.

Reference: UCI OpenCourseWare - GR 06 (Playlist)

Equation of free fall from Lagrangian

Last time we wrote down the equation of free fall:

$$ \frac{d^2x^\lambda}{d\tau^2} + \Gamma^\lambda{}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0\\ \\ \text{or}\\ \\ g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} + \Gamma_{\lambda\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0 $$

where $\Gamma$ is the Christoffel symbol of the first kind, which can be written as a function of metric tensor:

$$ \Gamma_{\lambda\mu\nu} = \frac 12 \left( \frac{\partial g_{\nu\lambda}}{\partial x^\mu} + \frac{\partial g_{\mu\lambda}}{\partial x^\nu} - \frac{\partial g_{\mu\nu}}{\partial x^\lambda} \right) $$

Today we are going to derive the equation of free fall from the variational principal.


$$ A = \int_A^Bd\tau\\ $$

the principal is that the equation of motion can be derived from:

$$ \delta A = 0 $$

By this, it means a particle goes from $A$ to $B$ from a trajectory $x^\mu(p)$ where $p \in [0, 1]$ is an parameter. If we perturb the trajectory with an arbitrary finite variation $x^\mu(p)$ multiply by an infinitesimal factor of $\Delta$ , then the action remains the same. We can write it down:

$$\begin{align*} \delta \int_A^Bd\tau &= \delta \int_A^Bdp \frac{d\tau}{dp}\\ \\ d\tau^2&=-g_{\mu\nu}dx^\mu dx^\nu \\ \delta d\tau&=\delta\sqrt{-g_{\mu\nu}dx^\mu dx^\nu} \\ &=\frac{1}{2d\tau}\delta(-g_{\mu\nu}dx^\mu dx^\nu) \\ &=\frac{1}{2d\tau}( -\delta g_{\mu\nu}dx^\mu dx^\nu -g_{\mu\nu}\left( d\delta x^\mu dx^\nu + dx^\mu d\delta x^\nu \right) ) \\ &=\frac{1}{2d\tau}( -\frac{\partial g_{\mu\nu}}{\partial x^\lambda}\delta x^\lambda dx^\mu dx^\nu -g_{\mu\nu}d\delta x^\mu dx^\nu -g_{\mu\nu}dx^\mu d\delta x^\nu ) \\ \end{align*}$$

The boundary term is $0$ and integrate by partial:

$$\begin{align*} -g_{\mu\nu}d\delta x^\mu dx^\nu &\to d(g_{\mu\nu}dx^\nu) \delta x^\mu \\ &= (dg_{\mu\nu}dx^\nu + g_{\mu\nu}d^2x^\nu) \delta x^\mu \\ &= (dg_{\lambda\nu}dx^\nu + g_{\lambda\nu}d^2x^\nu) \delta x^\lambda \\ &= \left( \frac{\partial g_{\lambda\nu}}{\partial x^\mu}dx^\mu dx^\nu + g_{\lambda\sigma}d^2x^\sigma \right) \delta x^\lambda \\ \\ -g_{\mu\nu}dx^\mu d\delta x^\nu &= \left( \frac{\partial g_{\lambda\mu}}{\partial x^\nu}dx^\mu dx^\nu + g_{\lambda\sigma}d^2x^\sigma \right) \delta x^\lambda \\ \end{align*}$$

Plug it in,

$$\begin{align*} \delta d\tau &\to \frac{1}{2d\tau}\left( -\frac{\partial g_{\mu\nu}}{\partial x^\lambda}dx^\mu dx^\nu + \frac{\partial g_{\lambda\nu}}{\partial x^\mu}dx^\mu dx^\nu + \frac{\partial g_{\lambda\mu}}{\partial x^\nu}dx^\mu dx^\nu + 2g_{\lambda\sigma}d^2x^\sigma \right)\delta x^\lambda \\ &= \frac{1}{d\tau}\left( \Gamma_{\lambda\mu\nu}dx^\mu dx^\nu + g_{\lambda\sigma}d^2x^\sigma \right)\delta x^\lambda \\ &= \left( \Gamma_{\lambda\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} + g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} \right)\delta x^\lambda \\ \end{align*}$$

$$\begin{align*} 0 = \delta \int_A^Bd\tau &= \int_A^Bd\tau \left( \Gamma_{\lambda\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} + g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} \right)\delta x^\lambda\\ \end{align*}$$

Since $\delta x^\lambda$ is arbitrary, we get again the equation of free fall:

$$ \Gamma_{\lambda\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} + g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} = 0 $$

By definition, this is trajectory of the stationary path (in terms of proper time) in 4 dimentional spacetime. This is the Geodesics. Therefore we replaced gravitational force with the geometry of spacetime.

There has a clear consequence with QM because of path integral:

$$ Amp(A\to B)\propto \int d \exp\left({\frac{i}{\hbar} \int d\tau mc^2A}\right) $$

Newtonian approximation

For slow moving particles:

$$\begin{align*} \frac{d^2x^\lambda}{d\tau^2} + \Gamma^\lambda{}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}&\approx \frac{d^2x^\lambda}{dt^2}\left(\frac{dt}{d\tau}\right)^2 + \Gamma^\lambda{}_{00}\left(\frac{dt}{d\tau}\right)^2\\ \frac{d^2x^\lambda}{dt^2} + \Gamma^\lambda{}_{00} &\approx 0 \end{align*}$$

For weak, stationary fields:

$$\begin{align*} g_{\alpha\beta} &=\eta_{\alpha\beta}+h_{\alpha\beta},\quad\quad|h_{\alpha\beta}| \ll 1\,,\dot h = 0\\ g^{\alpha\gamma} &=\eta^{\alpha\gamma}+h'^{\alpha\gamma} \\ g^{\alpha\gamma}g_{\alpha\beta}&= \delta^\gamma{}_\beta+\eta^{\alpha\gamma}h_{\alpha\beta}+\eta_{\alpha\beta}h'^{\alpha\gamma}+o(h^2)\\ 0 &= \eta^{\alpha\gamma}h_{\alpha\beta}+\eta_{\alpha\beta}h'^{\alpha\gamma}\\ &=h^\gamma{}_{\beta}+h'^\gamma{}_{\beta}\\ g^{\alpha\gamma} &= \eta^{\alpha\beta}-h^{\alpha\beta}\\ \\ \Gamma^{\lambda}{}_{00}&\approx\frac{1}{2}\eta^{\sigma\lambda}( \partial_0 g_{\sigma 0} + \partial_0 g_{\sigma 0} - \partial_\sigma g_{0 0} )\\ &=\frac{1}{2}\eta^{\sigma\lambda}( \partial_0 h_{\sigma 0} + \partial_0 h_{\sigma 0} - \partial_\sigma h_{0 0} )\\ &=-\frac{1}{2}\eta^{\sigma\lambda}\partial_\sigma h_{0 0}\\ &=-\frac{1}{2}(0, \vec\nabla h_{0 0})^\lambda\\ \end{align*}$$

If both conditions are met:

$$ \vec{a}= \frac{1}{2}\vec\nabla h_{0 0} $$

This looks a lot like the the Newtonian gravitation:

$$ \vec{a} = -\vec\nabla \phi $$

where $\phi$ is the gravitational potential (which is always negative),

$$ h_{00} = -2\phi + const $$

At the boundary, the gravitation field is $0$ , the $h_{00}$ should also be $0$ , therefore the constant should be $0$ as well.

$$ h_{00} = -2\phi $$

$$\begin{align*} \phi &= -\frac{mG}{|r|}\\ g_{00} &= -1-2\phi\\ &= \frac{2mG}{|r|}-1 \end{align*}$$

Remark: 1. $g_{00}$ goes to $0$ at the event horizon. 2. The boundary gravitational field may not be $0$ .

In the case of stronger gravitational field $\phi$ (which is dimensionless):

  1. $10^{-4}$ at the surface of a white dwarf;
  2. $10^{-6}$ at the surface of the sun;
  3. $10^{-9}$ at the surface of the earth;
  4. $10^{-39}$ at the surface of an proton;


Time Dilation

$$ d\tau = \sqrt{-\eta_{\alpha\beta} d\xi^\alpha d\xi^\beta} = \sqrt{-g_{\alpha\beta} dx^\alpha dx^\beta} $$

We can caluculate $dt/d\tau$ which is the factor of time dilation:

$$\begin{align*} \frac{dt}{d\tau}&=\left(-g_{\alpha\beta} \frac{dx^\alpha}{dt} \frac{dx^\beta}{dt}\right)^{-\frac 12}\\ \frac{dx^\mu}{dt} &= (1, \vec v)\\ \end{align*}$$

If the $\vec v=0$ :

$$ \frac{dt}{d\tau} = \frac{1}{\sqrt{-g_{00}}} $$

comparing two clocks:

$$ \frac{dt_1}{dt_2} = \frac{\sqrt{-g_{00}(x_2)}}{\sqrt{-g_{00}(x_1)}} $$

The frequency of the two stationary clock:

$$ \frac{dv_2}{dv_1} = \frac{\sqrt{-g_{00}(x_2)}}{\sqrt{-g_{00}(x_1)}} $$

In the previous conditions:

$$\begin{align*} \frac{dv_2}{dv_1} &\approx 1 + \phi(x_2) - \phi(x_1) \\ &= 1 + MG \frac{|r_1| - |r_2|}{|r_2r_1|} \\ \end{align*}$$

General Relativity 04
Last updated: 2016-08-14 03:35:13 PDT.

Reference: UCI OpenCourseWare - GR 05 (Playlist)

Starting with Newtonian gravity:

$$\begin{align*} \hat F &= - G \frac{m M}{r^2} \hat e_r,\quad G = 6.67\times 10^{-8} cm^2/g^2\\ \frac{\hbar G}{c^3} &= l_P^2(\text{Planck length}) \\ &= (1.61624(12) \times 10^{-33} cm)^2 \end{align*}$$

An easier way to describe this is through potential:

$$\begin{align*} \hat F &= -m \vec{\nabla}\phi \end{align*}$$

where $\phi$ is the gravitational potential, which only depends on the position. For point mass:

$$ \phi(\vec x) = - \frac{MG}{|\vec x -\vec x_0|}\,, $$

for mass distribution:

$$ \phi(\vec x) = -\int\,d^3\vec x'\,\frac{G\rho(\vec x)}{|\vec x -\vec x'|} $$

It means that we can introduce a gravitational field:

$$\begin{align*} \vec G &= -\vec \nabla\phi\\ \vec\nabla\cdot\vec G &= -4\pi G\vec\nabla \rho(\vec x)\quad(\text{Gauss})\\ \Delta\phi &= 4\pi G\rho\quad(\text{Poisson})\\ \end{align*}$$

Gravitational force is special:

  1. It is the oldest force.
  2. It couples to everything.
  3. Gravity cannot be screened.
  4. It is long range. (Mediated by massless particle if there is any.)
  5. It is the weakest force.

History of development of Einstein's GR

  1. ~1907
    • Gravitational fields have relative existence. (Gravitational field can be removed by coordinate transformation).
    • There's a need to extend SR to accelerating frames.
    • Newtonian gravity is not Lorentz invariant.
    • Leads to Equivalence Principal.
    • Gravitational red-shift (testable!).
    • Bending of light.
    • Geometry will be curved. By spinning disk.
  2. 1911
    • Bending of light by the sun is observable. His calculation was off by a factor of 2. ($E=h\nu$ )
  3. 1912 Equation for the "c-field", which is way off:

    $$\Delta c = \kappa \left(c \sigma + \frac{1}{2\alpha}\frac{(\vec\nabla c)^2}{c}\right)$$

    In the c-field, the speed of light is modified by the gravity.
  4. 1912 Start of Marcel Grossmann collaboration.
    • Metric tensor $g_{\mu\nu}(x)$ .
    • $g_{00}=-[1+2\phi(x)]$ .

Principal of Equivalence

Statement: If you have a particle in a gravitational field:

$$ m_I \vec a = m_G \vec g $$

where the inertial mass $m_I$ equals gravitational mass to $m_G$ . It is tested to be true in the level of $10^{-12}$ .


$$ m \frac{d^2x}{dt^2}= m \vec g + (\text{other force}) $$

By applying coordinate change:

$$ x = x - \frac 12 g t^2\\ t = t $$

The gravitational field is canceled.

General statement:

At everything spacetime point, it is possible to choose a "locally inertial frame" where gravity is canceled. In that frame of reference (freely falling frame), the law is the law of SR.

A more general description of gravitational force

Here we generalize by considering a free-falling particle in its own frame of reference $\xi^\mu$ , and transform the coordinate system to an arbitrary coordinate system $x^\mu$ .

In the freely-falling frame:

$$ \frac{d^2\xi^\alpha}{d\tau^2} = 0 $$

In the lab frame:

$$ x^\alpha = \left(x(\xi)\right)^\alpha $$

where $x$ is the function that maps $\xi$ to $x$ . From the first equation (in fact, 4 equations parameterized by $\alpha$ ):

$$\begin{align*} 0 &= \frac{d}{d\tau}\left(\frac{\partial \xi^\alpha}{\partial x^\mu}\frac{d x^\mu}{d \tau}\right)\\ &= \frac{\partial \xi^\alpha}{\partial x^\mu}\frac{d^2 x^\mu}{d \tau^2} + \frac{\partial^2 \xi^\alpha}{\partial x^\nu\partial x^\mu}\frac{d x^\mu}{d \tau}\frac{d x^\nu}{d \tau} \\ \end{align*}$$

multiply (contract) by ${\partial x^\lambda}/{\partial \xi^\alpha}$ (now we have 4 equations parameterized by $\lambda$ ):

$$\begin{align*} 0 &= \frac{\partial x^\lambda}{\partial \xi^\alpha}\frac{\partial \xi^\alpha}{\partial x^\mu}\frac{d^2 x^\mu}{d \tau^2} + \frac{\partial x^\lambda}{\partial \xi^\alpha}\frac{\partial^2 \xi^\alpha}{\partial x^\nu\partial x^\mu}\frac{d x^\mu}{d \tau}\frac{d x^\nu}{d \tau} \\ &= \frac{d^2 x^\lambda}{d \tau^2} + \frac{\partial x^\lambda}{\partial \xi^\alpha}\frac{\partial^2 \xi^\alpha}{\partial x^\nu\partial x^\mu}\frac{d x^\mu}{d \tau}\frac{d x^\nu}{d \tau}\\ \\ \text{Define(Affine Connection):}\quad\, \Gamma^\lambda{}_{\mu\nu} &\equiv \frac{\partial x^\lambda}{\partial \xi^\alpha}\frac{\partial^2 \xi^\alpha}{\partial x^\nu\partial x^\mu}\\ \\ 0 &= \frac{d^2 x^\lambda}{d \tau^2} + \Gamma^\lambda{}_{\mu\nu} \frac{x^\mu x^\nu}{d^2\tau} \end{align*} $$

This is called the "Equation of free fall" or "Geodesics Equation". This is apparently not the end of the manipulation because $\Gamma^{\lambda}{}_{\mu\nu}$ still depends on $\xi$ . There has to be a way to express the affine connection with properties related to $x$ .

Continue with $d\tau^2$ :

$$\begin{align*} d\tau^2 &= -\eta_{\alpha\beta} d\xi^\alpha d\xi^\beta\\ &= -\eta_{\alpha\beta} \frac {\partial \xi^\alpha}{\partial x^\mu} \frac {\partial \xi^\beta} {\partial x^\nu} dx^\mu dx^\nu\\ \text{Define(Metric):}\quad\,g_{\mu\nu} &\equiv -\eta_{\alpha\beta} \frac {\partial \xi^\alpha}{\partial x^\mu}\frac {\partial \xi^\beta} {\partial x^\nu}\\ d\tau^2 &= - g_{\mu\nu} dx^\mu dx^\nu \\ \end{align*}$$

To express $\Gamma^\gamma{}{}_{\mu\nu}$ in terms of $g_{\mu\nu}$ . Take the derivative respect to $x^\lambda$ , (64 equations parameterized by $\lambda, \mu, \nu$ ):

$$\begin{align*} \frac{\partial g_{\mu\nu}}{\partial x^\lambda} &= -\eta_{\alpha\beta} \left( \frac{\partial}{\partial x^\lambda} \frac {\partial \xi^\alpha}{\partial x^\mu} \frac {\partial \xi^\beta} {\partial x^\nu} + (\mu\leftrightarrow\nu) \right)\\ &= -\eta_{\alpha\beta} \left( \frac{\partial^2\xi^\alpha}{\partial x^\lambda \partial x^\mu} \frac {\partial \xi^\beta} {\partial x^\nu} + (\mu\leftrightarrow\nu) \right)\\ \end{align*}$$

$$\begin{align*} \frac{\partial^2\xi^\alpha}{\partial x^\lambda \partial x^\mu} &=\delta^\alpha{}_\rho\frac{\partial^2 \xi^\rho}{\partial x^\lambda\partial x^\mu} \\ &=\frac{\partial \xi^\alpha}{\partial x^\rho}\frac{\partial x^\rho}{\partial \xi^\beta} \frac{\partial^2 \xi^\beta}{\partial x^\lambda\partial x^\mu}\\ &=\Gamma^{\rho}{}_{\lambda\mu}\frac{\partial \xi^\alpha}{\partial x^\rho} \end{align*}$$

$$\begin{align*} \frac{\partial g_{\mu\nu}}{\partial x^\lambda} &= -\eta_{\alpha\beta} \left( \Gamma^{\rho}{}_{\lambda\mu}\frac{\partial \xi^\alpha}{\partial x^\rho} \frac {\partial \xi^\beta} {\partial x^\nu} + (\mu\leftrightarrow\nu) \right)\\ &= \Gamma^{\rho}{}_{\lambda\mu}g_{\rho\nu}+ \Gamma^{\rho}{}_{\lambda\nu}g_{\rho\mu} \\ \end{align*}$$

At this stage, we realize we can solve $\Gamma$ from just $x$ . This is indeed possible:

$$\begin{align*} \frac{\partial g_{\mu\nu}}{\partial x^\lambda} + \frac{\partial g_{\lambda\nu}}{\partial x^\mu} - \frac{\partial g_{\mu\lambda}}{\partial x^\nu} &= \Gamma^{\rho}{}_{\lambda\mu}g_{\rho\nu}+\cancel{\Gamma^{\rho}{}_{\lambda\nu}g_{\rho\mu}} + \Gamma^{\rho}{}_{\mu\lambda}g_{\rho\nu}+\cancel{\Gamma^{\rho}{}_{\mu\nu}g_{\rho\lambda}} - \cancel{\Gamma^{\rho}{}_{\nu\mu}g_{\rho\lambda}}- \cancel{\Gamma^{\rho}{}_{\nu\lambda}g_{\rho\mu}}\\ &= 2\Gamma^{\rho}{}_{\lambda\mu}g_{\rho\nu} \end{align*}$$

Now we need to define the inverse of $g_{\rho\nu}$ as $g^{\rho\nu}$ :

$$ g^{\nu\sigma}g_{\rho\nu} = \delta^\sigma{}_\rho\,, $$


$$ \Gamma^{\sigma}{}_{\lambda\mu}=\frac 12 g^{\nu\sigma} \{ \partial_{\{\lambda,} g_{\mu\}\nu} - \partial_\nu g_{\mu\lambda} \} $$

My further notes:

The affine connection $\Gamma^d{}_{ab}$ is also called Christoffel symbol of the second kind.

The definitions given below are valid for both Riemannian manifolds and pseudo-Riemannian manifolds, such as those of general relativity, with careful distinction being made between upper and lower indices (contra-variant and co-variant indices). The formulas hold for either sign convention, unless otherwise noted. Einstein summation convention is used in this article. The connection coefficients of the Levi-Civita connection (or pseudo-Riemannian connection) expressed in a coordinate basis are called the Christoffel symbols.

The first kind is defined as:

$$ \Gamma_{cab} = g_{cd}\Gamma^d{}_{ab} $$

Which is amazing since we don't need the inverse metric tensor to express it, as it is simply:

$$ \Gamma_{cab} = \frac 12(\partial_bg_{ca}+\partial_ag_{cb}-\partial_cg_{ab}) $$

This is so simple that we can write it down as: $[ab,c]$ .

My question will be, can we find such an consistent mapping $x$ in a curved spacetime?

General Relativity 03
Last updated: 2016-08-13 22:56:08 PDT.