# Bei's Study Notes

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

Reference: UCI OpenCourseWare - GR 05 (Playlist)

The energy-momentum tensor of a point particle is give as:

$$T^{\alpha\beta}(x) = mu^\alpha u^\beta \gamma^{-1}\delta^{(3)}(\mathbf x - \mathbf {\hat x} (t))$$

There will be 4 conserved quantities:

$$\partial_\alpha T^{\alpha\beta} = 0\,.$$

Plugging it in, we get:

\begin{align*} \partial_\alpha mu^\alpha u^\beta \delta^{(3)}(\mathbf x - \mathbf {\hat x} (t)) &=0\\ \end{align*}

In EM, we have only one conserved quantity $\rho = J_0$, which comes from:

$$\square A_\mu = - J_\mu$$

Since GR is about the energy momentum tensor, in order for it to be covariant, we are essentially looking for a equation that looks like this:

$$\square (?)_{\mu\nu} = GT_{\mu\nu}$$

Einstein didn't realize that this must be an equation of a 10-component tensor until 1912. This is the gravitational field. (In contrary, in Newtonian gravity, you can define everything with a one-component gravitational potential $\phi(x)$).

### Angular momentum and spin

Starting from energy-momentum conservation:

$$\partial_\alpha T^{\beta\alpha} = 0$$

we define an object $M^{\gamma\alpha\beta}(x)$ (angular momentum density):

$$M^{\gamma\alpha\beta}(x) = x^\alpha T^{\beta\gamma} - x^\beta T^{\alpha\gamma}$$

Then we take the partial derivative of it:

\begin{align*} \partial_\gamma M^{\gamma\alpha\beta} &= \delta^\alpha{}_\gamma T^{\beta\gamma} + x^\alpha{} \partial_\gamma T^{\beta\gamma} - \delta^\beta{}_\gamma T^{\alpha\gamma} - x^\beta \partial_\gamma T^{\alpha\gamma} \\ &= T^{\beta\alpha} - T^{\alpha\beta} \\ &= 0 \\ \end{align*}

This implies a conserved quantity (angular momentum) $J^{\alpha\beta} \equiv \int M^{0\alpha\beta}\,dx^3= - J^{\beta\alpha}$. The spatial components of the angular momentum tensor gives the classical angular momentum:

$$J^{23} = \int M^{023}\, d x^3 = \int x^2 T^{30} - x^3 T^{20}\, d x^3 \\ = \gamma \int (y p^z - z p^y) \, d x^3 = \gamma I_x$$

The time components of it gives mass moment:

$$J^{01} = \int M^{001}\, d x^3 = \gamma \int t p_x - x m\, d x^3$$

A special note is that this quantity contains both orbital part and spin part of angular momentum, the latter does not depend on the origin. (I should make another note on this issue). This is called the Pauli–Lubanski pseudovector.

### Perfect fluid

Perfect fluid in special relativity

## Lorentz Group

(This is essentially talking about the Lie algebra of the Lorentz Group as we discussed before in QFT notes.)

In QM, we discussed rotation group $R$ in 3 dimensional Euclidian space:

$$\mathbf x' = \mathbb R \cdot \mathbf x$$

where $\mathbb R$ is a 3x3 matrix.

it corresponds to the transformation of an indexed wave function:

$$\psi_n' = u_{nm}\cdot \psi_m$$

and $u$ is the rotation matrix on the state vectors. This mean while we need to change the coordinate to index into a field under rotation, we also need to change the components of the field at the same time given the field has more than one components.

The rotation matrix has a simple form:

$$u = e^{\frac{i}{\hbar} \alpha \cdot J}$$

where $J$ is the angular momentum matrices. It satisfies certain algebra (in terms of commutation relations):

$$[J_i, J_k] = i\hbar \epsilon_{ijk} J_k$$

one of the outcome of this algebra is that only certain eigenvalues of the spin is allowed:

$$J^2 = \hbar^2 j(j+1),\quad j = 0,\frac 12, 1, ...$$

Considering special relativity, the true transformation of the world is not only rotation, but also boost. In SR, $\mathbf x \to x^\alpha$, therefore, we need to redo the exercise to make sure we can still only get spin $0, \frac 12, 1, ...$ but not $\frac 23$ and such.

We write

\begin{align*} x'^\alpha &= \Lambda^\alpha{}_\beta x^\beta \\ \psi' &= D(\Lambda)\cdot\psi \end{align*}

which $D(\Lambda)$ is the transformation associated with $\Lambda$. They form a group:

$$D(\Lambda)D(\Lambda') = D(\Lambda \Lambda')$$

To find $D$, consider an infinitesimal Lorentz transformation looks like:

$$\Lambda^\alpha{}_\beta = \delta^\alpha{}_\beta + \omega^\alpha{}_\beta$$

The product of two Lorentz transformation is:

\begin{align*} \Lambda^\alpha{}_\beta\Lambda'^\beta{}_\gamma &= (\delta^\alpha{}_\beta + \omega^\alpha{}_\beta)(\delta^\beta{}_\gamma + \omega'^\beta{}_\gamma)\\ &= \delta^\alpha{}_\gamma + (\omega^\alpha{}_\beta\delta^\beta{}_\gamma + \delta^\alpha{}_\beta\omega'^\beta{}_\gamma) + o(\omega^2)\\ &= \delta^\alpha{}_\gamma + (\omega^\alpha{}_\gamma + \omega'^\alpha{}_\gamma) + o(\omega^2) \end{align*}

So the product of two infinitesimal Lorentz transformation is the infinitesimal Lorentz transformation with the displacement equals to the sum of the two transformations.

where $\omega$ is infinitesimal. From its being rotation:

$$\Lambda^\alpha{}_\gamma\Lambda^\beta{}_\delta\eta_\alpha{}_\beta =\eta_\gamma{}_\delta$$

Therefore:

\begin{align*} (\delta^\alpha{}\gamma + \omega^\alpha{}_\gamma)(\delta^\beta{}_\delta + \omega^\beta{}_\delta)\eta_\alpha{}_\beta &= \eta_\gamma{}_\delta \\ \eta_\gamma{}_\delta + (\delta^\alpha{}\gamma \omega^\beta{}_\delta + \omega^\alpha{}_\gamma\delta^\beta{}_\delta)\eta_\alpha{}_\beta + o(\omega^2) &= \eta_\gamma{}_\delta \\ (\delta^\alpha{}\gamma \omega^\beta{}_\delta + \omega^\alpha{}_\gamma\delta^\beta{}_\delta)\eta_\alpha{}_\beta &= 0 \\ \omega^\beta{}_\delta\eta_\gamma{}_\beta + \omega^\alpha{}_\gamma\eta_\alpha{}_\delta &= 0 \\ \omega_\gamma{}_\delta + \omega_\delta{}_\gamma &= 0 \\ \end{align*}

Therefore $\omega_\alpha{}_\beta$ is anti-symmetric.

For $D(\Lambda)$, we will pick an anti-symmetric $\sigma$ to make $D$ a group:

$$D(\Lambda) = D(1 + \omega) = I + \frac 12 \omega^{\gamma\delta} \sigma_{\gamma\delta}$$

These $\sigma$s, which we pick to be anti-symmetric., are called the "generators of the Lorentz group".

Lorentz group has 6 different parameters, therefore we need 6 of these 4x4 $\sigma$ matrices, denoted as $\sigma_{\gamma\delta}$.

If $\psi$ is a 4-vector, then the $D(\Lambda) = \Lambda$, and then:

\begin{align*} \left(I + \frac 12 \omega^{\gamma\delta} \sigma_{\gamma\delta}\right)^\alpha{}_\beta &= \delta^\alpha{}_\beta + \omega^\alpha{}_\beta \\ \left(\frac 12 \omega^{\gamma\delta} \sigma_{\gamma\delta}\right)^\alpha{}_\beta &= \omega^\alpha{}_\beta \\ \left(\frac 12 \omega^{\gamma\delta} \sigma_{\gamma\delta}\right)^\alpha{}_\beta &= \eta_\beta{}_\delta\omega^\gamma{}^\delta\delta_\gamma{}^\alpha \\ \end{align*}

If all components of $\omega^\gamma{}^\delta$ are independently arbitrary, then we can equalize the coefficients of the them (thus removing $\omega$ from the equation). Unfortunately $\omega$ is anti-symmetric. So we need to expand the summation:

\begin{align*} \left(\frac 12 \omega^{\gamma\delta} \sigma_{\gamma\delta}\right)^\alpha{}_\beta &= \left(\sum_{\gamma < \delta} \omega^{\gamma\delta} \sigma_{\gamma\delta}\right)^\alpha{}_\beta \\ \delta_\gamma{}^\alpha\eta_\beta{}_\delta\omega^\gamma{}^\delta &= \left(\sum_{\gamma < \delta}+ \sum_{\gamma > \delta}\right)\delta_\gamma{}^\alpha\eta_\beta{}_\delta\omega^\gamma{}^\delta \\ &= \sum_{\gamma < \delta} \left(\delta_\gamma{}^\alpha\eta_\beta{}_\delta\omega^\gamma{}^\delta + (\gamma \leftrightarrow \delta)\right) \\ &= \sum_{\gamma < \delta} \left(\delta_\gamma{}^\alpha\eta_\beta{}_\delta - (\gamma \leftrightarrow \delta)\right)\omega^\gamma{}^\delta \end{align*}

Now we can cancel $\omega$ and get:

\begin{align*} (\sigma_{\gamma\delta})^\alpha{}_\beta &= \delta_\gamma{}^\alpha \eta_{\beta\delta} - \delta_\delta{}^\alpha \eta_\alpha{}_\delta \end{align*}

(Last time, I derived this from $D(\Lambda^{-1})\sigma^{\mu\nu}D(\Lambda) = \Lambda^\mu{}_\rho\Lambda^\nu{}_\sigma\sigma^{\rho\sigma}$. This time its much simpler.)

The next step is to figure out the commutation relations. The result is:

\begin{align*} [\sigma_\alpha{}_\beta, \sigma_\gamma{}_\delta] &= (\eta_{\gamma\beta}\sigma_{\alpha\delta} - (\alpha\leftrightarrow\beta)) - ((\gamma\leftrightarrow\delta)) \end{align*}

Set(Pauli)

\begin{align*} a_i &= \frac 12 (-i \frac 12 \epsilon_{ijk} \sigma_{jk} + \sigma_{i0})\\ b_i &= \frac 12 (-i \frac 12 \epsilon_{ijk} \sigma_{jk} - \sigma_{i0})\\ [a_i, b_j] &= 0 \\ [a_i, a_j] &= i\epsilon_{ijk}a_k \\ [b_i, b_j] &= i\epsilon_{ijk}b_k \\ \end{align*}

So this breaks up into two rotation groups. So the eigenvalues of $a$ and $b$ should be:

\begin{align*} \hat a = a(a+1)\quad a=0,\frac 12, 1, \frac 32, ... \hat b = b(b+1)\quad a=0,\frac 12, 1, \frac 32, ... \end{align*}

Therefore all the representations of the Lorentz group should be classified by the a pair of numbers $(a, b)$.

(The relationship between dimension of the field/particle and the $(a,b)$ pair requires some further explanation).