Bei's Study Notes

General Relativity 02
Last updated: 2016-06-13 01:37:20 PDT.

Reference: UCI OpenCourseWare - GR 04 (Playlist)

Scalars, Vectors, and Tensors (Contd.)

A tensor is a physical value that transform in certain ways.

How to construct tensors

  1. Linear combination of tensors of the same kind
  2. Multiply tensors with different indices (direct product). This gives symmetric tensors;
  3. Contraction. This reduces the total number of indices by 1.
  4. Differentiation/Division. Similar to direct product.

Special tensors

  1. \eta_{\alpha\beta} = \Lambda^\gamma{}_\alpha\Lambda^\delta{}_\beta\eta_{\gamma\delta} (4-invariant)
  2. \eta_{\alpha\beta}\eta^{\beta\gamma} = \delta_\alpha{}^\gamma
  3. Levi-Civita symbol:

    \varepsilon_{ijk} = \begin{cases} +1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1) \text{ or } (3,1,2), \\ -1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2) \text{ or } (2,1,3), \\ \;\;\,0 & \text{if }i=j \text{ or } j=k \text{ or } k=i \end{cases}

(Wikipedia) ... Also, the specific term "symbol" emphasizes that it is not a tensor because of how it transforms between coordinate systems, however it can be interpreted as a tensor density.

Special topics

Current Densities and Electromagnetism

If we have a simple particle that follows \mathbf {\hat x}(t) .

\newcommand\Traj[0]{\mathbf {\hat x}(t)} \newcommand\DTraj[0]{\delta^{(3)}(\mathbf x-\Traj)} \mathbf J(\mathbf x,t) = e \DTraj \frac{\mathbf {\hat x}(t)}{dt} \\ \rho(\mathbf x,t) = e \DTraj

Defining x^0(t) = t , This generalizes to

J^\alpha = e \DTraj \frac{d\hat x^\alpha}{dt}

This does not manifestly look like a 4-vector. For: 1. the Dirac delta is (3) multiple, and 2. explicit use of t . Now we can turn it in to a manifestly covariant form by add a delta term and a integral

\begin{align*} \DTraj &= \int \,\delta^{(4)}(x-\hat x(t))\,dt\\ &= \frac{dt}{d\tau}\int \,\delta^{(4)}(x-\hat x(\tau))\,d\tau \,, \end{align*}

which also means that \gamma^{-1}\DTraj is a scalar. Then

\begin{align*} J^\alpha &= e \delta^{(3)}(\mathbf x-\Traj) \frac{d\hat x^\alpha}{dt} \\ &= e \left(\int \,\delta^{(4)}(x-\hat x(\tau))\,d\tau\right)\frac{dt}{d\tau} \frac{d\hat x^\alpha}{dt} \\ &= e \hat u^\alpha \left(\int \,\delta^{(4)}(x-\hat x(\tau))\,d\tau\right) \, .\\ \end{align*}

J^\alpha is indeed a 4-vector.


J^\alpha is conserved

\begin{align*} \mathbf \nabla \cdot \mathbf J &= \frac{\partial}{\partial x^i} \left(e \DTraj \frac{d\hat x^i}{dt}\right) \\ &= \frac{\partial}{\partial x^i} \left(e \DTraj\right) \frac{d\hat x^i}{dt}\\ \end{align*}

Observe that \partial_{x^i} \DTraj = -\partial_{\hat x^i} \DTraj ,

\begin{align*} \mathbf \nabla \cdot \mathbf J &= -\frac{\partial}{\partial \hat x^i}\frac{d\hat x^i}{dt} \left(e \DTraj\right) \\ &= -\frac{d}{dt}\left(e \DTraj\right) \\ &= -\dot \rho \\ \end{align*}

This can also be written as \partial_\alpha J^\alpha = 0 .


Electromagnetic tensor F^{\alpha\beta}

From Maxwell's equation:

Source equations:

\begin{align*} \mathbf \nabla \cdot \mathbf E &= \rho \\ \mathbf \nabla \times \mathbf B &= \mathbf J + \dot E \\ \end{align*}

Non-source equations:

\begin{align*} \mathbf \nabla \cdot \mathbf B &= 0 \\ \mathbf \nabla \times \mathbf E &= -\dot B \\ \end{align*}

We can write this down using 4-vectors/tensors. First introduce electromagnetic tensor F^{\alpha\beta} = -F^{\beta\alpha} defined as follows:

\begin{align*} F^{0j} &= E^j \\ F^{ij} &= \epsilon^{ij}{}_kB^k \\ \end{align*}

The source equations can be written as

\begin{align*} \partial_i E^i &= \rho \\ \partial_i F^{i0} &= -J^0 \\ \partial_\alpha F^{\alpha0} &= -J^0; \\ \epsilon^{ij}{}_k\partial_j B^k &= J^i + \dot E^i \\ \partial_i F^{ij} &= -J^i - \partial_0F^{i0} \\ \partial_\alpha F^{\alpha j} &= -J^j \\ &\implies \\ \partial_\alpha F^{\alpha\beta} &= -J^\beta \end{align*}

The non-source equation can be written as

\begin{align*} \partial_i B^i &= 0 \\ \frac{1}{2}\epsilon_{ijk}\partial_k {F^{ij}} &= 0 \\ \epsilon_{ijk}\partial_i E^j &= - \partial_0 B^k \\ \epsilon_{ijk}\partial_i F^{0j} &= \epsilon_{ijk} \partial_0 {F^{ij}} \\ \epsilon_{ijk}\partial_i F^{0j} - \epsilon_{ijk} \partial_0 {F^{ij}} &= 0 \\ \epsilon_{aijk}\partial_i F^{aj} &= 0 \\ \implies \\ \epsilon_{\alpha\beta\gamma\delta}\partial_\alpha F^{\beta\gamma} &= 0 \\ \end{align*}


Equation of motion for EM

\begin{align*} m \frac{d^2x^\alpha}{d\tau^2} &= e F^\alpha{}_\beta u^\beta \\ \frac{d\mathbf p}{d\tau} &= \gamma e (\mathbf E + \mathbf v \times \mathbf B) \\ \frac{d\mathbf p}{dt} &= e (\mathbf E + \mathbf v \times \mathbf B) \end{align*}


Energy-Momentum Tensor

The conservation of charse arises from the fact that current density is a 4-vector. In order for momentum to conserve, we need a order-2 tensor to correspond to the field.

The tensor of a particle looks like this:

\begin{align*} T^{\alpha\beta} &= \hat p^\alpha(t) \delta^{(3)}(\mathbf x - \mathbf {\hat x}(t)) \frac{\hat x(t)^\beta}{dt} \\ &= m u^\alpha u^\beta \gamma^{-1} \delta^{(3)}(\mathbf x - \mathbf {\hat x}(t))\\ \end{align*}

which is manifestly a Lorentz tensor.


Electromagnetic field itself also contains energy. This energy can be described in terms of energy-momentum tensor:

T^{00} = \frac{1}{2}(\mathbf E^2+ \mathbf B^2) \\ T^{0i} = (\mathbf E \times \mathbf B)^i \\

Then we can guess the relation between the tensors:

T^{\alpha\beta} = \eta_{\gamma\delta} F^{\alpha\gamma}F^{\delta\beta} - \frac{1}{4}\eta_{\alpha\beta}F^{\gamma\delta}F_{\gamma\delta}