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General Relativity - UCI OCW

General Relativity - UCI OCW

Perfect Fluid in Special Relativity
Last updated: 2016-06-08 11:28:33 PDT.

I was watching a OCW from UC Irvine on GR. It starts with some discussions about special relativistic. This one about perfect fluid caught me because is not obvious how the Euler equation is derived. So I did it here.

It is an interesting topic in astrophysics since most of the objects we study can be considered (general) relativistic perfect fluid. As a first step, perfect fluid in special relativity is studied.

Starting by write down the stress-energy tensor T^{\alpha\beta} at a point x_0 . First create the reference frame at that point, and the tensor looks like

\begin{align*} T^{00} &= \rho \\ T^{i0} &= 0 \\ T^{ij} &= p \delta^{ij}\quad. \\ \end{align*}

Then generalize it using Lorentz transformation:

\begin{align*} T^{\alpha'\beta'}&=\Lambda^{\alpha'}{}_\alpha\Lambda^{\beta'}{}_\beta T^{\alpha\beta} \\ \end{align*}

where consider only the boost

\begin{align*} \Lambda^{0}{}_0 &= \gamma \\ \Lambda^{0}{}_j &= \gamma v_j \\ \Lambda^{i}{}_0 &= \gamma v_i \\ \Lambda^{i}{}_j &= \delta^{i}{}_j + (\gamma-1)v_i v_j / v^2 \\ \end{align*}

where \gamma \equiv 1/\sqrt{1-v^2} is the Lorentz factor, then the tensor T^{\alpha\beta} is

\begin{align*} \newcommand{\Lij}[2]{\Lambda^{#1}{}_{#2}} \newcommand{\Lzz}[0]{Lij 00} \newcommand{\Lzj}[1]{Lij 0{#1}} \newcommand{\Liz}[1]{Lij {#1}0} \newcommand{\LUDzz}[0]{\gamma} \newcommand{\LUDzj}[1]{\gamma v_{#1}} \newcommand{\LUDiz}[1]{\gamma v_{#1}} \newcommand{\LUDij}[2]{\delta^{#1}{}_{#2}+(\gamma-1) v_{#1}v_{#2}/ v^2} T^{0'0'} &= {\Lij 00}{\Lij 00}T^{00}+ {\Lij 00}{\Lij 0j}T^{0j}+ {\Lij 0i}{\Lij 00}T^{i0}+ {\Lij 0i}{\Lij 0j}T^{ij} \\ &= \gamma^2 \rho + \gamma^2 v^2 p \\ T^{0'b'} &= \Lambda^0{}_0\Lambda^{b'}{}_\beta T^{0\beta} + \Lambda^0{}_i\Lambda^{b'}{}_\beta T^{i\beta} \\ &= \gamma\Lambda^{b'}{}_\beta T^{0\beta} + \Lambda^0{}_i \Lambda^{b'}{}_\beta p\delta^{i\beta} \\ &= \gamma\Lambda^{b'}{}_0 T^{00} + p \Lambda^0{}_i \Lambda^{b'}{}_i \\ &= \rho\gamma^2v_{b'} + p (\LUDzj{i})(\LUDij{b'}{i}) \\ &= \rho\gamma^2v_{b'} + p (\gamma v_{b'} + \gamma (\gamma -1 )v_{b'}) \\ &= \rho\gamma^2v_{b'} + p \gamma^2 v_{b'} \\ &= (\rho + p)\gamma^2v_{b'} \\ T^{a'b'} &= {\Lij {a'}0}{\Lij {b'}0}T^{00}+ {\Lij {a'}0}{\Lij {b'}j}T^{0j}+ {\Lij {a'}i}{\Lij {b'}0}T^{i0}+ {\Lij {a'}i}{\Lij {b'}j}T^{ij} \\ &= \rho{\Lij {a'}0}{\Lij {b'}0}+ p{\Lij {a'}i}{\Lij {b'}j}\delta^{ij} \\ &= \rho\gamma^2v_{a'}v_{b'}+ (\LUDij {a'}i)(\LUDij {b'}i)p \\ &= \rho\gamma^2v_{a'}v_{b'}+ p(\delta^{a'b'}+2(\gamma-1)v_{a'}v_{b'}/v^2 + (\gamma-1)^2v_{a'}v_{b'}/v^2) \\ &= \rho\gamma^2v_{a'}v_{b'}+ p(\delta^{a'b'}+(\gamma^2-1)v_{a'}v_{b'}/v^2) \\ &= \rho\gamma^2v_{a'}v_{b'}+ p(\delta^{a'b'}+\gamma^2v_{a'}v_{b'}) \quad\quad \text{ since } (\gamma^2 -1) / v^2 = \gamma^2 \\ &= \rho\gamma^2v_{a'}v_{b'}+ p\delta^{a'b'} \quad.\\ \end{align*}

Combine the equations:

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

To get the equation of motion, we can use the energy-momemtum conservation law:

\begin{align*} \partial_\alpha T^{\alpha\beta} &= \partial_\alpha p \eta ^{\alpha\beta} + \partial_\alpha (\rho + p)u^\alpha u^\beta \\ &=\left. (-\dot p, \nabla \mathbf p)^\beta + (\dot\rho + \dot p, \nabla \rho + \nabla p)_\alpha u^\alpha u^\beta + (\rho + p) \partial_\alpha(u^\alpha u^\beta) \right. \\ &= 0 \quad. \end{align*}

For \beta=0 ,

\begin{align*} \partial_\alpha T^{\alpha0} &= -\dot p + \gamma^2(\dot\rho + \dot p + (\mathbf v \cdot \nabla) (\rho + p)) + \gamma^2(\rho + p) \nabla \cdot \mathbf v \\ &=(\gamma^2-1)\dot p + \gamma^2(\dot\rho + (\mathbf v \cdot \nabla) (\rho + p)) + \gamma^2(\rho + p) \nabla \cdot \mathbf v \\ &= 0 \\ v^2\dot p + \dot\rho &= -(\mathbf v \cdot \nabla) (\rho + p) + (\rho + p) \nabla \cdot \mathbf v \\ &= -\nabla \cdot ((\rho + p)\mathbf v) \end{align*}

For \beta > 0 ,

\begin{align*} \nabla p + \gamma^2 (\dot\rho + \dot p + (\mathbf v \cdot \nabla) (\rho + p)) \mathbf v + (\rho + p) \partial_\alpha(u^\alpha u^\beta) = 0 \end{align*}


\begin{align*} \partial_\sigma{}(u^\alpha u^\beta) &\equiv \partial_\sigma(u^\alpha u^\beta) \\ \partial_\sigma{}(u^0u^0) &= 0 \\ \partial_\sigma{}(u^iu^0) &= \gamma^2 \partial_\sigma(v^i)\\ \partial_\sigma{}(u^0u^j) &= \gamma^2 \partial_\sigma(v^j)\\ \partial_\sigma{}(u^iu^j) &= \partial_\sigma(\gamma^2 v^iv^j)\\ &= \gamma^2 \partial_\sigma( v^iv^j) \quad. \end{align*}

\begin{align*} \partial_\alpha(u^\alpha \mathbf v) &= \gamma^2 \dot {\mathbf v} + \gamma^2 (\nabla \cdot \mathbf v) \mathbf v + \gamma^2 (\mathbf v \cdot \nabla) \mathbf v \end{align*}

\begin{align*} (\rho + p) (\gamma^2 \dot {\mathbf v} + \gamma^2 (\nabla \cdot \mathbf v) \mathbf v + \gamma^2 (\mathbf v \cdot \nabla) \mathbf v) &= -\nabla p - \gamma^2 (\dot\rho + \dot p + (\mathbf v \cdot \nabla) (\rho + p) \mathbf v \\ \end{align*}

\begin{align*} \dot {\mathbf v} + (\mathbf v \cdot \nabla) \mathbf v &= -\frac{1}{\rho+p}\left(\nabla p/\gamma^2 + (\dot\rho + \dot p + \nabla \cdot ((\rho + p)\mathbf v)\right) \mathbf v)\\ &= -\frac{1}{\rho+p}\left(\nabla p/\gamma^2 + (\dot\rho + \dot p - v^2\dot p - \dot\rho\right) \mathbf v)\\ &= -\frac{1}{\rho+p}\left(\nabla p/\gamma^2 + (\dot p - v^2\dot p\right) \mathbf v)\\ &= -\frac{1-v^2}{\rho+p}\left(\nabla p + \dot p \mathbf v \right) \\ \end{align*}

This is called relativistic Euler Equaltion for Fluid Dynamics.

General Relativity 01
Last updated: 2017-05-01 22:09:22 PDT.

Reference: UCI OpenCourseWare - GR 02 (Playlist)

Reminders of Special Relativity

Special Relativity is a thoery of translating coordinates in planar spacetime, which has a constant measure \eta_{\alpha\beta} .

Lorentz Transformation

Any two inertial reference frames can be related by a Lorentz Transformation (or Poincaré transformation). Consider a photon is shot from a point O , then we can move the origin of the two frame to that point. Then we can rotate the frames so that their relative speed lays on the x axis. The location of the photon in these two frames can be defined as x, y, z, t and x', y', z', t' , then the photon's motion forms a straight line in each frames of reference. Since the speed of light is constant, we have these relations:

x^2 + y^2 + z^2 = c^2t^2 \\ x'^2 + y'^2 + z'^2 = c^2t'^2 \\

In Newtonian transformion, frame of references are related by Galilean transformation:

x' = x - vt \\ y' = y \\ z' = z \\ t' = t \\

Thus we have the notion of "absolute" time flow. This does not satisfy \text{(LInv)} . Lorentz transformation multiplies the time coordinate by a factor \gamma , which depends on |\mathbf v| :

x' = \gamma (x - vt) \\ y' = y \\ z' = z \\ t' = \gamma (t - vx/c^2) \\

Pluggin it into \text{(LInv)} :

\begin{align*} \gamma^2 x ^2 - 2\gamma^2 xvt + \gamma^2v^2t^2+y^2+z^2&=c^2\gamma^2t^2-2\gamma^2tvx+\gamma^2v^2x^2/c^2 \\ \gamma^2 x^2 - \gamma^2 x^2 + \gamma^2v^2t^2+c^2t^2&=c^2\gamma^2t^2 + \gamma^2v^2x^2/c^2 \\ \gamma^2 (x^2 + v^2t^2 - c^2t^2 - v^2x^2/c^2) &= x^2 - c^2t^2 \\ \gamma^2 (x^2 - c^2t^2) (1 - v^2/c^2) &= x^2 - c^2t^2 \\ \gamma^2 (1 - v^2/c^2) &= 1 \\ \gamma &= \frac{1}{\sqrt{1 - v^2/c^2}} \\ \end{align*}

In general, the Lorentz transformation \Lambda^\alpha{}_\beta relates two coordinates in different reference frames by

x'^\alpha = \Lambda^\alpha{}_\beta x^\beta\,.

\Lambda has a important property:

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

where \eta_{00} = -1, \eta_{0i} = \eta_{i0} = 0, \eta_{ij} = \delta_{ij} . It is similar to rotation in Euclidean space. It is usually seen as rotation in Minkovsky space.

From now on, we set c=1 .

Define proper time \tau that

d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2 = - \eta_{\alpha\beta} dx^\alpha dx^\beta \\


dx^\beta = \Lambda^\beta{}_\alpha dx^\alpha

, so the proper time in another frame is

\begin{align*} d\tau'^2 &= -\eta_{\alpha\beta} dx'^\alpha dx'^\beta \\ &= -\eta_{\alpha\beta} \Lambda^\alpha{}_{\gamma} dx^\gamma \Lambda^\beta{}_{\delta} dx^\delta \\ &= -\eta_{\gamma\delta} dx^\gamma dx^\delta \\ &= d\tau^2 \\ \end{align*}

General form of boost-only Lorentz transformations:

\begin{align*} \Lambda^0{}_0 &= \gamma \\ \Lambda^i{}_0 &= -\gamma v^i \\ \Lambda^0{}_j &= -\gamma v^j \\ \Lambda^i{}_j &= \delta_{ij} + (\gamma - 1)\frac{v_iv_j}{|\mathbf v|^2} \\ \end{align*}

If we define v^0 = \gamma v / (1-v) , we then have a unified form:

\begin{align*} \Lambda^\alpha{}_\beta &= \delta^{\alpha}{}_{\beta} - v^{\alpha}v^{\varepsilon}\eta_{\varepsilon\beta}\frac{1}{1+\gamma} \end{align*}

We can check if this obeys the identity:

\begin{align*} \Lambda^{\alpha}{}_{\gamma}\Lambda^{\beta}{}_{\delta} &= & \\ \end{align*}

Some 4-vectors

4-force F^\alpha

Define 4-force F^\alpha in that

  1. If the particle is momentarily at rest, then dt = d\tau , then the 4-force is

    f = (0, \mathbf F)

  2. Then apply Lorentz boost

\begin{align*} f'^\alpha &= \Lambda^\alpha{}_\beta f^\beta \\ f'^0 &= \Lambda^0{}_j f^j \\ &= \gamma \mathbf v \cdot F \\ f'^i &= \Lambda^i{}_j f^j \\ &= (\delta^i_{j} + v^iv_j\frac{\gamma-1}{v^2})f^j \\ \mathbf F' &= \mathbf F + \mathbf v\frac{\gamma-1}{v^2} (\mathbf v \cdot \mathbf F) \\ \end{align*}

4-velocity u^\alpha

u^\alpha = \frac{dx^\alpha}{d\tau} = (\gamma, \gamma \mathbf v) \\ u_\alpha u^\alpha = -\gamma^2 + \gamma^2 v^2 = -1

4-momentum p^\alpha

It can be defined as

p^\alpha = m u^\alpha\,,

then we get:

p = (\gamma m, \gamma \mathbf p)

Another way to generalize momentum: the momentum at rest is p = (p^0, 0) , applying Lorentz boost:

\begin{align*} p'^0 &= \gamma p^0 \\ p'^i &= \gamma v^i p^0 \end{align*}

When |v| \to c , \gamma \to 1 :

\begin{align*} p'^i &= v^i p^0 \end{align*}

This should become the clasicall limit where \mathbf p' = m\mathbf v , therefore p^0 = m .

p_\alpha p^\alpha = -m^2\\ E^2 - \mathbf p^2 = m^2

Scalars, Vectors, and Tensors

Contravariant vectors transform by Lorentz transformation; covariant vectors transform by inverse Lorentz transformation:

V^\alpha \to V'^\alpha = \Lambda^\alpha{}_\beta V^\beta\\ V_\alpha \to V'_\alpha = \Lambda_\alpha{}^\beta V_\beta\\


\Lambda_\alpha{}^\beta \equiv \eta_{\alpha\gamma}\eta^{\beta\delta}\Lambda^\gamma{}_\delta\,.

To prove this is indeed inverse Lorentz transformation:

\begin{align*} \Lambda_\alpha{}^\beta \Lambda^\alpha{}_\rho &= \eta_{\alpha\gamma}\eta^{\beta\delta}\Lambda^\gamma{}_\delta \Lambda^\alpha{}_\rho \\ &=\eta_{\rho\delta}\eta^{\beta\delta}\\ &=\delta^{\beta}{}_{\rho}\\ \end{align*}

Derivative with respect to a contravariant vector is covariant.

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}

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

Reference: UCI OpenCourseWare - GR 05 (Playlist)

More about SR

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


\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*}


\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).

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) .