General Relativity in Matrix Formalism

Oct

Square Root of Relativity

It should be noted that the line element in general relativity is always written in a square format. This has an unfortunate consequence of loss of geometric information. The square root of this format may in fact be either positive or negative, but in fact can be written in an even more general way:

$ds = \sigma_{\mu}dx^{\mu}$

Here we have replaced the metric tensor with a set of 4 2×2 matrices. This still leaves a total of 16 unknowns. We can write the relationship between the metric and these matrices as follows:

$2g_{\mu \nu} = \hat{\sigma}_{\mu} \sigma_{\nu} + \hat{\sigma}_{\nu} \sigma_{\mu}$

The hat here represents the adjoint matrix. If we use the Pauli matrices as a basis, we can rewrite these equations as follows:

$ds = \gamma_{\mu}^{a} \sigma_{a} dx^{\mu}$
$g_{\mu \nu} = \gamma_{\mu}^{a} \gamma_{\nu}^{b} \eta_{ab}$

The gamma values are often referred to as “tetrads”. $\eta$ is just the Minkowski metric, though it is not in fact a tensor. The tetrads transform as vectors under a coordinate change, but the elements marked with regular letters don’t change under coordinate changes as they are in fact just constants. The 4 vectors represented by the tetrad gamma holds all the information for our geometry.

Let’s go ahead and look at what happens when we take the differential of the length of a vector, assuming the vector has zero covariant derivative (is a constant vector):

$dl = d(\gamma_{\mu}^{a} \sigma_{a} \psi^{\mu}) = \sigma_{a} dx^{\tau}(\frac{\partial \gamma_{\mu}^{a}}{\partial x^{\tau}} \psi^{\mu} + \gamma_{\mu}^{a} \frac{\partial \psi^{\mu}}{\partial x^{\tau}}) = \sigma_{a} dx^{\tau}(\frac{\partial \gamma_{\kappa}^{a}}{\partial x^{\tau}} + \gamma_{\mu}^{a} \Gamma_{\tau \kappa}^{\mu}) \psi^{\kappa}$

Since this vector is constant, we will assume it’s length is also constant, meaning the above value must be zero. This then gives us an equation relating the Christoffel symbols and the tetrads (we have dropped the tilda over the Pauli matrices, but they are just the static matrices).

$\frac{\partial \gamma_{\kappa}^{a}}{\partial x^{\tau}} + \gamma_{\mu}^{a} \Gamma_{\tau \kappa}^{\mu} = 0$

This is just the equation for 4 constant contravariant vectors. Plugging this back into our equation for the metric tensor gives us back our original equation relating it to the Christoffel symbols. Using the symmetry properties of the connection, we can conclude the following:

$\frac{\partial \gamma^a_{\kappa}}{\partial x^{\tau}} - \frac{\partial \gamma^a_{\tau}}{\partial x^{\kappa}} = 0$

Jul

On The Differential Geometry of C^4

Let us suppose we are working in a 4 dimensional complex space. We will define the metric tensor as e, and construct the space so that this metric a Hermitian matrix.

$\epsilon^{+} = \epsilon$ (1)

From this space, we will also define a projection to the real space using the following mapping (written as a matrix equation):

$\left \{ x_0, ix_j \right \}=Z^{+}\gamma_{\mu}Z$ (2)

We have chosen this particular projection so as to allow the four space-time variables to be both positive and negative. Dirac originally multiplied the gamma matrices with gamma 0 to force the time element to be positive, as this projection was used for the probability density. Here, we have yet to deduce anything quantum mechanical. Rather, we are only creating a single C^4 space and defining 4 real variables by the above equation. With this projection in mind, it is possible to write the derivatives with respect to Z and Z+ in terms of the 4 real variables x:

$\frac{\partial }{\partial Z} = \frac{\partial x_{\mu}}{\partial Z}\frac{\partial }{\partial x_{\mu}}=Z^{+}\gamma_{\mu}\frac{\partial }{\partial x_{\mu}}$ (3)

The presence of the i in front of the space coordinates cancels out in our derivative. Also, we have assumed the gamma matrices do not dependent on Z. This is not true in the more general case and will (as we will see later on) require us to introduce the C^4 affine connections. For now, let us assume our space is flat.

We can take a further derivative with respect to Z+, and assuming our tensors and metric in C^4 depend only on Z we get:

$\frac{\partial^2 }{\partial Z^{+} \partial Z} \equiv \bigtriangledown^2 = \gamma_{\mu}\frac{\partial }{\partial x_{\mu}}$ (4)

Jumping ahead a bit, lets go ahead and write a C^4 equivalent of a wave equation:

$\bigtriangledown^2 \psi = \gamma_{\mu}\frac{\partial \psi }{\partial x_{\mu}} = k\psi$ (5)

Sep

Extending Christoffel Symbols

Einstein was working on a non-symmetric field theory near the end of his life’s work in relativity. It really boiled down to removing certain symmetry assumptions in his field equations, in particular with the Christoffel Symbols $\Gamma^{\tau}_{\mu \nu}$

The equations:

$\Gamma^{\tau}_{\mu \nu} = \Gamma^{\tau}_{\nu \mu}$

$g_{\mu \nu} = g_{\nu \mu}$

Are removed in general to allow for non-symmetric elements. Let us propose the following general adjustment to the Christoffel Symbols:

$\Gamma ^{i}_{jk}\rightarrow \Gamma ^{i}_{jk} - ie\delta_{j}^{i} A_{k}$

Giving us a Reimannian Curvature Tensor as:

$R_{ijm}^{k}\rightarrow R_{ijm}^{k} - ie \delta_{i}^{k} F_{mj}$

Contracting this we get:

$R_{ij}\rightarrow R_{ij} + ie\left ( A_{j,i} - A_{i,j} \right )=R_{ij} - ieF_{ij}$

Hence, the electromagnetic anti-symmetric tensor F plays the part of the anti-symmetric part of the total curvature R. R is now a combination of a real symmetric component (gravity), and an imaginary anti-symmetric component (electromagnetism). The total curvature tensor acts as a hermitian operator. Let us G to represent the total curvature so as to keep these known elements clear in our minds:

$G_{\mu \nu} = R_{\mu \nu} - ieF_{\mu \nu} = G_{\nu \mu}^{*}$

Looking at the typical set of field equations for R and F:

$G_{\mu \nu ,\tau } + G_{\nu \tau ,\mu }+G_{\tau \mu ,\nu }=\left (R_{\mu \nu ,\tau } + R_{\nu \tau ,\mu } + R_{\tau \mu ,\nu } \right ) + ie\left (F_{\mu \nu ,\tau } + F_{\nu \tau ,\mu } + F_{\tau \mu ,\nu } \right ) = 0$

Jul

Alterations to 2×2 matrix equations for General Relativity

In a previous post, I showed how the metric of Minkowski space (special relativity) came naturally from the determinant of a specially designed 2×2 matrix. Converting a 4 vector into a 2×2 martix using the Pauli matrices did the trick, making a connection between the Pauli matrices and special relativity obvious.

To extend this concept to a general metrix g, we will need to generalize the Pauli matrices. As it turns out, however, any 2×2 matrix can be written as a linear combination of the original Pauli matrices, so this generalization is not too hard to write:

$\sigma _{\mu }=\epsilon _{\mu \nu}\tilde{\sigma }^{\nu}$

This then gives us the following generic definition for the matrix of a vector:

$A = \sigma _{\mu }A^{\mu }=\epsilon _{\mu \nu}\tilde{\sigma }^{\nu}A^{\mu }$

Looking at the determinant of this matrix, we get the following end result:

$det(A)=g_{\mu \nu }A^{\mu }A^{\nu }$

when:

$g_{\mu \nu }=\epsilon _{\mu \tau}\epsilon _{\nu \kappa} \eta ^{\tau \kappa}$

where:

$\eta ^{00}=1, \eta^{11}=\eta^{22}=\eta^{33}=-1, \eta^{\tau \kappa}=0 (\tau \neq \kappa)$

While this last part just looks like the Minkowski metric, I remind the reader that these values do not change under coordinate shift (it is NOT a tensor), but instead act more on the same level as the Kronecker Delta.

Let us now write some of the coordinate transform equations:

${\sigma}'_{\mu }={\epsilon_{\mu \nu}}'\tilde{\sigma }^{\nu}=\frac{\partial {x_\mu }}{\partial {{x}'_a}}\frac{\partial {x_\nu}}{\partial {x}'_b}\epsilon _{ab}\tilde{\sigma }^{\nu}=\frac{\partial x_{\mu}}{\partial {x}'_{a}}\epsilon_{ab}S\tilde{\sigma}^{b}S^{+}=\frac{\partial x_{\mu}}{\partial {x}'_{a}}S\sigma_{a}S^{+}$

where we use:

$\frac{\partial x_{\nu}}{\partial {x}'_{\mu}} \tilde{\sigma}^{\nu} = S\tilde{\sigma}^{\mu}S^{+}$

giving:

${A}'=SAS^{+}$

alos note that:

$S^{+} = S^{-1}$

Jun

More fun with Matrices – General Relativity

I keep harping on this, and I know for the most part it’s just fun with math, but I can’t help it. The simplicity of matrix equations draws me in. Even if most physicists read this and say: “and so? nothing new here”, I feel compelled to write it.

Let’s start with the simplest and most fundemental equation in differential geometry, the length of a vector:

$l^{2}=g_{\mu \nu }A^{\mu }A^{\nu }$

Here we are going to use Einstein summation notation. Written in matrix format this becomes:

$l^{2}=A^{+}gA$

where we are using the conjugate transpose. To ensure this length is a real number (something measureable) we must also set the following restriction on the matrix metric g:

$g^{+}=g$

To keep “covariant” and “contravariant” seperate and more obvious, I will use lower case for the covariant and upper case for the contravariant. Under a coordinate shift, we have the following:

$A{}'=\Lambda A$

$g{}'=\lambda^{+} g \lambda$

$\lambda \Lambda = \Lambda^{+} \lambda^{+} = 1$

Continuing down this track, we can rewrite the covariant derivative in matrix format as:

$D_{\mu}A = \frac{\partial A}{\partial x^{\mu}} - \Gamma_{\mu}A$

Under coordinate transformation we have:

$\Gamma_{\mu}{}'=\frac{\partial x^{\nu}}{\partial x^{\mu}{}'}\left ( \Lambda \Gamma_{\nu} + \frac{\partial \Lambda}{\partial x^{\nu}} \right )\lambda$

which gives:

$D_{\mu}{}'=\frac{\partial x^{\nu}}{\partial x^{\mu}{}'} \Lambda D_{\nu} \lambda$

For curvature, we just flip flop the covariant derivatives:

$R_{\mu \nu}A = D_{\mu}D_{\nu}A - D_{\nu}D_{\mu}A = \left ( \frac{\partial \Gamma_{\mu} }{\partial x^{\nu}} - \frac{\partial \Gamma_{\nu} }{\partial x^{\mu}} + \Gamma_{\mu} \Gamma_{\nu} - \Gamma_{\nu} \Gamma_{\mu} \right )A$

with finally having a coordinate shift:

$R_{\mu \nu}{}'=\frac{\partial x^{\alpha}}{\partial x^{\mu}{}'} \frac{\partial x^{\beta}}{\partial x^{\nu}{}'} \Lambda R_{\alpha \beta} \lambda$

Finally, we can write the relationship between the covariant derivative, curnvature tensor, and the metric matrix:

$\frac{\partial G}{\partial x^{\mu}} - \Gamma_{\mu}G - G \Gamma^{+}_{\mu} = 0$

where

$Gg = 1$

then

$\Gamma_{\mu} = \frac{1}{2}\frac{\partial G}{\partial x^{\mu}}g$

and finally:

$R_{\mu \nu}=\frac{1}{4}\left ( \frac{\partial G}{\partial x^{\mu}} \frac{\partial g}{\partial x^{\nu}} - \frac{\partial G}{\partial x^{\nu}} \frac{\partial g}{\partial x^{\mu}} \right )$

Now I must voice a bit of concern here at these equations. First of all, I have not yet figured out how to write Einstein’s field equation using matrices. Secondly, note that the solution for the Chrsitoffel matrices is only known to within an unknown imaginary field. Namely:

$\Gamma_{\mu} \rightarrow \Gamma_{\mu} + iA_{\mu}$

is also a solution to the equation relating the Chrisoffel matrices and the metric matrix. While this extra term makes total sense in the context of the covariant derivative when quantum fields are in question, it is not entirely clear to me yet how it plays into the overall sceme of things, and how / where it would plop itself into the original equations for the Chrisoffel symbols.

Being relatively anal, I worry too that these equations are “too simplistic” and that I’ve somewhere made a mistake in my calculations. This is most apparent in the equations connecting the metric and the Christoffel symbols, and the metric and the curvature matrices. Perhaps, when I have more time for simple algebra, I will go through and confirm these equations piece by piece…

Jun

Matrix Fun

Ok, so it’s been a damn long time since I’ve posted anything physics related. Lately, in my free time while wondering the halls, waiting for meetings, or just sitting at lunch, I’ve been plugging away at some re-representations of certain equations and notions in matrix formalism. Why? Simple: The Pauli matrices a friggen awesome.

Let’s define them quickly:

$\sigma _{\mu } \equiv \begin{Bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} & \begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix} & \begin{bmatrix}0 & -i \\ i & 0 \end{bmatrix} & \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} \end{Bmatrix}$

To see why these matrices are so darn awesome, we have to look at a few properties. First of all, they are all Hermitian. That means if we multiple any combination of them by real numbers and add them, we will get a 2×2 Hermitian matrix. Also, as there are 4 of them, ANY 2×2 matrix can be represented as a sum. Let us look at a generic sum of 4 numbers (they may be complex):

$A = \sigma _{\mu }A^{\mu } = \begin{bmatrix}A^{0} + A^{3} & A^{1} - iA^{2} \\ A^{1} + iA^{2} & A^{0} - A^{3} \end{bmatrix}$

This matrix has the following determinant:

$det(A) = (A^{0} + A^{3})(A^{0} - A^{3}) - (A^{1} - iA^{2})(A^{1} + iA^{2}) = (A^{0})^{2} - (A^{1})^{2} - (A^{2})^{2} - (A^{3})^{2}$

Look familiar? Well it should, it’s the inner product of the vector A with itself in Minkowski space! In otherwords, our choice of the Pauli matrices has given us the metric of special relativity out of the box. It’s pretty hard to ignore this and simply call it coincidence.

That said, let’s have some fun by changing some of our favorite equations over to using 2×2 matrices instead of vectors. We see from the above arrangement that any 4-vector can be replaced by a 2×2 matrix using the summation we performed for A. Of course, I chose “A” on purpose, since that is the electromagnetic field vector. We need to write out a few more 2×2 matrices of importance before we can really start having fun:

$D \equiv \sigma_{\mu }\frac{\partial }{\partial x_{\mu }} = \begin{bmatrix}\partial / \partial x_{0} + \partial / \partial x_{3} & \partial / \partial x_{1} - i\partial / \partial x_{2} \\ \partial / \partial x_{1} + i\partial / \partial x_{2} & \partial / \partial x_{0} - \partial / \partial x_{3} \end{bmatrix}$

with

$det(D) = \square ^{2}$

let’s also define:

$\bar{\sigma} _{\mu } = C\sigma _{\mu }^{*}C^{-1} = \begin{Bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} & -\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix} & -\begin{bmatrix}0 & -i \\ i & 0 \end{bmatrix} & -\begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} \end{Bmatrix}$

$C\equiv \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}$

Then we can write:

$\bar{D}\equiv \bar{\sigma _{\mu }}\frac{\partial }{\partial x_{\mu }}$

and:

$\bar{D}D = D\bar{D} = \square^{2}$

Now, I’m going to skip some simple Calculus here and give you some of the 2×2 matrix answers for Electromagnetism:

$F\equiv \frac{1}{2}\left ( D\bar{A}-\bar{D}A \right ) = \sigma _{\mu }\omega ^{\mu }$

$\omega ^{\mu } \equiv \left \{ 0 , \frac{E_{x}}{c} + iB_{x} , \frac{E_{y}}{c} + iB_{y} , \frac{E_{z}}{c} + iB_{z}\right \}$

$det(F) = \left | B \right |^{2} - \left | \frac{E}{c} \right |^{2}-\frac{2i}{c}\left ( E_{x}B_{x} + E_{y}B_{y} + E_{z}B_{z} \right )$

and finally, we have Maxwell’s equation:

$DF = \mu _{0}J=\mu _{0}\begin{bmatrix}c\rho + J_{z} & J_{x} - iJ_{y}\\ J_{x} + iJ_{y} & c\rho - J_{z} \end{bmatrix}$

For spinor fields, we can use a similar construction, and we will get the 2×2 matrix formatted Dirac equation as:

$\left ( \bar{D} - e\bar{A} \right ) \Psi + \frac{iE_{0}}{hc}\bar{\Psi}C=0$

After performing some simple modification, we can get the equivalent equation but with the bars reversed:

$\left ( D - eA \right ) \bar{\Psi} - \frac{iE_{0}}{hc}\Psi C=0$

We can then use these two equations to find the square:

$\left ( \bar{D} - e\bar{A} \right )\left ( D - eA \right ) \Psi + \frac{E_{0}^{2}}{h^{2}c^{2}}\Psi=0$

It is easy to see that this is just the Klein-Gordon equation in matrix format

Dec

Coordinate Transformations

Ok, so we have our metric and our covariant derivative. Now let’s look at what happens under coordinate transformations.

$x_{\mu}^{‘} = dx_{\mu}^{‘} / dx_\nu \ x_\nu = \Lambda_{\mu}^{\nu}x_\nu$

or
$X^{‘} = \Lambda X$

Since we are working with column matrices, we should not that the above equations are only true if the column matrix represents a vector in the space. In quantum mechanics, we are actually more interested in other constructions such as spinors and boson fields.

A spinor, for example, transforms via a different but related matrix S. Spinors can be thought of as the “square root” of a vector.

$\Psi^{‘} = S\Psi$

where
$\Lambda_{\mu}^{\nu} \gamma_{\nu} = S^{-1} \gamma_{\mu} S$
and
$\gamma_{\nu} \equiv$ 4×4 Dirac Matrices

We wish to extend the principal of relativity to spinor fields by saying that all laws of physics should be independent of S. It should be noted that S has more degrees of freedom than L. It should also be noted that while the format of the matrix version of the covariant derivative is the same for both vectors and spinors, the form of the Christoffel symbols is in fact different. It would not, therefor, be beyond the realm of reason to state that the metric matrix may also have a different form for spinors than for vectors.

For example, the metric for vectors transforms differently than the metric for spinors:

$g^{‘} = \Lambda^{\dagger-1} g \Lambda^{-1}$

vs.
$g^{‘} = S^{\dagger-1} g S^{-1}$

Given this format, I think it worthwhile to continue using indices for vector elements and matrices for spinors. Like the coordinate transform there is a relationship between the metric experienced by spinor fields and the metric tensor experienced by vectors:

Dec

Metric as a Matrix – Why?

At the surface it might seam relatively trivial. In a way, writing tensors as matrices doesn’t really change the math. The reason I want to explore this is that quantum mechanics uses fields and matrices (operators) as primary objects of interest. Also, I believe it worth investigating the idea that certain fields, namely spinor fields, experience geometry in a different way that as expressed in standard GR.

So lets dive into the basics:

We can write the inner product of a vector with itself in two different ways (Einstein summation notation used)

$ds^2 = g_{\mu \nu} dx^\mu dx^\nu$

or
$ds^2 = dX^\dagger\ g\ dX$

where the symmetry conditions on the metric can be written as:

$g_{\mu \nu} = g_{\nu \mu}$

or
$g^\dagger = g$

Here we can plainly see that the metric, written as a matrix, is in fact a Hermitian Operator. It’s eigenvalue is just the inner product of a column matrix (be it vector or spinor). Let’s look next at the covariant derivative: