Tensor Based Special Relativity Begins! EM II Notes 2014_08_25

Summary:  This one took awhile.  I got busy in the lab  These notes start with rotation matrix properties and the transpose products of matrices.  special relativity via tensors also begins.  Specifically, the Lorentz transformation tensor components are reviewed and the number of independent parameters are counted.


Did a few concrete checks that a matrix times its transpose is symmetric.  Sure enough, it is.
$\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}\begin{pmatrix}
a & c \\
b & d \\
\end{pmatrix} = \begin{pmatrix}
a^2 + b^2 & ca + bd \\
ca+bd & c^2+d^2 \\
\end{pmatrix}$

$\begin{pmatrix}a & b & c\\
d & e & f \\
g & h & i \\
\end{pmatrix}\begin{pmatrix}
a & d & g\\
b & e & h \\
c & f & i \\
\end{pmatrix} = \begin{pmatrix}
a^2 + b^2 + c^2 & ad + be + cf & ag + bh  + ci\\
ad + be + cf & d^2 + e^2 + f^2 & gd + eh + fi \\
ga + hb + ic & gd + he + if & g^2 + h^2 +i^2 \\
\end{pmatrix}$

Sure enough, symmetric!


We cover the Minkowski metric, written as

$\eta_{\mu\nu} = \begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1\\
\end{pmatrix}$


Using the Minkowski metric shown above, we can right the invariant interval as:\\
$x^2 + y+^2 + z^2 - t^2 = \eta_{\mu\nu}x^\mu x^\nu$

Lorentz transforms will now be denoted as\\
$x^{\prime \mu} = \Lambda_\nu^\mu x^\nu$


Here's an interesting bit.  The Lorentz transform tensor's components can be written as\\
$\Lambda_0^0 = \gamma$

$\Lambda_i^0 =\gamma v_i$

$\Lambda_0^i =\gamma v_i$

$\Lambda_i^j = \delta_{ij} + \dfrac{\gamma - 1}{v^2} v_i v_j$

The $i$, $j$ terms contain the same $\gamma - 1$ term that was in the earlier notes about the vector Lorentz transform.  See

http://goo.gl/Z9MGjK for a complete explanation
The pretext to Eqn. 1.70 is re-specifying the invariant time-space interval in terms of tensor notation

$\eta_{\mu\nu}x^\mu x^\nu = \eta_{\mu\nu}x^{\mu\prime} x^{\nu\prime}$

Moving to all unprimed variables we get

$\eta_{\mu\nu}x^\mu x^\nu = \eta_{\mu\nu}\Lambda^\mu_\rho \Lambda^\nu_\sigma x^\rho x^\sigma$

Then we have a good example of replacing dummy indices by any name we please including the dummy indices on the other side of the equation to get

$\left(\eta_{\rho\sigma} - \eta_{\mu\nu}\Lambda^\mu_\rho \Lambda^\nu_\sigma \right)x^\rho x^\sigma = 0$

and finally
$\eta_{\mu\nu}\Lambda^\mu_\rho \Lambda^\nu_\sigma= \eta_{\rho\sigma}$

Independent Lorentz Parameters

There are $4 \times 4$ parameters to begin with.  The transform is symmetric, so that eliminates the bottom triangle as redundant.  That leaves the upper triangle and the diagonal which gives 10 constraints, also known as equations.  That leaves 6 unspecified parameters.  Three of these are the velocities in each direction.  The other three are the spatial rotation parameters.

Question
How does the Lorentz group relate to the Conformal group?  I think it's a subgroup.

Questions

Does the Minkowski metric correspond to the identity tensor?\\
If so, then does the expression

$\eta_{\mu\nu}\Lambda^\mu_\rho \Lambda^\nu_\sigma = \eta_{\rho\sigma}$
Correspond to the transpose identity in the $O\left(3\right)$ group?

A This does correspond to the transpose identity, but the Minkowski metric doesn't have anything to do with it.  See \url{http://goo.gl/7zTnJV} for more details on how transposes look in index notation.

Q Why are the transformation in 1.68 only examples of pure boosts?  What extra parameters do they need to be completely general?

A: They need off diagonal terms in the three by three that's in the $i$, $j$, indices.  Two boosts in different directions will multiply to create terms in these positions giving a Thomas-Wigner rotation.  Any general boost can be decomposed to a pure boost and a rotation, but the terms do not commute.  Depending on the order of the decomposition, boost then rotation, or rotation then boost, you get different terms.

Q: Why do we need to take the transpose of the l.h.s. of 1.75 to show that it's symmetric?  Wouldn't showing the product of the transform and its transpose is symmetric work just as well?  I ask because the Minkowski metric that is subtracted out is diagonal and won't/can't affect the symmetry.

A: This was a bit of overkill in the notes.  Symmetric matrices are their own transposes.

Q: How does the expression $\left(4 \times 5\right)/2$ enter into the discussion?

A: This is a combinatoric expression for the number of independent parameters in a symmetric matrix.  The anti-symmetric counterpart is

$n\left(n-1\right)/2$

One way to look at it for the two dimensional case is to consider the antisymmetric case and keep in mind that the number of parameters for a completely general $n \times n$ matrix is $n^2$, and just subtract the anti-symmetric portion to arrive at the result above.

The above result also generalizes to higher dimensional tensors.  For example, in a three index tensor, we get

$\dfrac{\left(n+2\right)\left(n+1\right)n}{3!}$

Things to memorize

The rotation matrices with $det\left(M\right)$ do not form a group because the product of any two members will take you back to the space of $SO\left(3\right)$, i.e. matrices with a positive determinant.

The number of independent parameters is found by subtracting the number of independent equations, (which can be solved for parameters, making them therefore not independent), from the total number of parameters.

Matrix identity

$\left(AB\right)^T = B^T A^T$

Symmetric matrices by definition have a lower triangle's worth of redundant equations.

Picture of the Day:
The Tunnel Top Bar in San Francisco