Lecture 7

Handouts:
5. Solutions, Homework 1
6. Homework 2

We further discussed the transformations of observables and operators
under Lorentz boosts, en route to writing Maxwell's equations in a
Lorentz covariant way.

Reviewed how the four-vector x^\mu = (ct, \vec{x}) transforms as a
vector under Lorentz transformations: that is, the components ct and
\vec{x} mix with each other in such a way that the four-vector x^\mu
is just matrix-multiplied by a Lorentz tranformation matrix \Lambda.
(Later in lecture we clarified why our index summation conventions mean
that \Lambda^\mu_\nu x^\nu is just matrix multiplication of \Lambda on
the vector x^\nu .)

We then explored the consequences of Lorentz transformations for other
observables. Redefining space and time (as Lorentz transformations do)
induces redefinitions of other observables: for example, the velocity
depends explicitly on space and time, and changes when those change
due to a Lorentz transformation. Similarly, densities depend on the
definition of a volume element, and so change under Lorentz
transformations, as do currents which depend both on densities and
velocities.

We noted that we can group observables together to find objects which
change under a Lorentz transformation in exactly the same way as x^\mu
--- that is, they form four-vectors which get matrix-multiplied by
\Lambda under a Lorentz transformation. Such four-vectors include
p^\mu = (E/c, \vec{p}) : energy mixes with momentum in such a way that
the rest mass m^2 c^4 = E^2 - |\vec{p}|^2 c^2 is invariant. That j^\mu
= (c\rho, \vec{j}) and A^\mu = (\phi, \vec{A}) are also four-vectors
is shown in Appendix B of Rolnick.

We also noted that x^\mu = (ct, \vec{x}) determines a four-dimensional
gradient operator, \partial / \partial x^\mu, whose components are
c^{-1} \partial/ \partial t and \grad. We get tired of writing so many
\partial 's and label this operator \partial_\mu . Calculus tells you
that under a change of variables (and a Lorentz transformation is just
a specific change of variables) independent variables remain
independent.  That is, initially we know that the partial derivative
of x^\nu with respect to x^\mu is 1 if \mu and \nu are the same index,
and 0 otherwise. If the same thing is to be true in our
Lorentz-tranformed frame, multiplication of x^\nu by \Lambda must be
accompanied by multiplication of \partial_\mu by \Lambda^{-1}. Thus
\partial_\mu really transforms like a covector, and its lower index
makes sense.

Now that we have grouped the relevant observables and operators into
objects that transform in a well-defined way under Lorentz
tranformations (as vectors or covectors), we seek to write the laws of
electromagnetism in a Lorentz-covariant way. We focus on Maxwell's
equations; you will see a covariant formulation of the Lorentz force
law in your homework.

I will not reproduce the details of the derivation
here. Schematically, we started with Maxwell's equations in terms of
the potentials \phi and \vec{A}. These looked like

\Box \phi     - 1/c \partial_t  ( object 1 ) = \rho
\Box \vec{A}  +          \grad  ( object 1 ) = 1/c \vec{j}.

We worked as follows:

1) 

Given that A^\mu = (\phi, \vec{A}) and \partial_\mu = 
( 1/c \partial_t, \grad ), we identified their components in object 1
to show that object 1 = \partial_0 A^0 + \partial_i A^i = \partial_\mu
A^\mu. This object is a scalar: since it has 1 upper and 1 lower
index, it is multiplied by one factor of \Lambda and one factor of
\Lambda^{-1} under Lorentz transformations, which cancel to leave it
invariant. Thus the Lorentz condition, \partial_\mu A^\mu = 0, is
Lorentz covariant.

2)

We showed that \Box = \partial_\mu \partial^\mu . To show this, we
digressed on how raising \partial_\mu to determine \partial^\mu
introduces signs on the spatial components, due to matrix
multiplication by g^{\mu\nu} = diag(+1,-1,-1,-1). The dot product
\partial_\mu \partial^\mu then gives 1/c^2 \partial_t^2 - (\grad)^2 =
\Box. We also calculated the sum \partial^\mu \partial_\mu =
g^{\mu\nu} \partial_\nu \partial_mu explicitly, showing how the 4
nonzero diagonal entries fix the signs so that we get \Box.

3)

Having simplified those parts, we came back to the original equations,
and used the identities A^\mu = (\phi, \vec{A}), j^\mu = (c\rho,
\vec{j}) and \partial^\mu = ( 1/c \partial_t, - \grad ) --- the sign
on \grad comes from raising \partial_\mu ! We read off their
components in the Maxwell equations to get a simple, covariant
equation, in which a four-vector on the left-hand side equals another
four-vector on the right-hand-side.

---
KB