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\def\R{{\z9R}}
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\headline{\ifnum\pageno=1  
 \medtitle \noindent Math\. 489, Sec.\ 502    \hfill    Fall, 
2005
 \else \tenrm {}\hfill Page \folio\fi}
\nopagenumbers

\title \medbold 
 Special Relativity and Electromagnetism \endtitle


\noindent{\bf The following problems 
 (composed by Prof\. P. B. Yasskin)
 will lead you through the construction of the theory of
 electromagnetism in special relativity. 
 Please write your response as a connected essay,
 similar to a chapter of a textbook.
 If possible, use \TeX\ or a word processor so that you can 
make revisions easily.  
 The first draft is due on Thursday, October 13. 
 You may consult books and have discussions with other students,
 but outright copying 
 (except from the problems themselves, when appropriate)
 is not allowed.
}\bigbreak

  We
regard spacetime as the vector space $\R^4$ with a Lorentz-signature metric
(pseudo-inner product).  Thus, if we choose the orthonormal basis to be
$$
e_0 = (1,0,0,0), \quad
e_1 = (0,1,0,0), \quad
e_2 = (0,0,1,0), \quad
e_3 = (0,0,0,1)
$$
(so that all indices run from 0 to 3), and the dual basis to be
$\theta^\alpha$,  then the metric is
$$
\eta = \eta_{\alpha\beta} \theta^\alpha \otimes \theta^\beta, \qquad \zT{where}
\quad
\eta_{\alpha\beta} = \pmatrix-1&0&0&0\cr 0&1&0&0\cr 0&0&1&0\cr 0&0&0&1
 \endpmatrix,
$$
and the inverse metric is
$$
\eta^{-1} = \eta^{\alpha\beta} e_\alpha \otimes e_\beta,
 \qquad \zT{where} \quad
\eta^{\alpha\beta} = \pmatrix-1&0&0&0\cr 0&1&0&0\cr 0&0&1&0\cr 0&0&0&1
 \endpmatrix .
$$
Derivatives will be denoted by
$\,\displaystyle
\partial_\alpha = {\partial \over \partial x^\alpha}\,,
$
and indices will be raised and lowered using $\eta$ and $\eta^{-1}$.

\bigbreak
\hrule
\bigbreak
In problems 1--11, we will study the electromagnetic field, which is the 2-form
$$
F = F_{\alpha\beta}\, \theta^\alpha \otimes \theta^\beta
 \qquad \zT{where} \quad
F_{\alpha\beta} = \pmatrix 0 &-E_1&-E_2&-E_3\cr
   E_1&  0 & B_3&-B_2\cr
   E_2&-B_3&  0 & B_1\cr
   E_3& B_2&-B_1&  0 \endpmatrix ,
$$
the electromagnetic potential, which is the 1-form
$$
A = A_\alpha \theta^\alpha \qquad \zT{where} \quad
A_\alpha = (\phi, A_1, A_2, A_3),
$$
and the electromagnetic current, which is the vector
$$
J = J^\alpha e_\alpha \qquad \zT{where} \quad
J^\alpha = (\rho, J^1, J^2, J^3) .
$$

 \bigbreak\goodbreak
%\vfill\eject
\item{1.} Show that the rank-3 tensor
$$
S_{\alpha\beta\gamma} = \partial_\gamma F_{\alpha\beta}
      + \partial_\alpha F_{\beta\gamma}
      + \partial_\beta	F_{\gamma\alpha}
$$
is totally antisymmetric, and hence is a 3-form. ({\sl Hint:\/}
  There are three pairs of
indices to transpose.) Show that this implies that the components of $S$ are all
zero except for those for which $\alpha$, $\beta$, and $\gamma$ are distinct.

\bigbreak
\item{2.} Write out the components of the equations
$$
\partial_\gamma F_{\alpha\beta} + \partial_\alpha F_{\beta\gamma}
+ \partial_\beta  F_{\gamma\alpha} = 0
\qquad \zT{and} \qquad
\partial_\beta F^{\alpha\beta} = 4 \pi J^\alpha
$$
to see that these are the Maxwell equations
$$
\eqalign{&\del \cdot \vec B = 0, \cr
 &\del \times \vec E + \partial_t \vec B = 0, \cr}
\qquad \qquad
\eqalign{&\del \cdot \vec E = 4 \pi \rho, \cr
 &\del \times \vec B - \partial_t \vec E = 4 \pi \vec J . \cr}
$$

\bigbreak
\item{3.} Write out the components of the equations
$$
F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha
$$
to find expressions for $\vec E$ and $\vec B$ in terms of $\phi$ and $\vec A$.
({\sl Note:\/} $\phi$ is the negative of the usual scalar potential.)

\bigbreak
\item{4.} Show (in 4-dimensional notation) that if
$$
F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha
$$
is satisfied, then
$$
\partial_\gamma F_{\alpha\beta} + \partial_\alpha F_{\beta\gamma}
+ \partial_\beta  F_{\gamma\alpha} = 0
$$
is automatically satisfied. ({\sl Note:\/}
  The 3-dimensional version of these equations
is a pair of identities: $\del \cdot \del \times \vec A = 0$ and
$\del \times \del \phi = 0$.)
 Consequently, this subset of the Maxwell equations
is actually an identity; it is  sometimes
referred to as the electromagnetic Bianchi
identity.

\bigbreak
\item{5.} Substitute
$$
F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha
$$
into the remaining Maxwell equation
$$
\partial_\beta F^{\alpha\beta} = 4 \pi J^\alpha
$$
to obtain the Maxwell equation for $A_\alpha$.

\bigbreak
\item{6.} Let $\chi$ be a function.  Also let
$$
A'_\alpha = A_\alpha + \partial_\alpha\chi
\qquad \zT{and} \qquad
F'_{\alpha\beta} = \partial_\alpha A'_\beta - \partial_\beta A'_\alpha\, .
$$
This is called a gauge transformation.
 Relate $F'_{\alpha\beta}$ to
$F_{\alpha\beta}$ to see that the electromagnetic field is gauge-invariant.

\bigbreak
\item{7.} Given $A_\alpha\,$, show that there always exists a function $\chi$
such that $A'_\alpha$ satisfies
$$
\partial^\alpha A'_\alpha = 0 .
$$
({\sl Note:\/}
  You may assume that it is always possible to solve a wave equation with
an arbitrary but specified source.
 (Perhaps you can find a reference for this fact?))
 Write out the Maxwell equations for
$A'_\alpha$ to see that the Maxwell equations may always be solved for any
arbitrary but specified current $J^\alpha$.

\bigbreak
\item{8.} Write out the components of the equation
$$
\partial_\alpha J^\alpha = 0
$$
to see that this is conservation of electric charge.  Show that if
$$
\partial_\beta F^{\alpha\beta} = 4 \pi J^\alpha
$$
is satisfied, then
$$
\partial_\alpha J^\alpha = 0
$$
is automatically satisfied. This is sometimes referred to as an automatic
conservation law.

\bigbreak
\item{9.} Write out the function
$$
{\cal L} = - {\zZ2\frac14} F^{\gamma\delta} F_{\gamma\delta}
$$
in terms of $\vec E$ and $\vec B$.  This function is called the Lagrangian
density for the vacuum electromagnetic field.  It is sometimes interpreted as
the difference between the kinetic energy ${1\over2} |\vec E|^2$ and the
potential energy ${1\over2} |\vec B|^2$.  Also write out the Lagrangian density
in terms of $\phi$ and $\vec A$.

\bigbreak
\item{10.} Write out the components of the tensor
$$
T^{\alpha\beta} = {1 \over 4 \pi} \((F^{\alpha\gamma} F^\beta{}_\gamma
    - {\zZ2\frac14} \eta^{\alpha\beta} F^{\gamma\delta} F_{\gamma\delta}\))
$$
in terms of $\vec E$ and $\vec B$, to see that this consists of the
electromagnetic energy density, momentum density, energy current and momentum
current (or stress).  This tensor is called the Maxwell energy-momentum-stress
tensor.

\bigbreak
\item{11.} In 4-dimensional notation, compute the divergence of the
energy-momentum tensor and use the Bianchi identities and the Maxwell equations
to show that
$$
\partial_\beta T^{\alpha\beta} = - J_\beta F^{\alpha\beta} .
$$

\bigbreak
\hrule
\bigbreak
Problem 11 shows that the electromagnetic energy-momentum is not conserved if
the current is nonzero. The reason for this is that we have ignored the
energy-momentum of the charged particles producing the current.  In problem 12,
we study the motion of a charged particle. Then in problem 13 we study the
energy-momentum tensor of a fluid of charged particles.

\bigbreak
\item{12.} A particle of mass $m$ with electric charge $q$ is moving on the
parametrized path $x^\alpha(\tau)$ where $\tau$ is the proper time.
Consequently, it has unit timelike tangent vector
$$
U = U^\alpha e_\alpha, \qquad \zT{where} \quad
U^\alpha = {\partial x^\alpha \over \partial \tau}
	 = (\gamma , \gamma v^1 , \gamma v^2 , \gamma v^3 )
$$
and where
$$
\gamma = {1 \over \sqrt{1 - |\vec v|^2}} \,.
$$
Further, its 4-momentum is
$$
p^\alpha = m U^\alpha .
$$
Write out the components of the equations
$$
U^\beta \partial_\beta(p^\alpha) = q\, U^\beta F^\alpha{}_\beta
$$
to obtain the Lorentz force and power laws.  ({\sl Hints:\/}
 Don't expand $p^\alpha$.
Be careful with the factors of $\gamma$.)

\bigbreak
\item{13.} Consider a fluid of charged particles of rest mass $m$ and charge
$q$, with fluid velocity $U^\alpha$ and energy density $\rho$ in the
instantaneous local rest frame.  Then the charge density in the instantaneous
local rest frame is
$\,\displaystyle
{q \over m} \rho
\,$
and the electromagnetic current is
$$
J^\alpha = {q \over m}\, \rho\, U^\alpha .
$$
Assuming that the particles are non-interacting except for their electromagnetic
forces, then
\itemitem{(i)} the energy momentum tensor for the fluid is that for dust:
$$
T_\zT{fluid}^{\alpha\beta} = \rho\, U^\alpha U^\beta ,
$$
\itemitem{(ii)} each particle moves according to the Lorentz force equation:
$$
U^\beta \partial_\beta(m U^\alpha) = q\, U^\beta F^\alpha{}_\beta\, ,
$$
\itemitem{(iii)} the energy momentum tensor for the electromagnetic field is
$$
T_\zT{em}^{\alpha\beta} = {1 \over 4 \pi} \((F^{\alpha\gamma} F^\beta{}_\gamma
    - {\zZ2\frac14} \eta^{\alpha\beta} F^{\gamma\delta} F_{\gamma\delta}\)) ,
$$
\itemitem{(iv)} and the electromagnetic field satisfies the Bianchi identities
and the Maxwell equations with current $J^\alpha$.
\item{} Then, as seen in problem 8, the electromagnetic current is conserved:
$$
\partial_\alpha J^\alpha = 0,
$$
and, as seen in problem 11, the electromagnetic energy-momentum tensor
satisfies
$$
\partial_\beta T_\zT{em}^{\alpha\beta} = - J_\beta F^{\alpha\beta} .
$$
Now use the Lorentz force equation and the conservation of electromagnetic
current to show that the fluid energy-momentum tensor satisfies
$$
\partial_\beta T_\zT{fluid}^{\alpha\beta} =  J_\beta F^{\alpha\beta} .
$$
({\sl Hint:\/} Factor $T_\zT{fluid}^{\alpha\beta}$ as
$$
T_\zT{fluid}^{\alpha\beta} = \big( U^\alpha \big) \big( \rho U^\beta \big)
$$
and use the product rule.)  Thus the total energy-momentum is conserved:
$$
\partial_\beta \big( T_\zT{fluid}^{\alpha\beta}
    + T_\zT{em}^{\alpha\beta} \big) = 0 .
$$

\bigbreak
\hrule
\bigbreak
In problems 14 and 15, we study the behavior of the electromagnetic field under
rotations and Lorentz boosts.  Under a general Lorentz transformation,
$\Lambda^{\alpha'}{}_\gamma\,$, the electromagnetic field transforms
according to
$$
F_{\alpha'\beta'} = F_{\gamma\delta} (\Lambda^{-1})^\gamma{}_{\alpha'}
     (\Lambda^{-1})^\delta{}_{\beta'}\, .
$$
We then write
$$
F_{\gamma\delta} = \pmatrix 0 &-E^1&-E^2&-E^3\cr
    E^1&  0 & B^3&-B^2\cr
    E^2&-B^3&  0 & B^1\cr
    E^3& B^2&-B^1&  0 \endpmatrix
\quad \zT{and} \quad
F_{\alpha'\beta'} = \pmatrix  0   &-E^{1'}&-E^{2'}&-E^{3'}\cr
     E^{1'}&   0   & B^{3'}&-B^{2'}\cr
     E^{2'}&-B^{3'}&   0   & B^{1'}\cr
     E^{3'}& B^{2'}&-B^{1'}&   0\endpmatrix .
$$

\bigbreak
\item{14.} First assume that the Lorentz transformation is a rotation about the
$z$-axis:
$$
\Lambda^{\alpha'}{}_\gamma = \pmatrix1 &0 & 0 &0\cr 0\cr 0&&R\cr 0 
\endpmatrix
\qquad \zT{where} \quad
R^{i'}{}_j = \pmatrix \cos\theta & \sin\theta & 0 \cr
      -\sin\theta & \cos\theta & 0 \cr
    0  & 0     & 1 \endpmatrix .
$$
Show that $\vec E$ and $\vec B$ transform as vectors:
$$
E^{i'} = R^{i'}{}_j E^j
\qquad \zT{and} \qquad
B^{i'} = R^{i'}{}_j B^j .
$$
Show that this generalizes to arbitrary rotation matrices.

\bigbreak
\item{15.} Now assume that the Lorentz transformation is a boost in the
$z$-direction with velocity $\vec v = v \hat e_z\,$:
$$
\Lambda^{\alpha'}{}_\gamma
     = \pmatrix \cosh\lambda & 0 & 0 & \sinh\lambda \cr
      0       & 1 & 0 &      0     \cr
      0       & 0 & 1 &      0     \cr
 \sinh\lambda & 0 & 0 & \cosh\lambda \endpmatrix
$$
where
$
\zZ1\,\cosh\lambda = \gamma = {1 \over \sqrt{1 - v^2}}\,$
%\quad \zT{and} \quad
and
$\zZ1\,\sinh\lambda = \gamma v = {v \over \sqrt{1 - v^2}}\,
$.
Find expressions for $\vec E'$ and $\vec B'$ in terms of $\vec E$ and $\vec B$
and either $\lambda$ or $v$.



\bigbreak
\hrule
\bigbreak

%\noindent
%{\bf NOTE: DO NOT TRY THE REMAINING PROBLEMS UNTIL YOU HAVE COMPLETED ALL OF
%THE PREVIOUS PROBLEMS.}
%
%\vskip 15pt

In problems 16 and 17, we study the Lagrangian and Hamiltonian formulations of
electromagnetism. Each problem begins with a discussion of the analogous
formulation of classical mechanics and the situation for a general field theory
with fields $\psi^A$, for $A = 1, \ldots, N$.
 Then the special case of
electromagnetism is treated with $\psi^A$ replaced by $A_\alpha$.

\bigbreak
\item{16.} In classical particle mechanics, the Lagrangian is
$$
L = T - V = {\zZ2\frac12} m |\vec v|^2 - V(\vec x) .
$$
In discussing this Lagrangian, it is useful to regard  $\vec x$  and
 $\,\displaystyle
\vec v = {d\vec x \over dt}
\,$
as independent variables.  One then computes
$$
{\partial L \over \partial x^i} = - \partial_i V
\qquad \zT{and} \qquad
p_i = {\partial L \over \partial v^i} = m v_i\, .
$$
(The quantity $p_i$ is called the  momentum conjugate to $x^i$.)  Then the
Euler--Lagrange equations for this Lagrangian are
$$
{d \over dt}{\partial L \over \partial v^i}
    - {\partial L \over \partial x^i} = 0,
$$
or
$$
{d \over dt}(m v_i) + \partial_i V = 0,
$$
which is Newton's equation with the force identified as
$F_i = - \partial_i V$,  the gradient of the potential.

\item{}\indent Similarly, in field theory, in discussing a Lagrangian density,
${\cal L}$, it is useful to regard the fields $\psi^A$ and their derivatives
$\partial_\alpha\psi^A$ as independent variables.  One then computes
$$
{\partial{\cal L} \over \partial\psi^A}
\quad \zT{and} \quad
\pi_A{}^\alpha = {\partial{\cal L} \over \partial\partial_\alpha\psi^A}\, .
$$
(The quantity $\pi_A \equiv \pi_A{}^0$ is the conjugate momentum to
$\psi^A$, while the 4-vector $\pi_A{}^\alpha$ is sometimes
 called the conjugate
multimomentum to $\psi^A$ .) Then the Euler--Lagrange equations for the
Lagrangian density, ${\cal L}$, are
$$
\partial_\alpha \left( {\partial{\cal L} \over \partial\partial_\alpha\psi^A}
\right) - {\partial{\cal L} \over \partial\psi^A}  = 0 .
$$

\item{}\indent  Both sets of Euler--Lagrange equations
given above can be derived from appropriate variational principles.
   In this exercise, we apply
the field theory version to the vacuum Maxwell Lagrangian density,
$$
{\cal L} = - {\zZ2\frac14} F^{\gamma\delta} F_{\gamma\delta}\, .
$$

\item{}\indent Compute
$$
{\partial{\cal L} \over \partial A_\alpha}
\qquad \zT{and} \qquad
\pi^{\alpha\beta} = {\partial{\cal L} \over \partial\partial_\beta A_\alpha}
\, .
$$
Identify the conjugate momenta to $A_0 = \phi$ and to $A_i\,$:
$$
\pi^\alpha  \equiv \pi^{\alpha 0}
     = {\partial{\cal L} \over \partial\partial_0 A_\alpha}\, .
$$
Compute the Euler--Lagrange equations
$$
\partial_\beta \left( {\partial{\cal L} \over \partial\partial_\beta A_\alpha}
\right) - {\partial{\cal L} \over \partial A_\alpha}  = 0
$$
and verify that these are the vacuum Maxwell equations.

\item{}\indent {\sl Hints:\/}
   Explicitly write out all metrics in ${\cal L}$ (but do not
use  $\alpha$ or $\beta$ as dummy indices). 
At the first step,  do not expand $F$ in terms of
derivatives of $A$.  Use the chain rule.  Then compute the 
derivatives of $F$
using formulas such as
$$
{\partial\partial_\delta A_\gamma \over \partial\partial_\beta A_\alpha}
     = \delta^\delta_\beta \delta^\gamma_\alpha\, .
$$

\bigbreak
\item{17.} In classical mechanics, the Hamiltonian is
$$
H = p_i v^i - L(\vec x,\vec v)
$$
but expressed as a function of $\vec x$ and $\vec p\,$:
$$
H = {\vec p^{\,2} \over m} - \([{\vec p^{\,2} \over 2m} - V(\vec x)\)]
  = {\vec p^{\,2} \over 2m} + V(\vec x) = T + V .
$$
Similarly, for a field theory, the Hamiltonian density is
$$
{\cal H} = \pi_A \partial_0 \psi^A - {\cal L}(\psi^A ,
\partial_\alpha \psi^A),
$$
but expressed as a function of $\psi^A$, $\partial_i \psi^A$, and $\pi_A\,$.
  For
electromagnetism, express the Hamiltonian density
$$
{\cal H} = \pi^\alpha \partial_0 A_\alpha
   - {\cal L}(A_\alpha ,\partial_\beta A_\alpha)
$$
as a function of $A_\alpha$, $\partial_i A_\alpha$, and the non-zero components
of $\pi^\alpha$.  Then express $\cal H$ as a function of $\vec E$ and $\vec B$
to see that the Hamiltonian density, $\cal H$, is equal to the energy density,
$T^{00}$.




\bye


