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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \bigskip

{\bf Set 6:} \bigskip

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\noindent
{\bf 14.\ Charge conjugation, parity and time reversal in 
electrodynamics.}
\smallskip

\noindent (1) Using the given values $(-,+,+)$ for $\rho$ as input, 
our $(C,P,T)$ results are obtained as follows:
\begin{eqnarray} \nonumber
  1.~A^0 \sim \rho\ &\Rightarrow&\ (-,+,+)\,,\\ \nonumber
  2.~\vec{J} \sim \vec{v}\,\rho\ &\Rightarrow&\ (-,-,-)\,,\\ \nonumber
  3.~\vec{A} \sim \vec{J}\ &\Rightarrow&\ (-,-,-)\,,\\ \nonumber
  4.~E^i=F^{i0}=\partial^iA^0-\partial^0A^i\ &\Rightarrow&\ (-,-,+)\,,
  \\ \nonumber
  5.~B^k\sim F^{ij}=\partial^iA^j-\partial^jA^i\ &\Rightarrow&\  
     (-,+,-)\,,\\ \nonumber
  6.~\vec{p}=\int \vec{x'}\,\rho(\vec{x'})\,d^3x' \ &\Rightarrow&\  
     (-,-,+)\,,\\ \nonumber
  7.~\vec{m}=\frac{1}{2c} \int \vec{x'}\times \vec{J}(\vec{x'})\,d^3x'
     \ &\Rightarrow&\  (-,+,-)\,.
\end{eqnarray}
\noindent (2) From the explicit matrix forms we find
$$ F^{\alpha\beta}\,F_{\alpha\beta}=-2\,\vec{E}^{\,2}+2\,\vec{B}^{\,2}
   = - ^*F^{\alpha\beta}\ ^*F_{\alpha\beta} ~~~{\rm and}~~~ 
   ^*F^{\alpha\beta}\, F_{\alpha\beta} = -4\,\vec{E}\cdot\vec{B}\,.$$
Using the $(C,P,T)$ results above, we find $(+,+,+)$ for
$F^{\alpha\beta}\,F_{\alpha\beta}$ and $^*F^{\alpha\beta}\ 
^*F_{\alpha\beta}$ (proper scalars), and $(+,-,-)$ for 
$^*F^{\alpha\beta}\, F_{\alpha\beta}$ (pseudoscalar).
\bigskip


\noindent
{\bf 15.\ Proca Lagrangian.} \smallskip

\noindent (1) From the calculation of the script/lecture we have
$$ \partial_{\alpha}\,
  \frac{\partial\ {\cal L}}{\partial\,\partial_{\alpha}A_{\beta}} 
  = - \frac{1}{4\pi}\,\partial_{\alpha}\,F^{\alpha\beta}\ . $$ 
Now
$$ \frac{\partial\ {\cal L}}{\partial\,A_{\beta}} =
   \frac{m^2}{8\pi}\,\left( g^{\beta}_{~\alpha}A^{\alpha}+
   A_{\alpha}g^{\beta\alpha}-\frac{1}{c}J_{\alpha}g^{\beta\alpha}
   \right) = \frac{m^2}{4\pi} A^{\beta} -\frac{1}{c}J_{\beta} $$
and the Euler-Lagrange equation 
$$ \partial_{\alpha}\,
   \frac{\partial\ {\cal L}}{\partial\,\partial_{\alpha}A_{\beta}} 
   =  \frac{\partial\ {\cal L}}{\partial\,A_{\beta}}~~~{\rm reads}
   ~~~\partial_{\alpha}\,F^{\alpha\beta} + m^2 A^{\beta} =
      \frac{4\pi}{c}J^{\beta}\,. $$
(2) With the Lorentz gauge condition $\partial_{\alpha}A^{\alpha}=0$
we have
$$   \partial_{\alpha}\,F^{\alpha\beta} = 
     \partial_{\alpha}\partial^{\alpha}\,A^{\beta} -
     \partial_{\alpha}\partial^{\beta}\,A^{\alpha} =
     \partial_{\alpha}\partial^{\alpha}\,A^{\beta}+0 $$
so that the Euler-Lagrange equation becomes
$$   \partial_{\alpha}\partial^{\alpha}\,A^{\beta} + 
     m^2 A^{\beta} = \frac{4\pi}{c}J^{\beta}\,. $$
(3) The Proca Euler-Lagrange equation is not gauge invariant, but 
transforms under the gauge transformation $A^{\beta}\to A^{\beta}
+\partial^{\beta}\Lambda$ into
$$ \partial_{\alpha}\,F^{\alpha\beta} + m^2 A^{\beta} 
   + m^2 \partial^{\beta} \Lambda = \frac{4\pi}{c}J^{\beta}\,.$$
(4) Nevertheless the Lorentz gauge condition should hold: Taking the 
$\partial_{\beta}$ derivative of the Euler-Lagrange equation we find
$$ \partial_{\beta}\partial_{\alpha}\,F^{\alpha\beta} + 
   m^2\,\partial_{\beta} A^{\beta} = 
   \frac{4\pi}{c}\,\partial_{\beta}J^{\beta}\,. $$
On the left-hand side $\partial_{\beta}\partial_{\alpha}\,
F^{\alpha\beta}$ is zero due to the usual argument (contraction of
a symmetric with an antisymmetric tensor). On the right-hand side we 
have $\partial_{\beta}J^{\beta}=0$ due to charge conservation. So 
we end up with
$$ \partial_{\beta} A^{\beta} = 0\ . $$
\bigskip


\noindent
{\bf 16.\ Electromagnetic energy-momentum tensor.} \smallskip

\noindent (1) We have
\begin{eqnarray} \nonumber
  T^{00} &=& -{1\over 4\pi}\,\left( g^{0\mu}\,F_{\mu\lambda}\,
  \partial^0\,A^{\lambda} - \frac{1}{4}\, g^{00}\, 
  F_{\mu\lambda}\, F^{\mu\lambda} \right) \\ \nonumber
  &=& -{1\over 4\pi}\,F^0_{~\lambda}\,\partial^0\,A^{\lambda} 
   + \frac{1}{8\pi}\, \left( \vec{B}^{\,2}-\vec{E}^{\,2}\right)
\end{eqnarray}
and now
\begin{eqnarray} \nonumber
  - F^0_{~\lambda}\,\partial^0\,A^{\lambda} &=&
  + \sum_i F^{0i}\,\partial^0\,A^i \,=\, 
  - \sum_i \left[ E^i\,\left(\partial^0\,A^i 
  - \partial^i\,A^0\right) + E^i\,\partial^i\,A^0\right] \\ \nonumber 
  &=& \vec{E}^{\,2} + \nabla\cdot\left(\vec{E}\,\Phi\right)\ .
\end{eqnarray}
In the last step the relations $\Phi=A^0$, $\nabla\cdot\vec{E}=0$ and 
$\nabla = (-\partial^i)$ were used. Putting the above equations 
together, we have the desired result
$$ T^{00} = \frac{1}{8\pi}\,\left(\vec{E}^{\,2}+\vec{B}^{\,2}\right)
   + \frac{1}{4\pi}\,\nabla\cdot\left( \Phi\vec{E}\right)\ . $$

\noindent (2) We have
$$ \Theta^{00} = \frac{1}{4\pi}\, F^0_{~\lambda}\,F^{\lambda0} +
   \frac{1}{16\pi}\, F_{\mu\lambda}\, F^{\mu\lambda}\ . $$
The first term can be re-written as
$$ \frac{1}{4\pi}\,\sum_{i=1}^3 F^0_{~i} F^{i0} = 
   \frac{1}{4\pi}\,\sum_{i=1}^3 F^{0i} F^{0i} = 
   \frac{1}{4\pi}\,\vec{E}^{\,2} $$
and the second term is known to be $(\vec{B}^{\,2}-\vec{E}^{\,2})/
(8\pi)$ from our previous problem. So the result is
$$ \Theta^{00} = \frac{1}{8\pi}\, \left( \vec{E}^{\,2} +
   \vec{B}^{\,2} \right) $$
Next, as $g^{0i}=0$ we have 
$$ \Theta^{0i} = \frac{1}{4\pi} F^{0\lambda} F_{\lambda}^{~i}
    = - \frac{1}{4\pi} \sum_{j\ne i} F^{0j} F^{ji} $$
$i=3:$
$$ \Theta^{03} = - \frac{1}{4\pi} 
   \left( F^{01} F^{13} + F^{02} F^{23} \right) = 
   \frac{1}{4\pi} \left( E^1 B^2 - E^2 B^1 \right) = 
   \frac{1}{4\pi} \left( \vec{E}\times\vec{B} \right)^3 . $$
Repeating this evaluation for $i=1,2$, we find for all components
$$ \Theta^{0i} = \frac{1}{4\pi} F^{0\lambda} F_{\lambda}^{~i} =
   \frac{1}{4\pi} \left( \vec{E}\times\vec{B} \right)^i\ . $$
For $\Theta^{ij},\, i\ne j$ we find
$$ \Theta^{ij} = \frac{1}{4\pi} F^{i\lambda} F_{\lambda}^{~j}
    = \frac{1}{4\pi} \left( F^{i0} F^{0j} - \sum_{k\ne i,j}
    F^{ik} F^{kj} \right) = 
    - \frac{1}{4\pi} \left( E^i E^j + B^i B^j \right)\ . $$
For $\Theta^{ii}$ with {\bf no} summation over $i$ we have
$$ \Theta^{ii} = \frac{1}{4\pi} \left( F^{i0} F^{0i} - 
   \sum_{k\ne i} F^{ik} F^{ki} - \frac{1}{4}
    F_{\alpha\beta}F^{\alpha\beta} \right) = $$
$$ \frac{1}{4\pi} \left[ - E^i E^i + \vec{B}^{\,2} - B^i B^i - 
   \frac{1}{2} \left( \vec{B}^{\,2}-\vec{E}^{\,2} \right) \right] = $$
$$ - \frac{1}{4\pi} \left[ E^i E^i + B^i B^i - \frac{1}{2} 
   \left( \vec{E}^{\,2}+\vec{B}^{\,2} \right) \right]\ . $$
All $\Theta^{ij}$ together:
$$ \Theta^{ij} = - \frac{1}{4\pi} \left[ E^i E^j + B^i B^j +
   \frac{1}{2}\, g^{ij}\, \left(\vec{E}^{\,2}+\vec{B}^{\,2}\right) 
    \right]\ . $$
\bigskip

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