Physics 401 Assignment # 5:
Wed. 1 Feb. 2006 - finish by Wed. 8 Feb.

This is a relatively [pun intended] short Assignment, since the first Midterm Exam is on Monday February 6 (in class, 50 minutes). Nevertheless it will count the same as other Assignments, if you choose to tackle it. For the exam, you may bring your own 1-page summary sheet with any hard-to-remember equations etc. The exam will cover all of Chapters 7 and 8, sections 10.1 of Chapter 10, and all of Chapter 12. There will not be anything on the Midterm explicitly about $F^{\mu\nu}$ (second problem below) but applications of Eqs. (12.108) are fair game.

  1. If we stay in the Lorentz gauge ( $\partial_\mu A^\mu = 0$), $A^\mu$ is a genuine 4-vector and therefore $A_\mu A^\mu$ is a Lorentz scalar. Use this fact to show that, in any frame, $\vert\Vec{J}\vert^2 = c^2 (\rho^2 - \rho_0^2)$, where $\rho_0$ is the charge density in the frame where $\Vec{J} = 0$. ANSWER If $A_\mu A^\mu$ is really a Lorentz scalar, then it has the same value in any reference frame as it does in the frame where $\Vec{J} = 0$, namely (in Griffiths' notation) $-c^2\rho_0^2$. In general, $A_\mu A^\mu = -c^2\rho^2 + \vert\Vec{J}\vert^2$. Setting these equal and rearranging terms gives the desired result. This is just a warm-up exercise in "invariant thinking".

  2. Use the formal definition of the FIELD TENSOR $F^{\mu\nu}$ in Eq. (12.118)1 and the rule for its Lorentz transformation,2  $F'^{\mu\nu} = \Lambda^\mu_\alpha \Lambda^\nu_\beta F^{\alpha\beta}$, to derive the equations analogous to Eqs. (12.108) describing the transformation of $\Vec{E}$ and $\Vec{B}$ under a "boost" into a reference frame moving at velocity u (with the usual corresponding definitions of $\beta$ and $\gamma$) in the positive $\Hat{z}$ direction.3ANSWER This one drove me nuts until I broke the "matrix multiplication" habit I learned in High School. Here we don't find the $\mu\nu$th component of the product of two tensors by summing the products of the elements of the $\mu$th row of the first with the elements of the $\nu$th column of the second; instead we sum the products of the elements of the $\mu$th row of the first with the elements of the $\nu$th row of the second. Doing this twice [first with $\Lambda F$ and then with $\Lambda (\Lambda F)$] gives $ \Lambda F = $

\left( \begin{array}{cccc}
\gamma & 0 & 0 & -\beta\gamma  . . . 
 . . . ma e_y+\gamma B_x) & \beta\gamma e_z \cr
\end{array} \right)

\hbox{\rm and} \quad \Lambda (\Lambda F) =
\left( \begin{ . . . 
 . . . (-\beta\gamma^2e_z+\beta\gamma^2e_z) \cr
\end{array} \right)

= \left( \begin{array}{cccc}
0 & \gamma (e_x-\beta B_y) & . . . 
 . . . x+B_y) & \gamma (-\beta e_y-B_x) & 0 \cr
\end{array} \right)

    where $\Vec{e} \equiv \Vec{E}/c$ has been defined for compactness.

    That is, \fbox{ ${\displaystyle
E'_x = \gamma (E_x-uB_y) \;\;\; &  . . . 
 . . . ta E_y/c) & B'_y = \gamma (B_y-\beta E_x/c) & B'_z = B_z \cr
}$\ } .

  3. A capacitor made from two square parallel plates a on a side and d apart is given a charge -Q on the upper plate and +Q on the lower plate. Let the origin of coordinates be in the centre of the capacitor, the edges of the plates parallel to $\Hat{x}$ and $\Hat{y}$ and the gap in the $\Hat{z}$ direction.

    Using Lorentz transformations, find $\Vec{E}$ and $\Vec{B}$ inside the capacitor
    1. in a frame moving at a velocity u in the $\Hat{x}$ direction;
      ANSWER For this we can just plug $E_x=E_y=\Vec{B}=0$
      and $E_z=\epsz Q/a^2$ into Eqs. (12.108) to get the only nonzero
      components \fbox{
$E'_z = \gamma \epsz Q/a^2$\ } and $B'_y = \gamma (u/c^2) \epsz Q/a^2$ where as usual $\gamma \equiv \left(1-u^2/c^2\right)^{-1/2}$.
    2. in a frame moving at a velocity u in the $\Hat{z}$ direction. ANSWER Here we can use the analogous equations derived in the previous problem to get the only nonzero component $E'_z = E_z=\epsz Q/a^2$. That is, the fields have not changed. The only difference in geometry is that the plates are closer together due to Lorentz contraction; but that does not affect the electric field, as we know.
    3. For the former case, compare your result with what you would expect from simply transforming the dimensions of the plates into the moving frame and treating their motion as sheets of current. ANSWER Lorentz contraction shrinks distances parallel to the motion, but not perpendicular. Thus the plates are no longer square; the edge in the $\Hat{x}$ direction is a factor of $1/\gamma$ shorter in the primed frame while the other edge still has a length a. This compresses the net charge Q (an invariant) into an area smaller by the same factor, increasing $\sigma$' and consequently E'z by a factor of $\gamma$.  Meanwhile the plates are now sheets of current carrying $\Vec{K}' = \mp \sigma' u \Hat{x}$ for the top and bottom plates, respectively.4 As for the electric field, $\sigma' = \gamma Q/a^2$. We know that two opposite sheets of current produce a uniform magnetic field between them given by $\Vec{B} = (\muz/2) K \Hat{y}$ or $B_y = (\muz/2)(\sigma' u) = \gamma Q(\muz/2a^2)u$. Multiplying by $(\epsz/\epsz)$ and noting that $\muz\epsz = 1/c^2$ gives \fbox{ $B'_y = \gamma (u/c^2) \epsz Q/a^2$\ } .
      In a simple case like this, we get the right answer by simply taking Lorentz contraction into account. So one might wonder if all the elaborate formalism is ever any real help in solving practical problems. The answer is probably, "It depends on how flawlessy you can visualize all the effects without losing track of any." Equations like (12.108) allow you to transform into the frame of your choice "without much thought" even for rather complicated situations (e.g. when the fields vary with position). But without relativity we would have no way of making E&M obey NEWTON'S LAWS, of which we are understandably fond. It emerges naturally that magnetic fields are an obligatory extension of COULOMB'S LAW when we transform consistently into moving frames; and that knowledge is worth having!

Jess H. Brewer