THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 401 Assignment # 4:
 
POTENTIALS, GAUGES and RELATIVITY
 
SOLUTIONS:
 
Wed. 25 Jan. 2006 - finish by Wed. 1 Feb.

Please review Section 10.1 and Ch. 12.

1. (p. 420, Problem 10.3) - GIVEN V & $\Vec{A}$ . . . Find the $\Vec{E}$, $\Vec{B}$, $\rho$ & $\Vec{J}$ corresponding to
   
   $$V(\Vec{r},t) = 0, \qquad \Vec{A}(\Vec{r},t) = - {1\over4\pi\epsz} \, {qt\over r^2} \; \Hat{r} \; .$$
   
   
   ANSWER:   . $\Vec{B} = \Curl{A} = % {1\over r}\left[{1\over\sin\theta}\DbyD{A_r}{\phi}\right]\Hat{\theta} - {1\over r}\left[\DbyD{A_r}{\theta}\right]\Hat{\phi}$ but A_r varies only with r, not with $\phi$ or $\theta$, so  . These are the $\Vec{E}$ and $\Vec{B}$ of a stationary point charge at the origin, so and  .

2. POINT CHARGE:
   1. Find the $\Vec{E}$ and $\Vec{B}$ fields corresponding to a stationary point charge q situated at the origin. ANSWER:  This is a freebie.

   2. State the charge and current distributions of this situation. ANSWER:  See above.

   3. What are the electric and magnetic potentials? ANSWER:  We have learned to use  . Those potentials are certainly not wrong, but they don't look like the ones in Problem 10.3!

   4. Is there any relation between this situation and that described in Problem 10.3? ANSWER:  Well, obviously, since they give the same fields, they describe exactly the same physics! This is a vivid illustration of GAUGE INVARIANCE: an infinite variety of choices for V and $\Vec{A}$ will all give the same $\Vec{E}$ and $\Vec{B}$ as long as they differ only by adding the gradient of some scalar function to $\Vec{A}$ while subtracting the time derivative of the same function from V.

3. (p. 420, Problem 10.5) - GAUGE TRANSFORMATION: Use the gauge function
   
   $$\lambda = -{1\over4\pi\epsz} \; \left(q \, t\over r\right)$$
   
   
   to transform the potentials in Problem 10.3, and comment on the result. ANSWER:  $\Vec{A}'(\Vec{r},t) = \Vec{A}(\Vec{r},t) + \Grad{\lambda}$ and ${\displaystyle \Grad{\lambda} = {qt\over4\pi\epsz} {\Hat{r}\over r^2} = - \Vec{A}(\Vec{r},t) }$ so as in Problem 2. Meanwhile $V'(\Vec{r},t) = V(\Vec{r},t) - \dbyd{\lambda}{t}$ where $V(\Vec{r},t) = 0$ and ${\displaystyle \DbyD{\lambda}{t} = -{q\over4\pi\epsz r} }$ so $V'(\Vec{r},t)$ is just the potential due to a point charge q at the origin, again as in Problem 2. So $\lambda$ is just the gauge function required to transform between these two descriptions of the same thing.

4. WHICH GAUGE?
   1. In Problem 10.3 above, are the potentials in the Coulomb gauge, the Lorentz gauge, both, or neither? ANSWER:  With $V(\Vec{r},t) = 0$ we have $\dbyd{V}{t} = 0$ while ${\displaystyle \Div{A} = - {qt\over4\pi\epsz} \; \delta(\Vec{r}) = {t\over4\pi\epsz} \, \rho }$. This is neither zero (Coublob gauge) nor $-\dbyd{V}{t}$ (Lorentz gauge), so the answer is .

   2. In Problem 2 above, are the potentials in the Coulomb gauge, the Lorentz gauge, both, or neither? ANSWER:  In this case V is constant in time, so again $\dbyd{V}{t} = 0$; meanwhile, if $\Vec{A} = 0$ then $\Div{A} = 0 = - \dbyd{V}{t}$, so the answer in this case is  . It's important to note that "being in" one gauge does not preclude "being in" another!

5. NATURAL UNITS: Since c is now a defined quantity that keeps appearing in confusing places in our notation for 4-vectors etc., and since nanoseconds (ns) are perfectly handy units for distance, it seems silly to not just measure time and distance in the same units (seconds) and set c=1. While we're at it, why not set the ubiquitous constant in quantum mechanics to unity as well ($\hbar = 1$) so that all angular momenta are unitless and (because $E = \hbar \omega$) energies are measured in s^-1?
   1. In what units would we then measure velocities, momenta, masses, forces and accelerations? ANSWER:  Velocities are of course ( $v \equiv \beta$). Momenta are part of the same 4-vector as energies, and thus differ from them only by factors of c=1 (e.g. for massless particles E = pc), so momenta are also measured in  . Masses are also measured in  , since E = mc². Forces have the same units as the time rates of change of momenta, namely  . Accelerations are the time rate of change of velocities or forces divided by masses; either way, they are measured in  .

   2. Suppose we set the Coulomb force constant $k_E \equiv \kE = 1$ as well. In what units would we then measure charge, electric field, magnetic field, and potentials V and $\Vec{A}$? ANSWER:  COULOMB'S FORCE LAW is now written $\Vec{F} = q_1 q_2 \Hat{r} / r^2$, so [charge]² = [force] x [distance]² so Charge is  . Electric field is force per unit charge, so it now has the same units as force, namely . Alternatively, E² is proportional to energy per unit volume. Volume is in s³, so E² is in s^-4. Thus again E is in s^-2. Magnetic field is in the same as $\Vec{E}$, because B² is also proportional to energy per unit volume.¹ V is $\int \Vec{E} \cdot d\Vec{\ell}$ and $\ell$ is in seconds, since velocities are unitless; so V is in  . (This can also be deduced from U = q V.) $\Vec{B} = \Curl{A}$ so $\Vec{A}$ has one less s in its denominator: it is measured in just like V. (This should not be surprising.)

   3. Write out Maxwell's equations in this system of units. (Hint: We must have $\epsz \, \muz = 1$.)²ANSWER:  We have set $\epsz = 1/4\pi$ so we must set $\muz = 4\pi$ to ensure c=1. Then COULOMB'S LAW reads $\Vec{E} = q \Hat{r} / r^2$ which, integrated over a sphere centred on the charge, gives $\osurfint \Vec{E}\cdot d\Vec{a} = (4\pi r^2)\times(q/r^2) = 4\pi q$, so GAUSS' LAW has a $4\pi$ where the $(1/\epsz)$ used to be, just as expected. By the same token we expect to replace $\muz$ by $4\pi$ in the AMPÈRE/MAXWELL LAW; so Maxwell's equations become just
      
      $$\begin{array}{rclrcl} \Div{E} &\!\!\!=\!\!\!& 4\pi \rho & . . . . . . \!\!\!=\!\!\!& 4\pi\Vec{J} + \DbyD{\Vec{E}}{t} \cr \end{array}$$
      
      
      where both sides of each equation are measured in units of s^-3. These are of course the "microscopic" equations; if we want to use the "macroscopic average" fields $\Vec{H} = \Vec{B}/\mu$ and $\Vec{D} = \epsilon\Vec{E}$, we must include a dimensionless permittivity $\epsilon \equiv \kappa$ (where $\kappa$ is the dielectric constant) and dimensionless permeability $\mu \equiv 1 + \chi_m$ (where $\chi_m$ is the magnetic susceptibility).

      That was pretty simple. But before you get too enthusiastic, you might want to make a quantitative calculation of your own weight or the electron's charge in these units. For the former I get something like 10⁵⁸ s^-2. (I wouldn't want to have to paint the dials on bathroom scales if the world embraced these particular "natural units"!)³One can also set Newton's universal gravitational constant G=1 to define PLANCK UNITS, but I don't understand how that works. We already have everything covered, as far as I can see.

6. 4-POTENTIAL: In Eq. (12.131) on p. 541, Griffiths states that, "As you might guess, V and $\Vec{A}$ together constitute a 4-vector: $A^\mu = (V/c, A_x, A_y, A_z)$." This is a very strong statement with profound consequences. You can't just take any 3-vector and combine it with a convenient scalar in the same units to make a true 4-vector! Explain why we should believe this about $A^\mu$, and list any essential conditions that must be met for it to be true. ANSWER:  If, as also stated in Griffiths, $\partial_\mu \equiv \dbyd{}{x^\mu}$ is a covariant 4-vector, then its inner product with $A^\mu$ would be $\partial_\mu A^\mu = {1\over c^2}\DbyD{V}{t} + \Div{A}$. But the LORENTZ GAUGE explicitly sets the RHS of this equation to zero, and zero is certainly a Lorentz invariant scalar. This inner product is therefore invariant, so $A^\mu$ must be a bona fide 4-vector, as long as we stay in the LORENTZ GAUGE.⁴