THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 401 Assignment # 10:
 
RETARDED
POTENTIALS
 
SOLUTIONS:
 
Wed. 15 Mar. 2006 - finish by Wed. 22 Mar.

1. (p. 426, Problem 10.8) - Retarded Gauge: Confirm that the RETARDED POTENTIALS satisfy the LORENTZ GAUGE condition,
   

   $$\Div{A} = - {1\over c^2} \DbyD{V}{t} \quad \hbox{\rm or} \quad {\partial A^\mu \over \partial x^\mu} = 0$$ (1)

   
   
   
   

   $$\hbox{\rm where} \quad A^0 \equiv {V \over c} \quad \left( \hbox{\rm and} \quad J^0 \equiv c \rho \right) \, .$$ (2)

   
   
   ANSWER:  Following the hint, we first show
   

   $$\Vec{\nabla}\cdot\left(\Vec{J}\over\rr\right) = {1\over\r . . . . . . {J}) - \Vec{\nabla}^\prime\cdot\left(\Vec{J}\over\rr\right)$$ (3)

   
   
   where $\Vec{\rr} \equiv \Vec{r} - \Vec{r}^\prime$, $\Grad{}$ denotes derivatives with respect to $\Vec{r}$, and $\Vec{\nabla}^\prime$ denotes derivatives with respect to $\Vec{r}^\prime$:   The identity
   

   $$\Vec{\nabla}\cdot\left(f\Vec{v}\right) = f\left(\Div{v}\right) + \Vec{v}\cdot\Grad{f}$$ (4)

   
   
   and the (hopefully by now familiar) results
   

   $$\Grad{\left(1\over \rr\right)} = - {\Hat{\rr}\over \rr^2} = - \Vec{\nabla}^\prime\left(1\over \rr\right)$$ (5)

   
   
   
   

   $$\Longrightarrow \; \Vec{\nabla}\cdot\left(\Vec{J}\over\rr . . . . . . ight) - \Vec{J}\cdot\left(\Hat{\rr}\over \rr^2\right) \; \;$$ (6)

   
   
   
   

   $$\quad \& \quad \Vec{\nabla}^\prime\cdot\left(\Vec{J}\over\ . . . . . . \right) + \Vec{J}\cdot\left(\Hat{\rr}\over \rr^2\right) \, .$$ (7)

   
   
   Adding together Eqs. (6) and (7) gives Eq. (3).
   Next, noting that $\Vec{J}\left(\Vec{r}^\prime, t-\rr/c\right)$ depends on $\Vec{r}^\prime$ both explicitly and through $\rr$, whereas it depends on $\Vec{r}$ only through $\rr$, we confirm that
   

   $$\Div{J} = -{1\over c} \dVec{J} \cdot \left(\Grad{\rr}\right)$$ (8)

   
   
   
   

   $$\Vec{\nabla}^\prime\cdot\Vec{J} = -\dot{\rho} -{1\over c} \dVec{J} \cdot \left(\Vec{\nabla}^\prime\rr\right) \; :$$ (9)

   
   
   Derivatives of $\Vec{J}(\Vec{r}^\prime,t_r)$ with respect to $\Vec{r}$ (on which it does not depend explicitly) mix in the time derivative through the implicit dependence of t_r on $\Vec{r} = \Vec{\rr} + \Vec{r}^\prime$. That is,
   

   $$\Div{J}(\Vec{r}^\prime,t_r) = \left(\DbyD{\Vec{J}}{t_r}\right)\cdot\Grad{t_r} = - {\dVec{J}\cdot\Hat{\rr} \over c}$$ (10)

   
   
   because, for a given $\rr$, ${\displaystyle \DbyD{\Vec{J}}{t_r} = \DbyD{\Vec{J}}{t} = \dVec{J} }$, and ${\displaystyle\Grad{t_r} = - {1 \over c} \Grad{\rr} = - {\Hat{\rr} \over c} }$.
   However, $\Vec{J}(\Vec{r}^\prime,t_r)$ depends explicitly and implicitly upon $\Vec{r}^\prime$, and must locally satisfy the EQUATION OF CONTINUITY $\Vec{\nabla}^\prime\cdot{J} = -\dot{\rho}$ (i.e. charge conservation) at any instant of time in terms of the source coordinates $\Vec{r}^\prime$, so we have
   

   $$\Vec{\nabla}^\prime\cdot{J}(\Vec{r}^\prime,t_r) = -\dot{\rho} + {\dVec{J}\cdot\Hat{\rr} \over c}$$ (11)

   
   
   because ${\displaystyle \Vec{\nabla}^\prime t_r = - {1 \over c} \Vec{\nabla}^\prime\rr }$ ${\displaystyle = + {\Hat{\rr} \over c} }$.
   Finally we use this to calculate the divergence of $\Vec{A}$ in Eq. (10.19):
   

   $$A^\mu(\Vec{r},t) = {\muz\over4\pi}\VolInt {J^\mu(\Vec{r}^\prime,t_r) d\tau^\prime \over \rr}$$ (12)

   
   
   
   
   $$\begin{eqnarray*} \Div{A} &=& {\muz\over4\pi}\VolInt \left[ {1\over\rr}\left( . . . . . . ime\cdot\left(\Vec{J}\over\rr\right) \right] d\tau^\prime \; . \end{eqnarray*}$$
   
   The DIVERGENCE THEOREM tells us that
   
   $$\VolInt \left[\Vec{\nabla}^\prime\cdot\left(\Vec{J}\over\rr . . . . . . rime = \oSurfInt {\Vec{J}\cdot d\Vec{a}^\prime \over \rr} \; .$$
   
   
   Now, if the closed surface encloses all the charges and currents in the source volume, $\Vec{J} = 0$ over the whole surface and the surface integral is zero, leaving
   
   $$\Div{A} = {\muz\over4\pi}\VolInt \left(-\dot{\rho}\over\rr\right)d\tau^\prime$$
   
   
   
   
   $$= -\muz\epsz\DbyD{}{t}\left\{ {1\over4\pi\epsz} \VolInt \left(\rho\over\rr\right)d\tau^\prime \right\}$$
   
   
   or .

2. (p. 427, Problem 10.10) - Weird Loop:

   

   A piece of wire bent into a weirdly shaped loop, as shown in the diagram, carries a current that increases linearly with time:
   
   $$I(t) = k t \; .$$
   
   
   1. Calculate the retarded vector potential $\Vec{A}$ at the center.   ANSWER:  Choose the origin at the same place as the field point: the centre. Thus $\Vec{r}=0$ and $\Vec{\rr} = -\Vec{r}^\prime$. The source region is uncharged, so V = 0.
      
      $$\Vec{A}(0,t) = {\muz \over 4\pi} \int {\Vec{I}(\Vec{r}^\prime, t-r^\prime/c) \over -r^\prime} d\ell^\prime$$
      
      
      
      
      $$= -{\muz k \over 4\pi} \left[ 2 \int_a^b {(t-\ell/c) \Hat{x} d\ell \over \ell} \right.$$
      
      
      
      
      $$+ \int_0^\pi {(t-b/c) \Hat{\theta} b d\theta \over b}$$
      
      
      
      
      $$- \left. \int_0^\pi {(t-a/c) \Hat{\theta} a d\theta \over a} \right]$$
      
      
      where $\Hat{\theta} = -\Hat{x}\sin\theta +\Hat{y}\cos\theta$. Now, by symmetry there is as much current going "up" as "down" at the same $r^\prime$ and t_r, so the $\Hat{y}$ components cancel. This leaves
      
      $$\Vec{A}(0,t) = {\muz k \over 4\pi} {\cal I} \Hat{x}$$
      
      
      where
      
      $${\cal I} \equiv 2t \int_a^b {d\ell \over \ell} -{2\over c} \int_a^b d\ell$$
      
      
      
      
      $$- \left(t - {b \over c}\right) \int_0^\pi \sin\theta d\theta$$
      
      
      
      
      $$+ \left(t - {a \over c}\right) \int_0^\pi \sin\theta d\theta$$
      
      
      
      
      $$= 2t\ln\left(b\over a\right) - {2(b-a)\over c}$$
      
      
      
      
      $$- 2t + 2{b \over c} + 2t - 2{a \over c}$$
      
      
      or .

   2. Find the electric field at the center.

      ANSWER:  Since V=0 we have just
      .

   3. Why does this (neutral) wire produce an electric field?   ANSWER:  Because the vector potential is changing with time, "Doh!" I think this is meant as a retroactive hint in case you got hung up on the preceding question.

   4. Why can't you determine the magnetic field from this expression for $\Vec{A}$?

      ANSWER:  Finding $\Vec{B} = \Curl{A}$ requires knowledge of the dependence of $\Vec{A}$ on $\Vec{r}$; but we have calculated $\Vec{A}$ only at one point in space! If you want a differentiable $\Vec{A}(\Vec{r})$ you will have a far more difficult calculation to perform.

3. (p. 434, Problem 10.13) - Circulating Charge: A particle of charge q moves in a circle of radius a at constant angular velocity $\omega$. [Assume that the circle lies in the $x\,y$ plane, centered at the origin, and that at time t=0 the charge is at (a,0), on the positive x axis.] Find the LIÉNARD-WIECHERT POTENTIALS for points on the z axis.   ANSWER:  In general,
   
   $$V(\Vec{r},t) = {q\over4\pi\epsz} \left[c \over \rr c - \Vec{\rr}\cdot\Vec{v}\right]_{\rm ret}$$
   
   
   
   
   $$\Vec{A}(\Vec{r},t) = V(\Vec{r},t) \left[\Vec{v}\over c^2\right]_{\rm ret}$$
   
   
   where $\left[\cdots\right]_{\rm ret}$ means that the quantities in the square brackets are to be evaluated at the retarded time $t_r = t - \rr/c$. Relative to the origin, $\Vec{r}^\prime = a \Hat{s}$ $= a \left[\Hat{x}\cos(\omega t) + \Hat{y}\sin(\omega t)\right]$.
   For a point on the z axis, $\Vec{r} = z \Hat{z}$ and $\Vec{\rr} = z \Hat{z} - a \cos(\omega t) \Hat{x} - a \sin(\omega t) \Hat{y}$ so $\rr = \sqrt{z^2 + a^2}$, independent of time. We also have $\Vec{v} = a\omega\left[-\Hat{x}\sin(\omega t) + \Hat{y}\cos(\omega t)\right]$ and $v = a\omega$. Thus $\Vec{\rr}(t_r) = z \Hat{z}$ $- a \cos\theta_r \Hat{x}$ $- a \sin\theta_r \Hat{y}$ and $\Vec{v}(t_r) = a\omega\left[-\Hat{x}\sin\theta_r + \Hat{y}\cos\theta_r\right]$ where . Then $\Vec{\rr}(t_r)\cdot\Vec{v}(t_r) =$ $a^2\omega[\cos\theta_r\sin\theta_r$ $- \sin\theta_r\cos\theta_r] = 0$, leaving
   and
   .

4. (p. 441, Problem 10.19) - Sliding String of Charges: An infinite, straight, uniformly charged string, with $\lambda$ charge per unit length, slides along parallel to its length at a constant speed v.
   1. Calculate the electric field a distance d from the string, using Eq. (10.68):
      
      $$\Vec{E}(\Vec{r},t) = {q \over 4\pi\epsz} \; { 1 - v^2/c^2 . . . . . . 2 \sin^2 \theta / c^2 \right)^{3/2} } \; {\Hat{R} \over R^2}$$
      
      
      where $\Vec{R} \equiv \Vec{r} - \Vec{v} t$.

      

      ANSWER:  Suppose the field point is a perpendicular distance s from the string; measure z from the nearest point on the string, as shown in the diagram. Equation (10.68), in which we do not need to evaluate anything at a retarded time, gives the contribution to $\Vec{E}$ from a single charge q. We need to superimpose such contributions from all charge elements $dq = \lambda dz$ at positions $-\infty<z<+\infty$ down the string: for each of these we use $\Vec{R} = s \Hat{x} - z \Hat{z}$:
      
      $$\Vec{E}(\Vec{r},t) = {\lambda \over 4\pi\epsz} (1 - v^2/c^2) \; \Vec{\cal I} \qquad \hbox{\rm where}$$
      
      
      
      
      $$\Vec{\cal I} \equiv \int_{-\infty}^\infty {\Hat{R} \over R^ . . . . . . z \over \left( 1 - \beta^2 \sin^2 \theta \right)^{3/2} } \; .$$
      
      
      For each element dz at z there is an equal element dz at -z; thus the "horizontal" components cancel, leaving only the x component of $\Hat{R}$, namely $\Hat{x}\sin\theta$. Meanwhile, since $s = R\sin\theta$, $1/R^2=\sin^2\theta/s^2$; and since $-z=s\cot\theta$, $dz = s\csc^2\theta d\theta = s d\theta/\sin^2\theta$. So $dz/R^2 = d\theta/s$ and
      
      $$\Vec{\cal I} = \int_0^\pi {\Hat{x}\sin\theta d\theta \over s \left( 1 - \beta^2 \sin^2 \theta \right)^{3/2} } \; .$$
      
      
      Let $u = \cos\theta$ so that $\sin\theta d\theta = -du$ and $\sin^2\theta=1-u^2$:
      
      $$\Vec{\cal I} = {\Hat{x}\over s} \int_{-1}^1 {du \over \left( 1 - \beta^2 [1-u^2] \right)^{3/2} }$$
      
      
      
      
      $$= {\Hat{x}\over s\beta^3} \int_{-1}^1 {du \over \left( a^2 + u^2 \right)^{3/2} }$$
      
      
      
      
      $$= {\Hat{x}\over s\beta^3} \left[u \over a^2 \sqrt{a^2 + u^2}\right]_{-1}^1$$
      
      
      
      
      $$\hbox{\rm where} \qquad a^2 \equiv {1\over\beta^2} - 1 \; . \qquad \hbox{\rm Thus}$$
      
      
      
      
      $$\Vec{\cal I} = {\Hat{x}\over s\beta^3}\left[2 \over \left({1\over\beta^2} - 1\right)\sqrt{{1\over\beta^2} - 1 + 1}\right]$$
      
      
      
      
      $$= {\Hat{x}\over s}\left[2 \over 1 - \beta^2 \right] \, , \qquad \hbox{\rm so}$$
      
      
      
      
      $$\Vec{E}(\Vec{r},t) = {\lambda \over 2\pi\epsz} {1 - \beta^2 \over 1-\beta^2} {\Hat{x}\over s} \qquad \hbox{\rm or}$$
      
      
      
      just as for a line charge at rest!

   2. Find the magnetic field of this string, using Eq. (10.69):
      
      $$\Vec{B} = {1 \over c} \left( \Vec{\rr} \times \Vec{E} \right) = {1 \over c^2} \left( \Vec{v} \times \Vec{E} \right)$$
      
      
      where $\Vec{\rr} \equiv \Vec{r} - \Vec{r}^\prime$.   ANSWER:  Well, $\Vec{v} = v \Hat{z}$ and $\Hat{z}\times\Hat{x} = \Hat{y}$, so this is trivial:¹
      
      where $I = \lambda v$. (Again, the same result as for a steady current in magnetostatics.)