point charges as current
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Short physics note by Phil dated 1.28.03 in his E&M notes. It shows that J = q v delta(r - a) satisfies the continuity equation for a point charge, checks this by integrating over a box, then treats a mechanical dipole of charges +q and -q. It uses Jackson (9.13) to get the n=0 vector potential and recovers the dipole moment p = qd, agreeing with Jackson (9.16).
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Point charges as current: using delta functions PhL 1.28.03
Consider a point charge q moving at some slow constant velocity v. What is J and what is ?
(r) = q (r - a) //point charge located at position a
We know this is right because we integrate over a surrounding volume to get q. Next, let's guess for J:
J(r) = q v (r - a)
To verify this conjecture, consider (v is a constant vector)
J(r) = q (v)(r - a)
On the other hand, we can see that
t(r) = q t (r - a) = q (t a) a (r - a) = q v (-r) (r - a) = -q (v)(r - a)
Since this satisfies J = - t(r), we know it must be the right form for J. Remember that continuity just says that if there is a net outflow of current from a volume, the enclosed charge must decrease.
Example in the z direction
Let v = v . Then J(r) = q v (r - a) . Then J(r) = zJz = q v z (r - a), where z acts only on the z factor of the delta function. Now imagine an aligned box containing the charge. Integrate this divergence over the box and we get qv dz z (z - az) = only the parts = qv [ (h - az) - (az) ]. This is only non-zero when the charge is passing through the upper or lower face of the box. On the other hand, if we integrate the current density over the box, we get Q(az) = q (az) (h- az). That is, it is non-zero only if az is in the box. If we apply a time derivative to this Q, we use x (x) = (x) to get
t Q(az) = q t [( az) (h- az) ] = q az/t az [( az) (h- az) ] // chain rule
= q v az [( az) (h- az) ] // recognize v
= q v { ( az)[ -( h- az)] + (az) (h- az) } // product rule
= q v { ( h)[ -( h- az)] + (az) (h-0) } // use deltas
= q v { [ -( h- az)] + (az) }
= - q v [ (h - az) - (az) ]
So thing do work out. It just seems a little odd to say that J(r) = qv z (r - a) in this case, and generally
(r) = q (r - a) J(r) = q v (r - a) J(r) = - /t = qv (r - a)
Application to a mechanical dipole radiator
Imagine charges q and -q separated by distance a cos(t) . In other words, +q is located at
q is at: a(t) = (d/2) exp(-it) -q is at: - a(t) = - (d/2) exp(-it)
(r) = (+q) (r - a) + (-q) (r+ a) = q [(r - a) - (r + a) ]
The velocity of the +q charge is t a(t) = - i a(t), while that of -q is the opposite. So we have
The current density is
J(r) = (+q)(- i a(t) )(r - a) + (-q)(+ i a(t) )(r + a) = -iq a(t) [(r - a) + (r + a) ]
Now use Jackson (9.13) to compute the n=0 term of the vector potential! Note that the two currents are in the same direction! We get:
A(r,t) = J d3r = -ik [2 q a(t)] = -ik [2 q (d/2) exp(-it) ] = -ik {qd} exp(-it)
The thing {qd} is usually called p, the dipole moment,
p = q d A(r) = -ik p // agreeing with (9.16)
We could also compute p by integrating :
p = r d3r = q [(r - a) - (r + a) ] r d3r = q a - (q (-a)) = 2 q a = q d
Note that things are defined with and without the time phase in various places.