2
$\begingroup$

There is a problem in Griffiths "Introduction to Electrodynamics" whose available "solutions" online all seem to miss the point entirely. The exact problem I'm talking about is problem 5.12 (4th edition), or problem 5.38 (3rd edition).

I will recapitulate the problem here:

Since parallel currents attract each other, it may have occurred to you that the current within a single wire, entirely due to the mobile electrons, must be concentrated in a very thin stream along the wire. Yet, in practice, the current typically distributes itself quite uniformly over the wire. How do you account for this?

Okay, so the problem so far seems to be asking us to show that indeed, contrary to our initial guess, the mobile electrons spread out uniformly across the cross-section of the wire. But then the problem goes on to say the following:

If the uniformly distributed positive charges (charge-density $\rho_+$) are "nailed down", and the negative charges (charge-density $\rho_-$) move at speed $v$ (and none of these depends on the distance from the axis), show that $\rho_-=-\rho_+\gamma^2$, where $\gamma=1/\sqrt{1-(v/c)^2}$.

What?! In the boldfaced-text, Griffiths (seemingly) defeated the purpose of this problem!

My question is, can it be shown classically that, according to Maxwell's equations (and symmetry principles), the electron charge density and/or the velocity of those electrons is independent of the radial distance from the central-axis of the wire? I have tried to show it, but I can only end up with the following equation:

$$\frac{v(s)}{c^2}\int_0^s \rho(s')v(s')s'ds'=\int_0^s(\rho_++\rho_-(s'))s'ds'$$

From here, if I assume $\rho_-(s)$ and $v(s)$ are constant, then the result $\rho_-=-\rho_+\gamma^2$ easily follows.

NOTE: This is not a homework problem for a class of mine, but even if it was I think question is still valid.

EDIT:

If we assume the velocity is independent of the radial distance from the center, then we can show that this implies $\rho_-(s)=\rho_-$ (constant).

$$ \begin{align*} \frac{v(s)}{c^2}\int_0^s \rho(s')v s'ds'&=\int_0^s(\rho_++\rho_-(s'))s'ds'\\ \implies0&=\int_0^s\left(\rho_++\left(1-\frac{v^2}{c^2}\right)\rho_-(s')\right)2\pi s' ds'~~~~\textrm{(for all }s\textrm{)} \\ \implies 0&= \rho_++\left(1-\frac{v^2}{c^2}\right)\rho_-(s)\\ &\implies \boxed{\rho_-(s)=-\gamma^2\rho_+ }~~~~\textrm{(constant!)} \end{align*}$$

|cite|improve this question
$\endgroup$
  • $\begingroup$ Maybe I'm wrong, but I think the point of the problem is about showing radial density to be constant. If parallel currents attract, then there should be a larger density of it at the core. Speed being dependent on distance from the center shouldn't matter. Neither the positive charge distribution. $\endgroup$ – Diego Nov 11 '16 at 3:08
  • $\begingroup$ @Diego If that was the point of the problem, why would Griffiths then go right on to assume the radial density to be constant? In the end, it is claimed that the radial density doesn't depend on radial distance, and the same for the velocity. My question is, through what justification can this be shown to be true? See my edit. $\endgroup$ – Arturo don Juan Nov 11 '16 at 3:13
  • $\begingroup$ But he assumed only the positive charge density to be constant. Think of it as the atoms fixed on the copper wire, but the electrons are free to move in and out. $\endgroup$ – Diego Nov 11 '16 at 3:31
  • 1
    $\begingroup$ I have doubts that the link you post points to a legal copy of the book. If it is not, please delete the link. $\endgroup$ – garyp Nov 11 '16 at 3:39
  • $\begingroup$ @Diego He didn't assume only the positive charge density to be constant, and a quick look at the available solutions online proves that. Following garyp's advice, I won't link the actual solutions manual, but you can easily google it (e.g. try "griffiths introduction to electrodynamics solutions") $\endgroup$ – Arturo don Juan Nov 11 '16 at 4:07
3
$\begingroup$

The point of the problem (and this is not well explained in Griffiths) is this:

If you assume the densities $\rho_+$ and $\rho_-$ are uniform, you may derive that $\rho_-=-\gamma^2\rho_+$. Once you've found this, you can then prove that the net force on an electron travelling along the wire is zero. The magnetic attraction towards the center is exactly balanced by the electric repulsion caused by the fact that $\rho_-$ is larger than $\rho_+$. This shows that the arrangement $\rho_-=\text{constant}$ is an equilibrium arrangement, since in this arrangement the net force on any charge is zero. Furthermore, it's a stable equilibrium, since increasing the electron density in the center will cause electrons to be forced away from the center, and vice versa. So this problem is really only the first step in showing why the density is uniform, but it's not assuming its conclusion.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ What you say makes sense, but what about other possible equilibrium arrangements? How can one definitely say that no other equilibrium states exist? My physical intuition tells me that it is indeed the only (physical) equilibrium state, but my physical intuition doesn't always give me the right answer. $\endgroup$ – Arturo don Juan Nov 11 '16 at 20:48
  • $\begingroup$ And furthermore, if that was the purpose of the problem, then how come I'm able to prove it so easily (after assuming the velocity-profile is constant)? The assumption is superfluous and sidestepped. $\endgroup$ – Arturo don Juan Nov 11 '16 at 21:06
  • $\begingroup$ Yeah, that's a fair criticism. But you should have that criticism of the rigor in Griffiths in general. This problem isn't supposed to prove it's the only possible equilibrium, just give a plausibility argument. I think the real point is to get you to think about relativistic electrodynamics $\endgroup$ – Jahan Claes Nov 12 '16 at 5:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.