# Infinitely long wire carrying current

1. Aug 11, 2015

### fluidistic

1. The problem statement, all variables and given/known data
With respect to a frame of reference K, a conducting wire of infinite length and internal radius "a" carries a current I -due to a flow of electrons- and has no charge density. An observer in a reference frame K' is moving parallely with respect to the wire. From the viewpoint of that observer (in K'), calculate the charge density and the drift velocity of the electrons.

2. Relevant equations
Fields transformation formulae, E and B fields due to an infinite line of charge and infinite wire with current.

3. The attempt at a solution
From the point of view of the observer in K, the E field is worth 0 everywhere (since no charge density), and there's a magnetic field worth $\vec B =\frac{\mu_0I}{2\rho}\hat \theta$ where I used cylindrical coordinates, $\rho$ is the distance from the surface of the wire to a point ouside of the wire. Inside of the wire the B field vanishes since it's a perfect conductor. Let $\vec v=v_0 \hat z$ be the velocity of K' with respect to K.
Now, for the observer in K', there are both an E' and B' fields. I get that $\vec E'=\gamma v_0 \frac{\mu_0 I}{2\rho}\hat \rho$ and $\vec B'=\gamma \vec B = \gamma \frac{I\mu_0}{2\rho}\hat \theta$. The E' field allows me to calculate $\lambda'$, the charge density the observer can see or feel. Since for an infinite wire, $\vec E=\frac{\lambda}{2\pi \rho}\hat \rho$, I get that $\lambda ' =\gamma \pi v_0 I \mu_0$. Looks ok to me, when v_0 tends to 0, I get lambda' tends to 0 which is expected since the observer in K has lambda=0.

Now to get the drift velocity of the electrons, I use the fact that $\vec J =\lambda' \vec v_{\text{drift}}$. This is where I'm having a doubt, is $\vec J = i \hat z$? Or should I express \vec J in terms of a Dirac delta in cylindrical coordinates like so?: $\vec J=i\frac{\delta(\rho -a)}{2\pi a^2}\hat z$ If I use the former, I reach that $\vec v_{\text{drift}}=\frac{1}{\gamma\pi v_0\mu_0}\hat z$ but this doesn't seem to have units of speed.
Now maybe it's due to the units, when I got the E' field due to lambda' I may have forgotten a factor of $\varepsilon _0$ that I should get for both $\lambda '$ and the drift velocity.
And since c²=sqrt(mu_0 epsilon_0) I'd reach $\vec v_{\text{drift}}=\frac{c^2}{\gamma\pi v_0} \hat z$ which seems to have units of speed, but this speed seems gigantic to me, maybe even greater than c... What am I doing wrong?

2. Aug 11, 2015

### Orodruin

Staff Emeritus
You are taking a long way around by trying to go via the fields. The current density is a 4-vector and you can Lorentz transform it directly.

3. Aug 11, 2015

### fluidistic

Thanks for the help.
I have that $J^\alpha =\rho _0 \sqrt {1-v_0^2/c^2}U^\alpha$ where $U^\alpha$ is the four-velocity worth $(ct, 0,0,v_0)$ in my case. Is that what you meant by Lorentz transformation of J?
Also, how does this help me to get the drift velocity?

Edit: I see, with the Lorentz transformation under a 4x4 matrix I could get $J'^\alpha$ knowing $J^\alpha$. But then, I'm still not sure how to proceed further in order to get the drift velocity.

Last edited: Aug 11, 2015