Does a rotating point charge constitute a current ?

1. Jul 6, 2014

Vibhor

Does a point charge 'q' (say an electron) moving with constant angular speed ω in a circle of radius R constitute a current ?

A rotating ring of uniform charge density is treated as a current flowing in the ring .

On similar lines I think a point charge q is equivalent to current given by i = qω/2π .

Could someone help me understand this concept ?

Many Thanks

Edited : Replaced 'rotating' with 'moving'

Last edited: Jul 6, 2014
2. Jul 6, 2014

WannabeNewton

A point charge cannot rotate. At best you can talk about the magnetic field due to the intrinsic spin of electrons but this is not due to an actual rotation in space of the charge, it's just due to the usual quantum mechanical spin.

3. Jul 6, 2014

Vibhor

It seems you didn't like the word 'rotate' ,but I find it hard to believe that you didn't understand my question .

I have edited post#1 . Please have a look at the OP .

4. Jul 6, 2014

stevendaryl

Staff Emeritus
For an object to rotate means that parts are moving in different directions. But a point particle has no parts, so it's not clear what you mean by saying that it is rotating. Maybe you don't mean a point-particle, but a sphere, where the radius of the sphere is very small?

5. Jul 6, 2014

Vibhor

:surprised

I don't know how to respond . It is pretty clear I have already removed the word 'rotate' from the OP .

6. Jul 6, 2014

WannabeNewton

Ah I see, my apologies. I thought you were talking about a particle rotating in place, not a particle moving around in a circle. Well in that case yes there is certainly a current density associated with the charge and hence a current associated with it as well but it won't be what you wrote down. There are various ways to see this. One thing to note is that according to your expression for the current of such a particle, the current is stationary. This is obviously not true. Imagine just for a moment that the particle has a finite but very small radius $\epsilon$ so that the particle travels in a circular tube of radius $2\epsilon$, with its center of mass traversing the original circular trajectory (we can take $\epsilon \rightarrow 0$ in the end). At any point on this trajectory, take the cross section of the tube. Recall that current is the charge passing through this cross sectional area per unit time. Certainly if all we have is a single particle moving around in the tube, the charge passing through this (fixed) cross sectional area will not be the same per unit time. What you wrote down only works if we have a steady flow of current, which requires a persisting flux of charges. In fact we can make an overall guess. The amount of charge passing through any point of the circular trajectory goes like $q \delta(\phi - \omega t)$ where I have oriented the coordinates so that $\phi_0 = 0$. Then the current goes like $q\frac{ \delta(\phi - \omega t)}{t}$.

7. Jul 6, 2014

stevendaryl

Staff Emeritus
Okay. If you mean, does a point charge moving in a circle count as a current, then answer is yes.

8. Jul 6, 2014

Vibhor

So ,experts in Classical Physics section are of the opinion that point charge moving in a circle is equivalent to current .

How are experts in Introductory Physics section of the opinion that point charge moving in a circle does not constitute a current ?

Why is there a difference of opinion ?

Thanks

9. Jul 6, 2014

WannabeNewton

It's probably just a difference in terminology. Some people might define "current" to strictly mean "steady current". But in physics literature current is simply the flux of current density and a point charge moving in a circle certainly has a current density, one that is in fact very easy to write down.

10. Jul 7, 2014

vanhees71

Any moving charge implies a non-vanishing current density. Charge and current of any moving-charge distribution is decribed by the four-vector field
$$j^{\mu}(x)=\begin{pmatrix} c \rho(x) \\ \vec{v}(x) \rho(x) \end{pmatrix} = \rho_0(x) u^{\mu}(x),$$
where
$\rho(x)$ is the charge density as measured in the inertial reference frame under consideration (the "lab frame", where the observer is at rest) and $\vec{v}(x)$ is the velocity field of the charged fluid (this is NOT the spatial part of a four-vector). $\rho_0(x)$ is the charge density as measured in the local restframe of the heat bath and
$$u^{\mu}=\frac{1}{\sqrt{1-\vec{v}^2/c^2}} \begin{pmatrix} c \\ \vec{v} \end{pmatrix}$$
is the four-velocity of the fluid field, which is a four-vector field (and $\rho_0$ is a scalar field).

For a single point charge the expressions are
$$\rho(x)=q \delta^{(3)}[\vec{x}-\vec{y}(t)], \quad \vec{j}(x)=q \vec{v} \delta^{(3)}[\vec{x}-\vec{y}(t)], \quad \vec{v}=\dot{\vec{y}}.$$
Here $\vec{y}(t)$ is the trajectory of the particle measured in the lab frame.

To show that this gives a four-vector field one can write it as
$$j^{\mu}(x)=q \int \mathrm{d} \tau \frac{\mathrm{d} y^{\mu}}{\mathrm{d} \tau} \delta^{(4)}[x^{\mu}-y^{\mu}(\tau)],$$
which is obviously manifestly covariant (and independent of the parametrization of the world line with a parameter, $\tau$, for which you can choose the proper time of the particle. You come back to the original more intuitive expressions, if you choose the lab time, $t$ as the parameter in the integral.

11. Jul 7, 2014

Vibhor

I appreciate your response . But I am a High School student ,and what you have explained is beyond my level.

Could someone else help me understand this concept ?

12. Jul 7, 2014

stevendaryl

Staff Emeritus
The simple explanation is that current is charged particles in motion.

13. Jul 7, 2014

Staff: Mentor

Go with WbN's response in #9 of the thread - it's a difference in terminology. Sometimes the word "current" is used to mean any movement of electrical charge, in which case the moving electron counts as current. Sometimes the word is used to mean a steady flow, and then the moving electron wouldn't be such a good example.

14. Jul 7, 2014

Vibhor

15. Jul 7, 2014

Staff: Mentor

The guys in that other thread were using one of the accepted meanings of "current", we've been using another. No big deal, just shows that we could all be better at precisely defining the terms we use.

16. Jul 7, 2014

ChrisVer

Well, yes it does consist a current.
It generates the same magnetic field as you'd get from a ring flowed by some current I...

17. Jul 7, 2014

stevendaryl

Staff Emeritus
None of those posts say that a moving charge doesn't count as a current. They say that a moving charge is not a steady current. There is no difference of opinion about that.

18. Jul 7, 2014