# The current of a single electron

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1. Nov 1, 2015

### AuroralLight

I guess my question has multiple parts. Any help in understanding the questions is appreciated.

Assume a single electron in free space. The electron starts moving because of some force applied to it. The source of the force could be pretty much anything, lets say a uniform E-field in X direction. The electron will move on the X axis from left to right and passes the origin (X=0) at arbitrarily chosen time t=0. As we know current is a function of space and time. My question is:

1) What is the current as a function of time at origin? An observer (a sensor) is staring at the origin only.
2) Is the current at origin a multiple of Dirac delta function? This is my initial guess. Then, what would be the value of the multiplier to the Dirac delta function? Like in I = a * D(t), where D(t) is the Dirac delta function. What is multiplier "a" here?
3) If it is a Dirac delta function, it means it contains all frequency components (Fourier transform), isn't it? Including DC and visible frequencies.
4) If it includes DC, it means a portion of the current has been at the origin at all times even before electron reaches there (at negative times).
5) If it includes visible light, it means we should be able to see the light from the moving electron in free space if we stare at origin.
6) One might argue that the flash of light is so short that we cannot see it. What if we send a train of electrons at equi-time intervals (Dirac comb function). Then, we tune the time interval in such a way that one peak in frequency domain happens at a visible frequency. Then, we should see a constant light there.

If #2 above is wrong, then all other questions will not be valid. But, if it is right, then it is so strange to me.

2. Nov 1, 2015

### Staff: Mentor

Since you posted in the classical physics forum I assume that you are not actually interested in an electron as a quantum mechanical particle as described by quantum electrodynamics, but you are instead using "electron" as a shorthand for a "classical point charge". Is that correct?

Your 2) is essentially correct. If you have a point charge of total charge $q$ then its charge density would be $\rho=q \, \delta (\vec r - \vec r_q (t))$ where $\vec r_q(t)$ is the position of the charge at time t, and its current density would be $\vec j = q \, \vec v_q(t) \delta (\vec r - \vec r_q (t))$ where $\vec v_q(t)$ is the velocity of the charge at time t.

4) Sure, that is the way the Fourier transform works.

5) You don't see current, so I think that you are talking about the E-field. That would not be the same thing. Again, for the field you would calculate the Lienard Wiechert potentials.

6) See 5.

Last edited: Nov 1, 2015
3. Nov 1, 2015

### AuroralLight

Thanks a lot @DaleSpam . I am looking at this in classical terms. I guess Lienard Wiechert potential is what I was looking for.

4) I understand that the DC term is mathematically correct, but does it have any physical meaning? The current has been at origin and will be there forever, even though there is no charge there (except for a fraction of time). Well, the electromagnetic effect of this moving charge (the displacement current) exists at the origin forever, but how about current? I feel the definition of current as I = dq/dt somehow does not apply here, or I am missing something.

5 and 6) You are right. I think here the real question is not current density, but more electromagnetic field. And the wrong part of my question was that we cannot look at the electromagnetic field at one point (here origin) by ignoring the presence of the electron in the past somewhere in the negative X or in future in positive X. But, if the electron is created suddenly (somehow!) at the origin and vanishes suddenly (again at the origin) in a tiny fraction of time, then the electromagnetic field created by this hypothetical strange phenomena would create a light, isn't it?

Last edited: Nov 1, 2015
4. Nov 1, 2015

### Staff: Mentor

Yes, from your follow up response I think so. The LW potentials are very important for understanding electromagnetism, so it is well worth looking into at some point in your studies.

I wouldn't single out the DC term here. By definition, every single term in the Fourier transform has the same property. They each represent a sinusoid which has gone on forever and will continue to go on forever. This is just a fact of the basis functions used in the Fourier transform.

This also isn't a unique property of the delta function. Any function which has a Fourier transform will necessarily be represented in terms of "forever" functions. Functions which are time-limited will be frequency-unlimited, so they will have some component at each "forever" frequency.

This is just how the Fourier transform works. You can transform functions using many different sets of basis functions. If you choose a "forever" basis then any term in the expansion will be forever.

Maxwell's equations imply conservation of charge, so this scenario would violate Maxwell's equations. So the question would need to use some different theory of electromagnetism, but we only discuss mainstream theories here.

You could briefly create a charge dipole. That is kind of what Hertz did to detect EM waves.

5. Nov 1, 2015

### AuroralLight

Very true. Thanks again.

6. Nov 1, 2015

### Staff: Mentor

You are welcome, and welcome to PF!