# Computing light

1. Apr 3, 2009

### snoopies622

I want to know how to find the electromagnetic field at a given point around a moving charge and this has led me to a few questions. My first one is, do both the

$$\mu _0 \bf {J}$$

term and the

$$\mu_0 \epsilon _0 \frac {\partial}{\partial t} \bf E$$

term contribute to the magnetic component of the field even when the "current" is only a single charged particle?

2. Apr 3, 2009

### f95toli

If you are asking if a single moving charge can be considered a current, the answer is yes.
You can in principle (and in the future hopefully in practice) define current(the ampere) by simply counting the number of charges that passes by a certain point in a circuit per unit time, it doesn't matter if these charges are traveling inside a conductor or in vacuum.

3. Apr 3, 2009

### clem

I can answer that part of your question. The equations you wrote are only part of the full set needed. All four of Maxwell's equations must be used to get the wave equation for the EM potentials. These must be solved to get the Leinard-Wiechert fields of a moving point charge. The whole procedure is given in full detail in more advanced EM texts. There really is no short cut.

4. Apr 3, 2009

5. Apr 4, 2009

### clem

Yes, SR can be used to get the constant velocity fields, but it is harder for general motion.

6. Apr 4, 2009

### snoopies622

But must it be? Since any changing electric field is produced by a moving charge or charges somewhere, this implies that there are no circumstances in which one would use only the dE/dt term to find the curl of the magnetic field anywhere. Is this right?
Is there one you can recommend for me?
Thanks for the link! That looks like a good source.

7. Apr 4, 2009

### clem

I use "Classical Electromagnetism" by Franklin. It does the L-W fields in detail.
The curl B ~ dE/dt equation holds at any point, so it can give curl B if there is no current at that point.

8. Apr 5, 2009

### snoopies622

Thanks. I will see if the physics library closest to me has it.

In the meantime, the Wikipedia article on the Biot-Savart law includes a section called, "point charge at constant velocity" which includes the formula

$$\vec B = \vec v \times \frac {1}{c^2} \vec E$$

where

$$\vec E = \frac {1}{4 \pi \epsilon _0} \frac {q \vec r }{r^2}$$

and seems to imply that this accounts for both terms in Ampere's (Maxwell-modified) law for the magnetic field in the case of a moving charge. Is my interpretation of this correct?

9. Apr 5, 2009

### clem

As often happens Wiki is just dead wrong. If you follow its Griffith footnote to the end, you will find "but this is simply $$\it wrong$$" about that equation for B.
An earlier post and even Wiki has the correct formula, but only for constant velocity.

10. Apr 6, 2009

### snoopies622

I imagine that the answer I'm looking for will be in the Franklin text, but since I can't get to the library for a few days I'll try to reformulate it here. Suppose I have a charged particle that is moving according to

$$x= A sin (\omega t) \hspace {10 mm} y=0 \hspace {10 mm} z=0$$

and I want to find out about the electromagnetic wave at point (3A, A, 0) at some specific future time. Since the electrical field around the charge is changing due to the charge's motion, the Maxwell portion of Ampere's law will contribute to the magnetic field there. My question is: is that the only thing that contributes to it, or must I superimpose another magnetic field on top of that one due to the "current" that is the single moving charge?

11. Apr 6, 2009

### Bob S

Two things:
1) If a uniformly moving charge is moving faster than c/n in a medium with index of refraction n, then it radiates (Cerenkov (Chrenkov) radiation). See Schiff "Quantum Mechanics" (2nd ed) pgs 267 - 271

2) If the charge is accelerating or decelerating, there is a radiation field. See Panofsky and Phillips "Classical E & M" Chap. 19 "Fields from an Accelerated Charge."

12. Apr 8, 2009

### snoopies622

Doh! I just took a good look at what the Lienard-Wiechart potentials are and realized that they completely answer my questions. Thanks all.