# Curl of the Electric Field

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1. Dec 15, 2014

### ngendler

I am thinking about the curl of the electric field and want to make sure I have something straight:

Say you have a charged particle moving along some prescribed path. The electric field propagates outward at speed c, leading to a "retarded" time that you need to calculate in order to get the true electric field at any point.

My question: Of course, in this scenario, the curl of E will not be zero due to the changing magnetic field produced. But will there be an additional "curliness" due to including the time-delay of the field?

Thanks in advance(d time)! (har har)

2. Dec 15, 2014

### Simon Bridge

The "retarded " etc is just a calculation method - not cause and effect.

The instantaneous rate rate of change of the magnetic field tells you the curl of the electric field.

As a known charge travels on some path, then an observer someplace will observe a time varying electric field.
This means the observer will use the time rate of change of the electric field to work out the curl of the resulting magnetic field (we know the charge, remember).
But the whole thing is together at the same time part of the same phenomenon.

You may want to consider the imaginary case of a charge that suddenly shifts position a bit to see how it works out.
i.e. $x(t)=(h(t)-h(t-\delta t))vt : h(t)$ is the heaviside step.

3. Dec 15, 2014

### ngendler

That makes sense.

However, in the case of, say, a charge moving in a circle--taking into consideration the retarded time points part of the electric field in the direction of motion at any given moment. Doesn't this add a curl to the field?

4. Dec 15, 2014

### Simon Bridge

The "retarded" part of the calculation is not cause and effect - it's an artifact of the method chosen to do the calculation.
If you did not do that, then you'd have to chose another method to do the calculation or get a result that does not look much like reality.
The time delay does not cause the curl.

You don't like my example?
Try doing the math for a rotating dipole, with the observer a long way away.
What is the physical phenomena that gives the curliness?

5. Dec 15, 2014

### ngendler

It's not cause and effect, but it certainly changes what the field looks like at any given time...

Can you explain why my circling charge has a component of the field in the direction of motion?

6. Dec 15, 2014

### Simon Bridge

Because it is accelerating.

The method of calculation changes what numbers you get out the other end - it makes no difference to Nature.

7. Dec 15, 2014

### ngendler

Ah, okay! Thanks.

8. Dec 15, 2014

### ngendler

But my computer simulation doesn't know about the electric field generated by the magnetic field.... All it knows is that E=q \vec{r}/ r^3

9. Dec 15, 2014

### ngendler

Also, in any case the electric field must be different! Without the retarded time, the electric field around a charge that appears is there instantaneously--with it, there is no electric field for a short amount of time.

10. Dec 16, 2014

### vanhees71

I think the confusion is that you think the electric field (or part of it?) is "generated" by the magnetic field. This is a point of view you can sometimes read, but it's quite misleading, because it's complicating things. It leads to an apparently non-local description of electrodynamics, which in fact is the paradigmatic example for a local classical relativistic field theory.

The sources of the electromagnetic field (it's one field, which has electric and magnetic field components, and the split into electric and magnetic components is frame dependent) are the charge and current densities. Have a look at "Jefimenko's equation".