Electric Field of Moving Charge

[chickenz]

The electric field of a charge moving with uniform velocity points towards the charge's present location. You can either get that from a whole lotta math on the vector and scalar potentials. Or more easily from special relativity.

Is there a quick, minimally mathematical, and intuitive explanation for this otherwise miraculous coincidence?

 

[Dale]

Hi chickenz, welcome to PF,

The easy explanation is that Maxwell's equations are invariant under the Lorentz transform, so you can transform to a nice coordinate system, solve it there, and then transform back to your original system. The fact that choosing a nice coordinate system simplifies computations should not be a surprise.

 

[chickenz]

Hi chickenz, welcome to PF,

The easy explanation is that Maxwell's equations are invariant under the Lorentz transform, so you can transform to a nice coordinate system, solve it there, and then transform back to your original system. The fact that choosing a nice coordinate system simplifies computations should not be a surprise.

Oh, that wasn't what I was asking about. Sorry. Let me rephrase the question:

Is the fact that the electric field of a moving charge points to the charge's current location immediately obvious from fundamental principles, or does it just follow from the math?

 

[Dale]

I don't understand the distinction. The fundamental principles are Maxwells equations, which are mathematical.

 

[RedX]

The electric field of a charge moving with uniform velocity points towards the charge's present location. You can either get that from a whole lotta math on the vector and scalar potentials. Or more easily from special relativity.

Is there a quick, minimally mathematical, and intuitive explanation for this otherwise miraculous coincidence?

Basically the argument is like this: you can never know what's going on with the charge now, only how it was doing at a time=(distance/speed of light) ago. So say you are at x=3x10⁸, and you receive information at t=0 that the charge is at x=0 and going at speed v=1. Well that information was sent at t=-1, so that's old information, since right now is t=0. So where do you think the particle is now, at t=0? It's natural to think the particle is at x=1, which would be the case if there were no acceleration. If it did accelerate between t=-1 and t=0, then you wouldn't know it. But you have to define a value for the field now at t=0, because you just can't not have a field there just because you're unsure. So it's most natural to say that field at t=0 is as if the particle were at x=1 moving with velocity v=1. You have to choose a value of the field, as the electric field is not an indefinite quantity. And based on all available information, the field should be that created by the charge at x=1 with velocity v=1.

 

[cosmik debris]

The electric field of a charge moving with uniform velocity points towards the charge's present location. You can either get that from a whole lotta math on the vector and scalar potentials. Or more easily from special relativity.

Is there a quick, minimally mathematical, and intuitive explanation for this otherwise miraculous coincidence?

Still mathematical but I think this explanation is quite good. Note at the bottom he makes a comment about why the "force" of gravity on the Earth points at the present postion of the sun, this is for the same reason as your charge problem.

http://www.mathpages.com/home/kmath249/kmath249.htm"

 

[chickenz]

Ah. I think my question is still being misinterpreted. Here's my question, rephrased again:

The electric field of a uniformly moving charge points at the charge's present location, not to the retarded position. Is this obvious from Maxwell's Equations or Special Relativity? (And by obvious, I mean immediately clear from only a few lines of simple math)

 

[protazoa]

the intensity of an electric field is determined by the charge of the emitting particle- which I will presume is constant for your stated moving particle- and the distance away from said particle.

Since the distance from a point is inevitable linked to the points point location, this is aptly explained by Coulombs Law

Electric field is equal to:charge of the particle
___________________________________
(4*pi*[electric constant])*distance^2

 

[chickenz]

the intensity of an electric field is determined by the charge of the emitting particle- which I will presume is constant for your stated moving particle- and the distance away from said particle.

Since the distance from a point is inevitable linked to the points point location, this is aptly explained by Coulombs Law

Electric field is equal to:


charge of the particle
___________________________________
(4*pi*[electric constant])*distance^2

I'm afraid Coulomb's Law is wrong in the context of moving charges.

 

[Dale]

Ah. I think my question is still being misinterpreted. Here's my question, rephrased again:

The electric field of a uniformly moving charge points at the charge's present location, not to the retarded position. Is this obvious from Maxwell's Equations or Special Relativity? (And by obvious, I mean immediately clear from only a few lines of simple math)

Well, I guess it depends on if "simple math" includes four-vectors and boosts. If so, then I think that you can do it as follows.

Start in the rest frame of the central charge, the test charge is moving with some velocity, -v. Use Coulomb's law and the Lorentz force law in that frame to determine the four-force on the test charge. Calculate the displacement four-vector. Note that they are parallel meaning that the force points to where the charge is. Boost to the frame where the test charge is stationary and the central charge is moving with velocity v. Note that the four-force remains parallel to the displacement four-vector.

 

[chickenz]

Well, I guess it depends on if "simple math" includes four-vectors and boosts. If so, then I think that you can do it as follows.

Start in the rest frame of the central charge, the test charge is moving with some velocity, -v. Use Coulomb's law and the Lorentz force law in that frame to determine the four-force on the test charge. Calculate the displacement four-vector. Note that they are parallel meaning that the force points to where the charge is. Boost to the frame where the test charge is stationary and the central charge is moving with velocity v. Note that the four-force remains parallel to the displacement four-vector.

Yes, that's the type of explanation that I was looking for.

Unfortunately, I don't believe your argument. Let O be the origin of our coordinate system, and let A be the event where we measure the force on the test charge. O and A are simultaneous in the rest frame of the central charge.
1. The four-force and four-displacement are not parallel. In the rest frame of the central charge, the four-force has a nonzero timelike component, whereas the four-displacement has a zero timelike component.
2. When we do the boost into the rest frame of the test charge, the four-displacement gains a nonzero timelike component (and the four-force's timelike component becomes zero). That means O and A are no longer simultaneous. But we wanted to prove that the Electric Field points to the charge's current location, i.e., we want to measure the force when the central charge crosses the origin.

Nevertheless, thank you. I hadn't thought about four-vectors. If I choose to think about this further, I'll try to come up with an argument for why the Electric Field points to the charge's present location using four-vectors.

 

[Dale]

1. The four-force and four-displacement are not parallel. In the rest frame of the central charge, the four-force has a nonzero timelike component, whereas the four-displacement has a zero timelike component.

D'oh! you are right. I was trying to be too clever.

 

[Dale]

Try this. Start in the rest frame of the central charge. Use Coulomb's law to calculate the E-field in that frame at every point along the path of a test charge moving with velocity -v. There is no B-field so the force is in the direction of the E-field. Boost into the rest frame of the test charge and use the relativistic transformation laws in Einstein's original OEMB paper to calculate the E-field in that frame at every point along the path. The test charge is not moving so the force is again in the direction of the E-field.

I think that the only complication will be that you will need to carefully parameterize the test charge's path by the proper time.

 

[chickenz]

Try this. Start in the rest frame of the central charge. Use Coulomb's law to calculate the E-field in that frame at every point along the path of a test charge moving with velocity -v. There is no B-field so the force is in the direction of the E-field. Boost into the rest frame of the test charge and use the relativistic transformation laws in Einstein's original OEMB paper to calculate the E-field in that frame at every point along the path. The test charge is not moving so the force is again in the direction of the E-field.

I think that the only complication will be that you will need to carefully parameterize the test charge's path by the proper time.

Hmmm... This sounds like it could work. It's more mathematical than I would have hoped, but parametrizing worldlines is something that I ought to try, and maybe I'll gain some insight from it. Thanks for your help.