Do Accelerated Charges Always Radiate?

[shoestring]

Assume two charges, an electron and a proton, accelerate together. For example, let them start at (x,y)=(0,h) and (0,-h) and move in the +x direction along parallel trajectories (x(t),h) and (x(t),-h) while accelerating.

If they are far apart I assume they will each radiate on its own, but what happens if they are close, or even combine to form a hydrogen atom? When will they stop radiating?

Any neutral piece of materia made up of electrons, neutrons and protons consists of charges, so why doesn't a neutral piece of materia radiate under acceleration?

In short: When do accelerated charges radiate and when don't they?

 

[Fewmet]

Assume two charges, an electron and a proton, accelerate together. For example, let them start at (x,y)=(0,h) and (0,-h) and move in the +x direction along parallel trajectories (x(t),h) and (x(t),-h) while accelerating.

If they are far apart I assume they will each radiate on its own, but what happens if they are close, or even combine to form a hydrogen atom? When will they stop radiating?

Any neutral piece of materia made up of electrons, neutrons and protons consists of charges, so why doesn't a neutral piece of materia radiate under acceleration?

In short: When do accelerated charges radiate and when don't they?

This is not my strongest area, but I think it is that opposite charges, when accelerated, will produce oppositely directed magnetic fields, so there is no net field. I am eager, though, to see comments from someone with a stronger background.

 

[shoestring]

Thanks, that makes sense. What happens to the fields is perhaps more important than the acceleration of the charges itself.

A slightly different question: If a dipole is accelerated, will it radiate?

 

[Bill_K]

Yes, they will radiate. Although this is not obvious!

Let d be the position vector of the midpoint of the two particles, and h the vector from there to particle 1. The particles will be located at r₁ = d + h and r₂ = d - h. The dipole moment will be p = ∑ q_i r_i = e(d + h) - e(d - h) = 2eh. So what, you say? The point is that p is constant. You can't possibly get dipole radiation from the charges, since their dipole moment is constant.

Ok, now look at the quadrupole moment. Q = ∑ q_i r_ir_i = e(d + h)(d + h) - e(d - h)(d - h) = 2e(dh + hd). So what, you say? The point is that Q is *not* constant. Since you can choose d(t) to be anything you like, Q can also be made to vary in time in any way you like. So in general you can get quadrupole radiation. But there's a condition that must be imposed on d(t).

E & M books are so quick to Fourier transform everything, it's hard to find a radiation formula that still has t dependence in it. In Jackson, for example, the radiated power of an oscillating dipole is given as P = ck⁴/3 |p|². If he hadn't Fourier transformed it, this would have been P = c/3 |p⁽²⁾|², where ⁽²⁾ means the second time derivative. A few pages later, the power from an oscillating quadrupole is given as P = ck⁶/360 |Q|². Meaning P = c/360 |Q⁽³⁾|². In general, the radiation formula for the m^th multipole moment will have m+1 time derivatives on it.

The point here is that not just any time dependent p or Q will radiate. You need a p such that p⁽²⁾ is nonzero. Likewise you need a Q such that Q⁽³⁾ is nonzero. Solutions for which p ~ t or Q ~ t² do *not* radiate. These are called nonradiative motions.

In the present problem, if you use a constant acceleration d(t) = at²/2, then Q(t) ~ d(t) ~ t² is a nonradiative motion. You'll get quadrupole radiation if and only if the acceleration is not constant.

 

[cosmik debris]

Great post Bill.