Nope.

Redditor /u/Mimshot gave the following example:

If an observer is near the path of a small, moving charged particle (unless there’s some special quantum effect I’d love you to tell me about if it exists) the observer will see the

Efield increase and then decrease and will see theBfield ramp from baseline, then reverse direction, which is certainly wave-like. I’m not saying it radiates photons, but I’m wondering if “no, it must be accelerating” is a complete answer.Is there some quantum effect I’m missing?

I “know” immediately there is no radiation in this case, because the theory of relativity tells us we can use a frame of reference in which the particle is stationary. Hence, as a rule, only accelerating particles radiate and thus give rise to traveling waves. Nevertheless, this question did get me to think about what the fields would be like in such a situation. A passing electron would seem to have some time dependent magnetic fields because the “ramp” explanation above, but it cannot be the case since we should know, just “because”, only accelerating charges radiate.

After some thought I came up with the following proof that the magnetic field is static in this case.

Start here

**J**(**r**,t) = ρ(**r**,t)**v**(**r**,t) = *e* δ(**r – r’**,t)**v**(**r – r’**)

**v** has no time dependence.

The current *I* is ∫ **J** d^{2} **x’**

**I** = ∫ d^{2} **x’** *e* δ(**r – r’**,t)**v**(**r – r’**) = *e* **v** = a constant

To find **B** we use ampere’s law for some closed loop

∫ **B** d**x** = μ **I** = constant

If you’re concerned about the ∂**E**/∂t term lets look at the full maxwell equation

**∇** x **B** = μ **J** + μ ε ∂**E**/∂t

Applying the operation ∫ d^{2} **x** to both sides gives

∫ **B** d**x** = μ **I** + μ ε ∂/∂t ( ∫ d^{2} **x** **E** )

The RHS of the above equation is simpified using gauss law, the integral gives the charge enclosed by a surface

∫ d^{2} **x** **E** = q/ε

so

∫ **B** d**x** = μ **I** + μ ε ∂/∂t ( q/ε )

but ∂/∂t ( q ) = 0

so that term doesn’t change things.

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