# I Generation of EM radiation

1. Jun 7, 2017

### Pushoam

A clump of charged particles moving in circular motion emits em radiation.
But if I go on increasing no. of charged particles till the time the system becomes current in circular path, then since each particle is having same acceleration and velocity, the current will be uniform. So, now the system becomes uniform loop of current.

Now the electromagnetic field created by uniform loop of current doesn't change w.r.t. time, so this system won't emit em radiation.

But,

if I consider loop of current system as ,say, 5 clumps of charged particles, then each clump emits em radiation.
Since the net radiation is zero, does it mean that radiation created by clumps get canceled among themselves?
Each charged particle in the loop emits em radiation.
Radiation emitted by one particle gets canceled by radiation emitted by another particles using superposition principle. Is this correct?

But,
how to understand this thing in photon picture?
Each charged particle emits em radiation.⇒ Each charged particle emits photons.
Photons emitted by one particle doesn't get canceled by photons emitted by another particles .
Since the net radiation is zero, total no. of photons is also 0.
How to understand this?

2. Jun 7, 2017

### vanhees71

Better get first an understanding in terms of classical physics. Photons are a complicated subject and unnecessary to understand what's going on here.

It's in fact quite simple. If you have a single point particle running in a circle, the current density obviously is
$$\rho(t,\vec{x})=Q \delta^{(3)}[\vec{x}-\vec{y}(t)],$$
where
$$\vec{y}=R \cos \omega t \vec{e}_x + R \sin \omega t \vec{e}_y$$
is the trajectory on a circle. The current density is
$$\vec{j}(t,\vec{x})=Q \dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)].$$
Obviously, it's time dependent and you get a radiation field via the retarded Lienard-Wiechert solutions.

For a homogeneous line of charge running in a circle you have [CORRECTED!!!]
$$\rho(t,\vec{x})=\lambda \delta(\rho-R) \delta(z),$$
where $\lambda$ is the charge per unit length and
$$\vec{j}(t,\vec{x})=\vec{v} \rho(t,\vec{x})=R \omega \lambda \vec{e}_{\varphi} \delta(\rho-R) \delta(z),$$
where $(\rho,\varphi,z)$ are the usual cylinder coordinates. Here $\rho$ and $\vec{j}$ are time-independent, so that the retarded Lienard-Wiechert solutions lead to the usual electro- and magnetostatic solutions.

Last edited: Jun 14, 2017
3. Jun 7, 2017

### Pushoam

The loop could be accelerated ,too. So, ω will depend on time and so is current density.

4. Jun 8, 2017

### vanhees71

According to #1 the OP considers the case of a stationary current in the limit of the continuous charge distribution. So I assumed a stationary current for this case in my argument. Of course, if the current is time-dependent, you get a radiation field again.

5. Jun 13, 2017

### Pushoam

Integrating ρ(t,⃗ x) to get total charge,
$\int_V \frac λ R \delta (r-R) r^2 sinθ dr dθ dΦ = \int_{r=0}^ \inf \frac λ R \delta(r-R)r^2 dr \int_{θ=0}^{π}sinθ dθ\int_{Φ=0}^{2π} dΦ = 4πλ R$ instead of 2πλ R
Is this correct?

6. Jun 14, 2017

### vanhees71

First I have to correct myself (also corrected it in my original posting #2). The correct charge density is, of course,
$$\rho=\lambda \delta(\rho-R) \delta(z).$$
Further, we work in cylinder not spherical coordinates, i.e., we have
$$\mathrm{d}^3 \vec{x}=\mathrm{d} \rho \mathrm{d} \varphi \mathrm{d} z \rho$$
and thus
$$Q=\int_0^{\infty} \mathrm{d} \rho \int_0^{2 \pi} \mathrm{d} \varphi \int_{-\infty}^{\infty} \mathrm{d} z \rho \lambda \delta(\rho-R) \delta(z)=2 \pi R \lambda,$$
as it must be, because $2 \pi R$ is the length of the circle and thus $\lambda=Q/(2 \pi R)$.