What is the difference between cyclotron radiation and Larmor formula? On Wikipedia, it says 

Cyclotron radiation is electromagnetic radiation emitted by accelerating charged particles deflected by a magnetic field.

and 

The Larmor formula is used to calculate the total power radiated by a non-relativistic point charge as it accelerates. 

Looking only at these two sentences, the only difference seems to be that the description for cyclotron radiation specifically mentions that it is used for accelerating charged particles deflected by a magnetic field.


*

*What is the difference between those two formulas?

*When I'm looking at an accelerating charged particle in a magnetic field, is it more accurate to use cyclotron radiation formula?

*What does it mean by “total power” in the description for Larmor formula? What's the difference between just saying “power”?


Below is the formula for cyclotron radiation.
$$\frac{-dE}{dt}=\frac{\sigma_t B^2 v^2}{c \mu_0}$$
$\sigma_t$ is the total Thomson cross-section, $B$ is the magnetic field strength, $v$ is the charged particle's velocity perpendicular to the magnetic field, c is the speed of light, and $\mu_0$ is the permeability of free space. (Sorry for using words instead of proper notations. I'm not quite familiar with this place yet.)
 A: There is no difference between the formulas. The formula in the cyclotron radiation article is just the Larmor formula applied to the specific case of nonrelativistic charged particles moving in circles in a magnetic field.
For such particles, $F=ma$ combined with the Lorentz force law $F=qvB$ gives the acceleration as
$$a=\frac{qBv}{m}$$
so the Larmor formula
$$P=\frac{q^2a^2}{6\pi\epsilon_0c^3}$$
gives the radiated power as
$$P=\frac{q^2}{6\pi\epsilon_0c^3}\left(\frac{qBv}{m}\right)^2=\frac{q^4B^2v^2}{6\pi\epsilon_0m^2c^3}.$$
To see that this is the same as the cyclotron radiation formula,
$$-\frac{dE}{dt}=\frac{\sigma_tB^2v^2}{c\mu_0},$$
just substitute the formula for the Thomson cross section,
$$\sigma_t=\frac{8\pi}{3}\left(\frac{q^2}{4\pi\epsilon_0mc^2}\right)^2$$
and the formula for the vacuum permeability,
$$\mu_0=\frac{1}{\epsilon_0c^2}$$
to get
$$-\frac{dE}{dt}=\frac{8\pi}{3}\left(\frac{q^2}{4\pi\epsilon_0mc^2}\right)^2\frac{B^2v^2}{c}\epsilon_0c^2=\frac{q^4B^2v^2}{6\pi\epsilon_0m^2c^3}.$$
The total power $P$ given by the Larmor formula is the power radiated in all directions. The formula for the power per solid angle radiated in a specific direction is
$$\frac{dP}{d\Omega}=\frac{q^2a^2\sin^2\theta}{16\pi^2\epsilon_0c^3}.$$
Here $\theta$ is the angle between the acceleration vector and the direction in which the power is measured.
If you integrate this over all solid angles using $d\Omega=\sin\theta\,d\theta\,d\phi$, you'll get the Larmor formula for $P$ because
$$\int\sin^2\theta\,d\Omega=2\pi\int_0^\pi \sin^3\theta\,d\theta=2\pi\int_{-1}^1 (1-u^2)\,du=\frac{8\pi}{3}.$$
