Particles radiating energy when accelerating? Let us say we have a charged particle moving in the positive $X$ direction with velocity $v$. If we give the particle a nudge in the $-X$ direction causing it to decelerate. Then from the saying 'particles that accelerate, radiate' the particle will start to slow down as it radiates energy. Here is where I am confused. When will the particle stop accelerating in the $-X$ direction? The natural answer would be of course when it stops moving, but this gives an undue priority to the frame we are standing in, since in any other frame the particle would still be moving. The only other answer I can think of is that it doesn't and continues to accelerate in the $-X$ direction. This is of course nonsense as it would lead to a perpetual motion machine. So what would happen in this situation (i.e. when you cause a particle to accelerate briefly in the direction opposite to its motion)? 
 A: The question is: how do you "give the particle a nudge"?
The way you interact with electrically charged particles is to let them interact with photons: either real photons, as in the Compton effect, or the virtual photons of the electric and magnetic field.  When your particle interacts with your photon, the particle's momentum is changed and the photon's momentum is changed as well.  That scattered photon is your "accelerating" radiation.
A: 
Let us say we have a charged particle moving in the positive X direction with velocity v. If we give the particle a nudge in the −X direction causing it to decelerate. Then from the saying 'particles that accelerate, radiate'

Let us look at the classical description of this statement.
The classical formula for the radiated power from an accelerated electron is
$$P=\frac{2Ke^2}{3c^3}a^2$$
A "nudge" is a $\Delta t$ event, an impulse, so there is no continuous force ($F=m a$, Newtons second law) to give a continuous acceleration as you suppose. 

the particle will start to slow down as it radiates energy.

the electron will lose a $\Delta E$ from its energy , for deceleration as in your example, and continue motion with a lower momentum (Newton's first law)

Here is where I am confused. When will the particle stop accelerating in the −X direction?

At $t+\Delta t$

The natural answer would be of course when it stops moving,

If there is no new "nudge" it keeps on moving.
You are confusing acceleration with constant velocity . Energy is supplied by the "nudge" for a $\Delta t$ interval, energy is radiated by the formula above, and then the particle continues with constant velocity, lower or higher according to the direction of the impulse/nudge.
The quantum mechanical details are addressed in Rob's answer.
