Contradiction between energy conservation and radiation damping force

We know that if charged particle is accelerated, it will radiate. According to Larmor formula, the power is $$P = {2 \over 3} \frac{q^2 a^2}{ c^3}$$ in the Gauss units, where $a$ is acceleration.

And accelerated particle will experience radiation damping force(Abraham–Lorentz force) $$\mathbf{F}_\mathrm{rad} = { 2 \over 3} \frac{ q^2}{ c^3} \mathbf{\dot{a}}$$

The equation of motion of particle should be $$m \frac{d \mathbf{v}}{dt}= \mathbf{F}_{ext}+ { 2 \over 3} \frac{ q^2}{ c^3} \mathbf{\dot{a}} \tag{1}$$

But a contradiction arises when the external force is a constant force. In this case, the particle has constant acceleration ($\mathbf{\dot a}=0$). But it will still radiate energy ($P\neq0$). The work done by external force $\mathbf{F}_{ext}$ will be totally translated to particle's kinetic energy(due to uniform acceleration), but the particle will still radiate energy to infinity. Does this violate the energy conservation?

Note1: For constant force, the uniform acceleration solution $\mathbf{\dot a}=0$ is still the new equation(1) 's solution. That is, $\dot a=0$ is a solution for $a=b$. Certainly this solution($\dot a=0$ ) can also be the solution of $a= b+d \dot a =b$.

Note2: Even though you take the relativistic effect into consideration, you still cannot explain this contradiction. The relativistic generalization of radiation damping force is in Landau's The classical theory of fields (76.2) $$F_{rad}^\mu= \frac{2 e^2}{3 c}(\frac{d^2 u^{\mu}}{ds^2}-(u^\mu u^\alpha)\frac{d^2 u_\alpha}{ds^2})$$ So we see with constant external force $F_{ext}^\mu=\text{cosntant}$, equation of motion, $$m\frac{du^{\mu}}{ds}=F^{\mu}_{ext}+F^{\mu}_{rad}\tag{2}$$

Constant $4$-acceleration $\frac{du^{\mu}}{ds}=\frac{F^{\mu}_{ext}}{m}$, $\frac{d^2 u^{\mu}}{ds^2}=0$ is still the equation(2)'s solution. So there is still no radiation damping force.

• Your differential equation becomes of the form $a=b+d\dot a$ where $b$ and $d$ are constants. It is a first order linear differential equation. Apr 30, 2017 at 5:16
• @Farcher But original solution is still new equation's solution.
– user153663
Apr 30, 2017 at 5:17
• Where did you get the idea that the acceleration is constant.? Apr 30, 2017 at 5:19
• @Farcher $\dot a=0$ is a solution for $a=b$. Certainly this solution can also be the solution of $a= b+d \dot a =b$
– user153663
Apr 30, 2017 at 5:20
• Apparently, this is a hard question: physics.stackexchange.com/q/70915 Apr 30, 2017 at 11:29

1 Answer

The equation (1) is only an approximate equation of motion of a charged body of non-zero dimensions, because the expression

$$\frac{2}{3}\frac{q^2}{c^3}\dot{\mathbf a}$$

only approximates total internal EM force that is acting on such body due to non-rectilinear motion of its parts. Because of this approximation, this equation does not reflect the law of conservation of energy, it only gives approximate description of motion of the body as a whole.

The exact expression for net internal force would be much more complicated, it would involve internal degrees of freedom of the charged body and interaction of its parts. For highly simple models of such charged body (like uniformly charged sphere), the net force can be expressed as infinite series and the above term is just one term of this series.

In an exact model of the charged body where conservation of energy is present and can be proven, the energy that the system radiates is balanced by decrease of internal energy inside and near the charged body. That includes the EM energy in the vicinity of the body.

But if the internal degrees and dimensions of the body are neglected so it becomes a point, this internal energy becomes invisible and the remaining visible degrees of freedom manifest violation of energy conservation. It is just because of simplifying the model of the extended charged body so much, it is not an indication there is something wrong with the general theory.

See also my answer here: Does a constantly accelerating charged particle emit EM radiation or not?