How to solve Relativistic Lorentz Force equation if $\gamma$ is not constant?

Question

I am trying to numerically obtain the trajectory of an electron inside a periodic magnetic field $\mathbf{B}$, taking into account that the relativistic factor $\gamma$ is not a constant (the electron loses energy due to radiation). The energy loss is related to the instantaneous power radiated, given by the Liénard formula: $$ \frac{d\gamma}{dt}=\frac{1}{mc^2}\frac{dE}{dt}=\frac{e^2\gamma^6}{6\pi\varepsilon_0mc^3}\big(\dot{\mathbf{\beta}}^2-(\mathbf{\beta}\times\dot{\mathbf{\beta}})^2\big),\tag{1} $$ where $\beta=\mathbf{v}/c$ and $\dot{\beta}=\mathbf{a}/c$ ($\mathbf{v}$ and $\mathbf{a}$ are the usual velocity and acceleration vectors of the electron). Note that $d\gamma/dt$ is a positive quantity:

$$ (\mathbf{\beta}\times\dot{\mathbf{\beta}})^2=\Vert\mathbf{\beta}\times\dot{\mathbf{\beta}}\Vert^2=\Vert\beta\Vert^2\Vert\dot{\beta}\Vert^2\sin^2\theta, $$ where $\theta$ is the angle between the vectors $\beta$ and $\dot{\beta}$. Therefore,

$$ \dot{\mathbf{\beta}}^2-(\mathbf{\beta}\times\dot{\mathbf{\beta}})^2=\Vert\dot{\mathbf{\beta}}\Vert^2-\Vert\mathbf{\beta}\times\dot{\mathbf{\beta}}\Vert^2=\Vert\dot{\mathbf{\beta}}\Vert^2(1-\Vert\beta\Vert^2\sin^2\theta)\geq0, $$ where we have used that $\Vert\beta\Vert^2\leq1$ and $\sin^2\theta\leq1$. However, the electron is losing energy, so $d\gamma/dt$ should be negative. Thus, we shall write

$$ \frac{d\gamma}{dt}=-\frac{e^2\gamma^6}{6\pi\varepsilon_0mc^3}\big(\dot{\mathbf{\beta}}^2-(\mathbf{\beta}\times\dot{\mathbf{\beta}})^2\big). $$

After some research, I have come across Abraham-Lorentz force (see this Wikipedia article, for reference). This force accounts for the

"reaction force on an accelerating charged particle caused by the particle emitting electromagnetic radiation by self-interaction".

which I believe should be also included in the $\mathbf{F}=m\mathbf{a}$ equation if we consider that radiation is taking place. The expression for this reactive force, $\mathbf{F}_{\text{rad}}$, reads:

$$ \mathbf{F}_{\text{rad}}=\frac{\mu_0e^2}{6\pi c}\bigg(\gamma^2\dot{\mathbf{a}}+\frac{\gamma^4\mathbf{v}(\mathbf{v}\cdot\dot{\mathbf{a}})}{c^2}+\frac{3\gamma^4\mathbf{a}(\mathbf{v}\cdot\mathbf{a})}{c^2}+\frac{3\gamma^6\mathbf{v}(\mathbf{v}\cdot\mathbf{a})^2}{c^4}\bigg). $$

The magnetic field where the electron is moving is given by

$$ B_x=\frac{B_0k_x}{k_y}\sin{(k_{x}x)}\sinh{(k_{y}y)}\cos{(k_{z}z)} $$ $$ B_y=-B_0\cos{(k_{x}x)}\cosh{(k_{y}y)}\cos{(k_{z}z)} $$ $$ B_z=\frac{B_0k_z}{k_y}\cos{(k_{x}x)}\sinh{(k_{y}y)}\sin{(k_{z}z)}, $$ where $k_x$, $k_y$ and $k_z$ are just some constants.

Therefore, Lorentz Force equation in its relativistic formulation reads: $$ \frac{d}{dt}(\gamma m \mathbf{v})=-e(\mathbf{v}\times\mathbf{B})+\mathbf{F}_{\text{rad}} $$ $$ m\mathbf{v}\frac{d\gamma}{dt}+m\gamma\mathbf{a}=-e(\mathbf{v}\times\mathbf{B})+\mathbf{F}_{\text{rad}}.\tag{2} $$

I have tried solving the ODE system that stems from this last equation (where the variables are the electron's position, velocity, acceleration and $\gamma$). Derivatives of the acceleration are directly obtained from the $\mathbf{F}_{\text{rad}}$ expression.

The results I obtain are not consistent, since the $v_z$ component of the velocity increases as the electron moves through $\mathbf{B}$, and eventually becomes $v_z>c$, which makes no physical sense. I think I may be overlooking some relevant aspect in all this, since the electron dynamics I obtain make sense except for this $v_z>c$ issue.

I would highly appreciate any insight on this topic.

Answer 1

The power of the ALD force and the Liénard radiation formula are two separate quantities, they need not be equal. If you are interested in the time derivative of $\gamma$, it is the former that is relevant due to the relativistic work-energy theorem. In general, they differ by a temporal boundary term and you don't need to prescribe the extra minus sign by hand. From the power of the ALD, you can find: $$ \begin{align} mc^2\dot \gamma &= \frac{\mu_0e^2}{6\pi c}\gamma^2\left(\ddot \beta +\gamma^2(\ddot\beta\cdot\beta)\beta +3\gamma^2(\beta\cdot\dot\beta)\dot\beta+3\gamma^4(\beta\cdot\dot\beta)^2\beta\right)\cdot\beta \tag{1} \\ &= \frac{\mu_0e^2}{6\pi c}\gamma^4\left((\beta\cdot \ddot\beta) +3\gamma^2(\beta\cdot\dot\beta)^2\right) \\ &= \frac{\mu_0e^2}{6\pi c}\left(-\gamma^4\dot\beta{}^2-\gamma^6(\beta\cdot\dot\beta)^2+\frac{d}{dt}[\gamma^4(\beta\cdot\dot \beta)]\right) \\ &= -\frac{\mu_0e^2}{6\pi c}\gamma^6\left((1-\beta^2)\dot\beta^2+(\beta\cdot\dot\beta)^2\right)+\frac{\mu_0e^2}{12\pi c}\frac{d^2}{dt^2}\gamma^2 \\ &= -\frac{\mu_0e^2}{6\pi c}\gamma^6\left(\dot\beta^2-(\beta\times\dot\beta)^2\right)+\frac{\mu_0e^2}{12\pi c}\frac{d^2}{dt^2}\gamma^2 \end{align} $$ using: $$ \gamma = \frac1{\sqrt{1-\beta^2}} \quad \dot\gamma = \gamma^3(\beta\cdot\dot\beta) \quad \beta^2\dot\beta{}^2 = (\beta\times\dot\beta)^2+(\beta\cdot\dot\beta)^2 $$ You can recognise the first term as the Liénard formula, but the second term is the advertised boundary term, which is typically not zero. Notice that the minus signs appears from the math, no need to impose it. For periodic trajectories, the boundary term will cancel in time averages, but in general it will be important.

In your example, since the magnetic force does not work, the previous formula still applies. Back to your question, there is quite a lot of literature about the runaway solutions due to ALD force. It is therefore hardly surprising that for generic initial conditions, you obtain FTL solutions which do not make sense physically.

Answer 2

As stated in answers (see Ján Lalinský answer) to other similar questions the way to go is to evaluate the ALD force on the trajectory without the radiation particle reaction.

Following this idea,the ALD force: \begin{equation} \label{reactiveforce} \mathbf{F_\text{rad}}=\frac{\mu_0e^2\gamma^2}{6\pi}\bigg(\mathbf{\ddot{\beta}}+\gamma^2\mathbf{\beta}(\mathbf{\beta}\cdot\mathbf{\ddot{\beta}})+3\gamma^2\mathbf{\dot{\beta}}(\mathbf{\beta}\cdot\mathbf{\dot{\beta}})+3\gamma^4\mathbf{\beta}(\mathbf{\beta}\cdot\mathbf{\dot{\beta}})^2\bigg), \end{equation}

can be evaluated using the equations of motion without radiation emission:

\begin{equation} \label{simplelorentz} \begin{aligned} \dot{\mathbf{\beta}}&=-\frac{e}{m\gamma}(\mathbf{\beta}\times\mathbf{B})\\ \mathbf{\ddot{\beta}}&=-\frac{e}{m\gamma}\mathbf{\dot{\beta}}\times\mathbf{B}\\ \mathbf{\beta}\cdot\mathbf{\dot{\beta}}&=0 \end{aligned} \end{equation}

Using The previous approximations will simplify $\mathbf{F_\text{rad}}$ to a point that it only depends on $\mathbf{\beta}$ and the magnetic field $\mathbf{B}$ which makes the ODE system much simpler and well behaved.

I think there is no such inconsistency in the EM theory. The problem is that we are using the Lienard formula (from which the ALD is derived using the relativistic work-energy theorem) which is an approximation. It is derived using Maxwell's equation assuming some field sources (the charge and its movement) that are not affected by the electromagnetic field. One should re-derive it without that assumption and solve the equations of motion together with Maxwell's equations to obtain the exact solution. I guess this is only really important in high magnetic field cases like neutron stars or something like that.