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In the classical formulation of electrodynamics given an initial electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ a particle moving through this field experiences a force $\mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B})$.

  1. If I understand correctly when the particle accelerates its movement itself affects the magnetic and electric field $\mathbf{B}$ and $\mathbf{E}$ through electromagnetic radiation. So the movement of the particle is affected by $\mathbf{B}$ and $\mathbf{E}$ and the movement of the particle is affecting the field $\mathbf{B}$ and $\mathbf{E}$. Is my understanding correct here?

  2. If so, then when looking at the Lagrangian formulation of electrodynamics it seemed to me that the scalar and vector potential $\phi$ and $\mathbf{A}$ are specified off the bat and then the motion of the particle is computed as if this motion did not affect $\phi$ and $\mathbf{A}$ ? For instance this is the way things are presented in 2.2 of "Principles of Quantum mechanics" by Shankar.

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Your understanding of 1. is a correct summary of radiation-reaction.

For 2, you are correct that if we use ONLY $L = \frac{1}{2} mv^2 -q\phi +q/c \vec{v}\cdot \vec{A},$ then we are assuming that $\vec{A}$ and $\phi$ are fixed and that the motion of the particle doesn't cause $\vec{E}$ and $\vec{B}$ to change.

However, the complete electromagnetic Lagrangian with dynamical fields will be $L_{kin} + \int d^3r \left[\frac{1}{2}(E^2-B^2) + \vec{J}\cdot{A} - \rho \phi \right]$ where $L_{kin}$ describes the kinetic term for the particle of interest, and can be chosen to be either relativistic or non-relativistic, such as $L_{kin}=-mc/\gamma$ or $L_{kin}=\frac{1}{2}mv^2$ for the case of a single particle.

To find the equations of motion of the electromagnetic fields (i.e. Maxwell's equations), you will vary the potentials $\phi(\vec{r},t)$ and $\vec{A}(\vec{r},t)$ to find the stationary action, and to find the motion of the particle (the Lorentz force), you will vary $x(t)$. For the latter case, note that $x(t)$ is implicit in $\rho$ and $\vec{J}$ as $\rho(\vec{r}, t) = q \delta(\vec{r}-\vec{x}(t))$ and $\vec{J}(\vec{r}, t) = q \dot{\vec{x}} \delta(\vec{r}-\vec{x}(t))$. You can in principle find the effects of the particle on the field by simultaneously solving Maxwell's equations and the Lorentz force equation, but all of the coupled equations and infinities make it a still-unsolved problem. Thus, it's most natural in beginning problems to only consider the effect of some external fields on the particle, not how the particle also contributes to the fields with associated backreaction.

Thus, the answer is that the Lagrangian considered in Shankar in 2.2 only captures the dynamics of the particle and assumes external fields; i.e. fields are not affected by the motion of the particle. Similar but not quite the same things happen when you solve the hydrogen atom in introductory QM; the full complexity of Maxwell's equations is avoided.

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  • $\begingroup$ Thank you very much! $\endgroup$ – blaber Nov 2 '20 at 6:17
  • $\begingroup$ Is there a method by which you can vary $\vec{E}$ and $\vec{B}$ to get Maxwell's equations? The standard method involves varying $\phi$ and $\vec{A}$ instead. $\endgroup$ – Michael Seifert Nov 2 '20 at 14:09
  • $\begingroup$ @MichaelSeifert This is a new question, but no there is not. This is one reason to consider $(\phi , \vec A )$ the physical fields of electromagnetism. $\endgroup$ – my2cts Nov 2 '20 at 18:35
  • $\begingroup$ @my2cts: I didn't think so, but I wanted to make sure there wasn't some alternate method I was unaware of. I've edited the answer to reflect this, but the OP can roll it back if they see fit. $\endgroup$ – Michael Seifert Nov 2 '20 at 18:51
  • $\begingroup$ @MichaelSeifert Good catch! Thanks! $\endgroup$ – user196574 Nov 2 '20 at 20:02
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If I understand correctly when the particle accelerates its movement itself affects the magnetic and electric field $\mathbf{B}$ and $\mathbf{E}$ through electromagnetic radiation. So the movement of the particle is affected by $\mathbf{B}$ and $\mathbf{E}$ and the movement of the particle is affecting the field $\mathbf{B}$ and $\mathbf{E}$. Is my understanding correct here?

Yeah, you are correct. When particle accelerates, it radiate electromagnetic radiation which will affect the fields already present in region. The potential is given by so called Liénard–Wiechert potential.The Liénard–Wiechert potentials $\phi$ (scalar potential field) and $A$ (vector potential field) are for a source point charge $q$ at position $\mathbf{r}_s$ traveling with velocity $\mathbf{v}_s$. $$ \phi(\mathbf{r},t)=k\left(\frac{q}{(1-\mathbf{n}_s\cdot\mathbf{\beta}_s)|\mathbf{r}-\mathbf{r}_s|}\right)$$ and $$\mathbf{A}(\mathbf{r},t)=\frac{\mathbf{\beta}_s(t_r)}{c}\phi(\mathbf{r},t)$$

where $\mathbf{\beta_s(t_r)}$ is the velocity of the source expressed as a fraction of the speed of light. Now for small velocity, the force of recoil due to electromagnetic radiation is given by Abraham–Lorentz force:

$$\mathbf{F}_{rad}=\frac{q^2}{6\pi\epsilon_0 c^3}\dot{\mathbf{a}}$$

You can see the effect is very small due to a factor of $c^3$ on the denominator (please go though the detail criteria for it's applicability). That's the reason we don't take it into account. But under specific circumstances, it may affect the motion,in that cases, we take it into account. But at elementary level of physics courses, We don't care.

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  • $\begingroup$ Perhaps you should mention that self acceleration is an unsolved problem in classical electrodynamics. $\endgroup$ – my2cts Nov 2 '20 at 18:37

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