In classical electro dynamics, the Lorentz force is $qE + qv\times{}B$ where $E$ is electric field, $B$ magnetic field, $v$ velocity, and $q$ is charge. So, the force when $E=0$ is $qv\times{}B$ which is orthogonal to the velocity, and so cannot change the speed of the electron. The electron, under this force goes around in circles. To explain the spiral path of the electron, Lamor radiation formula is used, radiating power proportional to $a^2$ where $a$ is acceleration. This is effectively another force on the electron, tangential to the path. Call it the Abraham force. The electron slows under the Abraham force and spirals in under the Lorentz force.

In general relativity, the electromagnetic stress energy (or Maxwell) tensor is the spacetime extension of the classical electromagnetic stress tensor as was derived by Maxwell from effectively the Lorentz force. So, in general relativity, using the standard electromagnetic stress energy tensor, does the electron go in circles or spirals in a pure magnetic field?

The crucial part of this question is whether there is a braking action of a magnetic field on an electron in some reasonable reference frame. I asked the question looking for general principles - but would be happy with a specific case such as the path of an electron, or even the dynamic equations for the path, in a specific reference frame in a specific spacetime such as Bertotti-Robinson.

I asked because I heard someone say that in GR the dynamics of the electron were determined by the stress energy tensor, making it distinct from CEM and SR. The answers so far lead me to believe that there is no definitive known answer. If so, a reference to any serious attempt would answer my question.

  • $\begingroup$ What magnetic field are we assuming? For a homogeneous magnetic field, the associated spacetime is the Bertoti-Robinson spacetime (aka the Levi-Civitta spacetime) $\endgroup$ – Slereah Jul 17 '18 at 7:11
  • $\begingroup$ I would expect that an answer to what is the path of an electron in the Bertotti-Robinson spacetime would be fine. I will edit the question to reflect this point. Thanks. $\endgroup$ – Ponder Stibbons Jul 17 '18 at 11:07
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    $\begingroup$ I feel that you feel that this question is not going anywhere, so I have asked the question in the negative sense explicitly as an alternative [physics.stackexchange.com/questions/418054/… $\endgroup$ – Ponder Stibbons Jul 17 '18 at 11:53

So, in general relativity, using the standard electromagnetic stress energy tensor, does the electron go in circles or spirals in a pure magnetic field?

General relativity reduces to special relativity in the limit of scales small enough so that curvature doesn't matter. This is one way of stating the equivalence principle. Therefore switching from SR to GR doesn't change the answer to the question, provided that the motion is on a small enough scale in spacetime.

  • $\begingroup$ Thanks. However, the distinction between SR and GR was not really the question. Using the Lorentz force the electron goes in circles, but in a cyclotron it goes in spirals. So, in SR, using the standard electromagnetic stress energy tensor - is it circles or spirals? To get spirals do we need to explicitly include radiation as in classical mechanics? $\endgroup$ – Ponder Stibbons Jul 17 '18 at 3:11
  • $\begingroup$ @PonderStibbons please be specific about which part of physics your question is about. Among the tags I see "general-relativity", and GR is also mentioned in the original question. Thus this answer explains why general relativity effects are not important for small electromagnetic fields. Was your question really about special relativity instead? $\endgroup$ – Prof. Legolasov Jul 17 '18 at 8:05
  • $\begingroup$ My question is about GR. If the answer is that it behaves, in this respect, like SR - that is fine. But, saying that does not answer the GR question. And I would prefer an answer that uses the mathematical machinery of GR. An answer in terms of SR might be illuminating. But, I think I understand the SR answers reasonably well, and I am wondering if GR has anything new to add. If the answer to that is "no" with explanation, then I will consider the question answered. $\endgroup$ – Ponder Stibbons Jul 17 '18 at 11:00
  • $\begingroup$ After thinking, I upvoted this because I think it was actually relevant. But, it also does not answer the question for the following reason. Imagine that in SR electrons go round in precise circles. But, in GR, when the curvature is taken into account, the electron goes in a spiral. As the limit is taken from GR to SR, the spiral could become tighter until it becomes a circle. So, the fact that GR limits to SR, and SR electrons go around in circles, does not imply that GR electrons go around in circles. $\endgroup$ – Ponder Stibbons Jul 23 '18 at 13:04

After some hesitation and having checked the Stack Exchange policy as indicated on Meta Stack Exchange, I will attempt to answer my own question and hope to clarify what I was asking. Putting this together did help my own appreciation for the topic. Perhaps it might be of interest to others. It goes without saying that if I have made an error then I trust that someone will point it out to me.

My answer does perhaps involve a re-framing: Does General Relativity add any clarification to the dynamics of the electron, in particular pertaining to the radiation reaction?

In the following, I have left out some details, including some physical constants.

The direct answer is in the negative. The analysis of the dynamics of the electron does not particularly depend, other than on some technical details, on whether the background mechanics is classical electrodynamics, special relativity, or general relativity. More over, whether the path can actually be determined depends on decisions about the properties of the electron, decisions that are outside the scope of every one of these theories as given.

In a space-mode classical approach, the power associated with charged particle behaviour is determined by the power associated with the Coulomb $qE$ and Lorentz $qv\times{}B$ forces as well as the radiation power $\nabla\cdot{}S$, where $E$ is the electric field, $B$ is the magnetic field, $q$ is the charge on the particle, $v$ is the particle velocity, and $S=E\times{}B$ is the electromagnetic energy flux vector. The magnetic field vector $B$ can be replaced by an anti symmetric matrix making the Maxwell equations and the Lorentz force dimension agnostic. The practical difficulty in determining the actual power absorbed or emitted by the electron lies in the difficulty of solving the field equations.

The spacetime-modes of classical electromagnetic and special relativistic mechanics are essentially the same as each other. The velocity and charge of the particle are replaced by a 4-vector $J=(q,qv)$, and the electric and magnetic fields are packed into a 4-tensor.

$F = \left[\matrix{0 & E \cr E & -B\times}\right]$

The product $FJ = (E\cdot qv,\ qE + qv\times B)$, a force-power vector, which is the electromagnetic power and the Lorentz force. This is the rate of change of energy with time, and with space. The stress-energy tensor packs in the stress associated with the Lorentz force and the energy density, as well as the flux:

$T = \left[\matrix{ E^2+B^2 & E\times B \cr E \times B & E \otimes E + B \otimes B}\right] - \frac{1}{2}(E^2+B^2)I $

It follows that $\nabla\cdot T = FJ$, that is, the gradient of the stress tensor is a power-force. The temporal component of this is clearly the Poynting theorem $\nabla\cdot S + E\cdot J - \frac{\partial}{\partial{}t}e$, which is zero from the Maxwell equations on which this is all based.

So, the divergence of the stress-energy tensor is a force-power tensor. Hence, it does state both how much force is on the electron and how much power is being emitted - that is, the rate of change of energy of the particle with time and the rate of change of energy with position.

But, this is for virtual displacements. While the divergence of the stress-energy tensor is a force-power vector, it is not immediate that it is the force-power vector of the electron. The equating of the divergence of the stress-energy tensor with the force-power vector of the particle itself is a matter of mechanical assumption.

However, even if this is taken as axiomatic, and so the momentum-energy exchange between the electron and the field is known. To turn this into a path, which is kinematic, requires knowledge of the relation between velocity and momentum. In classical mechanics this is taken to be $p=mv$ and in special relativity $p=\gamma{}mv$, and there is an identical electromagnetic expression for an effective momentum.

But, this contains assumptions about the structure of an electron filled only by a further theory of matter, such as Quantum Mechanics, or more radically, stochastic electrodynamics. If the electron has, for the sake of argument, excited states, then the electron might become excited, like a partially ionised atom, on the absorption of a photon, and change its momentum and energy without changing its velocity.

In general relativity, the derivatives and connections required for curved spacetime make the analysis more sophisticated, but does not essentially change the structure of the discussion.

  • $\begingroup$ The assumption about the electron comes close to the assertion simply that it is a fundamental particle. I am not against this assumption, I am simply saying that it is required, and that GR does not require it any less. Which was the thrust of the original question. $\endgroup$ – Ponder Stibbons Jul 22 '18 at 9:27
  • $\begingroup$ I will mark my own answer as the answer, since it seems to resolve my own conundrum. And to close this matter off. It appears not to have invoked much interest in this forum. Sorry about that. $\endgroup$ – Ponder Stibbons Jul 29 '18 at 0:35

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