- #1
victorvmotti
- 155
- 5
We know that a charged particle will have a drift velocity in both a curved magnetic field and when there is a transverse spatial gradient in the magnitude of the magnetic field.
This drift velocity is added to the rotation velocity around the the field line.
In both cases the force vector on the particle, $$\mathbf{F}$$ is perpendicular to the velocity vector.
$$\mathbf{v_D}=\frac {c}{q} \frac{\mathbf{F} \times \mathbf{B}}{B^2}$$
Now the question is what the "reaction force", $$-\mathbf{F}$$ is applied on, in particular in the case of gradient drift?
Is it the "source" of the magnetic field?
But what can we say about a radiation field that spans throughout spacetime, far from the source?
I mean we know that there is nothing like a free electromagnetic field even though for practical purposes we treat radiation like a dynamical entity "unconnected to the source."
Is the reaction force due to either curvature or gradient drift irrelevant here?
Have we assumed a time-independent $$\mathbf{B}$$ for the derivation of $$\mathbf{v_D}$$?
This drift velocity is added to the rotation velocity around the the field line.
In both cases the force vector on the particle, $$\mathbf{F}$$ is perpendicular to the velocity vector.
$$\mathbf{v_D}=\frac {c}{q} \frac{\mathbf{F} \times \mathbf{B}}{B^2}$$
Now the question is what the "reaction force", $$-\mathbf{F}$$ is applied on, in particular in the case of gradient drift?
Is it the "source" of the magnetic field?
But what can we say about a radiation field that spans throughout spacetime, far from the source?
I mean we know that there is nothing like a free electromagnetic field even though for practical purposes we treat radiation like a dynamical entity "unconnected to the source."
Is the reaction force due to either curvature or gradient drift irrelevant here?
Have we assumed a time-independent $$\mathbf{B}$$ for the derivation of $$\mathbf{v_D}$$?