# How does a charged particle behave in a vector potential?

I know that a charged particle interacts with a magnetic field through the Lorentz force, thus knowing how it behaves in a given magnetic field.
However, I don't understand how a charged particle (be it at rest or moving) interacts with the vector potential.
If I insert $$\mathbf{B}= \nabla× \mathbf{A}$$ into the Lorentz force I obtain: $$\mathbf{F} = q\mathbf{E} + q\mathbf{v}×\mathbf{B} = q\mathbf{E} + q\mathbf{v} × (\nabla×\mathbf{A}) = q\mathbf{E} +q((\mathbf{v} \cdot \mathbf{A})\nabla - (\mathbf{v} \cdot \nabla)\mathbf{A})$$ I am not sure how to interpret this.

• Have you tried substituting $\bf{B} = \nabla \times \bf{A}$ in the Lorentz force? Jun 25, 2020 at 10:41
• Be careful : you expand $$\mathbf v \boldsymbol \times (\boldsymbol \nabla \boldsymbol \times\mathbf A) \tag{01}\label{01}$$ as if $\nabla$ is a 3-vector. But this is an operator. Your expansion is wrong, that's why you can't interpret it. Jun 25, 2020 at 12:06

First of all if the description with the vector potential is desired to be complete, one should consider that (under neglection of the scalar potential, as the question is only about the vector potential)

$$\mathbf{E} = -\frac{\partial \mathbf{A}}{\partial t}$$

Secondly, according to

$$\nabla(\mathbf{a}\mathbf{b}) = (\mathbf{a}\nabla\mathbf{ b}) + (\mathbf{b}\nabla \mathbf{a}) + \mathbf{b}\times(\nabla \times\mathbf{a}) + \mathbf{a}\times (\nabla\times \mathbf{b}) = (\mathbf{b}\nabla \mathbf{a}) + \mathbf{b}\times(\nabla\times \mathbf{a})$$

where the second equal sign can be used if it is assumed that $$\mathbf{b=v}$$ is independent of location. With this relationship we get the following expression from the original Lorentz force:

$$\mathbf{F}=\frac{d\mathbf{p}}{dt} + q(\mathbf{v} \nabla \mathbf{A}) = -q\frac{\partial \mathbf{A}}{\partial t} + q\nabla(\mathbf{A} \mathbf{v})$$

Putting $$-\frac{\partial \mathbf{A}}{\partial t}$$ on this left side and using

$$\frac{d \mathbf{A}}{dt} = \frac{\partial \mathbf{A}}{\partial t} + (\mathbf{v} \nabla \mathbf{A})$$

the equation can be "simplified" by using the canonical (or generalised) momentum $$\mathbf{P} = \mathbf{p} + q \mathbf{A}$$ in to order to write

$$\frac{d\mathbf{P}}{dt} = q\nabla(\mathbf{A} \mathbf{v})$$

which is more compact and also can be easily connected with the Lagrangian description $$L$$ of a charged particle in the vector potential (neglecting the scalar potential as at the beginning)

$$L = \frac{mv^2}{2} + q\mathbf{v}\mathbf{A}$$

For simplicity here only the non-relativistic Lagrangian was used.

A general remark should be made. The description of the motion of a charged particle in a vector potential field is questionable as the vector potential is not uniquely defined, it can always be changed according to the gauge transformation

$$\mathbf{A'} = \mathbf{A} + \nabla f$$

with the arbitrary function $$f$$ dependent on $$\mathbf{r}$$. Only the field strengths $$\mathbf{E}$$ and $$\mathbf{B}$$ can be really observed. Therefore for a description of a charged particle in an electromagnetic field the description with the field strengths should be clearly preferred. It is also easier to understand and does not contain complicated vector analytical expressions nor the generalised and rather abstract canonical momentum $$\mathbf{P}$$.

• What does $\mathbf{A} \mathbf{v}$ denote here? A scalar product? A dyadic tensor product? Something else? How does $\nabla$ act on it? This notation means different things in different places and cannot be used without explanation. Jun 25, 2020 at 13:52
• It is the scalar product $\sum_i A_i v_i$. The effect of the $\nabla$-operator is given by the second expression in my post. Jun 25, 2020 at 13:55
• Following @Emilio Pisanty's comment I suggest to use a bold centerdot for the inner product between 3-vectors or between 3-operators and 3-vectors, so your second equation (I think it's a copy from Jackson's "Classical Electrodynamics") would expressed as :.... Jun 25, 2020 at 18:31
• $$\boldsymbol{\nabla}\left(\mathbf a\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{=}\left(\mathbf a\boldsymbol{\cdot}\boldsymbol{\nabla}\right)\mathbf b\boldsymbol{+}\left(\mathbf b\boldsymbol{\cdot}\boldsymbol{\nabla}\right)\mathbf a\boldsymbol{+}\mathbf a\boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf b\right)\boldsymbol{+}\mathbf b\boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf a\right)$$ Jun 25, 2020 at 18:49
• Also what exactly are the terms $\mathbf b \nabla \times\mathbf{a}$, $\mathbf{a} \nabla\times \mathbf{b}$ ??? It would be very confusing if you mean that they are the right ones $\mathbf b\boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf a\right)$ , $\mathbf a\boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf b\right)$ and not the wrong ones $\mathbf b\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf a\right)$ , $\mathbf a\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf b\right)$ according to your notation. Jun 25, 2020 at 18:49

I think the confusion comes from the fact that you applied the bac-cab formula for $$\mathbb{R}^3$$ elements:

$$\mathbf{a} \wedge (\mathbf{b} \wedge \mathbf{c}) = \mathbf{b} \, (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} \, (\mathbf{a} \cdot \mathbf{b})$$

regardless of the fact that $$\nabla$$ is a differential operator, not a triple in $$\mathbb{R}^3$$. You cannot apply that formula in this case, for which you have:

$$\mathbf{v} × (\nabla×\mathbf{A}) = \left[\sum_{i} v_i (\nabla A_i)\right] - (\mathbf{v} \cdot \nabla)\mathbf{A}$$

Note that the thing you wrote, $$(\mathbf{v} \cdot \mathbf{A})\nabla$$ cannot appear as a contribution to a force: it is a differential operator and lives in a totally different vector space.