The Hamiltonian for a relativistic charged particle moving in a static electromagnetic field is the well known: $$H=c\sqrt{\left(\mathbf{P}-q\mathbf{A}\right)^{2}+m^{2}c^{2}}+q\phi$$ where,\begin{align*} \mathbf{B} & =\nabla\times\mathbf{A},\\ \mathbf{E} & =-\nabla\phi. \end{align*} Now let's suppose that one wants to write the magnetic field in terms of the magnetic scalar potential, $\phi_M$, rather than of $\mathbf{A}$, that is for a magnetic field written as: $$\mathbf{B}=-\nabla\phi_{M}.$$ What would the Hamiltonian look like, in terms of $\phi_M$ ?

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    $\begingroup$ Comments to the post (v1): 1. The formula for ${\bf E}$ seems incomplete. 2. Why should $\phi_M$ exist? $\endgroup$ – Qmechanic Apr 13 '18 at 5:55
  • $\begingroup$ 1. The formula for $E$ is complete -- they are all static fields in vacuum. 2. Because it does. E.g. as the numerical solution of the Laplace equation. $\endgroup$ – Hans Castrop Apr 13 '18 at 7:09
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    $\begingroup$ Those static assumptions seem to violate the spirit of relativity. $\endgroup$ – Qmechanic Apr 13 '18 at 7:14
  • $\begingroup$ What assumptions? I reckon the formula for the Hamiltonian is OK. How do I introduce $\phi_M$ in that formula? $\endgroup$ – Hans Castrop Apr 13 '18 at 7:17
  • $\begingroup$ If I am in the same reference frame where the fields are static, those equations are fine. I'd really like to introduce $\phi_M$ in the picture. $\endgroup$ – Hans Castrop Apr 13 '18 at 7:20

One way would be to express the vector potential as function[al] of the scalar potential and substitute in the Hamiltonian.

Magnetic field (induction) can be expressed as gradient of a potential function only in a limited region of space where the integral

$$ \int_1^2 \mathbf B\cdot d\mathbf s $$ between two points does not depend on the path; it may depend only on the endpoints.

Suppose we have such a region and such a potential function so magnetic field is a gradient.

Then, we can ( in principle ) solve the equation

$$ - \nabla \phi = \nabla \times \mathbf A $$

for unknown function $\mathbf A(\mathbf x)$. Suppose we found a solution which is expressed as function[al] of $\phi$ and space coordinates. Then we can substitute for $\mathbf A$ in the standard Hamiltonian you mentioned and thus obtain Hamiltonian that refers to $\phi$ and space coordinates only.

For example, if the magnetic field is uniform, so we can express it as $\mathbf B=[0,0,B_0]$, the potentials that would give the magnetic field correctly could be defined as

$$ \phi = -B_0z $$

$$ \mathbf A = \bigg[-\frac{1}{2}B_0 y,\frac{1}{2}B_0 x,0\bigg] $$

However, it is easy to see that vector potential at some point is not simply a function of scalar potential at the same point; if we are to relate the two, we need to involve also the spatial coordinates explicitly:

$$ \mathbf A = \bigg[-\frac{\phi}{2} \frac{y}{z},\frac{\phi}{2}\frac{x}{z},0\bigg ] $$

Now we can substitute for $\mathbf A$ in the Hamiltonian.

Finding $\mathbf A$ was simple here but for general magnetic field, the relation would be more complicated, probably integral: $\mathbf A$ would be a functional of $\phi$. Such complicated expression would be hard to work with in the Hamiltonian. Better use directly the vector potential.


As far as I know, each vector field can be unambiguously split into a gradient and a rotational part:

$$\mathbf{F} = \nabla \phi + \nabla \times \mathbf A$$

This is called the Helmholtz decomposition: https://en.wikipedia.org/wiki/Helmholtz_decomposition

Due to your assumption $\mathbf{B} = \nabla \times \mathbf A$ it follows that the gradient part is identically zero.

  • $\begingroup$ You are right. A generic vector field could feature both div-free and curl-free components. I should have been more precise. Now let's imagine that I have a field $\mathbf{B}$ featuring such grad of $\phi_M$ component: How does $\phi_M$ enters $H$ ? ... $\endgroup$ – Hans Castrop Apr 13 '18 at 8:07
  • $\begingroup$ My point is that in this case the magnetic field will have no divergence free component $\nabla \times \mathbf{A}$. If this is fine with you, then the field will enter the Hamiltonian exactly in the same way as the electric field does, namely with the potential term $q_M\phi_M$. But it would behave exactly the same way as an electric field (in particular have a "magnetic charge" $q_M$). Not sure if it is justified to call it a magnetic field in this case. $\endgroup$ – Photon Apr 13 '18 at 11:18
  • $\begingroup$ I am not sure it is that simple. Assuming there is not vector potential, and $\mathbf{B}$ is indeed $\mathbf{B}=-\nabla\phi_M$, then the force acting on the charged particle sees $\phi$ (electric) and $\phi_M$ (magnetic) in two very different roles: $$\mathbf{F} = q\left(-\nabla\phi + \mathbf{v}\times\left(-\nabla\phi_M\right)\right)$$ Therefore I doubt that the two $\phi$'s appear in the Hamiltonian with the same functional form... $\endgroup$ – Hans Castrop Apr 13 '18 at 13:14

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