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 $\phi_M$ ?

  • 1
    $\begingroup$ Comments to the post (v1): 1. The formula for ${\bf E}$ seems incomplete. 2. Why should $\phi_M$ exist? $\endgroup$
    – Qmechanic
    Commented Apr 13, 2018 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. $\endgroup$ Commented Apr 13, 2018 at 7:09
  • 2
    $\begingroup$ Those static assumptions seem to violate the spirit of relativity. $\endgroup$
    – Qmechanic
    Commented Apr 13, 2018 at 7:14
  • $\begingroup$ What assumptions? I reckon the formula for the Hamiltonian is OK. How do I introduce $\phi_M$ in that formula? $\endgroup$ Commented Apr 13, 2018 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. $\endgroup$ Commented Apr 13, 2018 at 7:20

3 Answers 3


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 $H$ ? ... $\endgroup$ Commented Apr 13, 2018 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
    Commented Apr 13, 2018 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... $\endgroup$ Commented Apr 13, 2018 at 13:14

I created account now so can't comment, maybe someone else can comment this and delete this answer:

Helmholtz decomposition is only valid for fields that decay fast enough at infinity. Many typical static magnetic fields do not.

I think the original question is interesting, and would add that one clue is that the magnetic vector potential does appear together with the electric scalar potential in the potential 4-vector, while the magnetic scalar potential does not, maybe suggesting that it is quite tricky to write down what OP wants.

One could ask the same question but for an electric field written as the curl of an electric scalar potential, again in the static source free case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.