# Hamiltonian formulation of single particle in an electromagnetic field

Consider a charged particle in a static electromagnetic field. Suppose that the domain is simply connected so that the second law of Newton's dynamics reads: $$m\frac{\mathrm{d}^2\vec{x}}{\mathrm{d}t^2}=-e\left(\vec{\nabla}\Phi+\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}\times(\vec{\nabla}\times\vec{A})\right)$$ where:

• $m$ is the mass of the particle,
• $\vec{x}$ is position vector,
• $\Phi$ the (scalar) electric potential,
• $\vec{A}$ the (vector) magnetic potential.

My question is how to derive the Hamiltonian formulation from here. The Hamiltonian is: $$H(\vec{x},\vec{p})=\frac{1}{2m}\left\|\vec{p}-e\vec{A}\right\|^2+e\Phi(\vec{x})$$ and the Hamilton equations look like: $$\begin{cases} \frac{\mathrm{d}x_i}{\mathrm{d}t}=\frac{1}{m}(p_i-eA_i)\\ \frac{\mathrm{d}p_i}{\mathrm{d}t}=\frac{e}{m}\sum_{j=1}^3(p_j-eA_j)\frac{\partial A_j}{\partial x^i}-e\frac{\partial\Phi}{\partial x^i} \end{cases}$$ Another question related to this one is why the magnetic potential is "included" in the "generalized" moment $\vec{p}-e\vec{A}$. I undertand why from the Lagrangian of the system but perhaps there is a physical explanation.

EDIT:

Consider a single particle in a conservative force field $F=-\vec{\nabla}V$. Then the Newton's equations are: $$m\frac{\mathrm{d}^2\vec{x}}{\mathrm{d}t^2}=-\vec{\nabla}V$$ This is a second order differential equation, which we can transform into a system of first order differential equations, by adding a new variable which is usually the speed $v$ in mathematics but here we will take the momentum $p=mv$. The system I am talking about is: $$\begin{cases} \frac{\mathrm{d}\vec{x}}{\mathrm{d}t}=\frac{\vec{p}}{m}\\ \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}=-\vec{\nabla}V(x) \end{cases}$$ which is just the hamiltonian form of the Newton's equations. Can we do the same kind of trick in the case of the system above?

• Since you're aware of the Lagrangian formulation and how the Hamiltonian formulation can be arrived at through that formulation (via Legendre transform), what sort of a derivation beyond this are you looking for? Perhaps you'd like some way of motivating the Hamiltonian without appealing to Lagrangians? Sep 15, 2013 at 8:02
• Ginzburg "Theoretical Physics & Astrophysics" spends it's first 26 pages examining the Hamiltonian approach to electrodynamics & parallel's Landau & Lifshitz's development pretty closely if you're interested in a reference. Sep 15, 2013 at 8:19
• This is not an homework question by the way.
– bc87
Sep 15, 2013 at 9:19
• Starting from Newton's equations you can use the principle of Virtual work to arrive at Lagrange's equations, c.f. Lanczos Variational Principles of Mechanics Chapters 3 to 5. Then the Hamiltonian is merely the Lagrangian after a change of variables via the Legendre transform (c.f. Lanczos or Gelfand) to make your equations of motion more symmetric & to enable more freedom in your coordinates. Maybe that will be of some use to you. Sep 15, 2013 at 9:49
• I had a look at the references but did not find the answer to my question. Again, the goal is not to use the Lagrangian formalism. For instance in the case of a particle in conservative force field, adding a variable (the momentum) to the system of Newton's (2nd order) equations enable us to convert it to its hamiltonian form, which is a system of first order equations. I would like to have the same kind of trick here.
– bc87
Sep 15, 2013 at 9:55

Brief layout of strategy:

1. Write down Lorentz force ${\bf F}$.

2. Find (velocity-dependent) potential $U$ for ${\bf F}$.

3. Write down Lagrangian $L=T-U$.

4. Define the canonical (as opposed to kinetic/mechanical) momentum ${\bf p}:=\frac{\partial L}{\partial{\bf v}}$.

5. Perform Legendre transformation ${\bf v}\to {\bf p}$.

6. Write down Hamiltonian $H={\bf v}\cdot{\bf p}-L$.

• Thanks for your answer but I really would like to derive the Hamiltonian formulation from the Newton's equation.
– bc87
Sep 15, 2013 at 9:08
• @Benjamin: It is not possible to systematically define the canonical momentum ${\bf p}$ without (what amounts to essentially) swing by the Lagrangian formalism first. On the other hand, if you already know Hamilton's eqs. of motion for the system, it is possible to just eliminate the momentum ${\bf p}$ to get to Newton's 2nd law. Sep 15, 2013 at 9:58
• I understand better why it is not systematically possible to do so now, thank you.
– bc87
Sep 16, 2013 at 18:56