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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?

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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? –  joshphysics Sep 15 '13 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. –  bolbteppa Sep 15 '13 at 8:19
    
Thanks for your answer but I really would like to derive the Hamiltonian formulation from the Newton's equation, not the Lagrangian formalism. I will have a look at Landau and Lifshitz. –  Benjamin Sep 15 '13 at 9:10
    
This is not an homework question by the way. –  Benjamin Sep 15 '13 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. –  bolbteppa Sep 15 '13 at 9:49

1 Answer 1

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 $H={\bf v}\cdot{\bf p}-L$.

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Thanks for your answer but I really would like to derive the Hamiltonian formulation from the Newton's equation. –  Benjamin Sep 15 '13 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. –  Qmechanic Sep 15 '13 at 9:58
    
I understand better why it is not systematically possible to do so now, thank you. –  Benjamin Sep 16 '13 at 18:56

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