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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?

Qmechanic
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bc87
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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 08:02
  • 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 08:19
  • This is not an homework question by the way. – bc87 Sep 15 '13 at 09:19
  • 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 09:49
  • 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 '13 at 09:55
  • Have you got a reference for that trick? It looks to me like you're doing a Legendre transformation on the Lagrange equations of motion in disguise right there. Another reference is Mandelstam - Variational Principles in Dynamics & Quantum Theory, section 8, however I'd say none of these will be of help until you clear up this issue about adding variables. – bolbteppa Sep 15 '13 at 09:59
  • @bolbteppa: Thank you for perseverance. I edited my question in order to explain what I said before. More generally, is it always possible to get the hamiltonian of a system from the Newton's equations without resorting to the lagrangian formalism? – bc87 Sep 16 '13 at 08:19
  • No problem, but notice that once you start with Newton's equations you are already assuming the Lagrangian formalism. The Euler-Lagrange equations of motion just end up deriving F = ma, thus when you start changing your variables you are basically just converting your Lagrangian formalism into the Hamiltonian formalism AFTER you've calculated the EOM from your Lagrangian, instead of first finding the Hamiltonian via a Legendre transform (after which you can then extremize to get Hamilton's equations). Coincidentally this process you're following now is one of the motivations for using – bolbteppa Sep 16 '13 at 08:30
  • the Hamiltonian formalism in the first place, because the equations of motion look more symmetric, it just amounts to choosing whether you change variables before or after extremizing really. Note that the Lagrange EOM can be broken into a system of equations just as Hamilton's equations are (check my thread on the Legendre transform to see what I'm talking about: http://physics.stackexchange.com/questions/75305/motivating-the-legendre-transform-mathematically ), so it can all be made to look very homogeneous theoretically, thus Hamiltonians are basically just Lagrangians in new coordinates. – bolbteppa Sep 16 '13 at 08:34
  • @bolbteppa: Thank you, your answered my question. But are you sure that in this particular simple case, it is not possible to derive hamiltonian equations directly with some trick? Could you turn your comments into an answer so that I accept it please? – bc87 Sep 16 '13 at 18:55

1 Answers1

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

Qmechanic
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  • Thanks for your answer but I really would like to derive the Hamiltonian formulation from the Newton's equation. – bc87 Sep 15 '13 at 09:08
  • @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 09:58
  • I understand better why it is not systematically possible to do so now, thank you. – bc87 Sep 16 '13 at 18:56