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Yet another question on Lagrangians

  1. Jan 9, 2010 #1
    I know this is getting really ridiculous but I have yet another question on Lagrangians...

    This is our Lagrangian:

    L=\frac{1}{2}m\dot{\vec{x}}^{2}+e\vec{A}.\dot{\vec {x}}

    Using the fact that:

    [tex]\vec P= \frac{\partial L}{\partial \dot{\vec{x}}}=m \dot{\vec{x}} + e\vec A[/tex]

    and substituting P for [tex]\dot{\vec{x}}[/tex] *

    We get this Hamiltonian:

    [tex]H=\frac{1}{2m}(\vec P - e\vec A )^2[/tex]

    (From * [tex]\dot{\vec{x}} = \frac{1}{m}(\vec P - e\vec A )[/tex], so presumably, the above Hamiltonian is equal to [tex]\frac{m}{2} \dot{\vec{x}}^2[/tex])

    The question is to find the Equations of Motion in the Hamiltonian Formalism, ie. we need to determine [tex]\dot{\vec{x}}[/tex] and [tex]\dot{P}_i[/tex]:

    We know [tex]\dot{\vec{x}} = \frac{1}{m}(\vec P - e\vec A )[/tex] from above.


    [tex]\dot{P}_i = \frac{\partial H}{\partial x^i} = \frac{1}{2m}\frac{\partial [(\vec P - e\vec A )^2]}{\partial x^i} = \frac{m}{2}\frac{\partial [(\dot{\vec{x}} ^2]}{\partial x^i}=0[/tex]

    HOWEVER the answer seems to be this:

    [tex]\dot{P}_i = \frac{e}{m}(P-eA)_j \epsilon_{jki} B_k[/tex]

    Can someone please explain, why P_dot is non zero?

  2. jcsd
  3. Jan 9, 2010 #2


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    The short answer would be that it's because of the vector potential... if this is anything like your previous questions,
    It's actually [itex]x[/itex] and [itex]p[/itex] that are the independent variables, not [itex]x[/itex] and [itex]\dot{x}[/itex]. In the simple case, when there's no vector potential, both statements are equivalent, but that's not the case when the momentum has that spatial dependence in it.
  4. Jan 9, 2010 #3
    Thanks for the reply diazona. I must admit, I don't know very much about vector potentials - I will try reading up on this. I guess the key thing to take away from your post is that x_dot and x aren't independent in the case of vector potentials (a revelation to me!).

    Can I ask whether:


    \dot{\vec{x}}^{2} = \sum_{i,j}\delta_{ij} \dot{\vec{x}}_i \dot{\vec{x}}_j


    and whether we can write:

    (\vec P - e\vec A )^2=(\vec P.\vec P - 2e\vec P.\vec A + e^2 \vec A.\vec A)

    Where [tex]\vec P.\vec P[/tex]

    and [tex]\vec P.\vec A[/tex]
    and [tex]\vec A.\vec A[/tex]

    are just scalar products?

    Because if I assume this, I get that:

    \frac{\partial H}{\partial x^i}=-2e\epsilon_{kji}B_j(P_k-eA_i)[/tex]

    -not the expression I should get.

  5. Jan 9, 2010 #4


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    It's nothing specific about vector potentials - the vector potential [itex]\vec{A}[/itex] is just one of many possible things that can be in a Lagrangian or Hamiltonian. It might be better to say that when you switch from the Lagrangian to the Hamiltonian formalism, [itex]x[/itex] and [itex]\dot{x}[/itex] stop being completely independent of each other, and [itex]x[/itex] and [itex]p[/itex] become the independent variables.

    If you find this a little confusing, you're not alone... I'll see if I can come up with a better explanation.
    Yep, that all seems reasonable to me...

    hmm, how exactly did you get that? There's definitely something wrong if you're getting a sum that involves different unsummed indices in different terms.
  6. Jan 9, 2010 #5
    Thanks for reassuring me of the basics there!

    I kind of see this. Ofcourse the first thing I was taught in this crash course was that H=H(p,q) and L=L(q,q_dot).

    I just spent ages typing out my solution, but as per usual I made I tiny mistake. And indeed, I got the right answer in the end.
  7. Jan 12, 2010 #6


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    Cool, glad you got it.

    Here's something you might be interested in: hopefully you know that the state of a physical system is described by a point in phase space. For a 1D system, you have a 2D phase space, so the state of the system is described by two coordinates. But you can choose those coordinates to be either [itex]x[/itex] and [itex]\dot{x}[/itex] (the Lagrangian way) or [itex]x[/itex] and [itex]p[/itex] (the Hamiltonian way). Or you could choose some set of generalized coordinates that are combinations of those.

    The point is, basically what you're doing by picking coordinates is choosing a basis for a vector space. A particular state of a system corresponds to a vector in the space, which you can express as either
    [tex]x\hat{x} + \dot{x}\hat{\dot{x}}[/tex]
    (Lagrangian) or
    [tex]x\hat{x} + p\hat{p}[/tex]
    (Hamiltonian). I'm using letters with hats to denote basis vectors and letters without hats to denote the coordinates. Visually, it looks like this:
    (okay, well that's kind of small, but the full-size version is clear)

    When you take a partial derivative, like [itex]\frac{\partial}{\partial x}[/itex], physically that corresponds to shifting the state of the system by some infinitesimal amount [itex]\mathrm{d}x[/itex] in the direction of increasing [itex]x[/itex] - that is, the [itex]\hat{x}[/itex] direction, which is to the right in these diagrams. You'll notice that if you take the Lagrangian view (right), [itex]\hat{x}[/itex] and [tex]\hat{\dot{x}}[/tex] are orthogonal, which means that when you adjust the state by that infinitesimal amount in the [itex]\hat{x}[/itex] direction, the [itex]\dot{x}[/itex] coordinate doesn't change at all. But the [itex]p[/itex] coordinate does. This is what it means to say that [itex]x[/itex] and [itex]\dot{x}[/itex] (but not [itex]x[/itex] and [itex]p[/itex]) are the independent variables in the Lagrangian formulation. Conversely, in the Hamiltonian view (on the left), when you adjust the state by [itex]\mathrm{d}x[/itex], the [itex]p[/itex] coordinate doesn't change, but [itex]\dot{x}[/itex] does. Thus [itex]x[/itex] and [itex]p[/itex] are the independent variables in the Hamiltonian formulation. (There are in fact two different meanings of "independent" being thrown around here, so it can easily get confusing!)
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