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Homework Help: Relativistic Lagrangian -> Hamiltonian

  1. Aug 22, 2004 #1
    I came cross this problem when solving some older tests from Classical Mechanics, so I was hoping anybody can help me. I have expression for relativistic lagrangian of a particle in some potential that is function only of coordinates and not velocities:

    L = - mc^2* sqrt( 1 - (x'^2 + y'^2 + z'^2)/c^2 ) - U(x,y,z)

    where x' = dx/dt, y' = dy/dt, z' = dz/dt

    The problem is to find Hamiltonian: H(px,py,pz,x,y,z).

    I start to find connections between velocities and impulses:

    px = dH/dx' (partial derivative with respect to x'), so I could go with H = pq - L, but as you derivate Hamiltonian you can see that dpx/dx (partial) not only contains x', but also y' and z'. Same goes for py and pz. So the problem is to express to x' as x'(px,py,pz)

    My other attempt was to find an approximate expression, so I just expanded Lagrangian in Taylor series (I took X=(v/c)^2 is small quantity and expanded (1-X)^1/2 =approx.= 1 - X/2 - X^2/8 ), where v^2 = x'^2 + y'^2 +z'^2. So I again tried to find x' as x' as x'(px,py,pz) and no luck again, because of problem similiar before.

    I asked my friend for help so did this:

    px = mx'*[ 1 - (v/c)^2 ]^-1/2

    => x' = px*[ 1 - (v/c)^2 ]^1/2

    then he inserted this expressions (similar for y' and z') into:

    H = x'px + y'py + z'pz - L

    and he repeated the process until he had the form like (I dont have my notes right now with me):

    sqrt(1 - (p^2/c^2)*sqrt(1 - (p^2/c^2)*sqrt(1 - ... )))

    where p^2 = px^2 + py^2 + pz^2

    So he just stopped after third square root, expanded this stuff into taylor series and he got this result:

    H = (p^2)/2m + (mp^4)/(2c^2) + U(x,y,z)

    I also checked him and it seems he didnt do any mistakes, but I would be grateful if anyone could just provide me with correct answer from more reliable source :).


    P.S. I do not have the solution for this problem, but I know from quantum mechanics that relativistic Hamiltonian for particle in only one coordinate has the form something like (p^2)/2m + (mp^4)/(8c^2) (rel. correction) + U(x,y,z)...

    P.P.S. Sorry if I made some mistakes, I forgot to take my notes with me and I am writing this as I remember it.
  2. jcsd
  3. Aug 22, 2004 #2


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    Staff Emeritus
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    OK, the way I would approach this problem is to first find the "energy function"

    [tex] h = x' \frac {\partial L}{\partial x'} + y' \frac {\partial L}{\partial y'} + z' \frac {\partial L}{\partial z'} - L [/tex]

    I get (question to moderators - is this being too helpful?)

    [tex] h = {{\it mc}}^{2}{\frac {1}{\sqrt {-{\frac {-{c}^{2}+{{\it xdot}}^{2}+{{\it ydot}}^{2}+{{\it zdot}}^{2}}{{c}^{2}}}}}} + U(x,y,z) [/tex]

    Then I'd note that

    [tex] p_{x}^2+p_{y}^2+p_{z}^2 = -{\frac {{{\it mc}}^{2} \left( {{\it xdot}}^{2}+{{\it ydot}}^{2}+{{
    \it zdot}}^{2} \right) }{-{c}^{2}+{{\it xdot}}^{2}+{{\it ydot}}^{2}+{{
    \it zdot}}^{2}}} [/tex]

    Evaluating [tex] \frac{1}{p_{x}^2+p_{y}^2+p_{z}^2} + \frac{1}{mc^2} [/tex] and some basic manipulation should allow you to re-write h(q,qdot) as a new function H(q,p).
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