Relativistic Lagrangian -> Hamiltonian

[Igor_S]

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 similar 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 don't 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 :).


Thanks...


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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.

 

[pervect]

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:

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

$$h = x' \frac {\partial L}{\partial x'} + y' \frac {\partial L}{\partial y'} + z' \frac {\partial L}{\partial z'} - L$$

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

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

Then I'd note that

$$p_{x}^2+p_{y}^2+p_{z}^2 = -{\frac {{{\it mc}}^{2} \left( {{\it xdot}}^{2}+{{\it ydot}}^{2}+{{<br /> \it zdot}}^{2} \right) }{-{c}^{2}+{{\it xdot}}^{2}+{{\it ydot}}^{2}+{{<br /> \it zdot}}^{2}}}$$

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

 

[M98Ranger]

The process of going from the Lagrangian to the Hamiltonian in classical mechanics is known as the Legendre transformation. In the relativistic case, it becomes more complicated due to the presence of the square root term in the Lagrangian.

To find the Hamiltonian, we first need to find the momenta conjugate to the coordinates. In this case, the momenta are given by:

px = dL/dx' = -mc^2*x'/(c^2-x'^2-y'^2-z'^2)^(1/2)

py = dL/dy' = -mc^2*y'/(c^2-x'^2-y'^2-z'^2)^(1/2)

pz = dL/dz' = -mc^2*z'/(c^2-x'^2-y'^2-z'^2)^(1/2)

Next, we use these expressions to eliminate the velocities (x', y', z') from the Hamiltonian, using the relation:

H = Σp_i * x_i' - L

Substituting in the expressions for the momenta and the Lagrangian, we get:

H = px * x' + py * y' + pz * z' + mc^2 * (c^2 - x'^2 - y'^2 - z'^2)^(1/2) + U(x,y,z)

Now, we can rearrange this expression to get x' in terms of the momenta and the coordinates:

x' = px * (1 - (p^2/m^2c^2))^(1/2)

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

Similarly, we can express y' and z' in terms of the momenta and coordinates.

Substituting these expressions back into the Hamiltonian, we get:

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

This is the same result that your friend obtained, and it is correct. It is the correct relativistic Hamiltonian for a particle in a potential that depends only on the coordinates. Note that the correction term (mp^4)/(2c^2) is a relativistic correction and is only significant when the particle's momentum is large compared to its rest mass (i.e. when p^2 is much larger than m^2c^2).