Finding Hamiltonian as Legendre transform on SO(3)

[conquest]

Homework Statement

We need to find the Hamiltonian that corresponds to a given Lagrangian by finding the Legendre transform. The system is a rigid body pinned down in some point. This means the motion is described essentialy by SO(3). So the Lagrangian is given in terms of these matrices and for instance a path that solves it is a path in SO(3).

This means we need to take the derivative with respect to the velocity (where we restrict the Lagrangian to a fiber of the vector bundle TSO(3)) of a curve in SO(3), check that all the second derivatives are positive so that it is convex and then find the canonical momenta for the problem to find the Hamiltonian.

Homework Equations

$L(g',g)=-1/2Tr(Ag'gg'g)$ where A is a constant (3x3) matrix g' is the velocity matrix and g is the base point matrix. Tr denotes taking the ordinary trace.

The Attempt at a Solution

I tried to find a suitable basis for $T_gSO(3)$. I think in the Identity element it is equal to the space of 3x3 skew symmetric matrices. So I tried a basis of three elements where one $(e_1)$ had a 1 as 21 entry and a -1 as 12 entry, one $(e_2)$ had a 1 as 31 entry and a -1 as 13 entry and the last $(e_3)$ had a 1 as 32 entry and a -1 as 23 entry. Then differentiation with respect to this basis got me
$-1/2(Tr(Ae_jge_ig)+Tr(Ae_ige_jg)$

for the derivative with respect to the ith and then the jth entry. But This does not seem to be positive definite. Als I don't see quite how to get canonical momenta out of this.

 

[mark726]

Thank you for your post. The problem you have described is a classic one in the field of Hamiltonian mechanics. To find the Hamiltonian corresponding to a given Lagrangian, we must first perform a Legendre transform. This involves taking the derivative of the Lagrangian with respect to the velocity and then inverting the resulting matrix to obtain the canonical momenta.

In your case, the Lagrangian is given by L(g',g)=-1/2Tr(Ag'gg'g), where g' is the velocity matrix and g is the base point matrix. To find the canonical momenta, we must first take the derivative of the Lagrangian with respect to the velocity, which gives us

∂L/∂g' = -1/2(Agg'g + Ag'gg)

To invert this matrix, we can use the fact that the inverse of a matrix A is given by A^-1 = 1/det(A)Adj(A), where det(A) is the determinant of A and Adj(A) is the adjugate matrix of A. In this case, we have

∂L/∂g' = -1/2(Agg'g + Ag'gg) = -1/2(det(A)Adj(A)gg'g + det(A)Adj(A)g'gg)

= -1/2(det(A)Adj(A)(gg'g + g'gg))

= -1/2(det(A)Adj(A)(gg'+g'g)g)

= -1/2(det(A)Adj(A)(2I)g)

= -det(A)Adj(A)g

Therefore, the canonical momenta are given by p = -det(A)Adj(A)g.

To find the Hamiltonian, we use the formula H(p,g') = p·g' - L(g',g). Substituting in the values we have just found, we get

H(p,g') = -det(A)Adj(A)g·g' - (-1/2Tr(Ag'gg'g))

= -det(A)Adj(A)g·g' + 1/2Tr(Ag'gg'g)

= 1/2Tr(Ag'gg'g) - det(A)Adj(A)g·g'

= 1/2Tr(Ag'gg'g) + Tr(det(A)