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In classical mechanics kinetic part of Hamiltonian is the Legendre transform of kinetic part of Lagrangian.

On the other hand kinetic part of Lagrangian is a metric on the configuration space. At first I though kinetic energy in Hamiltonian setting would be an inverse metric tensor.

However it's not true. In 1D:

$$T_L(\dot x, \dot x) = \frac{1}{2} m \dot x^2$$

$$T_H(p_x, p_x) = \frac{1}{2m} p_x^2$$

where as the inverse of $T_L$ would be

$$T_L^{-1}(p_x, p_x) = \frac{2}{m} p_x^2$$

Is in this case inverse metric tensor related to Legendre transform at all? Can Legendre transform be expressed in coordinate independent way in the case above?

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up vote 2 down vote accepted

The inverse transformation that you are thinking of works pointwise: you can take the metric on the tangent space at one fixed point, and compute with it.

But this is not how the Legendre transform works. It involves (though not explicitly) the derivative of the thing being transformed. If you know the original function at one point only, you don't know its transform at any point.

On the level of derivatives there is indeed an inverse relationship: gradient maps are inverses of each other. And for quadratic forms such as the kinetic energy this indeed makes the transformation into a linear-algebraic calculation.

For example, the gradient of $\sum_{i,j} a_{ij} x_i x_j$ (defined on Euclidean space) can be thought of as a map $(x_i)\mapsto (2\sum_j a_{ij} x_j)$. This is a linear map with the matrix $2A$. Its inverse has the matrix $A^{-1}/2$. The corresponding quadratic form is $1/4$ of what you would get if you simply inverted pointwise without doing the necessary calculus.

Executive summary: you can get the Legendre transform by inverting the matrix, if you keep the factor of $1/2$ separately - from $m(\dot x^2/2)$ you get $m^{-1}(p_x^2/2)$, etc.

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