reparameterization invariance

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Often (especially in relativistic theories) the action does not depend on the choice of parameterization. For eqample, given two parameters s\,, \tilde{s}\,, we have

ds\, L\left(q,\frac{dq}{ds}\right) = d\tilde{s}\,L\left(q,\frac{dq}{d\tilde{s}}\right)\,.

If \frac{d\tilde{s}}{ds} = \epsilon\,,

ds\, L\left(q,\frac{dq}{ds}\right) - \frac{d\tilde{s}}{ds} ds\,L\left(q,\frac{dq}{ds}\frac{ds}{d\tilde{s}}\right)=0\,,

or

ds\, L\left(q,\dot{q}\right) - \epsilon\,ds\,L\left(q,\frac{1}{\epsilon}\dot{q}\right)=0\,.

Differentiating with respect to \epsilon\, and subsequently setting \epsilon = 1\, gives

\frac{\partial L}{\partial \dot{q}} \dot{q} - L = 0\,.

Recognizing p = \frac{\partial L}{\partial \dot{q}}\, as the conjugate momentum of q\, we see that the Hamiltonian vanishes:

H = p \dot{q} - L = 0\,.

The equations of motion have not been used, so this is an identity. Differentiating with respect to \dot{q}_i\, (generalizing to many variables)

0\, = \frac{\partial}{\partial{\dot{q}_i} } \left( \frac{\partial L}{\partial \dot{q}_j} \dot{q}_j - L \right)\,,
= \delta_{ij} \frac{\partial L}{\partial \dot{q}_j}- \frac{\partial L}{\partial \dot{q}_i} + \frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j } \dot{q}_j\,,
= \frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j } \dot{q}_j \,.

The matrix \frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j }\, therefore has zero eigenvalues (alternatively, it has a non-trivial null-space or kernel) so that the Jacobian determinant for transforming from p_i\, to \dot{q}_i\, is zero:

\det\left|\frac{\partial p_i}{\partial \dot{q}_j} \right| = 0\,,

I.e. the Legendre transformation is not invertible (which can be seen from the fact that H\, vanishes) and we say that the Lagrangian is degenerate.

Finally, differentiating the first equation with respect to \dot{q}_i\, leads to a constraint in phase space:

\frac{\partial}{\partial \dot{q}_i }\left [ L(q_j,\dot{q}_j) - \epsilon\,L\left(q_j,\frac{1}{\epsilon}\dot{q}_j\right) \right ] = 0\,,

or

p_i(q_j, \dot{q_j}) - p_i\left(q_j, \frac{1}{\epsilon} \dot{q_j}\right) = 0\,,

so that the image of \dot{q}_i\,-space under p_j(q_i, \dot{q}_i)\, is a surface in p_j\,-space (depending on q_i\,) and

\Phi(q_i, p_i) = 0\,,

for some function \Phi(q_i, p_i)\, of the phase space coordinates. Since the Hamiltonian is zero, the time evolution is completely determined by this constraint, which we may add to the Hamiltonian using a Lagrange multiplier.

References

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