How do we explain the fact that if we want to get Hamilton's equations out of Lagrangian:
L=-mc\sqrt{(\dot x^\mu \dot x_\mu)},
where dot is noting derivative in proper time, we get that Hamilton's function equals zero ?
This is a general feafure of the COVARIANT Hamiltonian formulation of a theory. By taking the time PARAMETER as an EXTRA DYMANICAL VARIABLE, we can transform any theory into the covariant form in which the evolution parameter is arbitrary. As I will show you bellow, the introduction of an arbitrary time into Hamiltonian theory implies the existence of one 1st class constraint and a vanishing Hamiltonian. So, if the action (like yours) is "already covariant", i.e., it treats time as coordinate from the very begining, then the Hamiltonian vanishes identically as a result of parametrization-invariance.
To explain the problem of covariance, let us consider the very simple dynamical system of free non-relativistic particle
L(t) = \frac{1}{2} ( \frac{dx}{dt})^{2}
Notice that t is an evolution parameter, i.e., it is not a dynamical variable! well, peopel like myself do not like this fact, so we introduce a new evolution parameter (new time) \delta as a monotonic function of t, and take t \equiv x_{0} as an extra coordinate, i.e., we consider the new Lagrangian
L( \delta ) = \frac{1}{2} \frac{(dx/d \delta)^{2}}{dx_{0} /d \delta} \equiv \frac{1}{2} \frac{\dot{x}^{2}}{\dot{x}_{0}}
as a function of the coordinates (x_{0},x) and time \delta, so that
\int L(t) \ dt = \int L( \delta ) \ d \delta
By introducing the corresponding momenta
<br />
P^{0} = \frac{ \partial L( \delta)}{ \partial \dot{x}_{0}} = - (1/2) ( \dot{x} / \dot{x}_{0})^{2}<br />
<br />
P = \frac{ \partial L( \delta)}{ \partial \dot{x}} = \dot{x} / \dot{x}_{0}<br />
we find that the canonical Hamiltonian vanishes identically;
H_{c} \equiv \dot{x}_{0} P^{0} + \dot{x} P - L( \delta) = 0
This means that there exists at least one 1st class primary constraint. Indeed, we do have it;
\phi = P_{0} + \frac{1}{2} P^{2} \approx 0
so, the total Hamiltonian is of the form
H_{T} = f \phi
where f is an arbitrary multiplier, and the equations of motion are
<br />
\dot{x}_{0} = \{x_{0}, \phi \} f = f, \ \ \dot{P}_{0} = 0<br />
<br />
\dot{x} = \{x, \phi \} f = P f, \ \ \dot{P} = 0<br />
Notice that this system of equation is form invariant (covariant) under the arbitrary infinitesimal transformations
x_{0} \rightarrow x_{0} + \epsilon , \ \ P^{0} \rightarrow P^{0}
x \rightarrow x + \epsilon P , \ \ P \rightarrow P
To go back to time t, we choose the "gauge" x_{0} - \delta \approx 0. In this gauge the equations take the standard form
f = 1 , \ \ \dot{x} = P
In short; parametrization-invariance ( which shows up because of the way in which we formulate the action) leads to off-shell mathematical identities, i.e., the identities hold wether or not the action is extremized. It is this fact that makes the identity H = 0 different from the conserved quantity (1/2)P^{2} which holds only on-shell.
regards
sam
It is of great use since in that way people are catching those errors that I became blind to in the derivation. They are extremely few though.
