Lagrangian mechanics

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We wish to find an action that reproduces the equation of motion describing a free relativistic point particle, i.e.,

\frac{d p^\mu}{d\tau}=0\,.

Our action should be independent of any choice of inertial reference frame. Let us parameterize the trajectory by some parameter \lambda\, using the function x^\mu(\lambda)\,. The simplest invariant is the particle's proper time:

\tau = \int\!d\lambda \sqrt{ -\eta_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} }\,,

so we make the following guess as to the action:

S = - m c \int\!d\lambda \sqrt{ -\eta_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} }\,.

Denoting \tfrac{dx^\mu}{d\lambda} = \dot{x}^\mu\,, the equations of motion that result are

m c \frac{d}{d\lambda} \left( \frac{\dot{x}^\mu }{ \sqrt{-\dot{x}^\nu \dot{x}_\nu}} \right) = 0\,.

We may further divide both sides by \sqrt{-\dot{x}^\nu \dot{x}_\nu}\, which we realise is \tfrac{d\tau}{d\lambda}\,:

m \frac{d\lambda}{d\tau} \frac{d}{d\lambda} \left( \frac{d\lambda}{d\tau} \frac{dx^\mu}{d\lambda} \right) = 0\,,

or

\frac{d}{d\tau} \left( m \frac{dx^\mu}{d\tau} \right) = \frac{dp^\mu}{d\tau}= 0\,.

The choice of the overall factor in the action will become evident when we find the Hamiltonian. The canonical momentum is

P_\mu = m c \frac{\dot{x}_\mu }{ \sqrt{-\dot{x}^\nu \dot{x}_\nu}}\,,

so that

H = P_\mu \dot{x}^\mu - L = m c \frac{\dot{x}_\mu \dot{x}^\mu}{ \sqrt{-\dot{x}^\nu \dot{x}_\nu}} + m c \sqrt{ -\dot{x}^\nu \dot{x}_\nu} = 0\,.

The Hamiltonian is zero, and the Legendre transformation is not invertible. This is a consequence of the reparameterization invariance of the action. We can break reparameterization invariance by choosing \lambda = x^0 = c t\,. Then

S = - m c^2 \int\!dt\sqrt{ 1 - \frac{\vec{v}^2}{c^2} }\,,

and

p^i = \gamma m v^i\,,

whence

H\, = \gamma m v^2 + m c^2 \gamma^{-1}\,,
= \gamma^{-1}m c^2( \gamma^2 \frac{v^2}{c^2} + 1)\,,
= \gamma^{-1}m c^2\left( \frac{ \frac{v^2}{c^2} }{1- \frac{\vec{v}^2}{c^2} } + 1 \right)\,,
= \gamma^{-1}m c^2\left( \frac{ 1 }{1- \frac{\vec{v}^2}{c^2} }\right)\,,
= \gamma m c^2\,,

which is the usual expression for energy. Therefore, the original action gives the correct equations of motion and values for energy and momentum for the parameterization choice \lambda = c t\,. On the other hand, it doesn't seem general enough to handle massless particles such as the photon.

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