Recall that the Lagrangian path to the dynamics of a particle (which is most
easily made covariant, since it uses
as its variables)
is based on the **Action**

(17.3) |

(17.4) |

The simplest way to make this relativistic is to express it in terms of the
proper time and then require that the action be extremal in all frames.
Then,

(17.5) |

- If the extremal condition is to be invariant with respect to LT's, then must be invariant (and hence a 4-scalar);
- Therefore (since is invariant) must be invariant;
- Finally, since is just a number, must be a 4-scalar.

Note well that it is not clear that this argument (from Jackson) is
valid. , while a number, is not invariant under boosts -
indeed, it is parametrically related to the boost parameter! It is also
perfectly clear that the first statement is false - while it is true
that if is a four-scalar with respect to boosts that its extremums
must be preserved, it is equally true that this condition is not
necessarily unique - all that is *necessary* is that a boost
monotonically scale the action in such a way that the extremal property
is preserved!

A weaker (but still sufficient) argument might then be:

In my opinion this is clearer and still adequate for our purposes. being a 4-scalar (0th rank tensor in 4-space w.r.t. the Lorentz transformation) isIfis a 4-scalar, and is a monotonic function independent of the 4-coordinates, 4-velocities, and , then the property of a given trajectory resulting in an extremum of the action is preserved.

Either way, we will now assert that the Lagrangian of a free particle must be a 4-scalar (and hence must be formed out of the full contraction of tensors of higher rank), and will remain alert in the work below for any sort of inconsistency that might be related to .

Obviously, we want it to reproduce the classical non-relativistic
theory in the appropriate limit, that is, a free particle should have
constant energy and momentum, or, equivalently, 4-velocity. The
simplest (not ``only'' as Jackson states) Lorentz invariant function of
the 4-velocity is it's quadratic form:

(17.6) |

(17.7) |

If we now crunch through the Euler-Lagrange equation we find that this choice
for the constant leads to

(17.8) |

If one chooses a frame where the particle is initially at rest, a trajectory where it remains at rest will yield the least action (you should check this). This is because is maximal when (and the Lagrangian has a minus sign).

Now, suppose that the charged particle is in a electromagnet potential. If it were moving slowly in a scalar potential only, its potential energy would be . The non-relativistic Lagrangian in this case should be just (where is the free particle Lagrangian). The interaction part of the relativistic Lagrangian must therefore reduce to in this non-relativistic limit.

We must find a Lorentz invariant (scalar) form for that reduces to for non-relativistic velocities. Since is the time component of a four vector potential, we can guess that the correct generalization must involve the four vector potential . Since it must be a scalar, it must involve the scalar product of with some four vector(s). The only ones avaliable are and .

The correct
depends only on the . If it
depended on the coordinates as well, then the physics would not be
translationally invariant and the results of our calculation might well depend
on where we chose the origin. This does not seem reasonable. Once again,
this does not uniquely determine it, it only determines the simplest (linear)
form to within a sign and a constant:

(17.9) |

(17.10) |

The complete relativistic Lagrangian for a charged particle is thus

(17.11) |

(17.12) |

The canonical momentum conjugate to the position coordinates is
obtained (as usual) from

(17.13) |

(17.14) |

We make the Hamiltonian out of the Lagrangian and the canonical momentum via

(17.15) |

(17.16) |

(17.17) |

From this result, Hamilton's equations of motion should reproduce the Lorentz
force law. See that it does (although the relationship between the EL
equations and Hamilton's equations makes the result obvious). Note that if we
interpret the Hamiltonian (as usual) as the total energy of the particle,
this result is related to the free particle energy by
and the addition of the scalar potential energy . This is actually just a single change in the four-vector momentum:

(17.18) |

(17.19) |

It is really annoying to obtain the invariance properties of things after the fact. It is also annoying (although perhaps useful) to have the three vector coordinates hanging around at this point. So let us rederive these results using only four-vectors and suitable scalar reductions.