In , the Lagrangian for a charged particle is given as (12.9) as
Let’s work in detail from this to the Lorentz force law and the Hamiltonian and from the Hamiltonian again to the Lorentz force law using the Hamiltonian equations. We should get the same results in each case, and have enough details in doing so to render the text a bit more comprehensible.
We need the conjugate momenta for both the Euler-Lagrange evaluation and the Hamiltonian, so lets get that first. The components of this are
In vector form the canonical momenta are then
Completing the Euler-Lagrange equation evaluation is the calculation of
On the left hand side we have
and on the right, with implied summation over repeated indexes, we have
Putting things together we have
the first two terms are recognizable as the electric field. To put some structure in the remainder start by writing
The remaining term, with is now
We are left with the momentum portion of the Lorentz force law as expected
Observe that with a small velocity Taylor expansion of the Lagrangian we obtain the approximation
If that is our starting place, we can only obtain the non-relativistic approximation of the momentum change by evaluating the Euler-Lagrange equations
That was an exercise previously attempting working the Tong Lagrangian problem set .
Having confirmed the by old fashioned Euler-Lagrange equation evaluation that our Lagrangian provides the desired equations of motion, let’s now try it using the Hamiltonian approach. First we need the Hamiltonian, which is nothing more than
However, in the Lagrangian and the dot product we have velocity terms that we must eliminate in favor of the canonical momenta. The Hamiltonian remains valid in either form, but to apply the Hamiltonian equations we need , and not .
We can rearrange 2 for
but also has a dependence, so this is not complete. Squaring gets us closer
and a bit of final rearrangement yields
Writing , we can rearrange and find
Also, taking roots of 16 we must have the directions of and differ only by a rotation. From 14 we also know that these are colinear, and therefore have
This and 17 can now be substituted into 13, for
Dividing out the common factors we finally have the Hamiltonian in a tidy form
Hamiltonian equation evaluation.
Let’s now go through the exercise of evaluating the Hamiltonian equations. We want the starting point to be just the energy expression 20, and the use of the Hamiltonian equations and none of what led up to that. If we were given only this Hamiltonian and the Hamiltonian principle
how far can we go?
For the particle velocity we have no dependence and get
This is 18 in coordinate form, one of our stepping stones on the way to the Hamiltonian, and we recover it quickly with our first set of derivatives. We have the gradient part of the Hamiltonian left to evaluate
This looks nothing like the Lorentz force law. Knowing that is of significance (because we know where we started which is kind of a cheat), we can subtract derivatives of this from both sides, and use the convective derivative operator (ie. chain rule) yielding
The first and last terms sum to the electric field, and we seen evaluating the Euler-Lagrange equations that the remainder is . We have therefore gotten close to the familiar Lorentz force law, and have
The only untidy detail left is that doesn’t look much like , what we recognize as the relativistically corrected momentum. We ought to have that implied somewhere and 23 looks like the place. That turns out to be the case, and some rearrangement gives us this directly
This completes the exercise, and we’ve now obtained the momentum part of the Lorentz force law. This is still unsatisfactory from a relativistic context since we do not have momentum and energy on equal footing. We likely need to utilize a covariant Lagrangian and Hamiltonian formulation to fix up that deficiency.
 JD Jackson. Classical Electrodynamics Wiley. 2nd edition, 1975.
 Dr. David Tong. Classical Mechanics Lagrangian Problem Set 1. [online]. http://www.damtp.cam.ac.uk/user/tong/dynamics/mf1.pdf.