Primary constraint of electrodynamics

Question

I have some problems understanding the transition from the Lagrangian to Hamiltonian formalism of electrodynamics. I will use the metric $(-+++)$.

I want to start from the Lagrangian which is invariant for parameterization for a free particle: $$ L=- m \sqrt{-\frac{dx^\mu}{d\lambda}\frac{dx_\mu}{d\lambda}}. $$ The conjugated momenta of $\dot x_\mu$ is

\begin{equation} (*) \hspace{1cm}p_\mu= \frac{\partial L}{\partial \dot x_\mu} = m \frac{\dot x_\mu}{\sqrt{-\dot x_\mu \dot x^\mu}}. \end{equation}

Performing Legendre transformation we get $$ H=\dot x_\mu p^\mu -L = 0. $$ So, fixing the gauge $x^\mu=( \tau , \vec x)$, we can write $$ p^\mu=\begin{cases} p^0= m\frac{1}{\sqrt{1- \dot x_i \dot x^i}}\\ p^i = m\frac{\dot x^i}{\sqrt{1-\dot x_i \dot x^i}} \end{cases} \Rightarrow \boxed{\dot x^i =\frac{p^i}{p_0}} $$ And using it in the definition of $p^\mu$ (*) we can rewrite and obtain

$$ p_\mu p^\mu + m^2=0 $$ Which is a primary constraint, and it appears in the Hamiltonian as

$$ H= \xi (p_\mu p^\mu + m^2). $$ Where $\xi$ is a Lagrange multiplier. To conciliate this equation with the motion equation via least principle action we find that the form of $\xi$ must leads to $$ H= \sqrt{p_\mu p^\mu + m^2}. $$

My problem now is that I can't perform the same arguments for the total Lagrangian of the electrodynamics. $$ L=- m \sqrt{-\frac{dx^\mu}{d\lambda}\frac{dx_\mu}{d\lambda}} - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} - e \frac{dx^\mu}{d\lambda} \delta^{(4)}(x-x(\lambda))A_\mu(x(\lambda)). $$ I can't include the potential vector $A^\mu$ and I don't know how to treat the Lagrange multiplier for the Hamiltonian.

Answer 1

The Lagrangian you wrote is not quite a Lagrangian. When mixing fields and degrees of freedom, it's better to just write the action: $$ S = -\int\left[m\sqrt{-\dot x^\mu \dot x_\mu}+e\dot x^\mu A_\mu(x)\right]d\lambda -\frac{1}{4}\int F_{\mu\nu}(y)F^{\mu\nu}(y)d^4y $$ and vary $x$ and $A$.

If you're assuming the EM field to be fixed, you can disregard the second term and the bracketed term is you Lagrangian.

Btw, your gauge fixing perhaps you meant $x^0=\lambda$ (and not $\tau$)?

You don't need the gauge fixing to see the constraint. From: $$ p_\mu = m\frac{\dot x_\mu}{\sqrt{-\dot x^\nu \dot x_\nu}} $$ you get directly by squaring: $$ p_\mu p^\mu = -m^2 $$

Similarly, with minimal coupling, you get: $$ p_\mu = m\frac{\dot x_\mu}{\sqrt{-\dot x^\nu \dot x_\nu}}+eA_\mu(x) $$ You can reduce to the previous by considering: $$ p_\mu - eA_\mu(x) = m\frac{\dot x_\mu}{\sqrt{-\dot x^\nu \dot x_\nu}} $$ and squaring: $$ (p_\mu - eA_\mu(x))(p^\mu - eA^\mu(x)) = -m^2 $$

If you really want to fix the same gauge, you can just vary the spatial coordinates and this gives the Hamiltonian: $$ H = p_0 = -eA_0+\sqrt{m^2+(\mathbf x+e\mathbf A)^2} $$

Btw, if you also want to vary the EM field, you can rewrite the action as: $$ S = -\int m\sqrt{-\dot x^\mu \dot x_\mu}d\lambda-\int\left[\int e\dot x^\mu A_\mu(y)\delta(y-x)d\lambda +\frac{1}{4}F_{\mu\nu}(y)F^{\mu\nu}(y)\right]d^4y $$ You can disregard the first term and the bracket gives you the Lagrangian density. So you get the usual Maxwell equation in covariant form with the four current density: $$ j^\mu (y) = \int e \delta(y-x)dx^\mu $$

Hope this helps.