A charged relativistic free particle has the Hamiltonian in general:

$$ \mathcal{H} = \sqrt{{\bf p}^2c^2+m^2c^4}.$$

I read somewhere that says, it is possible to go further and say that the EoM are Hamilton's equations. But it is not done as there is "less interest" in such a discussion.

Is there something deeper to this? Like another formalism is ''better''.

(My guess is a more trivial one though. That is, it is not useful because the equations just get very cluttered and ugly)

A relativistic free particle has the Hamiltonian in general:

$$ \mathcal{H} = \sqrt{{\bf p}^2c^2+m^2c^4}.$$

I read somewhere that says, it is possible to go further and say that the EoM are Hamilton's equations. But it is not done as there is "less interest" in such a discussion.

Is there something deeper to this? Like another formalism is ''better''.

(My guess is a more trivial one though. That is, it is not useful because the equations just get very cluttered and ugly)

A charged relativistic free particle has the Hamiltonian in general:

$$ \mathcal{H} = \sqrt{p^2c^2+m^2c^4}.$$

I read somewhere that says, it is possible to go further and say that the EoM are Hamilton's equations. But it is not done as there is "less interest" in such a discussion.

Is there something deeper to this? Like another formalism is ''better''.

(My guess is a more trivial one though. That is, it is not useful because the equations just get very cluttered and ugly)

A charged relativistic free particle has the Hamiltonian in general:

$$ \mathcal{H} = \sqrt{{\bf p}^2c^2+m^2c^4}.$$

I read somewhere that says, it is possible to go further and say that the EoM are Hamilton's equations. But it is not done as there is "less interest" in such a discussion.

Is there something deeper to this? Like another formalism is ''better''.

(My guess is a more trivial one though. That is, it is not useful because the equations just get very cluttered and ugly)