Tuesday, 3 September 2019

electromagnetism - Is there a Hamiltonian for the (classical) electromagnetic field? If so, how can it be derived from the Lagrangian?


The classical Lagrangian for the electromagnetic field is


L=14μ0FμνFμνJμAμ.


Is there also a Hamiltonian? If so, how to derive it? I know how to write down the Hamiltonian from the Lagrangian where derivatives are taken only with respect to time, but I can't see the obvious way to generalize this.



Answer




Yes. There is a standard way to generalize to field theory.


Let a theory of n1 fields ϕi with a Lagrangian density L=L(ϕi,μϕi) be given. Here we use that standard abuse of notation in which ϕi denotes the vector whose components are the fields; ϕi=(ϕ1,,ϕn).


To obtain the corresponding hamiltonian density, one first defines the following canonical momentum corresponding to the field ϕi: πi(x)=L˙ϕi(ϕi(x),μϕi(x)),˙ϕi:=tϕi

Then, the Hamiltian density is H=πi˙ϕiL
where a sum over i is implied. Note that as in classical mechanics, on the right hand side of this expression, ˙ϕi should be replaced with its expression in terms of πi,ϕi so that the Hamiltonian is a function of (πi,ϕi) only, namely H(πi,ϕi)=πj˙ϕj(πi,ϕi)L(ϕi,˙ϕ(πi,ϕi)).
Again we have abused notation slightly here in writing ˙ϕi as a function of πi and ϕi. What we mean is the expression for ˙ϕi is obtained by solving the definition (1) of the canonical momentum for ˙ϕi in terms of πi and ϕi.


In your case, the fields are Aμ with corresponding momenta πμ.


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