Physics Asked by AFG on April 15, 2021
In this paper, Toms discusses the method that Faddeev and Jackiw proposed for quantization of constrained theories. In section III.B, he applies this method to a field theory, but I have several doubts about how this transition is done.
A first-order Lagrangian for a mechanical system can be written as
begin{equation}tag{2.1}
L=A_alpha(xi)dot{xi}^alpha+L_v(xi),
end{equation}
whose Euler-Lagrange equations are
begin{equation}tag{2.2}
F_{alphabeta}dot{xi}^beta=-frac{partial L_v}{partialxi^alpha},
end{equation}
with
begin{equation}tag{2.3}
F_{alphabeta}=frac{partial A_beta}{partialxi^alpha}-frac{partial A_alpha}{partialxi^beta}
end{equation}
being the symplectic two-form. So far so good.
Later, on section III.B, he says that the traduction of expresion (2.3) to field theory is
begin{equation}tag{3.41}
F_{alphabeta}=frac{delta A_beta}{deltaxi^alpha}-frac{delta A_alpha}{deltaxi^beta},
end{equation}
but I can’t see how this araises from Euler-Lagrange equations, i.e.
begin{equation}tag{EL}
frac{delta S}{deltaphi^alpha}=frac{partialmathcal{L}}{partialphi^alpha}-partial_muBigg(frac{partialmathcal{L}}{partial({partial_muphi^alpha})}Bigg)=0,
end{equation}
with
begin{equation}
S[phi]=int d^4{x} mathcal{L}(phi,partialphi).
end{equation}
On a field theory, I suppose that a first-orden lagrangian density would be written as
begin{equation}tag{LD}
mathcal{L}=A_alpha(phi,nablaphi)dot{phi}^alpha+mathcal{L}_v(phi,nablaphi),
end{equation}
where $phi^alpha$ are the fields and $nablaphi$ indicates dependence in spatial derivatives. However, if I insert that lagrangian density into (EL) I can’t get an expression equivalent to (2.2). Moreover, the "$A$"s in expresion (3.41) must be functionals. Which is the relation between those "$A$"s and the ones in (LD)?
OP is essentially asking the following.
How does the point-mechanical Faddeev-Jackiw formulas from chapter II generalizes to the field-theoretic formulas in section III.B?
Well, let's see. The field-theoretic first-order Lagrangian is $$ L~=~int!d^3{bf x}~A_{alpha}({bf x}) dot{xi}^{alpha}({bf x})~-~H.tag{2.1'}$$ An infinitesimal variation reads $$begin{align} delta L ~sim~& int!d^3{bf x}left( delta A_{alpha}({bf x}) dot{xi}^{alpha}({bf x}) -frac{dA_{alpha}({bf x})}{dt} deltaxi^{alpha}({bf x})right) ~-~delta H cr ~=~&int!d^3{bf x} int!d^3{bf x}^{prime} left( deltaxi^{alpha}({bf x}) frac{delta A_{beta}({bf x}^{prime})}{delta xi^{alpha}({bf x})} dot{xi}^{beta}({bf x}^{prime}) ~-~ dot{xi}^{beta}({bf x}^{prime}) frac{delta A_{alpha}({bf x})}{delta xi^{beta}({bf x}^{prime})} deltaxi^{alpha}({bf x}) right) cr &~-~ delta Hcr ~=~&int!d^3{bf x} int!d^3{bf x}^{prime} deltaxi^{alpha}({bf x}) F_{alphabeta}({bf x},{bf x}^{prime})dot{xi}^{beta}({bf x}^{prime}) ~-~ int!d^3{bf x}~ delta xi^{alpha}({bf x}) frac{delta H}{delta xi^{alpha}({bf x})}. end{align}$$ Here the $sim$ symbol means equality modulo total time derivative terms. Also we have defined the components of symplectic 2-form $$ F_{alphabeta}({bf x},{bf x}^{prime})~=~frac{delta A_{beta}({bf x}^{prime})}{deltaxi^{alpha}({bf x})}-frac{delta A_{alpha}({bf x})}{deltaxi^{beta}({bf x}^{prime})}.tag{2.3'/3.41'} $$ (Judging from eq. (3.42) it becomes clear that eq. (3.41) is bi-local.)
The Euler-Lagrange equations are Hamilton's equations $$ int!d^3{bf x}^{prime} F_{alphabeta}({bf x},{bf x}^{prime})dot{xi}^{beta}({bf x}^{prime}) ~=~ frac{delta H}{delta xi^{alpha}({bf x})}.tag{2.2'} $$
References:
Correct answer by Qmechanic on April 15, 2021
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