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Poisson Brackets of EM Field

  1. Dec 28, 2012 #1
    Since I couldn't find any reference on the subject of Poisson bracket formalism of classical field theory, I'm posting a few question here:

    A) What are the Poisson brackets of the source-less EM field?
    B) Does the law that the Poisson brackets between a dynamical variable and its conjugate momentum equals unity still hold? Is this law gauge invariant?
    C) In classical particle mechanics, the definition of Poisson brackets includes a sum over all dynamical variables. Is this sum replaced with an integral over space when talking about EM field? Is there a useful notion of "Poisson bracket density" which excludes the integral?

    Thanks
     
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  3. Dec 28, 2012 #2

    Jano L.

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    That is an interesting question. Goldstein in his Classical Mechanics (3rd ed.) says that the formalism of Poisson brackets is not much developed for field theory (like the theory of elastic material or EM field). The difficulty is that in Poisson bracket, we differentiate with respect to coordinates and momenta, but for field there are no such things. Instead, the state of the field is described by field function of space coordinates, which is quite different mathematical object.
     
  4. Dec 28, 2012 #3

    Jano L.

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    Actually, I read about it at 2nd edition, chapter 12, see page 567.
     
  5. Dec 28, 2012 #4

    samalkhaiat

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    The field functions are the coordinates in field theory. And yes you can (at least for unconstraint systems) formulate field theory on Poisson manifold.

    Sam
     
  6. Dec 29, 2012 #5
    So what is the formulation? Or where can I find it?

    When I tried to write down the Poisson bracket formulation myself, I ran into problems. Here is my attempt:

    The coordinates are [itex]{A^\mu }({\bf{x}})[/itex] where [itex]\mu [/itex] runs over the space-time indices and [itex]{\bf{x}}[/itex] runs over all space points.

    The Lagrangian density is [tex]{\cal L} = \left[ {{\partial _\mu }{A_\nu }({\bf{x}}) - {\partial _\nu }{A_\mu }({\bf{x}})} \right]\left[ {{\partial ^\mu }{A^\nu }({\bf{x}}) - {\partial ^\nu }{A^\mu }({\bf{x}})} \right][/tex] where I use summation notation. The Lagrangian density depends on the coordinates [itex]{A^\mu }({\bf{x}})[/itex], and their time derivative [itex]{{\partial _0}{A^\mu }({\bf{x}})}[/itex] evaluated at a specific space point [itex]{\bf{x}}[/itex]. (the space derivatives [itex]{{\partial _i}{A^\mu }({\bf{x}})}[/itex] can be obtained by taking the difference of very close coordinates. This can't be said for time derivatives since the coordinates are functions of a space point [itex]{\bf{x}}[/itex] and not a space time point [itex]x[/itex]. The idea is to find the time dependence of the coordinates (compare to the mechanics of particles)).

    The Lagrangian is [itex]L = \int {{d^3}{\bf{x}}\;} {\cal L}[/itex].

    The canonical conjugate momentum to [itex]{A^\mu }({\bf{x}})[/itex] is:
    [tex]{P_\mu }({\bf{x}}) = \frac{{\delta L}}{{\delta \left( {{\partial _0}{A^\mu }({\bf{x}})} \right)}} = \frac{{\partial {\cal L}({\bf{x}})}}{{\partial \left( {{\partial _0}{A^\mu }({\bf{x}})} \right)}} = {\partial ^0}{A_\mu }({\bf{x}}) - {\partial _\mu }{A^0}({\bf{x}})[/tex]
    (I used the "en.wikipedia.org\wiki\Functional_Derivative" [Broken]. was this correct? It seems that the conjugate momentum is equal to the conjugate momentum density...)

    Now I take the definition of Poisson brackets to be (compare to particle mechanics)
    [tex]\left\{ {F\;,\;G} \right\} = \int {{d^3}{\bf{x'}}\frac{{\delta F}}{{\delta {A^\alpha }({\bf{x'}})}}\frac{{\delta G}}{{\delta {P_\alpha }({\bf{x'}})}} - \frac{{\delta F}}{{\delta {P_\alpha }({\bf{x'}})}}\frac{{\delta G}}{{\delta {A^\alpha }({\bf{x'}})}}} [/tex]where [itex]F[/itex] and [itex]G[/itex] are functionals of [itex]{A^\alpha }[/itex] and of [itex]{{\partial _0}{A^\alpha }}[/itex], and are typically in a form of an integral over these.

    Now, I want to find [itex]\left\{ {{A^\mu }({\bf{x}})\;,\;{P_\nu }({\bf{y}})} \right\}[/itex]. First I write these in a functional form:
    [tex]F \equiv {A^\mu }({\bf{x}}) = \int {{d^3}{\bf{x'}}\;{A^\mu }({\bf{x'}})\delta ({\bf{x}} - {\bf{x'}})} [/tex]
    [tex]G \equiv {P_\nu }({\bf{y}}) = \int {{d^3}{\bf{y'}}\;{P_\nu }({\bf{y'}})\delta ({\bf{y'}} - {\bf{y}})} [/tex]

    Hence:
    [tex]\begin{array}{l}
    \frac{{\delta F}}{{\delta {A^\alpha }({\bf{x'}})}} = \delta ({\bf{x}} - {\bf{x'}})\delta _\alpha ^\mu \\
    \frac{{\delta F}}{{\delta {P_\alpha }({\bf{x'}})}} = 0 \\
    \frac{{\delta G}}{{\delta {A^\alpha }({\bf{x'}})}} = 0 \\
    \frac{{\delta G}}{{\delta {P_\alpha }({\bf{x'}})}} = \delta ({\bf{y}} - {\bf{x'}})\delta _\nu ^\alpha \\
    \end{array}[/tex]

    So finally we get:
    [tex]\left\{ {{A^\mu }({\bf{x}})\;,\;{P_\nu }({\bf{y}})} \right\} = \left\{ {F\;,\;G} \right\} = \int {{d^3}{\bf{x'}}\delta ({\bf{x}} - {\bf{x'}})\delta _\alpha ^\mu \delta ({\bf{y}} - {\bf{x'}})\delta _\nu ^\alpha } = \delta ({\bf{y}} - {\bf{x}})\delta _\nu ^\mu [/tex]

    Oops, seems like I got what I wanted to get... When I wrote it down on a paper I didn't work out. I tried to be as consistent and precise as I can with the mathematical definitions (specifically functional derivatives), and it actually turned out to be okay.

    But actually, there is a problem. As I read in Bjorken-Drell (Fields), the Poisson brackets in the Coulomb gauge are taken to be something else, with "transverse delta" and not "dirac delta". So how is this consistent with my derivation?


    I proceed to the Poisson brackets of the Electric and Magnetic fields:
    [tex]{E_i}({\bf{x}}) = {\partial _i}{A_0}({\bf{x}}) - {\partial _0}{A_i}({\bf{x}}) = {P_i}({\bf{x}}) = \int {{d^3}{\bf{x'}}\;{P_i}({\bf{x'}})\delta ({\bf{x'}} - {\bf{x}})} [/tex]
    [tex]\begin{eqnarray*}
    {B_j}({\bf{y}}) = {\varepsilon _{jkl}}\left[ {{\partial _k}{A_l}({\bf{y}}) - {\partial _l}{A_k}({\bf{y}})} \right]
    & = & {\varepsilon _{jkl}}\int {{d^3}{\bf{y'}}\;\left[ {{\partial _k}{A_l}({\bf{y'}}) - {\partial _l}{A_k}({\bf{y'}})} \right]\delta ({\bf{y'}} - {\bf{y}})}\\
    & = & {\varepsilon _{jkl}}\int {{d^3}{\bf{y'}}\;\left[ {{\partial _k}{A_l}({\bf{y'}})} \right]\delta ({\bf{y'}} - {\bf{y}})} - {\varepsilon _{jkl}}\int {{d^3}{\bf{y'}}\;\left[ {{\partial _l}{A_k}({\bf{y'}})} \right]\delta ({\bf{y'}} - {\bf{y}})} \\
    & = & {\varepsilon _{jkl}}\int {{d^3}{\bf{y'}}\;{A_l}({\bf{y'}}){\partial _k}\delta ({\bf{y'}} - {\bf{y}})} - {\varepsilon _{jkl}}\int {{d^3}{\bf{y'}}\;{A_k}({\bf{y'}}){\partial _l}\delta ({\bf{y'}} - {\bf{y}})}
    \end{eqnarray*} [/tex]
    So
    [tex]\begin{array}{l}
    \frac{{\delta {E_i}({\bf{x}})}}{{\delta {A^\alpha }({\bf{x'}})}} = 0 \\
    \frac{{\delta {E_i}({\bf{x}})}}{{\delta {P_\alpha }({\bf{x'}})}} = \delta ({\bf{x}} - {\bf{x'}})\delta _i^\alpha \\
    \frac{{\delta {B_j}({\bf{y}})}}{{\delta {A^\alpha }({\bf{x'}})}} = 2{\varepsilon _{jk\alpha }}{\partial _k}\delta ({\bf{y}} - {\bf{x'}}) \\
    \frac{{\delta {B_j}({\bf{y}})}}{{\delta {P_\alpha }({\bf{x'}})}} = 0 \\
    \end{array}[/tex]
    Hence
    [tex]\left\{ {{E_i}({\bf{x}})\;,\;{B_j}({\bf{y}})} \right\} = \int {{d^3}{\bf{x'}}\frac{{\delta {E_i}({\bf{x}})}}{{\delta {A^\alpha }({\bf{x'}})}}\frac{{\delta {B_j}({\bf{y}})}}{{\delta {P_\alpha }({\bf{x'}})}} - \frac{{\delta {E_i}({\bf{x}})}}{{\delta {P_\alpha }({\bf{x'}})}}\frac{{\delta {B_j}({\bf{y}})}}{{\delta {A^\alpha }({\bf{x'}})}}} = - \int {{d^3}{\bf{x'}}\delta ({\bf{x}} - {\bf{x'}})\delta _i^\alpha 2{\varepsilon _{jk\alpha }}{\partial _k}\delta ({\bf{y}} - {\bf{x'}})} = - 2{\varepsilon _{ijk}}{\partial _k}\delta ({\bf{y}} - {\bf{x}})[/tex]

    So it seems that the Poisson brackets are gauge invariant, since both sides of the equation are gauge invariant.

    Is any of this correct?
     
    Last edited by a moderator: May 6, 2017
  7. Dec 29, 2012 #6

    dextercioby

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    The source-less/free EM field is a constrained dynamical system, so that a correct Hamiltonian formulation would be developed with the ideas of Dirac.

    Because of the existing 1st class constaints, the fundamental (non-zero) Poisson brackets between the potentials [itex] A_{\mu}(\vec{x},t) [/itex] and their conjugate momentum densities [itex] \pi^{\alpha}(\vec{x},t) [/itex]

    [tex] [A_{\mu}(x), \pi^{\nu}(y)]_{x_0 = y_0} = \delta_{\mu}^{\nu} \delta^{3}(\vec{x}-\vec{y}) [/tex]

    bear no significance for the Hamiltonian dynamics. One needs to impose some gauge conditions on the dynamical variables to obtain a second class system which has the Dirac bracket as the right paranthesis on a reduced phase space. With the Dirac bracket one writes the correct Hamiltonian equations of motion and later obtain a quantization.
     
  8. Apr 18, 2013 #7

    Jano L.

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    I think your derivation of the Poisson brackets is correct, except for the case where ##P_0## occurs. Your Lagrangian does not contain ##\dot A_0##, so ##P_0## is zero. It does not make sense to use it as a canonical coordinate then, at least not in an easy manner. I think one easy way around this problem is to discard the scalar potential entirely and use only the vector potential ##\mathbf A##. Then the resulting Poisson bracket seems right.

    The reason Bjorken&Drell give for the transverse delta is interesting. They say that the commutator

    $$
    [E_r(\mathbf x), A_s(\mathbf y)] = k\delta_{rs} \delta(\mathbf x-\mathbf y)
    $$

    can't be right, because the divergence of the left-hand side in ##\mathbf x## can give zero for field which has ##\nabla \cdot \mathbf E = 0##, while the right-hand side is always non-zero distribution ##k\partial_s \delta(\mathbf x-\mathbf y)##. This argument presumes that one can smuggle in the differentiation inside the commutator. I do not know the justification for this.

    In the classical case, it works, but the fact that ##\nabla \cdot \mathbf E\ = 0## does not mean that the functional derivatives will come out zero. The reason is that the actual field may very well be divergence-less, but the functional derivatives do not preserve this when variations of the field are performed, the same way partial differentiation does not preserve consistency of the variables with the actual motion in ordinary mechanics.

    Craig&Thirunamachandran, sec. 3.5, say that the transverse delta occurs in the commutator because the vector potential is restricted by the Coulomb gauge

    $$
    \nabla \cdot \mathbf A = 0.
    $$

    I do not understand why this is so in their presentation, but perhaps it can be seen after the commutation relation is cast into Fourier representation - E, A are perpendicular to k, so the right-hand side should be too.


    I would like to admit a change of my view above, I thought about this a little bit and now I think Goldstein's reservations towards the Poisson bracket in field theory are perhaps not that serious problem. Still one point of his discussion I do not understand. He says in sec 124., p. 567

    "
    Further, if x_i plays the role of continuous indices on the mechanical variables, then fundamental Poisson brackets should involve functions at different values of x_i, which is not easily brought into the present formulation. For this reason there has been little exploration of canonical transformations for classical fields, a subject that for discrete systems proved to be so rich and consequential.
    "

    Can anybody explain?

    EDIT: I edited minor mistake above.
     
    Last edited: Apr 18, 2013
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