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Canonical conjugate momenta of EM Field Lagrangian density

Physics Asked by sawan kt on May 30, 2021

I have the EM Field Lagrangian density given as

$
mathcal{L} =- frac{1}{4} F_{mu nu} F^{mu nu}
$

where $F^{mu nu}$ is the Field strength tensor defined as $F^{mu nu} = partial^mu A^nu- partial^nu A^mu$, where $A^nu = (phi/c, A^x,A^y A^z)$

I need to find the canonical conjugate momenta ($pi^alpha$) w.r.t. this lagrangian density to find the primary constraint.

$pi^alpha = frac{partial L}{partial (partial_0 A_alpha)} $

I am unable to solve it. please help me. ( I tried by expanding the field strength tensor and then solving it, but that didn’t help)

The solution is $pi^alpha = – F^{0 alpha}$

One Answer

I'll do it step by step. First lower all the indices:

begin{equation} -frac{1}{4}F_{mu nu}F^{munu}=-frac{1}{4}F_{mu nu}g^{murho}g^{nusigma}F_{rhosigma} end{equation}

then expand the products,

begin{equation} -frac{1}{4}(partial_{mu}A_{nu}partial_{rho}A_{sigma}-partial_{mu}A_{nu}partial_{sigma}A_{rho}-partial_{nu}A_{mu}partial_{rho}A_{sigma}+partial_{nu}A_{mu}partial_{sigma}A_{rho})g^{murho}g^{nusigma} end{equation}

Using the symmetry of the metric, I can rename the indices and then switch them, to get

begin{equation} -frac{1}{4}(2partial_{mu}A_{nu}partial_{rho}A_{sigma}-2partial_{mu}A_{nu}partial_{sigma}A_{rho})g^{murho}g^{nusigma} end{equation}

Now let us take a functional derivative with respect to $partial_{lambda}A_{alpha}$

begin{equation} frac{partial}{partial(partial_{lambda}A_{alpha})}(partial_{mu}A_{nu}partial_{rho}A_{sigma})g^{murho}g^{nusigma}=(delta^{lambda}_{mu}delta^{alpha}_{nu}partial{rho}A_{sigma}+delta^{lambda}_{rho}delta^{alpha}_{sigma}partial_{mu}A_{nu})g^{murho}g^{nusigma}=2partial^{lambda}A^{alpha} end{equation}

Similarly for the other term. Hence, begin{equation} frac{partial}{partial(partial_{lambda}A_{alpha})}left(-frac{1}{4}F_{mu nu}g^{murho}g^{nusigma}F_{rhosigma}right)=-left(partial^{lambda}A^{alpha}-partial^{alpha}A^{lambda}right)=-F^{lambdaalpha} end{equation}

Now setting $lambda=0$ gives the desired result.

Correct answer by Ruben Campos Delgado on May 30, 2021

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