An Introduction To Quantum Field Theory - M. Peskin, D. Schroeder (Perseus, 1995) · Problem 2.1

The Klein-Gordon Field — M. Peskin, D. Schroeder (Perseus, 1995)

Physics — QFT · classical-electromagnetism-action

Problem

Classical electromagnetism (with no sources) follows from the action S=∫d4x(−14FμνFμν),where Fμν=∂μAν−∂νAμ.S = \int d^4x \left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right), \qquad \text{where } F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu. (a) Derive Maxwell's equations as the Euler-Lagrange equations of this action, treating the components Aμ(x)A_\mu(x) as the dynamical variables. Write the equations in standard form by identifying Ei=−F0iE^i = -F^{0i} and ϵijkBk=−Fij\epsilon^{ijk}B^k = -F^{ij}. (b) Construct the energy-momentum tensor for this theory. Note that the usual procedure does not result in a symmetric tensor. To remedy that, we can add to TμνT^{\mu\nu} a term of the form ∂λKλμν\partial_\lambda K^{\lambda\mu\nu}, where KλμνK^{\lambda\mu\nu} is antisymmetric in its first two indices. Such an object is automatically divergenceless, so T^μν=Tμν+∂λKλμν\widehat{T}^{\mu\nu} = T^{\mu\nu} + \partial_\lambda K^{\lambda\mu\nu} is an equally good energy-momentum tensor with the same globally conserved energy and momentum. Show that this construction, with Kλμν=FμλAν,K^{\lambda\mu\nu} = F^{\mu\lambda}A^\nu, leads to an energy-momentum tensor T^\widehat{T} that is symmetric and yields the standard formulae for the electromagnetic energy and momentum densities: E=12(E2+B2);S=E×B.\mathcal{E} = \frac{1}{2}(E^2 + B^2); \qquad \mathbf{S} = \mathbf{E} \times \mathbf{B}.

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