Landau’s gauge free solutions

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Carlo Maria Becchi and Camillo Imbimbo (2008), Scholarpedia, 3(11):7135. revision #44660 [link to/cite this article]

Curator: Dr. Carlo Maria Becchi, Genoa University, Italy
Curator: Dr. Camillo Imbimbo, Genoa University, Italy

Landau’s gauge free solutions

Decomposing the vector potential in its physical and unphysical parts, A_\mu(x)=A^{(ph)}_\mu(x)+A^{(u)}_\mu(x), the general solution of electrodynamic equations in Landau’s gauge reads as follows

A^{(ph)}_\mu(x)=\int {d^4 k\over(2\pi)^{3/2}}e^{-ik\cdot x}\theta(k_0)\left[\delta(k^2)\sum_{h=\pm 1}\epsilon_\mu(\vec k, h) a(\vec k,h)\right]+ c.-c.\ ,
A^{(u)}_\mu(x)=i\int {d^4 k\over(2\pi)^{3/2}}e^{-ik\cdot x}\theta(k_0)\left[\delta(k^2)\left(k_\mu\alpha(\vec k) -\bar k_\mu {\beta(\vec k)\over k\cdot\bar k}\right)- k_\mu\delta'(k^2)\beta(\vec k)\right]+ c.-c.\ ,
b(x)=\int  {d^4 k\over(2\pi)^{3/2}}e^{-ik\cdot x}\theta(k_0)\delta(k^2)\beta(\vec k)+ c.-c.\

where:

The polarization vectors define the unpolarized photon density matrix

\sum_{h=\pm}\epsilon _\mu(\vec k, h)\epsilon^*_\nu(\vec k, h)=-g_{\mu\nu}+{k_\mu\bar k_\nu+\bar k_\mu k_\nu\over k\cdot\bar k}\ .

It is easy to verify, using the identity x\delta'(x)=-\delta(x) and x\delta(x)=0, that for a generic choice of the functions a\ ,\alpha\ ,\ \beta the above equations give the general solution to the Landau's gauge free field equations.


Invited by: Dr. Eugene M. Izhikevich, Editor-in-Chief of Scholarpedia, the peer-reviewed open-access encyclopedia
Invited by: Dr. Riccardo Guida, Institut de Physique Théorique; CEA, IPhT; CNRS; Gif-sur-Yvette, France
Action editor: Dr. Riccardo Guida, Institut de Physique Théorique; CEA, IPhT; CNRS; Gif-sur-Yvette, France
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