Massive Vector Boson E.o.M.: Minimize Action & Lorentz Gauge?

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SUMMARY

The discussion focuses on deriving the equations of motion (E.o.M.) for a massive vector boson from the Lagrangian density L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{m^{2}}{2} B_{\mu}B^{\mu}. The E.o.M. is established as ( \partial^{2} + m^{2}) B^{\mu} - \partial^{\mu} (\partial B) = 0. A key point raised is the challenge of eliminating the term \partial B, which is linked to the Lorentz Gauge condition. The solution involves taking the derivative of the E.o.M., leading to the conclusion that \partial B = 0, thereby confirming the necessity of this condition in minimizing the action.

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ChrisVer
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Starting from the Lagrangian density:
[itex]L= -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{m^{2}}{2} B_{\mu}B^{\mu}[/itex]
we can derive the E.o.M. for the field [itex]B[/itex] which read:

[itex]( \partial^{2} + m^{2}) B^{\mu} - \partial^{\mu} (\partial B) = 0[/itex]
In the case of a massive field, I am not sure how I can kill out the partial of B through the field equations...
[itex]\partial B=0[/itex]
Does this come as a constraint/boundary condition of minimizing the action? or is there something I cannot see? In most cases they state it's a Lorentz Gauge, however I am not sure how this can be indeed shown...
 
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http://www.theory.nipne.ro/~poenaru/PROCA/proca_rila06.pdf
P.180 , eq 19,20 and the 1st paragraph of p.181 gave the answer... One has to take the derivative of the Equations of Motion [itex]\partial_{\mu}[/itex] and [itex]\partial B[/itex] comes out zero... (unfortunately for me I came out with the wrong EoM missing a minus sign and I couldn't even think of doing it)
 

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