I am aware of Maxwell Equations in Curved Spacetime. But how do these equations change if the cosmological constant is not assumed to vanish?


Adding a cosmological constant $\Lambda$ won't change Maxwell's equations (which are obtained when you vary the action with respect to the field $A_\mu$), but it will change the equations of motion in the gravity sector. See Supersymmetric, cold and lukewarm black holes in cosmological Einstein-Maxwell theory, for instance.

Also this wikipedia article.

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  • $\begingroup$ The metric appears in Maxwell’s equations, and the cosmological constant helps determine the metric. $\endgroup$ – G. Smith Jul 15 '19 at 21:41
  • $\begingroup$ @G.Smith Right. But we're talking at the level of the equations of motion, not solutions of it. These are 2 very different things. $\endgroup$ – Avantgarde Jul 15 '19 at 21:45
  • $\begingroup$ I have no idea what you mean. The metric appears in the equations, and therefore you get different solutions for different metrics. $\endgroup$ – G. Smith Jul 15 '19 at 22:13
  • $\begingroup$ @G.Smith I believe Avantgarde means that, indeed, the solutions are different for different metrics; but the form of the equation itself (i.e., $dA=d^\dagger A=0$) is the same, whether you turn on $\Lambda$ or not. $\endgroup$ – AccidentalFourierTransform Jul 16 '19 at 0:25
  • $\begingroup$ @AccidentalFourierTransform Yes, but that because the form of those equations has obscured the presence of the metric. When you solve Maxwell’s equations in curved spacetime, the metric matters. $\endgroup$ – G. Smith Jul 16 '19 at 0:29

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