- #1
Terilien
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I can't type in latex so in this post d^2a is the secpnd derivative of a, while (da)^2 is the square of the derivative.
This equation arose from the G_thetatheta compinent of the einstein tensor. Iwas solving tfor the shcwarzchildmetric where where the cosmos constant is nonzero.
the equation is:
e^2a(d^2a +2(da)^2 +2/r(da))= L where L is the cosmological constantand r is the radial coordinate. Umm how do we solve this?
This equation arose from the G_thetatheta compinent of the einstein tensor. Iwas solving tfor the shcwarzchildmetric where where the cosmos constant is nonzero.
the equation is:
e^2a(d^2a +2(da)^2 +2/r(da))= L where L is the cosmological constantand r is the radial coordinate. Umm how do we solve this?