
#1
Jun807, 03:12 PM

P: 140

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? 



#2
Jun807, 08:07 PM

P: 1,705

is that right? 



#3
Jun807, 08:59 PM

P: 140

Yes it is. can you helpme?
essentially I determined from G_tt and G_rr that b=a just like with the ordinary metric. When I plugged it into the G_thetathata equation, it was still fairly ugly, and icouldnt find a solution by inspection. Non linears are annoying. 



#4
Jun1007, 08:18 PM

P: 66

non linear differential equation
Is it not
[tex]a^{\prime \prime} + 2 a^{\prime 2} + \frac{2a}{r} = L e^{2a}[/tex] and, presumably the derivative is with respect to r, yes? 



#5
Jun1007, 08:43 PM

P: 140

yes it is. It may be that. It is possible that I've made a small mistake in my calculations of miswrote someon paper. How do we solve that? Can it be done by inspection? non linearity scares me.
could someone help me solve it? 


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