A Derivation of Equation 26 in Bardeen's Four Laws of Black Hole Thermodynamics

  • A
  • Thread starter Thread starter thatboi
  • Start date Start date
  • Tags Tags
    Derivation Paper
thatboi
Messages
130
Reaction score
20
Hi all,
I am currently reading Bardeen's Paper on The Four Laws of Black Hole Thermodynamics: https://projecteuclid.org/journals/...ws-of-black-hole-mechanics/cmp/1103858973.pdf
and am struggling with the derivation of equation 26. Specifically, I do not see how he uses the Lie derivative statement just above the equation to make the 2 terms disappears in the second equality because one of the terms involves the differential of n as opposed to l.
 
Physics news on Phys.org
Hm, that's indeed quite a headache. The little comment about the Lie derivative looks like it contains a typo (!), ought to be ##(L_{l} \delta l)^a = l^b \nabla_b (\delta l)^a - (\delta l)^b \nabla_b l^a = 0##, no?
 
  • Like
Likes robphy and vanhees71
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top