Robert McNees
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 Nov 12 comment Cubic term in gauge theories Adding F^3 to the standard YM Lagrangian gives a non-renormalizable theory; you should only consider it as part of a low-dimensional effective theory. F^3 is the "least irrelevant" gauge-invariant operator for pure YM. So the low-energy expansion of some theory that holds at higher energies might have F^2 as the leading term and F^3 term as the first "correction". Nov 9 answered Cubic term in gauge theories Nov 7 awarded Yearling Nov 7 answered Einstein tensor in Friedmann equations : where is the missing $c^2$? Sep 3 answered How do I calculate the induced metric in the Gibbonsâ€“Hawkingâ€“York boundary term? Aug 12 awarded Citizen Patrol May 28 awarded Critic Aug 12 comment Does the lack of modular nuclearity in string theory mean anything? You don't even need quantum gravity to see that extensivity breaks down when gravity is important. Solve the Tolman-Oppenheimer-Volkov equation for a self-gravitating perfect fluid with equation of state $p=\kappa\rho$. This gives an energy density $\rho \sim r^{-2}$. Cut off the solution at some radial size $R$ and join it on to a solution that goes to zero at a finite distance. Relate the entropy density to the energy density using the EOS, integrate, and you get entropy $\sim R^{\frac{1+3\kappa}{1+\kappa}}$. The entropy of the self-gravitating object never grows faster than the area. Aug 2 comment Cheat sheet of elementary particles Here is a handy Eightfold Way reference chart from an undergrad class I taught on particle physics: cl.ly/8yhy . Of course, it will only be useful if you understand what it represents! There is a good review in "Introduction to Elementary Particles" by Griffiths. Jun 7 answered 2nd order variation of Hilbert-Einstein action + Gibbons-Hawking-York boundary term Jun 3 comment Explicit Variation of Gibbons-Hawking-York Boundary Term You're right -- simply preserving $n^{\mu}n_{\mu} = 1$ would only determine $\delta n^{\mu}$ up to a vector orthogonal to $n^{\mu}$. Instead, you should consider the definition of $n^{\mu}$. Let $\partial M$ be an isosurface of some coordinate $r$. Then $\alpha_{\mu} = \nabla_{\mu} r$ is orthogonal to the surface. Now normalize that vector to obtain: $$n_{\mu} = \frac{\alpha_{\mu}}{\sqrt{g^{\nu\lambda}\alpha_{\nu}\alpha_{\lambda}}}$$ Consider how this expression behaves under a small variation of the metric and you obtain the result for $\delta n_{\mu}$. May 31 awarded Editor May 31 revised Explicit Variation of Gibbons-Hawking-York Boundary Term added 108 characters in body May 31 answered Explicit Variation of Gibbons-Hawking-York Boundary Term Apr 21 answered Supergravity calculation using computer algebra system in early days Mar 31 comment Space-like Killing vector of Robertson-Walker metric? Both of you are writing metrics on a portion of the full de Sitter spacetime. Four-dimensional de Sitter has topology $R^1 \times S^3$ and is conveniently realized as a hyperboloid in $R^5$. Various combinations of the coordinates on the hyperboloid give the metrics above, but will only cover part of the hyperboloid. There may be timelike Killing vectors on the patches associated with those metrics, but there is no global timelike Killing vector. Feb 24 awarded Supporter Feb 4 comment Treatment of boundary terms when applying the variational principle The solution -- which goes back to the Regge and Teitelboim paper I mentioned -- is to add additional surface terms to the action until you have a functional that doesn't require $\delta g$ to fall off faster than whatever the appropriate asymptotics are. As long as the new surface terms are intrinsic to the boundary you can still impose Dirichlet boundary conditions. In AAdS spacetimes, the surface terms are the famous "boundary counterterms". The fundamental motivation for them is establishing a good variational formulation of the theory; finiteness of the action is a consequence of that. Feb 4 comment Treatment of boundary terms when applying the variational principle The problem isn't with GHY. It's the claim that adding GHY to the EH action leads to a "well-defined variational problem". Take the EH action and add the GHY term. The Schwarzschild solution should be a stationary point of this action, right? Now consider the change in the action due to a small variation of the metric. You will find that the surface terms in the first-order change in the action only vanish if $\delta g$ falls off faster than 1/r. You have an 'action' that allows Dirichlet boundary conditions, but it can't tell between Schwarzschild and other metrics with 1/r asymptotics. Feb 4 comment Treatment of boundary terms when applying the variational principle I'll let you know next time I head down, which will hopefully be soon.