Recent content by StateOfTheEqn

January 13, 2016 will be the one hundredth anniversary of Karl Schwarzschild's paper on the solution to the GR field equations around a spherically symmetric gravitating body. You can find an English translation here: http://arxiv.org/pdf/physics/9905030v1.pdf I urge everyone to read it. You...

Are you wondering about the convergence of the series ##\Sigma_{k=0}^\infty \frac{1}{k!}## We can set ##e^x=\Sigma_{k=0}^\infty \frac{x^k}{k!}##. Then ##e^1=\Sigma_{k=0}^\infty \frac{1^k}{k!}=\Sigma_{k=0}^\infty \frac{1}{k!}## So, the series converges and equals ##e##.

We can assume the mass of a star is concentrated at a single point and a body of mass ##m## is a distance ##R## from the point. The binding energy is the energy required to move the body of mass ##m## from ##R## to infinity. Essentially I am asking what is the GR counterpart to the Newtonian...

What is the gravitational binding energy in GR in the spherically symmetric case? I calculate ##E=mc^2(1-\frac{1}{\sqrt{1-\frac{r_s}{R}}})## where ##m## is the mass of the body, ##r_s## is the Schwarzschild radius, and ##R## is the area radius as in the Birkhoff theorem.

The error in the notion of Black Hole 'event horizons' at r=2GM has been exposed back in 1989. The error began with Hilbert. See the paper Black Holes:The Legacy of Hilbert's Error. See also Schwarzschild's original 1916 paper in English.

##exp(\textbf{H})=\textbf{R}^+\times S^3## where ##\textbf{H}## is the Lie algebra of Quaternion space. This can be proven as follows: Let ##\textbf{q}=x+y\textbf{i}+z\textbf{j}+w\textbf{k}## ##x## commutes with ##y\textbf{i}+z\textbf{j}+w\textbf{k}## so...

The vector space of Hamiltonians forms a Lie algebra H. If you exponentiate the Lie algebra H you get the emergence of space-time as well as energy-gravity from absolute nothing. There is no t=0 so there is no first instant of time and no first cause.

Consider the spatial manifold ##\mathbb{R}^+\times S^2##. Suppose there are two metric on ##\mathbb{R}^+\times S^2##, one Euclidean and the other non-Euclidean. Define ##R=Area(S^2)/4\pi## as the area radius for the non-Euclidean and ##r=Area(S^2)/4\pi## the area radius for the Euclidean. For...

Yes. I'm currently working on a more detailed reply and I hope it won't take too long.

http://www.mathpages.com/rr/s8-07/8-07.htm is a good read and brings out the issues pretty clearly. In Schwarzschild's 1916 paper he used ##R=(r^3+r_s^3)^{1/3}## in the metric. The question that arose historically was (and is) where is ##r=0##? If ##r## originates at the central singularity then...

I think I have already answered this question. ##4\pi R^2/4\pi r^2=R^2/r^2=(r^3+r_s^3)^{2/3}/r^2 \rightarrow \infty## as ##r \rightarrow 0## and this implies the negative spatial curvature grows arbitrarily large near the central singularity.

I have rethought this a bit. First, I think that substituting ##(r^3+(2m)^3)^{1/3}## for ##R## gives the desired result ##48m^2/r^6##. But from the point of view of g(1916), ##{R}^{3}-8\,{m}^{3}## is just ##r^3##, the Euclidean distance from the origin cubed. So your little ##r## is just the...

I do not think this is true. In his original paper at http://de.wikisource.org/wiki/%C3%9Cber_das_Gravitationsfeld_eines_Massenpunktes_nach_der_Einsteinschen_Theorie (English translation at http://arxiv.org/pdf/physics/9905030v1), Schwarzschild defines little ##r## to be ##r=\sqrt{x^2+y^2+z^2}##...

Did you mean substituting ##(r^3+(2m)^3)^{1/3}## for ##R## ? Then I get ##48m^2/r^6##. This could mean we should understand the ##r## in g(current) to really be ##R=(r^3+(2m)^3)^{1/3}## where ##r## is the Euclidean radius with origin at the central singularity as in g(1916).

Both g(current) and g(1916) have rotational isometries in (at least) one of the angles.