Nigel Cook made an interesting comment about the three times. In particular, he pointed out that “physically, every body which has gained gravitational potential energy, has undergone contraction and time dilation, just as an accelerating body does. This is the equivalence principle of general relativity.”

Note that by considering stationary objects in the Sparling metric, in the approximation of one $c$ value, we obtain

$s^2 = c^2 (t^2 + u^2 + v^2)$

which is basically Louise’s equation, provided we set a time parameter $T$ to be the length of the time 3-vector. Unfortunately, Lunsford does not appear to have a blog, but here is a link to Nigel’s discussion on this matter.

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## Anonymous said,

April 28, 2007 @ 10:06 pm

Shucks. Lubos Motl might regret missing his chance to win a Nobel prize.

## nige said,

April 30, 2007 @ 1:33 am

Kea, thanks for this, which is encouraging.

In thinking about general relativity in a simple way, a photon can orbit a black hole, but at what radius, and by what mechanism?

The simplest way is to say 3-d space is curved, and the photon is following a curved geodesic because of the curvature of spacetime.

The 3-d space is curved because it is a manifold or brane on higher-dimensional spacetime, where the

time dimension(s) create the curvature.Consider a globe of the earth as used in geography classes: if you try to draw Euclidean triangles on the surface of that globe, you get problems with angles being bigger than on a flat surface, because although the surface is two dimensional in the sense of being an area, it is curved by the third dimension.

You can’t get any curvature in general relativity due to the 3 contractable spatial dimensions:

hence the curvature is due to the extra dimension(s) of time.This implys that the time dimension(s) are the

sourceof the gravitational field, because the time dimension(s) produce all of the curvature of spacetime. Without those extra dimension(s) of time, space is flat, with no curved geodesics, and no gravity.This should tell people that the mechanism for gravity is to be found in the role of the time dimension(s). With the cosmic expansion represented by recession of mass radially outward in three

timedimensions t = r/c, you have a simple mechanism for gravity since you have outward velocity varying specifically withtimenot distance, which impliesoutward accelerationof all the mass in the universe, using Hubble’s empirical law, dr/dt = v = Hr:a = dv/dt

= dv/(dr/v)

= v*dv/dr

= v*d(Hr)/dr

= vH.

Thus outward force of universe F=Ma = MvH. Newton’s 3rd law tells you there’s equal inward force. That inward force predicts gravity because it (which is corce-mediating gauge boson exchange radiation, i.e., gravitational field) exerts pressure against masses from all directions except where shielded by local, non-receding masses. The shielding is simply caused by the fact that non-receding (local) masses don’t cause a reaction force, so they cause an asymmetry, gravity. There are two different working sets of calculations for this mechanism which predict the same formula for G (which is accurate well within observational errors on the parameters) using different approaches (I’m improving the clarity of those calculations in a big rewrite).

Back to the light ray orbiting the black hole due to the curvature of spacetime: Kepler’s law for planetary orbits is equivalent to saying the radius of orbit, r is equal to 2MG/v^2, where M is the mass of the central body and v is the velocity of the orbiting body.

This comes from: E = (1/2)mv^2 = mMG/r, as Dr Thomas R. Love has explained.

Light, however, due to its velocity v = c, is deflected by twice as much by gravity than the slow moving objects (v << c). Instead of (1/2)mv^2 = mMG/r giving an orbital radius of 2MG/v^2, for light you have mc^2 = 2mMG/r The factor of two on the right hand side comes from the fact that light (moving perpendicular to gravitational field lines) is deflected twice as much by gravity than predicted by Newton’s law. The lack of the factor of (1/2) on the left hand side is due to the fact that the mass-energy equivalence for velocity c is E=mc^2 not the low velocity kinetic energy E = (1/2)mv^2. However, E = (1/2)mv^2 = mMG/r and E = mc^2 = 2mMG/r both lead to the same formula for radius or orbit: (1/2)v^2 = MG/r implies r = 2MG/v^2 mc^2 = 2mMG/r implies r = 2MG/c^2 So there are two relativistic factors involved, and each exactly offsets the other. This is clearly the reason why, paradoxically, Newtonian gravity gives the same formula for the event horizon radius of a black hole that general relativity does, and is not wrong by a factor of 2. In general relativity, the metric tensor (for the Schwarzschild metric) is g_{00} = [(1 – GM/(2rc^2)/(1 + GM/(2rc^2)]^2 g_{11} = g_{22} = g_{33} = -[1 + GM/(2rc^2)]^4 When each of these are expanded with Maclaurin’s series to the first two terms, g_{00} = 1 – [2GM/(rc^2)] g_{11} = g_{22} = g_{33} = -1 – [2GM/(rc^2)] In each case, the major effect of the gravitational field is that you reduce the metric tensor by an amount equal to 2GM/(rc^2), which is the ratio of the event horizon of a black hole to the distance r. Put another way, g_{00} = 1 – (1/n) g_{11} = g_{22} = g_{33} = -1 – (1/n) where n is simply distance as measured in units of

black hole event horizon radii(similar to the way that the Earth’s Van Allen belt’s are plotted in units of earth radii).For the case where n = 1, i.e., one event horizon radius, you get g_{00} = g_{11} = g_{22} = g_{33} = 0.

That’s obviously wrong because there is severe curvature. The problem is that in using Maclaurin’s series to the first two terms only, the result only applies to small curvatures, and you get a strng curvature at event horizon radius of a black hole.

So it’s vital at black hole scales to

notuse Maclaurin’s series to approximate the basic equations, but to keep them intact:g_{00} = [(1 – GM/(2rc^2)/(1 + GM/(2rc^2)]^2

g_{11} = g_{22} = g_{33} = -[1 + GM/(2rc^2)]^4

where GM/(2rc^2) = (2GM/c^2)/(4r) = 1/(4n) where as before n is the distance in units of event horizon radii. (Every mass consititues a black hole at a small enough radius, so this is universally valid.) Hence:

g_{00} = [(4 – 1/n)/(4 + 1/n)]^2

g_{11} = g_{22} = g_{33} = -(1/256)*[4 + 1/n]^4.

So for one event horizon radius (n = 1),

g_{00} = (3/5)^2 = 9/25

g_{11} = g_{22} = g_{33} = -(1/256)*5^4 = -625/256.

The gravitational time dilation factor of 9/25 at the black hole event horizon radius is equivalent to a velocity of about 0.933c.

It’s pretty easy for to derive the Schwarzschild metric for weak gravitational fields just by taking the Lorentz-FitzGerald contraction gamma factor and inserting v^2 = 2GM/r, on physical arguments, but then we have the problem that Schwarzschild’s metric only applies to weak gravity fields because uses only the first two terms in the Maclaurin’s series’ for the metric tensor’s time and space. It’s an interesting problem to try to get a completely defensible, simple physical model for the maths of general relativity. Of course, there is no real physical need to work beyond the Schwarzschild metric since it all the tests of general relativity apply to relatively weak gravitational fields within the domain of validity of the Schwarzschild metric. There’s not much physics in worrying about things that can’t be checked or tested.