"greywolf42" <***@sim-ss.com> wrote in message news:***@corp.supernews.com...
|
| Steve Carlip <***@dirac.ucdavis.edu> wrote in message
| news:bk7rqr$7p4$***@woodrow.ucdavis.edu...
| > greywolf42 <***@sim-ss.com> wrote:
| >
| > > Perion <***@hotmail.com> wrote in message
| > > news:ks-***@comcast.com...
| > >> Can anyone describe (or point me to some information) how Einstein
| > >> originally arrived at 8pi for the proportionality constant in G=8piT?
| >
| > > Quite simple. It was a backfit to Newton's equation of gravity -- in
| the
| > > weak limit. That is, it was assumed by Einstein as one of the
boundary
| > > conditions in determining constants of GR.
| >
| > This is right, for a sufficiently loose definition of ``backfit.''
| >
| > You should first as the question of where Newton's constant G
| > came from in Newton's law of gravity, F=GMm/r^2. (Warning:
| > I'm using G to mean Newton's constant, *not* the Einstein
| > tensor.)
|
| Newton's constant was derived by application of the calculus to Kepler's
| observational "laws" of planetary motion. Newton determined the equation
of
| force necessary to duplicate Kepler's laws. The formula F= GMm/r^2
resulted
| from this analysis. This is why Newton never attempted to explain 'why'
his
| gravitational equation was what it was ("Hypothesis non fingo.")
|
| > The answer is that given a system of units for mass,
| > length, and time, G was an undetermined coupling constant,
| > that had to be fixed by measurement. Note that this did not
| > significantly weaken the predictive power of Newtonian gravity:
| > one careful laboratory measurement determines G, and that
| > value can then be used to predict the results of millions of other
| > measurements.
|
| The *VALUE* of the constant, G, is given by the laboratory measurements
| (because we can use known masses). The equation of gravitational
attraction
| and the identification of the constant G, however, was derived elsewhere.
|
| > Now write the Einstein field equations as G_{ab} = kT_{ab}. The
| > constant k is again an unknown coupling constant, which has
| > to be measured. One can again set up a laboratory experiment
| > like the Cavendish experiment to make this measurement. The
| > main difference is that you now need to know the GR prediction
| > for the result of such an experiment. In principle, you can just
| > sit down with the field equations and work out the answer, to
| > whatever accuracy you need.
| >
| > In practice, it's easier to go via the Newtonian limit. The Einstein
| > field equations predict that for weak fields (and a careful definition
| > of coordinates, time, etc.) a pair of masses behave as if there were
| > a Newtonian gravitational attraction F = kc^4Mm/8 pi r^2, where
| > c is the speed of light. By comparing this to Newton's expression,
| > you can read off the value k=8 pi G/c^4. You can think of this as
| > ``backfitting'' to Newton's equation, but it's probably better to
| > think of it as fitting a single instance of Nwton's equation -- say,
| > Newton's equation applied to a particular Cavendish experiment.
| > As in Newtonian gravity itself, once the constant k is fixed by one
| > experiment, you have no more freedom left.
|
| But Einstein didn't do that ("Foundation of the General Theory of
| Relativity," 1916, section 21). Adding additional constants (8, pi, c)
to
| the mix only changes the units. Not the strength of the interaction
| (obviously). Einstein explicitly adjusted to the Newtonian *equation*.
Not
| to an arbitrary experiment.
|
| And the experiments were first done prior to GR, so the issue was moot.
| Experiments don't care if you're using GR to analyse the data or Newton.
|
| > Why the pi? Basically because the Einstein field equations are
| > differential equations. If you write Newton's gravitational force
| > law as a differential equation, the Poisson equation, you'll find
| > that the right-hand side is not just density, but 4pi(density).
| > This is the same 4pi you get when you go from Coulomb's law
| > to Gauss's. If you trace it back far enough, it comes from the
| > area of a sphere.
| >
| > Finally, note that k, like G, is a dimensionful constant, that is, it
| > depends on your choice of units. In Newtonian gravity, you could,
| > in principle, decide to use meters and seconds and to set G=1;
| > this would then simply determine what units you were using to
| > measure mass. In this sense, the whole 8pi G/c^4 can also be
| > viewed as defining a choice of units.
|
| Correct. The choice of units does not determine the physical strength of
| the constant (or couple).
How about if the 8pi comes from rotation?
FrediFizzx