Re: Neutrino Mass and Supernovae

carlip-nospam_at_physics.ucdavis.edu
Date: 10/09/04 

Date: Sat, 9 Oct 2004 21:21:01 +0000 (UTC)

Ken S. Tucker <dynamics@vianet.on.ca> wrote:
> carlip-nospam@physics.ucdavis.edu wrote in message
> news:<ck4dlf$1b6$1@skeeter.ucdavis.edu>...
> > Ken S. Tucker <dynamics@vianet.on.ca> wrote:
  
> > [...]
> > > But now we have more confidence that Einstein's Law *should*
> > > be G_uv = T_uv and NOT G_uv = G*T_uv.
 
> > That's all very well. But G_uv has units of 1/(length)^2,
> > and T_uv has units of mass/(length)^3. So you need some
> > conversion constant, or some choice of units such that
> > mass and length are measured in the same units. Otherwise
> > you fail at the freshman physics level -- you've written
> > an equation whose dimensions don't match.

> That's more complex...sorry, needs "word salad"...

> Suppose we use to define unit energy density, a 1 meter
> length metal bar, with a mass of 1 kg, and a sectional
> area A, (located in Paris of course), and define

> Energy Density = 1 kg/(1 meter * A) == M / (L*A)

> Beginning with SR, relatively to a CS moving parallel
> to L, Area A is invariant, since it is perpendicular to
> the direction of motion, so the energy density transforms
> to the moving FoR as,

> M' / (L'*A') = gamma^2 * M / L*A

> (used M' = M*gamma, L'=L/gamma, A' = A).

> Please note that 1/L^2 transforms like

> 1/L^2 == M/(L*A) == Energy Density

Yes, that's right -- the stress-energy tensor transforms the
same as the Einstein tensor. That does *not* mean that their
units are the same.

> so that the relativistic treatment of the units in

> G_uv = T_uv

> is satisfied using Energy Density with units M/L.

This makes no sense. Energy density is *defined* to be energy
per unit volume. It therefore has units of energy/volume. This
is high school physics, for goodness sakes!

> Area "A" in this application is an invariant scalar, it has
> no units,

This makes no sense. Area has units of area, that is, (length)^2.

> it's a pure number, just as one can set c =1, one
> can set A=1.

This makes no sense. The transverse area used to determine
mass density or energy density is different for every different
object whose density you are looking at. You can't just declare
that it's always one -- it observationally isn't.

> That said, one can balance the units on each side by using

> G_uv = (1/h)*T_uv

> where "h" is Plancks constant and has units of (ML). Now
> we find,
>
> G_uv = (1/h)*T_uv and in units is,

> 1/L^2 = (1/ML)* M/L

> balances, by replacing the conventional "G" in Einstein's
> Law by the reciprocal of "h", and one may set the proven
> invariant h=1.

This makes no sense. T_uv has units of M/L^3, not M/L.
(1/h)T_uv has units 1/L^4, not 1/L^2.

[...]
> > > So Steve, we can keep the 10 equations G_uv =T_uv provides,
> > > together with it's "predictive power", and I hope be more
> > > confident in that relationship.
> > > To boot, we have another toy from the analysis,
> >
> > > R = T = K^uv T_uv
> >
> > > that is now in accord with Einstein's Law, G_uv=T_uv, (no "G").

> > Well, if you accept that G_uv = T_uv, then by contracting both
> > sides with the metric and using the definition of the Einstein
> > tensor, you find that

> > R = -T

> > in four spacetime dimensions, where I am using the standard notation
> > that R is the scalar curvature (R = g^{ab}R_{acb}^d) and T is the
> > contracted stress-energy tensor (T = g^{ab}T_{ab}). So clearly
> > the claim R = T is wrong.

> Agreed, R == T when invariant "K" is unitized, I treated
> the (-1) as a constant, thanks for detailing that. ###

This makes no sense. You can't set -1 equal to 1, and you can't
arbitrarily drop factors from an equation just because they're
constants.

Steve Carlip

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