Re: Neutrino Mass and Supernovae

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 10/10/04 

Date: 10 Oct 2004 13:44:07 -0700

carlip-nospam@physics.ucdavis.edu wrote in message news:<ck9knt$lbt$1@skeeter.ucdavis.edu>...

Steve,
thank you for your comments and please pardon the (temporary)
top post. I'd like to answer but I'm a bit confused (at least!).

I do not quite understand the meaning of "volume" in the context
of G_uv = T_uv.

Here's my understanding...
Since G_uv conventionally defines curvature at a "point",
the "volume" should occupy a point given by either,

1) dV = (small Volume) = Area*(dL = small length)

or

2) dV = dx dy dz

I get nervous with (2) because of the density that may occur
within the transform of a 3D volume element, ie.

  dV' sqrt g' = dV sqrt g g=det|g_uv|

Using (1), I need to unitize an invariant Area, so which do you
use/accept or is there another choice.
Regards
Ken S. Tucker

> 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