Re: Neutrino Mass and Supernovae
carlip-nospam_at_physics.ucdavis.edu
Date: 10/09/04
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Date: Sat, 9 Oct 2004 21:47:28 +0000 (UTC)
Ken S. Tucker <dynamics@vianet.on.ca> wrote:
> carlip-nospam@physics.ucdavis.edu wrote in message
> news:<ck1k83$enq$1@skeeter.ucdavis.edu>...
> > Ken S. Tucker <dynamics@vianet.on.ca> wrote:
> > > carlip-nospam@physics.ucdavis.edu wrote in message
> > > news:<cjuv9o$2t3$4@skeeter.ucdavis.edu>...
> Steve says below,
> >I don't know what to say, beyond, "That makes no sense."
> Ok, I'll be more stringent and not use too much "word salad".
> > > The difficult accessablity of "G" to experiment and concensus
> > > is well known, so I chose to investigate it's *invariance* on the
> > > theoretical grounds of dimensionality.
> > This is a bad idea. Dimensionality has nothing at all to do with
> > coordinate invariance.
> That is a conjecture.
No, it's not. It's a fact, illustrated by hundreds of examples.
[...]
> > You are confusing two completely different meanings of the
> > symbols L and M. In the equation L'= L/gamma, L means "coordinate
> > distance." In the statement "the dimensionality of G is L^2,"
> > L means "having units of length." The two are unrelated. For
> > example, proper distance has dimension L,
> Correction, I presume you mean "proper length" is equivalent to
> "invariant" length, like (in 3D),
By proper length I mean proper length. Look it up.
> L^2 = g_ij L^j L^k
> in that case invariant "L" has no units. That's because the
> Kronecker Delta
> Delta^u_v = {0,1} has no units
That's true...
> Recall, that Delta is defined by the scalar product of
> "reciprocal" unitary base vectors as
> -> ->
> Delta^u_v = u^u . u_v
> In all CS's L^2 = L_u L^u, and the units like "meters" or
> "inches" must cancel, leaving a scalar with no units.
That's patently false. L^j has units of length; g_ij L^j L^k has
units (dimensionless number)x(length)x(length), or (length)^2.
There is no ambiguity here; there is nothing more complicated
than high school physics.
> You want real physics: To define mass and length use a
> metal bar (at rest) 1 meter long with a mass 1 kg. Use
> that as a CS basis, ((located in Paris)).
> Relative to a moving observer the bar has a length of
> L' = (L=1m)/gamma and M' = (M=1kg)*gamma
> All moving observers agree M'L' = ML is an invariant,
> and all do agree Planck's h is defined by ML = invariant.
> The rest system also defines the Gravitational Constant
> using the metal bar to be,
> G = L/M * (a number)
No, it doesn't. Why would you think this?
> however the moving systems find
> G' = G/gamma^2
No, it doesn't. Why would you think this?
[...]
> >For that matter, if m denotes rest mass, mv^7/c^7
> > has dimensions M, but does not transform even remotely like
> > M'= M*gamma.
> That's a conjecture.
No, it's not. While I don't know off hand of an experimental
test of the transformation of mv^7/c^7, there are *millions*
of tests of the transformation properiries of mv/c and mv^2/c^2.
It is a completely unambiguous experimental *fact* that m (rest
mass), mv/c (momentum/c), and mv^2/c^2 (twice kinetic energy/c^2)
all transform diffeerntly, even though each has dimension M.
[...]
> > > >Since T=0 for electromagnetic radiation,
> > > > are you proposing that light doesn't gravitate?
> > > Yes, light exchanges momentum, it does not gravitate.
> > > An in flight photon (floton) has by definition momentum
> > > but not mass.
> > So, you are saying that general relativity is completely wrong.
> On the contrary, I'm regarding a floton (flying photon) as a
> massless (non-gravitating) particle, that has momentum, and
> exchanges momentum when the floton is decaccelerated by some
> interacting field, in which a momentum exchnage can take place.
> For example, a g-field perturbs the "global" velocity of light
> and exchanges momentum with the deflecting particle without
> and need to presume the floton has gravitation.
That is, in fact, swating that general relativity is completely
wrong.
> > You
> > are also saying that tests of the equivalence principle, which show
> > that electromagnetic binding energy contributes E/c^2 to gravitational
> > mass, are completely wrong, despite the fact that they have been
> > carried out to an accuracy of a part in 10^10.
> Again contrary. Set up two charged sphere's of nil radius
> having charge q and Q and separated by a radar ranged distance
> of "s". The system energy (discounting q and Q alone) is,
> Energy = qQ/s
> If you like, we can define a "energy density" like,
> Energy Density = Energy/Volume = Energy/s^3
> = (qQ/s) /s^3 = qQ/s^4.
No, you can't. You need to go look up energy density. This definition
has nothing whatsoever to do with the term as it's used in physics.
> Then we have an Energy Density given by,
> Energy Density = E(q)*E(Q) where E(q) = q/s^2, etc. and
> the E(q) and E(Q) are the *invariant* Electric fields of q and Q.
> Using the quantity "T" to be "invariant" energy density, then
> T = qQ/s^4
That's not what T means. T is the contraction of the stress-energy
tensor, T = g^{ab}T_{ab}. You can calculate it for the situation
you describe -- it comes out zero.
If you are going to invent new, completely different meanings for
standard terms, you ought to at least warn us: "By T I mean something
comp[letely different from what physicists mean when they write T."
[...]
> > "Can't be differentiated" is *completely* different from "has
> > a derivative of zero."
> Ok, we require the following solution for T,
> dT = 0, T =/= constant
> (field variation) (quantum variation)
> I'm open to suggestions.
There is no "solution." If dT=0, then T is constant. Period. No
ifs, ands, or buts. No wiggle room.
This does not, by the way, contradict anything in quantum theory,
although it might contradict something in your imaginary reconstruction
of what you think quantum theory ought to be.
> > > > > Suppose we have a massive (m>0) small (no tidals) satellite
> > > > > orbiting Earth elliptically. The satellite will follow a
> > > > > geodesic defined T;w = 0, and DELTA T = 0.
> > > > First of all, T;w = 0 does not define a geodesic. Why do you
> > > > think it does?
> >
> > > Newton's 1st law of motion, together with General Covariance,
> > > I only need to find one CS where the invariant "T" is constant
> > > then find T;w=0 is true in all CS's. That's obviously true in
> > > all free-falling states to begin with.
> > It is certainly not "obviously true" that T is constant in all, or
> > any, freely falling state. In fact, it's generically not true.
> > Furthermore, even if it were true, that would not determine a
> > geodesic.
> In a previous post to Dr. Carlip, I used some "word salad"
> to explain how a g-field cannot transmit, quantum variations.
> Given the definition from above, T = qQ/s^4 we have good
> experimental evidence that T will remain constant in any
> state of free-fall, that is,
> T;w =0 in freefall.
This makes no sense. The stress-energy tensor is not a number, it's a
function of space and time. As I said previously, your definition of
T has nothing whatsoever to do with what everyone else means by T.
> > > >Second, unless the satellite is a continuous
> > > > ring around the Earth, T;w is certainly not zero
> > > Disagree, recall in the FoR of the satellite the PoE holds,
> > > and T;w=0 in that weightless state.
> > T;w is not equal to zero in a weightless state.
> That's a conjecture.
No, it's not. It is observably true that the density of a satellite
is not constant -- the outer metal shell, for instance, is more dense
than the empty space between components on the inside.
Have you somehow decided that the derivative in the equation T;w=0
is only a time derivative? It's not -- that equation says that
both the time and the spatial derivatives are zero.
> >Why in the world do you think it is?
> Explained above, T is invariant.
"Invariant" is not the same as "constant."
> > T is the trace of the stress-energy tensor. For ordinary slowly
> > moving matter, it is approximately the mass density. Are you
> > claiming that a freely falling object necessarily has constant
> > density?
> Yes, if I understand GR correctly, and in particular General
> Covariance, I can attach a FoR (Frame of Reference) to an
> object and in the free falling system expect to find the
> density remains constant (assuming tidal force notwithstanding).
This makes no sense. The Earth is in free fall around the Sun. Do you
claim that the density of the Earth's core is the same as the density
of the mantle and the crust, that these all have the same density as
the oceans, that the oceans have the same density as the atmosphere,
and that the atmosphere has the same density as the near-vacuum of
interplanetary space?
[...]
> >For T;w to be zero
> > everywhere, density has to be constant everywhere, throughout the
> > Universe.
> We might consider "gauge invariance". You are free to add
> (or subtract) to "T" a constant "t" (t>0) and make T=0 anywhere
> and T;w=0 still holds. Meaning there is no absolute value of "T",
> an no absolute derivation.
This makes no sense. It would imply that the gravitational field
of a massive object is the same as that of a massless object. It
also has nothing whatsoever to do with gauge invariance -- do you
just like the sound of the term?
[...]
> > I will be happy to continue if you start to talk about real physics.
> > At the moment, though, you are making claims so remote from either
> > physics or math that I don't know what to say, beyond, "That makes
> > no sense."
Still true.
Steve Carlip
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