Re: Neutrino Mass and Supernovae

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 10/11/04 

Date: 11 Oct 2004 12:32:53 -0700

Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckdkr6$6hp$1@news.urz.uni-heidelberg.de>...
> Ken S. Tucker wrote:
> > Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ck8ele$kub$1@news.urz.uni-heidelberg.de>...
...
> > google "w.w.sawyer" >1000 hits.
> > ((worked together from 1968))
>
> Thanks. Giving his name would have been enough.

Well if you have a chance you should read his ideas,
I didn't fully appreciate his help, I was a 15 year
old brat. I was using balloon membranes and drawing
graphs on them and then streching them to study Grad
Div and Curl, in LT and GR. We had some very interesting
conversations.
  He was particularily keen on what was wrong with
things, rather than taking an authoritarian approach.
  He was hard as hard on presumption as I am.
 
> >>And did this Prof also say this stuff about "u_j=0 in GR",
> >>and that this means that "absolute velocity vanishes", and that
> >>a Lorentz boost is a translation, and that Lorentz boosts are not
> >>allowed in GR?
>
> I notice that you did not answer this question below.

No, I was detailed a problem, that is my solution.

> > Either you do not understand the problem or you don't take it
> > seriously, let me explain.
> >
> > From the standpoint of mathematics a displacement like
> >
> > ds^2 = g_uv dx^u dx^v
> >
> > does NOT diffentiate between relative and absolute motion,
> > and neither do mathematicians, it's a an equation without
> > regard to the nature of the relativity imposed on FoR's.
>
> You do not make any sense. The equation above has nothing to
> do with motions! It only defines the *geometry* of a space - i.e. it
> tells you how lengths and angles are defined!

In geometry, ds^2 is regarded as a displacement in the
context of spacetime.
 
> > In fact relative motion is a subset (or group) of all
> > possible motions in all "proper" CS's. I take the ds^2
> > above to be the generic representitive set defining all
> > possible displacements, including absolute and relative
> > displacements.
>
> Care to explain what exactly the difference is? What do you *mean* when
> you say "absolute displacement"?

Generally speaking, any CS that permits dx_i dx^i =/=0
allows "absolute displacement", and that is NOT a
permissable CS where relativity is imposed.
 
> > In physics the u,v are conventionally set to integer
> > values 0,1,2,3, where "0" is messaged to being the
> > dimension of time.
> >
> > The ISU has agreed L=ct,
>
> ISU? From the context, perhaps International System Unit?

Yup, I work strictly in ISU guidelines.
...
 
> > Problem...
> >
> > define the *relative* displacement subset of
> >
> > ds^2 = g_uv dx^u dx^v.
> >
> > within the above conditions expressed generally
> > covariantly, and prove.
> >
> > Solution U_i =0 defines the theory of relativity,
> > in all FoR's.

> > Proof, Invariant Mass = (rest mass)*gamma.
> > experimentally verified by nuclear tests.
>
> 1) Invariant mass = rest mass, exactly. (rest
> mass)*gamma = *relativistic* mass, not invariant mass!

Simply put, your suggestion that rest mass, applicable
in one CS is invariant for all CS's is wrong.

> 2) That has nothing at all to do with U_i = 0 defining the theory
> of relativity.

It certainly does! All momentum covariant tensors like p_i
vanish and P_0 is rest mass/energy. The relativity condition
  
   U_i =0

is extremely powerful. Try it, tell me where it fails.
I've used it thoughout unified field theories, including
those involving nonsymmetrical tensors, and it always holds.
 
> >>>>>>The equations I gave above were not translations.
> >>>>>
> >>>>>
> >>>>>Of course they are,
> >>>>
> >>>>No, they aren't. A translation would be
> >>>> x' = x + x_0
> >>>> t' = t + t_0
> >>>>with constant x_0 and t_0. Try understanding the difference between a
> >>>>linear and an affine coordinate transformation, please.
> >>>
> >>>
> >>>Good, then solve &t'/&x when the relatively moving point
> >>>is at x=0, t=0, x'=0 and t'=0.
> >>
> >>Do you suggest here that I first set x=0 etc. and then carry out the
> >>derivative, or that I first carry out the derivative and then set x=0
> >>etc.? I hope for you that you meant the second...
>
> Care to say what you meant?

Yes, the second.
 
> >>BTW, what has the above to do with my point that a Lorentz boost is not
> >>a translation?
> >
> >
> > Well, a translation will create artifacts if the origins are not
> > coincident.
>
> A *spatial* translation will *ensure* that the origins are not
> coincident!!! But a Lorentz boost is *not* a translation!
>
>
> > The Lorentz boost is meaningless in GR.
>
> Please support that assertion.

Well, if you end up with a partial like &t'/&x
dependant upon a finite spacetime displacement you've
imported a translation into the transformation.
In simple tensor analysis transformations are meaningful
only at a point. More advanced treatments that include
nonsymmetrical g_uv require a finite in accord with the
ISU.
 
> >>Anyway, doing the second method (because the first makes no sense, as
> >>you hopefully realize): I already pointed out that for a Lorentz boost,
> >>&t'/&x = -beta gamma. This is indepedent of x, t, x' and t', and hence
> >>it is entirely irrelevant what I insert for these coordinates.
> >
> >
> > The ISU is quite specific on this point. Length is defined by time.
>
> Yes. So what? What on earth has that to do with my paragraph above???
>
> > You're attempting to import obsolete (1905) developemental equations
> > into the final refined decision.
>
> Huh?????
>
> > Bjoern, if you retain your 1905 thinking in 2004 you are a kook,
> > get an education, learn why the ISU standard is L=cT.
>
> What on earth has that to do with what I wrote above?????

Because Time is independent of length, &T'/&x =0 just
as &T/&x =0, I work within ISU standards.
 
> Oh, BTW: if I am a kook, how did I manage to obtain a PhD in physics?
> Working in Quantum Field Theory, where a good understanding of SR and
> tensor analysis is required?

Bjoern, even a superficial acquaintance to QFT requires
some analysis of Shapiro's experiments, as it relates to
length in fields.
...

> Bye,
> Bjoern

Alot of this is new to me
Ken S. Tucker

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