🚀 go-pugleaf

On Jul 18, 5:18 pm, rbwinn <rbwi...@juno.com> wrote:
> > > n'=t(1-v/w)
>
> > > n' is time on a cesium clock in a moving frame of reference, t is time
> > > on a cesium clock in a frame of reference at rest.
>
> > BTW, you made a mistake on your formula. v and w need to be squared,
> > and whole (1 - v^2/w^2) needs to be square-rooted (http://
> > en.wikipedia.org/wiki/Time_dilation, note the formula). I won't make a
> > big deal out of the fact that you used an ... unconventional letter
> > for speed of light. It's usually c.
>
> These are not the Lorentz equations.  They are equations derived from
> the Galilean transformation equations, the equations scientists

Oh, I'm sorry. I didn't realize you were discussing "unconventional"
physics. In that case (having looked at it a little more closely),
here is the problem I see. You say:
   x=wt
   x'=wn'
But I see absolutely no justification for this. First you adopt
Galilean transformation, which is incompatible with constant speed of
light in all inertial reference frames, and then to express a single
point in two reference frames, you say x = wt and x' = wn' (why the
prime, if you've already adopted a different letter, n, rather than
t'?), rather than x' = w'n', where w' is, of course, w - v (v
containing the sign). It is, on the face of it, self-contradictory.

Could you provide a physical picture (a Gedankenexperiment, as it
were) which these expressions are supposed to correspond to?

> Velocity of light is w, not c, because the velocity of a photon
> traveling on the x axis of a set of Cartesian coordinates in the
> negative direction is -c, not c.  The Lorentz equations ignore this
> fact by leaving the equations in an unreduced form so that c is always
> squared in those equations.   (-c)(-c)=c^2

Irrelevant. It is not the photons that are moving, it is the observer.
The term "speed of light" is very misleading in this sense (and I do
empathize with your confusion), because the constant value of "c" is a
property of space-time. In a flat space, in the formalism of general
relativity, this is expressed by the metric (I may be using
unconventional letters here myself, it's been a while since I've done
anything GR, oh, and the sign convention differs depending on whom you
talk to ) dr^2 = (c dt)^2 - (dx^2 + dy^2 + dz^2).

The presumption is, even if no photons actually existed (i.e. no
electromagnetic fields anywhere at all), the properties derived in
theories of relativity would hold. As for the "speed" part, we just
tend to measure it with photons because it's the only massless
particle we know.

And as for the sign of the observer moving in different directions,
it's expressly written into x' = x - vt, so the fact that you would
end up with v^2/c^2 elsewhere is irrelevant.

>     Scientists do not like these equations because they cannot
> disprove them, and they give answers almost identical to Lorentz
> equation answers with regard to anything that can be measured
> directly, such as the orbit of Mercury, which was the first proof used
> to show that the times given by the Lorentz equations agreed with
> scientific measurements better than Newton's equations for
> gravitation.

Er, I sincerely hope that you check these statements with a real
scientist real soon.

First, special relativity (and these modified Galilean transformations
that you propose to replace it with) has nothing to do with orbit of
Mercury. The theoretical prediction (or rather, explanation) for the
orbit of Mercury is an evidence for GENERAL RELATIVITY (Newtonian
gravity could not explain the precession of perihelion of Mercury, as
1/r potential gives rise to a closed orbit---only with corrections to
that 1/r potential (e.g. with GR effects) one would predict anything
like that), which must take into account the effects of gravity.
Proposing to do the same (or similar) with anything like Galilean
transformation (or even the Lorentz transformation!) is equivalent of
proposing to put the shuttles in earth's orbit using nothing but
airplane propellers---your tool is not powerful enough.

Nonetheless, do you actually have detailed calculation of these
effects that prove your claim?

Remember the burden is on the challenger to prove his own claim, not
on the incumbents to disprove the challenger's claim (as a mathematics
professor said once, if he had to read what every crank puts up on,
e.g., arXiv.org, he wouldn't have the time for anything else). Also,
to replace the orthodoxy, it's not sufficient only to duplicate what
the orthodoxy has shown and proven (after all, why take one over the
other in that case?)---you must also explain/predict something that
the existing theory hasn't.