Re: h bar and Lorentz transformation

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 03/20/05 

Date: 20 Mar 2005 00:38:32 -0800

Bilge wrote:
[snip agreebly]

> The number `hbar' is again, the result of choosing units rather
than
> what `hbar' represents. What it represents is 1 unit of angular
momentum.

Bilge, I argue, Plancks "h" is a unit of action,
it is a scalar with 1 single positive magnitude,
6.624x10^-27 erg x seconds.

A unit of angular momentum can have +/- values,
since that is, by convention, reversible by
rotation, and we use a vector normal to the
plane of rotation to describe angular momentum.

But the magnitude of that vector can be negative,
but "h" is never negative.

So I argue "h" is the ACTION quantum, and is
invariant.

If you have the time, let's consider that from
the standpoint of GR.

Since h is non-zero, only non-zero components
can be used.

Begin with

h = p_u x^u (u=0,1,2,3}

where p_u is the 4-momentum and x^u is a
length is spacetime.

In accord with Minkowski I'll use

U_i=0 {i=1,2,3}

then we find,

h= p_0 x^0 ,

where p_0 is ergs, and x^0 is seconds,
in accord with the definition of "h".

((p_u = p * U_u))

It important to notice that "h" results
from a product of relative components,
i.e. erg seconds == p_0 x^0, using

U_i =0.

The U_i=0 is required to make absolute
motion vanish. Relative motion is alive
and well in the U^i being non zero.

So we can connect, the requirement of
relative motion (that is to say, the
vanishing of absolute motion U_i=0 ),
to predict Plancks "h", using 4D.

Hence we can find the basis of "h" and
QT from relativity.

Regards
Ken S. Tucker
kxsxt