Re: Note on Kronecker delta (kst).
- From: Igor <thoovler@xxxxxxxxxx>
- Date: Sat, 19 Apr 2008 08:07:08 -0700 (PDT)
On Apr 18, 3:05 pm, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
On Apr 18, 8:41 am, Igor <thoov...@xxxxxxxxxx> wrote:
On Apr 18, 2:37 am, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
To Mike, Charles, Salviati and all.
This post is going to be weird. It's based partially
on Charles and Salviati's discussions about infinity.
On Apr 16, 4:06 pm, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
Thank you for your reply Mike.
I studied your post and links.
On Apr 15, 12:15 pm, Mike <mj...@xxxxxxxxx> wrote:
On Apr 5, 2:28 am, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
From my superficial understanding, the author (Dr. Kleinert)Yes, I was instructed to read chapter 10, 11, and 19 of his book, Path
is using "truly" curved spaces, (to his credit IMHO).
His "quantum equivalence principle" is excellent though it's
presented in close mathematical detail.
Probably best to contact the author himself for clarification,
and if you do, I'd appreciate you post his reply.
Regards
Ken S. Tucker
Integrals in Quantum Mechanics, Statistics, Polymer Physics, and
Financial Markets. I bought it and am waiting for it to come in. It
should do a lot to clearify the on-line article I'm asking about. I'll
let you know.
In the mean time, if anyone should happen to already know, my question
is this:
I've started to derive non-relativistic quantum mechanics from only
the definition of the Dirac delta function as shown in the following
link:
http://hook.sirus.com/users/mjake/delta_physics.htm
Mike, it looks to me you are really good at math!
But now I wonder if General Relativity and/or Quantum Gravity can be
derived from the equivalent of the Dirac delta function except in
curved spacetime.
I've been studying something similiar, maybe you
could help me out. Pardon the crappy ascii, I'll
write the best I can.
Begin with a contravariant finite length, "X^a", then
perform a covariant derivative w.r.t "X^b", to give,
X^a;b = &X^a / &X^b + GAMMA ^a_bc X^c.
where "&" is a partial, GAMMA is a Christoffel.
Let's denote the Kronecker delta by,
g^a_b = &X^a / &X^b,
and note that " X^a;b " and " g^a_b " are both tensors.
The difference between two tensors is a tensor, then,
GAMMA ^a_bc X^c = X^a_b - g^a_b ,
is a mixed tensor of rank two.
Although GAMMA ^a_bc is not a tensor, the product,
GAMMA ^a_bc X^c , apparently is.
For a ref, I'm using Richmyer, Kennard & Coopers,
"Intro to Modern Physics", that looks standard.
I'll use their Eq.(13.31), the integral,
(psi_n (X) and psi_j (X) be any two normalized
nondegenerate eigenfunctions of Schrodinger
equation...), ($ is integral),
oo
$ psi*_j psi_n dX = d_jn , Eq.(RKC 13.31)
-oo
I'll rewrite that with some improvement,
for covariance,
oo
$ psi^a psi*_b dX = g^a_b , Eq.(13.31a)
-oo
The infinity boundaries are difficult to deal with,
so I use the following. I'll use the deduction below,
oo
$ psi^a psi*_b dX -
-oo
X/2
$ psi^a psi*_b dX = (1/2) g^a_b .
-X/2
That's way of saying the particle has a 50%
probabiltiy of existing between X/2 and -X/2.
I'll enter the following solution to the last Eq.
(A Poincare transform, moving the origin),
X
$ (psi^a psi*_b) dX = (Psi^a Psi*_b) X , Eq.($)
0
= (1/2) g^a_b.
(I see no need for a constant of integration,
but that's pending).
From the far above post, we now have,
((GAMMA ^a_bc X^c = X^a_b - g^a_b))
GAMMA ^a_bc X^c = X^a_b - 2*$ .
Now I'll use 2 operations.
1) X^a_b =0 , (generally vanishes).
2) GAMMA ^a_bc X^c = GAMMA ^a_b X
(a contraction).
Those, with a bit of algebra, yield,
GAMMA ^a_b = (Psi^a Psi*_b), Eq.(GP).
Often the Christoffel (GAMMA) is called a connection,
well it looks like " (Psi^a Psi*_b) " is a connection too.
There is latitude of interpretation of Eq.(GP), I'll
provide IMHO...
A single particle cannot provide a measurement of a
g-field that is characterized by GAMMA. That g-field
measurement requires 2 particles. The RHS of Eq.(GP),
requires 2 particles, Psi and Psi*, and if either is zero,
the GAMMA is zero.
From the standpoint of Newton, we can rewrite the RHS,
(m/r) (m*/r) = Newtonian gravitation,
which is the product of the two potentials, Psi and Psi*.
It's costumary to set m=1 such that we cheat to
describe m*/r^2 = acceleration applied to a unit
mass "m", yet our common GR geodesic needs
a unit mass in,
dU^i/ds = - GAMMA ^i _uv U^u U^v.
For example,
m dU^i/ds = - m GAMMA ^i _uv U^u U^v.
GR uses a unit mass but the equation blows up
if m=0, because no measurement is possible.
Regards
Ken S. Tucker- Hide quoted text -
- Show quoted text -
Nobody expects the Spanish Inquisition. And these days, with your
posts, not even physics is immune.
If you have a look at AE's Guv=Tuv, the Tuv
is usually called "energy density" as in
mass/volume, expressed as Guv as well.
OTOH, the probability of finding a mass within
a volume is in units of mass/volume.
That looks the same to me.
Regards
Ken S. Tucker
Then I guess the choice is up to you: thumb screws or water boarding.
.
- References:
- Note on Kronecker delta (kst).
- From: Ken S. Tucker
- Re: Note on Kronecker delta (kst).
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