Re: Fw: Potential energy and relativity

On Jul 12, 3:36 am, Bilge <dubi...@xxxxxxxxxxxxxxx> wrote:
On 2007-07-12, Pmb <some...@xxxxxxxxxxxxx> wrote:

If you were to take a look at P_0 then you'd find
that the potential energy is part of P^0.

No, you wouldn't. The p_u and A_u satisfy the relation
p_u A^u = 0,

I made a mistake on that and if you truly knew the theory then you'd
have caught the error. Since you didn't then it only goes to
demonstrate my point. You're not that good at this.


Besides, who cares if that is true since its about a specific
quantity, i.e. a photon, and does not apply in general. This is yet
another example of what Eric Gisse failed to see. He must have just
read it and assumed you were right because, as it appears, you're his
hero who is impossible of making mistakes. Sad. Really and truly sad.
As for whetheer that is true you'd have to take into account the
correction I posted on an error I made where I posted the 4-mmomentum
and meand to post the canonical 4-momentum. I warned you about this
condition but all you were able to do is to focus on the term
"canonical" and misuse the term to "prove" I was in error somewhere,
which was an error in your part.


This concept of the potential energy idea holds in Nuclear Physics too where
the potential energy resides in the nuclear binding energy and the electric
potential energy energy. See
http://www.geocities.com/physics_world/sr/nuclear_energy.htm

Don't even go there. Nuclear physics is a morass of approximations,
a number of which, I am intimately familiar. Writing down a general
model, like an optical potential (which contains up to 11 potential
energy terms) is most certainly not a fundamental model of a nucleus
by any stretch of the imagination. Go look at the shell model and you'll
see a term called the ``residual interaction,'' which is exactly what
it sounds like - all of the stuff left over after attempting to include
whatever happens to be relevant to a particular experiment.

The concept of potential energy also is alive in the relativistic version of
the work energy theorem. See
http://www.geocities.com/physics_world/sr/sr_work_energy.htm
http://www.geocities.com/physics_world/sr/work_energy.htm

Get over it. It's certainlt possible to pick a particular coordinate
system and a particular gauge, (which is explicitly non-covariant
and non-gauge invariant) and write a potential energy for that
particular choice. But relativity is a theory about invariance,

Wrong. I see now that this is the root of your problem in that
everything which isn't covariant is meaningless in your twisted little
world. Relativity is about the laws of nature. That is what relativity
is about. To be more precise it means that the laws of physics must be
able to be expressed in the same form in all inertial frames of
reference. There is no demand that the laws be expressed in
geometrical form - I see you didn't read that part of MTW huh?

As far as approximations go then so what? If this is the case then go
tell it to someone in the relativity/quantum newsgroup to those people
who actually care about what you have to say. *This* is not a
newgroups on relativistic quantum mechanics or QFT. *This* newsgroup
is about classical relativity. That means, as I said in another post

In the new text "Classical Electrodynamics - Second Edition," by Hans.
C. Ohanian. On page 241 Ohanian writes
---------------------------------------------
Eq. (73) E_0 = mc^2
.......

As an example, let us apply Eq. (73) to a bound system such as an atom
or nucleus consisting of particles of masses m1, m2, m3, .... with
velocities v1, v2, v3, ...... The total C.M. (center-of-momentum)
energy of the system is the sum of the particle energies and potential
energies, and therefore the mass of the system is

(74) M = sum(i) gamma(vi) mi + U/c^2

where the summation is over all the particles of the system.
---------------------------------------------
where the U is the total potential energy of the neucleon. This is
almost exactly what I said in my web page on nuclear fission. Although
I set the initial velocities to zero for convenience.

Please Bilgey old man. Go back to the basics so we don't have to
correct you all the time and learn how to use classical relativity
without goinf outside its domain since this is sci.physics.relativity!

Best regards

Pete

.

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