Sujet : Re: Energy?
De : ross.a.finlayson (at) *nospam* gmail.com (Ross Finlayson)
Groupes : sci.physics.relativityDate : 29. Jul 2024, 00:30:26
Autres entêtes
Message-ID : <HVudnVs62uEdSTv7nZ2dnZfqn_cAAAAA@giganews.com>
References : 1 2
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On 07/28/2024 12:36 PM, J. J. Lodder wrote:
Stefan Ram <ram@zedat.fu-berlin.de> wrote:
>
In a chapter of a book, the author gives this relation for a
system with mass m = 0:
>
E^2/c^2 = p^"3-vector" * p^"3-vector"
>
. Then he writes, "This implies that either there is no particle
at all, E = 0, or we have a particle, E <> 0, and therefore
p^'3-vector' <> 0.".
>
So, his intention is to kind of prove that a particle without mass
must have momentum.
>
But I wonder: Does "E = 0" really mean, "there is no particle."?
>
It is nonsense, from a practical point of view.
Maybe you need a better textbook?
>
E = 0 merely means an energy which is too low to be detectable.
We know that there must be infinitely many of those photons.
See under 'infra-red catastrophe', in which infinitely many IR photons
still have a finite energy,
>
Jan
>
>
The word "catastrophe" is also "perestroika" or "opening" in
the relevant mathematics of the singularity theory which is
a principal branch of what's a multiplicity theory.
Then, the "ultra-violet catastrophe" brings it back to how
the conceit of particle mechanics is rehabilitated in what's
a continuum mechanics, with regards to Rayleigh-Jeans, blackbody
radiation, Millikan measurement, and about how, for example,
sodium has at least three lines in spectroscopy.
See "hobbled".