Re: Energy?

On 07/28/2024 02:37 AM, Stefan Ram 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."?
>
   300 years ago, folks would have said, "m = 0" means that there is
   no particle! Today, we know that there are particles with no mass.
>
   Can we be confident that "E = 0" means "no particle", or could there
   be a particle with "E = 0"?
>
   Here's the Unicode:
>
E²/c² = p⃗ · p⃗
>
   and
>
|This implies that either there is no particle at all, E = 0, or we
|have a particle, E ≠ 0, and therefore p⃗ ≠ 0.
>

It's about conservation law and continuity law, conservation law
that there's an invariant and symmetry or Noether's theorem,
continuity law that it adds up to zero so for a quasi-invariance
and a super-symmetry, then what that would mean would be that
that "point" "particle" is a "resonance system" of, "waves",
that in the algebraic derivation for that derived quantity
because it's built into the derivation, "equals zero".
I.e., the quantities are derivations themselves, and have
"implicits" that parameterize them.
Of course a "sum-of-histories sum-of-potentials" ticks all
the boxes about least action or Maupertuis, that the potential
fields are the real fields, that conservation laws are continuity
laws, for a total sort of theory and not just a dimensionless
linear impulse.
Then the mechanics of cube-wall and inverse-square and long-linear,
are pretty much a usual thing.