On 2/14/2014 11:05 AM, Bruno Marchal wrote:
On 14 Feb 2014, at 04:19, Russell Standish wrote:
On Thu, Feb 13, 2014 at 06:07:00PM +0100, Bruno Marchal wrote:
On 13 Feb 2014, at 16:40, Quentin Anciaux wrote:
2014-02-13 16:31 GMT+01:00 Bruno Marchal <[email protected]>:
On 13 Feb 2014, at 12:36, Quentin Anciaux wrote:
hence F=ma cannot be universaly true if comp is true.
...
Even F=m*a cannot be universal as I've shown,
It might be. I think it is (I mean the Feynman generalisation, which
is already close to comp-physics, but that's out of the topic).
...
The computation interfere below the substitution level, but the
artificial simulation with F≠ma, bring an artificial physics, which
does not result from the interference below the subst. level.
If qZ1* proves F=ma, and if my environment does not obeys F=ma, it
will looks "dreamy" to me, I will see that I am not in a real (comp)
physical reality, I will see the discrepancy.
F=ma is more of a definition actually, than a logical constraint. It
is how we define (and operationally measure) "force".
No problem with that, and that is why a answered with F = KmM/r^2, but that was not much
relevant.
If you have a copy of Vic Stenger's "Comprehensible Cosmos", he
discusses this from page 48.
No problem. I appreciate the argument.
I read it online, and it was taught by some physicists.
Actually, the correct relativistic form is F=dp/dt, where p is the 3
momentum of the object under consideration. F=ma is its low velocity
approximation.
Sure. Even F = dp/dt is a classical approximation deducible from Feynman
integral.
So I would be surprised if COMP fails to prove Newton's second law -
it would mean someone was using terminology inconsistently.
F= ma is like H phi = E phi. All is in F, or H. Those equality should be laws indeed,
and deducible from deeper laws. It might be more doubtful for F or H, except that the
Turing universality of the vacuum suggest some "H = 0", à-la Dewitt-Wheeler. But we are
not yet there ..
But this seems to point to a deeper problem. If we elaborate H and E as operators and psi
as a ray in a Hilbert space and if we further define the Hilbert space, we will still have
a symbolic expression which we can related ostensively to some apparatus. But we will
never get down to an arithmetical computation.
Brent
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