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Randy Poe said:
>
> Han de Bruijn wrote:
> > Randy Poe wrote:
> > > Han de Bruijn wrote:
> > >>Start of quotes:
> > >>
> > >>"consider an INFINITELY small element of volume v" and "rays that come
> > >>from the points of an INFINITELY small element of area d(sigma)" & "We
> > >>choose d(sigma) so that its linear dimensions are small compared with
> > >>those of v and consider the cone of rays which, starting at a point of
> > >>d(sigma), meets the volume v. This cone consists of an INFINITE number
> > >>of conical elements" and "The solid angle of such a conical element is
> > >>f/r^2 where f denotes the area of cross-section normal to the axis of
> > >>the cone at a distance r from the vertex."
> > >
> > > Ah, so when you said "physicists use infinitesimals" what you
> > > really mean was "physicists use integrals".
> >
> > No. Physicists use "infinitely small element"s and "an infinite number
> > of elements".
>
> In the form of integrals.
>
> > How can you possibly deny? Can't you just READ the above?
>
> Yes, I've read a thousand such passages, which are always deriving
> an integral equation or differential equation.
>
> > > No, something like dV that appears under an integral is
> > > not what is meant by "infinitesimals" in non-standard analysis.
> >
> > It _doesn't_ appear "under an integral" in the first place
>
> The integral is being defined. When it becomes time to actually
> come up with a useful equation, it will be an integral.
>
> You deny and deny this, insisting that Planck intends to use
> the naked infinitesimal he is describing here, but then...
>
> > Right. Of course it becomes a Riemann integral, at the very end of the
> > chapter.
>
> I rest my case.
>
> > But the point is that Max Planck is talking about "infinitely
> > small quantities" all the time, until he arrives there.
>
> It's a derivation.
>
> > And _before_ he
> > has arrived there, there is _no_ such integral.
>
> And before he arrives there, it's not a thing he's using. It's just
> part of a derivation whose point is to arrive at the integral.
>
> As I said, I'm perfectly happy to admit physicists use integrals.
> I can show you a thousand such paragraphs like the above.
> For instance, in deriving the Coulomb force from a static charge
> distribution, you'll always see an argument like this:
>
> "Let rho(r) = the charge density at position r. An infinitesimal
> volume element dV has charge rho(r) dV. Therefore the
> total field at point R is integral(all space) k*rho(r) (R-r)
> dV/|R-r|^3"
>
> The useful result is that integral. Nobody is going to "use"
> dV or the infinitesimal amount of charge, except as something
> under the integral sign.
>
> The fact that at step 1 of the proof it isn't under an integral
> sign yet does not mean it isn't being introduced just as something
> to PUT under an integral sign. You're reading a proof. It's
> intended to go somewhere. Individual steps in proofs do not
> in general stand by themselves as useful results. Especially
> physics proofs where intermediate steps are non-rigorous
> and everybody knows it.
>
> Physicists "use" delta functions many places as well, and you
> will frequently see them with no integral anywhere in sight. For
> instance, in quantum mechanics, there are theorems about
> certain commutators being "equal to delta(x)".
>
> Why is that OK? Because everybody knows that when you
> want to actually use that, to derive a measurable quantity.
> it's going to appear under an integral.
>
> Your infinitesimals are only used under integral signs.
>
> - Randy
>
>
I think that Han's whole point is that, while the final solution leads to an
integral, the derivation of the integral itself rests on the very notion of
infinitesimals that he's trying to put forth, as the limit of 1/n as n->oo. So,
if calculus itself is derived using this notion, the question becomes why this
notion of the infinitesimal cannot be discussed in its own right. In other
words, why is the concept of an infinitesimal probability for each of an
infinite number of equally likely mutually exclusive outcomes subject to
outright rejection? I really see no reason why this concept cannot coexist with
standard probability. If the answer is that it just doesn't, well then, maybe
it should, eh? :)
--
Smiles,
Tony
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