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From: achille on 14 Jan 2010 02:01
On Jan 14, 1:38 pm, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
> "achille" <achille_...(a)yahoo.com.hk> wrote in message
> OK, here is my reply, in the file linked below:
>
> http://jayryablon.files.wordpress.com/2010/01/dq.pdf
>
> I believe that I am doing this correctly, but am willing to keep an open
> mind, and if I am doing something wrong here, I hope you will able to
> pinpoint exactly what operation I am doing that is illegal for one
> reason or another.
>
> Fundamentally, it seems to me that when you say dq  "has nothing to do
> with difference of q between nearby time slices and hence completely
> unrelated to momentum p," you are saying that momentum p is (mass times)
> the "*difference of q between nearby time slices*" (of delta t).  That
> is a pre-calculus statement, and not quite accurate.  Newton and
> Leibnitz taught us how to define momentum *on a single time slice*, just
> as the derivative of a function can be defined *at a single point* using
> calculus.  It is calculus that gets us from differences in a function
> between nearby points divided by that difference, to exact slopes *at a
> single point.*
>
No, in regular calculus following Newton/Cauchy's treatment,
one can talk about the limit of dq/dt as dt, dq -> 0. However,
you cannot talk about dq, dt individually. They does not
exist independently.

Of course, one can switch to use non-standard analysis
where dq, dt does exists as infinitesimal quantities.
Expression like dq/dt and sqrt(dq) is also well defined.
However,

your m dq/dt still have nothing to do with momentum.

the dq in your expression is a difference on two hyperreals


Finally,
From: achille on 14 Jan 2010 02:10
On Jan 14, 3:01 pm, achille <achille_...(a)yahoo.com.hk> wrote:
>
>    your m dq/dt still have nothing to do with momentum.
>
> the dq in your expression is a difference on two hyperreals
>
> Finally,

***Damn, google send it before I complete my reply again***

okay, back to my reply.

the dq in your expression is the difference between two q
infinitesimal close on same t. while the proper m dq/dt
appear in a momentum is the difference between two q on
two infinitesimal close t.

Even if you use the proper dq to define your momentum.
It won't help. In the context of path integral, 'momentum'
is not a constant of motion at all, so sqrt( p dq ) does
not mean it is proportional to sqrt(dq).

Period. This is my last post on this thread.
From: Jon Slaughter on 14 Jan 2010 02:39
Jay R. Yablon wrote:
> In the process of exploring path integration I have come across an
> integral of the form:
>
> ${-oo,+oo} F(x) (dx)^.5 (1)
>
> Is there any body of literature about how one does such integrals?
> Please note, this does not appear to be a fractional integral d^.5x
> which is done via the Gamma function. Not does this appear to be a
> d(x^.5) which can easily be converted to .5 x^-.5 dx and thus give us
> a whole-integer power for dx.
>
> This seems more akin to a differential equation (x' = first
> derivative, e.g., dx/dt) which contains an x'^.5 rather than an x'^2.
>
> Thanks,
>

Such integrals are not useful because they either are 0 or infinity assuming
a generalization from the riemann integral at least.

The standard riemann integral is defined as as the limit of the riemann
sums(which we all know about)

so by analogy

sum(f(a + i*dx)*g(dx)) -> int(f(x)*g(dx))

for a uniform partition

let a = 0 and b = 1 then we have

g(1/n)*sum(f(i/n))

if the order of g is larger than that of the sum then the sum will be 0 else
it will be infinite.

to seet his note that the sum is approximated by the standard riemann
integral so that

int(f(x)*g(dx)) ~= g(dx)/dx*int(f(x))

as dx->0 we see that, the sum is only bounded when g(dx) is proportional to
dx.

Of course in general this may not always be true but for continuous or
piecewise continuous functions this seems likely(a more rigorous proof is
needed of course.

The point is that such integrals are most likely useless in most cases and
hence why they are not used.








From: Androcles on 14 Jan 2010 02:42

"achille" <achille_hui(a)yahoo.com.hk> wrote in message
news:b0c199b0-7d63-4d25-8015-5235976f86b8(a)h9g2000yqa.googlegroups.com...
On Jan 14, 3:01 pm, achille <achille_...(a)yahoo.com.hk> wrote:
>
> your m dq/dt still have nothing to do with momentum.
>
> the dq in your expression is a difference on two hyperreals
>
> Finally,

***Damn, google send it before I complete my reply again***


=============================================
Yeah, it's someone else's fault, not yours of course.

Oh dear, the rest of your message is snipped. The cat did it.







From: Androcles on 14 Jan 2010 02:46

"Jon Slaughter" <Jon_Slaughter(a)Hotmail.com> wrote in message
news:himhnn$lsh$1(a)news.eternal-september.org...
> Jay R. Yablon wrote:
>> In the process of exploring path integration I have come across an
>> integral of the form:
>>
>> ${-oo,+oo} F(x) (dx)^.5 (1)
>>
>> Is there any body of literature about how one does such integrals?
>> Please note, this does not appear to be a fractional integral d^.5x
>> which is done via the Gamma function. Not does this appear to be a
>> d(x^.5) which can easily be converted to .5 x^-.5 dx and thus give us
>> a whole-integer power for dx.
>>
>> This seems more akin to a differential equation (x' = first
>> derivative, e.g., dx/dt) which contains an x'^.5 rather than an x'^2.
>>
>> Thanks,
>>
>
> Such integrals are not useful because they either are 0 or infinity
> assuming a generalization from the riemann integral at least.
>
> The standard riemann integral is defined as as the limit of the riemann
> sums(which we all know about)
>
> so by analogy
>
> sum(f(a + i*dx)*g(dx)) -> int(f(x)*g(dx))
>
> for a uniform partition
>
> let a = 0 and b = 1 then we have
>
> g(1/n)*sum(f(i/n))
>
> if the order of g is larger than that of the sum then the sum will be 0
> else it will be infinite.
>
> to seet his note that the sum is approximated by the standard riemann
> integral so that
>
> int(f(x)*g(dx)) ~= g(dx)/dx*int(f(x))
>
> as dx->0 we see that, the sum is only bounded when g(dx) is proportional
> to dx.
>
> Of course in general this may not always be true but for continuous or
> piecewise continuous functions this seems likely(a more rigorous proof is
> needed of course.
>
> The point is that such integrals are most likely useless in most cases and
> hence why they are not used.
>
The point is that such replies as yours are most likely useless in most
cases and
hence why they are ignored.





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