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From: achille on 12 Jan 2010 10:21
On Jan 12, 10:59 pm, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
>
> Replicating what I posted elsewhere in the thread, I'll show you where.
> Look at Zee's derivation posted athttp://jayryablon.files.wordpress.com/2009/12/sakurai-and-zee.pdf.
>
> Look at $Dq just before equation (4).  Following some manipulation and
> in the limiting case, one can arrive for a given fixed time slice, at
> the expression:
>
> (-2pi i)^.5 $ (m/dt)^.5 dq = (-2pi i)^.5 $ p^.5 dq^.5   (1)
>
> if one uses the momentum:
>
> p = m dq/dt   (2)
>
> to absorb the m/dt term.  One is left in this event, at any fixed time

dq in $Dq is an integration variable over a fixed time slice.
It has nothing to do with difference of q between nearby
time slices and hence completely unrelated to momentum p.




From: Jay R. Yablon on 12 Jan 2010 12:11

"achille" <achille_hui(a)yahoo.com.hk> wrote in message
news:5ab67858-4d40-467b-9f50-5f342cd38c10(a)s31g2000yqs.googlegroups.com...
On Jan 12, 10:59 pm, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
>
> Replicating what I posted elsewhere in the thread, I'll show you
> where.
> Look at Zee's derivation posted
> athttp://jayryablon.files.wordpress.com/2009/12/sakurai-and-zee.pdf.
>
> Look at $Dq just before equation (4). Following some manipulation and
> in the limiting case, one can arrive for a given fixed time slice, at
> the expression:
>
> (-2pi i)^.5 $ (m/dt)^.5 dq = (-2pi i)^.5 $ p^.5 dq^.5 (1)
>
> if one uses the momentum:
>
> p = m dq/dt (2)
>
> to absorb the m/dt term. One is left in this event, at any fixed time

>dq in $Dq is an integration variable over a fixed time slice.
>It has nothing to do with difference of q between nearby
>time slices and hence completely unrelated to momentum p.

This seems to be the most sensible answer I have gotten so far. I
understand what you are saying, and want to think about this some more.
I'll be back to you.

Thanks,

Jay.

From: Ken S. Tucker on 13 Jan 2010 10:47
On Jan 12, 3:07 am, David C. Ullrich <ullr...(a)math.okstate.edu> wrote:
> On Mon, 11 Jan 2010 11:08:18 -0500, "Jay R. Yablon"
>
> <jyab...(a)nycap.rr.com> wrote:
> >In the process of exploring path integration I have come across an
> >integral of the form:
>
> >${-oo,+oo} F(x) (dx)^.5 (1)
>
> Where did you run across this? The notation makes very little
> sense.

Jays doing just fine, David I think you need to take a basic
calculus course, let me explain (using Jay's example), that I
frequently encounter.

V = dX/dT = Velocity.

sqrt(V) = sqrt (dX/dT) , sqrt(dX) = sqrt(V)*sqrt(dT).

That's SOP in calculus, from that a bit of algebraic massage
produces *generally* a means to solve.
Regards
Ken S. Tucker
....
From: eric gisse on 13 Jan 2010 20:16
Ken S. Tucker wrote:

> On Jan 12, 3:07 am, David C. Ullrich <ullr...(a)math.okstate.edu> wrote:
>> On Mon, 11 Jan 2010 11:08:18 -0500, "Jay R. Yablon"
>>
>> <jyab...(a)nycap.rr.com> wrote:
>> >In the process of exploring path integration I have come across an
>> >integral of the form:
>>
>> >${-oo,+oo} F(x) (dx)^.5 (1)
>>
>> Where did you run across this? The notation makes very little
>> sense.
>
> Jays doing just fine, David I think you need to take a basic
> calculus course, let me explain (using Jay's example), that I
> frequently encounter.

Wow, how does one quantify the spectacular arrogance of YOU telling David
Ulrich that *HE* needs to take a basic calculus course?

I notice you haven't offered one meaningful comment yet.

>
> V = dX/dT = Velocity.
>
> sqrt(V) = sqrt (dX/dT) , sqrt(dX) = sqrt(V)*sqrt(dT).
>
> That's SOP in calculus, from that a bit of algebraic massage
> produces *generally* a means to solve.
> Regards
> Ken S. Tucker
> ...

From: Jay R. Yablon on 14 Jan 2010 00:38

"achille" <achille_hui(a)yahoo.com.hk> wrote in message
news:5ab67858-4d40-467b-9f50-5f342cd38c10(a)s31g2000yqs.googlegroups.com...
On Jan 12, 10:59 pm, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
>
> Replicating what I posted elsewhere in the thread, I'll show you
> where.
> Look at Zee's derivation posted
> athttp://jayryablon.files.wordpress.com/2009/12/sakurai-and-zee.pdf.
>
> Look at $Dq just before equation (4). Following some manipulation and
> in the limiting case, one can arrive for a given fixed time slice, at
> the expression:
>
> (-2pi i)^.5 $ (m/dt)^.5 dq = (-2pi i)^.5 $ p^.5 dq^.5 (1)
>
> if one uses the momentum:
>
> p = m dq/dt (2)
>
> to absorb the m/dt term. One is left in this event, at any fixed time

dq in $Dq is an integration variable over a fixed time slice.
It has nothing to do with difference of q between nearby
time slices and hence completely unrelated to momentum p.

[JRY]
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.*

The details of how and why I believe my calculation stands up are in the
file linked above, and I hope we can continue our discussion until we
both see this the same way, one way or the other.

Thanks,

Jay



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