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From: Jay R. Yablon on 14 Jan 2010 11:01

"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***

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.

JRY: I know you said it was your last post. Perhaps I am being
pedantic, but if you are correct you should be able in equation (9) of
http://jayryablon.files.wordpress.com/2010/01/dq.pdf to identify the
precise point at which I make an illegal step and why, which is why I
wrote out (9) in four steps. Is the wrong step in your view simply the
fourth term, where I substitute a momentum? Or, is it earlier. I would
appreciate if you could please do so.

Jay

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