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Message from discussion Probability in an infinite sample space
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Robert Low  
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 More options Mar 28 2006, 6:47 pm
Newsgroups: sci.math
From: Robert Low <mtx...@coventry.ac.uk>
Date: Tue, 28 Mar 2006 17:47:32 +0100
Local: Tues, Mar 28 2006 6:47 pm
Subject: Re: Probability in an infinite sample space

Robert Low wrote:
> Han de Bruijn wrote:

>>I still feel rather uncomfortable with this comment. Because there is
>>some truth in it. By this I mean the way infinitesimals are _actually_
>>done in i.e. physics. I have an example of this and would be glad if
>>someone feels like scrutinizing it. It's in the following document:
>>http://hdebruijn.soo.dto.tudelft.nl/jaar2006/RobertLow.PDF
>>The problem is in the assumption at the bottom of page 1, where it is
>>assumed that two angles are 90 degrees, _before_ the limit is taken,
>>while this can be motivated only _after_ the limit has been taken.

>>It seems weird, but I don't know how to arrive at the result in another
>>manner. I would be thankful if somebody points me out how to accomplish
>>the same, but without being "sloppy" in this way.

> I'll just consider the 2-d case, where everything lies
> in a plane. I think you're computing the light intensity
> at B due to a unit source at F, so that's what
> the following will work with. You can recover the full
> 3-d case by rotating about AF if you want to.

Bunch of stuff deleted.

I should probably have mentioned that I don't think any
mathematician would bat an eyelid at what was essentially
your argument namely: Let dS be an element of surface at B
tangent to a sphere centred at F. Then the projection
of dS to the image plane has area dS/cos(alpha), so
the intensity at B is just  cos(alpha)/(4 pi R^2).

Or even shorter: the intensity vector E at B is given by
1/(4pi R^2) * the unit radial vector, so the
component perpendicular to that is just the dot product
of E with the perpendicular to the image plane,
which is the same as above.

(And I belatedly realised that you were using 1/2pi
there so that the total flux through the image
plane would be 1, which I would call a source
of strength 2, rather than 1.)


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