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MathGroup Archive 2005

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Re: Can you solve my gravity-maths problem in Mathematica?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg62856] Re: Can you solve my gravity-maths problem in Mathematica?
  • From: dh <dh at metrohm.ch>
  • Date: Tue, 6 Dec 2005 23:10:34 -0500 (EST)
  • References: <dn373m$32r$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Paul,
your function does not have a minimal value for r>0.
To show this, we first simplify your function by defining a new variable 
x= m r and we use a definite integral from zero:
f[r] == F[x] == m Exp[x] Integrate[Exp[-2x1]/x1^2,{0,x}]
We may drop m as it does not change the position of a minimum.
If this function  would have a minimum, a x0 would exists so that

F'[x0] = Exp[x0] Integrate[Exp[-2x1]/x1^2,{0,x0}] +Exp[x0] 
Exp[-2x0]/x0^2== 0

from this we get:
Integrate[Exp[-2x1]/x1^2,{0,x0}] = - Exp[-2x0]/x0^2

For positive x0 the left side is positive (positive integrand), but the 
right side is negative. Therefore, no x0 exists.

Daniel

Paul wrote:
> Hello, I have been working on a simple model of quantum gravity. Like
> atoms have a minimum electron shell distance caused by the quantisation
> of the electromagnetic force my approximation for the average distance
> between particles held together by a graviational force, using my
> model, becomes :
> 
> "Find r at the maximum value of:
> 
> f(r)   =  exp( m*r )  *  integrate(  exp(-2*m*r)/r^2 , r )
> 
> (r>0) "
> 
> Where we use constants such that c=h=G=1. In this system for example
> electron mass is something like 10^(-23) I believe. (The graviational
> bond enegy in this model becomes m^5 in my units.)
> 
> So I need to find f'(r)=0.
> 
> Can anyone find a good way to solve this for r, maybe as a series in
> terms of m? Also, I would be interested in seeing a graph of f(r).
> 


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