How do i find minimum for this potential?

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    Minimum Potential
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Homework Help Overview

The discussion revolves around finding the minimum of a potential function given by u(r) = L²/2*m*r² - b*exp(-ar)/r, where b and a are constants, L is angular momentum, and m is mass. Participants are exploring the critical points of the function through its derivative.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivative of the potential function and its critical points, with some questioning the correctness of expressions for u'(r). There are attempts to clarify the differentiation process and explore series expansions for simplification.

Discussion Status

The conversation is ongoing, with participants providing corrections to each other's expressions and discussing the complexity of solving for r. Some suggest using series expansions for approximations, while others express uncertainty about the necessity of such methods. There is no explicit consensus on the best approach yet.

Contextual Notes

Some participants mention that the minimum occurs close to zero and that the problem may lead to a third-order equation if not simplified properly. There is also mention of differing opinions among classmates regarding the use of series expansion.

wormhole
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how do i find minimum for this potential:

u(r)=L^2/2*m*r^2 - b*exp(-ar)/r ?

b,a - constants
L - angular momentum
m - mass

u`(r)=L^2/m+b*exp(-ar)*r(a*r-1)

(u`(r*):=0 => r*=r*(a,b,L,m) )

i plotted the function for some values of b,a,L and m and i see that for certain values there is a minimum but i just can't figure out how do i find the right r.

can anyone help/give some hint?
 
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wormhole said:
how do i find minimum for this potential:
u(r)=L^2/2*m*r^2 - b*exp(-ar)/r ?
b,a - constants
L - angular momentum
m - mass
u`(r)=L^2/m+b*exp(-ar)*r(a*r-1)
(u`(r*):=0 => r*=r*(a,b,L,m) )
i plotted the function for some values of b,a,L and m and i see that for certain values there is a minimum but i just can't figure out how do i find the right r.
can anyone help/give some hint?

If
[tex]U(r) = \frac{L^2}{2mr^2} - \frac{b e^{-ar}}{r}[/tex]
then, what you have written for U'(r) is not correct.
 
ok, the correct expression for u`(r):

u`(r)=L^2/m-b*r*(a*r+1)exp(-a*r)

still i don't quite know how to proceed...
 
wormhole said:
ok, the correct expression for u`(r):
u`(r)=L^2/m-b*r*(a*r+1)exp(-a*r)
still i don't quite know how to proceed...

How is that the correct expression?
What is
[tex]\frac{d}{dr} (\frac{1}{r^2})[/tex]?

and do you know how to find
[tex]\frac{d}{dr} (\frac{e^{-ar}}{r}) ?[/tex]
 
Last edited:
siddhart , so sorry
i simply already did some rearrangemnet:

[tex] \frac{du}{dr}=\frac{L^2}{mr^3} + \frac{b(are^{-ar} + e^{-ar})}{r^2}[/tex]

the expression i wrote is when i do
u`(r):=0
and i multiply by r^3
so sorry for confusion
 
Last edited:
Ok, by solving U'(r)=0, you can find the critical points and then you can find the points of minima with the second derivative test.On second thought, the equation you have to solve to find r looks really tough . I can find no obvious way to solve it.

For values of (ar)<<1, you could approximate by using the series expansion of [tex]e^{ar}[/tex]
 
Last edited:
i'll try to do that...
from the plots i did the r where u(r) is at minimum is very close to zero
i have to take only first term in [itex]e^{-ar}[/itex] expansion series otherwise i get an third order equation
 
Last edited:
wormhole said:
i'll try to do that...
from the plots i did the r where u(r) is at minimum is very close to zero
i have to take only first term in [itex]e^{-ar}[/itex] expansion series otherwise i get an third order equation

Can you post your solutions for r when you get it? This is an interesting problem and I want to check that I got the right answers so I can show my classmates.
 
Last edited:
  • #10
i asked my classmates about the solution and some say that there is no need to use expansion series...
i'm not sure what they did is right..so when i get the official solution(next week) i will give you a link to it
 

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