How do i find minimum for this potential?

In summary, the conversation is about finding the minimum for a potential function using the equation U'(r)=0 and the second derivative test. There is some confusion about the correct expression for U'(r) and the use of series expansion to approximate the solution. It is mentioned that some classmates do not think series expansion is necessary for this problem. The person speaking plans to share the official solution when it becomes available.
  • #1
wormhole
29
0
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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  • #2
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.
 
  • #3
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...
 
  • #4
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:
  • #5
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:
  • #7
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:
  • #8
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:
  • #9
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
 

1. What is the definition of a minimum potential?

A minimum potential is the lowest possible value of energy that a system can have. It is the point of equilibrium where the net force on a system is zero, and any small changes to the system will result in an increase in energy.

2. How do I determine the minimum potential of a system?

To find the minimum potential of a system, you need to find the point where the derivative of the potential function is equal to zero. This point is known as the critical point and corresponds to the minimum potential of the system.

3. What is the significance of finding the minimum potential in a system?

Finding the minimum potential in a system is important because it allows us to determine the stable equilibrium point of the system. It also helps us understand the behavior of the system and make predictions about its future state.

4. What are some common methods for finding the minimum potential?

There are several methods for finding the minimum potential, including graphical methods, algebraic methods, and numerical methods such as gradient descent. The choice of method depends on the complexity of the potential function and the availability of computational resources.

5. Can the minimum potential of a system change over time?

Yes, the minimum potential of a system can change over time if external forces or variables are applied to the system. This can cause the system to reach a new equilibrium point with a different minimum potential.

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