Solving a Parabolic Function Near the Minimum of the Morse Potential

Click For Summary
SUMMARY

The discussion centers on demonstrating that the Morse potential, defined by the equation U = E_0 (1 - exp(-a(r - r_0)))^2, behaves as a parabolic function near its minimum. Participants confirm that a Taylor expansion is the appropriate method to show this, noting that both the potential and its first derivative vanish at r = r_0, while the second derivative does not. Consequently, the potential energy can be approximated as U(r) ≈ U''(r=r_0)/2 * r^2, confirming its parabolic nature.

PREREQUISITES
  • Understanding of Morse potential and its mathematical formulation
  • Familiarity with Taylor series expansion
  • Knowledge of derivatives and their significance in function analysis
  • Basic concepts of potential energy in physics
NEXT STEPS
  • Study the application of Taylor series in physics problems
  • Explore the properties of the Morse potential in quantum mechanics
  • Learn about second derivatives and their role in determining function behavior
  • Investigate other potential energy functions and their approximations near minima
USEFUL FOR

Students and professionals in physics, particularly those focusing on molecular dynamics, potential energy analysis, and mathematical modeling of physical systems.

capslock
Messages
17
Reaction score
0
I have the equation for the Morse potential, U = E_0 (1-exp(-a(r-r_0))^2. I'm asked to show that near the minimum of the curve the potential energy is a parabolic function. I've tried to play around with the taylor series with no hope! :( :(

Many thanks, James
 
Physics news on Phys.org
capslock said:
I have the equation for the Morse potential, U = E_0 (1-exp(-a(r-r_0))^2. I'm asked to show that near the minimum of the curve the potential energy is a parabolic function. I've tried to play around with the taylor series with no hope! :( :(

Many thanks, James

It is indeed just a simple Taylor expansion! Can you show your work?
The potential vanishes at r=r_0 and the derivative of the potential also vanishes at r=r_0. The second derivative does not vanish at that point so you get that U(r) is approximately U''(r=r_0)/2 r^2 so a parabolic function.

Pat
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
Replies
5
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
14K
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K