Does the Generalised Laplacian satisfy a certain relation?

  • Thread starter Thread starter spaghetti3451
  • Start date Start date
  • Tags Tags
    Laplacian
Click For Summary
SUMMARY

The discussion confirms that the relation involving the Generalised Laplacian does not hold universally. Specifically, the left-hand side (LHS) simplifies to \(\left( \frac{D-1}{r} \frac{\partial}{\partial r} + \frac{\partial ^2}{\partial r^2} \right) \psi\), while the right-hand side (RHS) includes an additional term \(\left( \frac{D-1}{2} \right) \left( \frac{D-3}{2} \right) \frac{1}{r^2} \psi\). The relation is valid only under the condition that \(\left( \frac{D-1}{2} \right) \left( \frac{D-3}{2} \right) \frac{1}{r^2} \psi = 0\).

PREREQUISITES
  • Understanding of differential operators in mathematical physics
  • Familiarity with the concept of the Generalised Laplacian
  • Knowledge of dimensional analysis in the context of partial differential equations
  • Proficiency in manipulating mathematical expressions involving derivatives
NEXT STEPS
  • Research the properties of the Generalised Laplacian in various dimensions
  • Study the implications of boundary conditions on differential equations
  • Explore the role of dimensional analysis in solving partial differential equations
  • Investigate the conditions under which specific relations hold in mathematical physics
USEFUL FOR

Mathematicians, physicists, and researchers working with differential equations and mathematical modeling, particularly those focusing on the Generalised Laplacian and its applications in various dimensions.

spaghetti3451
Messages
1,311
Reaction score
31
Hi, I was wondering if the following relation holds:

$$ \frac{1}{r^{D-1}} \frac{\partial}{\partial r} \left( r^{D-1} \frac{\partial}{\partial r} \right) \psi = \frac{1}{r^{\frac{D-1}{2}}} \frac{\partial ^2}{\partial r^2} \left( r^{\frac{D-1}{2}} \right) \psi $$

I've seen that the LHS evaluates to:

$$\left( \frac{D-1}{r} \frac{\partial}{\partial r} + \frac{\partial ^2}{\partial r^2} \right) \psi $$

while the RHS evaluates to:

$$ \left( \frac{D-1}{r} \frac{\partial}{\partial r} + \frac{\partial ^2}{\partial r^2} + \left( \frac{D-1}{2} \right) \left( \frac{D-3}{2} \right) \frac{1}{r^2} \right) \psi $$

Am I correct?
 
Physics news on Phys.org
That relation holds only if ##\left( \frac{D-1}{2} \right) \left( \frac{D-3}{2} \right) \frac{1}{r^2} \psi = 0##.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K