Proving \Phi = 1 / r satisfies \nabla^2\Phi=0 in Cartesian coordinates

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SUMMARY

The discussion focuses on proving that the scalar potential function \(\Phi = \frac{1}{r}\) satisfies the Laplace equation \(\nabla^2\Phi = 0\) in Cartesian coordinates, specifically for \(r \neq 0\). Participants emphasize the importance of correctly defining \(r\) as \(r = \sqrt{x^2 + y^2 + z^2}\) and applying the appropriate differentiation techniques. The initial attempt at differentiation was flawed, leading to incorrect conclusions about the divergence of the gradient. Correct application of the Laplacian operator confirms that \(\nabla^2\Phi = 0\) holds true.

PREREQUISITES
  • Understanding of vector calculus, specifically the Laplacian operator.
  • Familiarity with Cartesian coordinates and their application in three-dimensional space.
  • Knowledge of partial derivatives and their computation.
  • Concept of scalar potential functions in physics and mathematics.
NEXT STEPS
  • Study the derivation of the Laplacian operator in Cartesian coordinates.
  • Learn about the properties of scalar potential functions in electrostatics.
  • Explore examples of solving Laplace's equation in different coordinate systems.
  • Investigate the implications of \(\nabla^2\Phi = 0\) in physical applications, such as fluid dynamics.
USEFUL FOR

This discussion is beneficial for students and professionals in mathematics, physics, and engineering, particularly those focusing on vector calculus and potential theory.

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Homework Statement



Using Cartesian coordinates show that [tex]\Phi = 1 / r[/tex] satisfies [tex]\nabla^2\Phi=0[/tex] for r not equals to 0.

Homework Equations


The Attempt at a Solution



I tried to let [tex]r = x+y+z[/tex], then [tex]\nabla \Phi = ( \frac{1}{r^2} , \frac{1}{r^2}, \frac{1}{r^2})[/tex], but [tex]\nabla . \nabla \Phi[/tex]is obviously not zero
 
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You didn't do the differentiation correctly. What is the partial derivative of 1/r with respect to x?
 
..and [itex]r^2=x^2+y^2+z^2[/itex] would work better too...
 

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