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Physical Interpreation of the Laplace Operator

  1. Feb 25, 2010 #1
    I am wondering if there is a physical interpretation of the Laplace operator (also known as Laplacian, Δ, ∇2, or ∇·∇).

    From my impression a gradient of a function is the vector field in the direction with the greatest change. Also a divergence is volume density of flux from a point source.

    Can I think of the Laplacian as the volume density of the flux in direction of greatest change from a point source? I am trying to use the identity that the Laplace operator is the divergence of the gradient. I am not sure what I am proposing makes sense. I am a new student to vector calculus so I am trying to understand the physical meaning of these terms. Please correct me if anything that I have stated is incorrect.

    Thanks.
     
    Last edited: Feb 25, 2010
  2. jcsd
  3. Mar 15, 2010 #2
    I've taken a deeper look into this question and for a scalar Laplacian as in the heat equation, the Laplacian is the divergence of the gradient,

    [tex] \nabla^2 T = \nabla \cdot \nabla T [/tex].

    But for cases when the function is not a scalar such as in the Navier-Stokes Equation, the Laplacian is the gradient of the divergence of V minus the curl of the curl of V.

    [tex] \nabla^2 V = \nabla (\nabla\cdot V) - \nabla \times (\nabla \times V) [/tex].

    Can I physically think of the Laplacian as the volume density of the flux (divergence) in direction of greatest change (gradient) from a point source? Is there a physical view available or should I see this as a mathematical construct?

    Thanks.
     
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