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Why does this say ∇-∇Φ=∇^2Φ?

  1. Jan 16, 2015 #1
    Hey all, in this line of a proof it went straight from
    ∇-∇Φ=-4πGρ to
    ∇ is divergence, Φ is supposed to be potential energy.
    G is gravitational constant and p is density so both are scalars, Any help apreciated.
  2. jcsd
  3. Jan 16, 2015 #2


    Staff: Mentor

  4. Jan 16, 2015 #3
    "gradient of the function Phi is the laplacian of the function Phi" So ∇Φ=∇^2Φ? but then why the ∇-∇Φ=∇^2Φ?
  5. Jan 16, 2015 #4


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    You misread the previous post. It says "divergence of the gradient is the Laplacian". Don't omit "divergence of".
  6. Jan 17, 2015 #5


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    I believe you are mis-reading. A "[itex]\nabla[/itex]" by itself does not have any meaning and [itex](\nabla- \nabla)\phi[/itex] would be equal to 0.

    I suspect that what you are reading as "-", a subtraction, is really "[itex]\cdot[/itex]", a dot product. [itex]\nabla^2 \phi[/itex] is defined as [itex]\nabla\cdot \nabla \phi[/itex].
  7. Jan 28, 2015 #6
    Brilliant! thank you so much, wow I cannot believe I didn't see that, wow.
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