Why does this say ∇-∇Φ=∇^2Φ?

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

Answers and Replies

  • #3
146
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"gradient of the function Phi is the laplacian of the function Phi" So ∇Φ=∇^2Φ? but then why the ∇-∇Φ=∇^2Φ?
 
  • #4
mathman
Science Advisor
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"gradient of the function Phi is the laplacian of the function Phi" So ∇Φ=∇^2Φ? but then why the ∇-∇Φ=∇^2Φ?
You misread the previous post. It says "divergence of the gradient is the Laplacian". Don't omit "divergence of".
 
  • #5
HallsofIvy
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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].
 
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  • #6
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Brilliant! thank you so much, wow I cannot believe I didn't see that, wow.
 

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