What does the equation ∇-∇Φ=∇^2Φ represent in this proof?

[NotASmurf]

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.

 

[jedishrfu]

This is just an application of a vector calculus identity that the div of the gradient of the function Phi is the laplacian of the function Phi:

http://en.wikipedia.org/wiki/Laplace_operator

 

[NotASmurf]

"gradient of the function Phi is the laplacian of the function Phi" So ∇Φ=∇^2Φ? but then why the ∇-∇Φ=∇^2Φ?

 

[mathman]

"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".

 

[HallsofIvy]

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

I suspect that what you are reading as "-", a subtraction, is really "\cdot", a dot product. \nabla^2 \phi is defined as \nabla\cdot \nabla \phi.

 

[NotASmurf]

Brilliant! thank you so much, wow I cannot believe I didn't see that, wow.