# Why does this say ∇-∇Φ=∇^2Φ?

1. Jan 16, 2015

### 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.

2. Jan 16, 2015

### Staff: Mentor

3. Jan 16, 2015

### NotASmurf

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

4. Jan 16, 2015

### mathman

You misread the previous post. It says "divergence of the gradient is the Laplacian". Don't omit "divergence of".

5. Jan 17, 2015

### 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 $\nabla\cdot \nabla \phi$.

6. Jan 28, 2015

### NotASmurf

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