Let $F$ be a function of $x,y,z$, namely $F(x,y,z)$.

My question:

What are the necessary and sufficient conditions for $\triangledown$$^2$$F(x,y,z)$=$0$, what does it signify?

I am aware that if $d$ is a differential operator, then $d$ $(\triangledown$$F(x,y,z))$$=0$, where $i, j, k$ are $dx,dy,dz$ respectively. That is, $d(\frac{\partial F}{\partial x}dx,\frac{\partial F}{\partial y}dy,\frac{\partial F}{\partial z}dz)=0$.

I know this is true if $F(x,y,z)$ is the force associated with a conservative potential function or in other words conservative vector field. However, I can not seem to relate it with the Laplacian. Are the two related, or they just seem related to me?

Thank you.

  • $\begingroup$ I think you mean $F(x,y,z)$ is the potential associated with a conservative force. $\endgroup$ Oct 22, 2018 at 3:09
  • $\begingroup$ @spaceisdarkgreen No, $F(x,y,z)$ is the force associated with a conservative vector field. That is, $\exists$$U(x,y,z)$ such that $\triangledown$$U(x,y,z)=-F(x,y,z)$ $\endgroup$ Oct 22, 2018 at 3:16
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    $\begingroup$ Well, you describe it as a function, not a vector field. And in the case of a conservative field/potential, both the potential and the vector field have zero laplacian. $\endgroup$ Oct 22, 2018 at 3:22
  • $\begingroup$ @spaceisdarkgreen Sorry for the confusion, I have seen vector fields referred to as function as well. It is generally done for ease of notation. In fact, a vector field is a function of $x, y$, and $z$. Thanks for the input that potential and the vector field has zero Laplacian. $\endgroup$ Oct 22, 2018 at 3:33
  • $\begingroup$ Yeah, I think that a vector field is a function, in the same sense that a scalar field is also a function (both with $>1$ variables). $\endgroup$
    – manooooh
    Oct 22, 2018 at 3:52

1 Answer 1


Remember that for any vector field $\mathbf{F}$, $\nabla^2 \mathbf{F} = \boldsymbol\nabla(\boldsymbol\nabla\cdot \mathbf{F}) - \boldsymbol\nabla\times(\boldsymbol\nabla\times \mathbf{F}$). So $\nabla^2 \mathbf{F} = 0$ is equivalent to $\boldsymbol\nabla (\boldsymbol\nabla\cdot \mathbf{F}) = \boldsymbol\nabla\times(\boldsymbol\nabla\times \mathbf{F})$.

If $\mathbf{F}$ is conservative, then $\boldsymbol\nabla\times\mathbf{F} = 0$, and so we must have $\boldsymbol\nabla(\boldsymbol\nabla\cdot\mathbf{F}) = 0$. A scalar function with vanishing gradient is a constant.

Thus, if $\mathbf{F}$ is a conservative vector field, $\nabla^2 \mathbf{F} = 0$ if and only if $\boldsymbol\nabla\cdot \mathbf{F} = C$ for some constant $C$.

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    $\begingroup$ Perfect, thank you very much. Can you please also answer the last part of my question? $\endgroup$ Mar 28, 2019 at 7:20
  • $\begingroup$ @BertrandWittgenstein'sGhost Which part is that? $\endgroup$ Mar 29, 2019 at 4:05
  • $\begingroup$ The part where I ask about the relation between differential of a gradient, and the Laplacian. It starts from: "I am aware if $d$ is a differential operator..." Regard. $\endgroup$ Mar 29, 2019 at 4:25
  • $\begingroup$ @BertrandWittgenstein'sGhost I'm not sure what you're getting at with it. Certainly it is true that the curl of a gradient is 0, and the curl is a differential operator. However, This is not true in general of differential operators. Could you link me a source for where you're getting this? $\endgroup$ Mar 29, 2019 at 4:50
  • $\begingroup$ its from differential geometry, wedge product. en.m.wikipedia.org/wiki/One-form. The problem is I know that differential of a gradient is $0$ if the gradient is a conservative field, but I don't know why that is. $\endgroup$ Mar 29, 2019 at 4:57

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