0
$\begingroup$

My understanding of this topic is that the Laplacian operator can be applied to both scalar fields as well as vector fields. The formula

$$\nabla^2 \equiv \frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2}$$ works for either a scalar or a vector.

1) Is it true that Laplacian can be applied to vectors(which I think is a yes)?

2) If yes,then just as we call the Laplacian of a scalar as "Divergence of a gradient",can we also call Laplacian of a vector field as "Gradient of Divergence"?`

(I am new to text editors so please if someone could edit it correctly)

PS:I did some research online but it rather confuses me as to whether or not Laplacian is applicable to a vector.

$\endgroup$
2
  • 1
    $\begingroup$ Have you checked the wiki article? $\endgroup$
    – ares
    Commented Sep 11, 2019 at 11:30
  • $\begingroup$ @ares thanks,doubt is cleared.Not grad of divergence only but with an extra term. $\endgroup$
    – DonGod
    Commented Sep 11, 2019 at 11:56

1 Answer 1

1
$\begingroup$

As you said the laplacian works for both scalar and vector. That being said -scalar laplacian=div(grad) -vector laplacian=grad(div) -curl(curl) that means that vector laplacian is not really grad(div) but grad(div) minus curl of curl.

$\endgroup$
1
  • $\begingroup$ Thanks for that! $\endgroup$
    – DonGod
    Commented Sep 27, 2019 at 5:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .