2
$\begingroup$

This is a question about notation. I know that for a function $\psi(x,t)$ $$\nabla\psi=\left(\frac{\partial\psi}{\partial x},\frac{\partial\psi}{\partial t}\right)$$ (the gradient of $\psi$) and $$\nabla^2 \psi=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial\psi^2}{\partial t^2} $$ (the laplacian of $\psi$).

By what does $(\nabla \psi)^2$ mean?

Notes

  • If $(\nabla \psi)^2$ is another way of representing the laplacian than why didn't they just write $\nabla^2 \psi$ or $\Delta \psi$?
  • $K$ is a scalar constant.
  • If context is needed see below

enter image description here

which is from "Novel Methods in Soft Matter Simulations" chapter 29 pg 21

$\endgroup$
9
  • $\begingroup$ It is a short-hand notation for the (square of the) norm of $\nabla\psi$. Physicists, right? $\endgroup$ – AccidentalFourierTransform Jan 31 '18 at 2:43
  • $\begingroup$ Then why not use $||\nabla \psi||$? $\endgroup$ – AzJ Jan 31 '18 at 2:44
  • 1
    $\begingroup$ beats me.${}{}$ $\endgroup$ – AccidentalFourierTransform Jan 31 '18 at 2:46
  • $\begingroup$ I've seen it used in math texts also but yeah it is definitely common in physics as it just means $\vec{v}^2 = \vec{v} \cdot \vec{v}$ $\endgroup$ – Triatticus Jan 31 '18 at 2:55
  • $\begingroup$ So wouldn't it just mean that $(\nabla \psi)^2$=$\nabla^2 \psi$? $\endgroup$ – AzJ Jan 31 '18 at 2:58
2
$\begingroup$

$\nabla \psi \cdot \nabla \psi=(u_x \hat i+u_y \hat j) \cdot(u_x\hat i+u_y\hat j)=(u_x)^2+(u_y)^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.