I am a undegraduate math major student. I have met a problem about gradient and I am comfused about it.
Compute the Gradient of the following functions:
(a) s(r) = sin (kr)/r
(b) s(r) = cos (kr)/r
(Here r is the absolute value of r.)
What is the difference between this one and
(a) s(r) = sin (k·r)
(b) s(r) = cos (k·r)?
Thank you very much for your answers!
In $s(\vec{r}) = \frac{\sin(kr)}{r}$ the right hand side is a function of $r=|\vec r|$ alone. So it takes the same value at all points on a circle (or sphere if you are working in three dimension) around the origin. Hence its gradient must always be radial - the gradient will take the form $\nabla s= f(r)\vec r$.
However in $s(\vec r) = \sin(\vec k \cdot \vec r)$ the right hand side is a function of $r$ and of the angle between $\vec r$ and $\vec k$. Even if we keep $r$ constant then $\vec k \cdot \vec r$ varies in value from $|\vec k||\vec r|$ to $-|\vec k||\vec r|$.