# A Gradient of scalar product

1. Apr 24, 2017

### LagrangeEuler

It is very well known result that $grad[e^{i\vec{k}\cdot \vec{r}}]=i\vec{k}e^{i\vec{k}\cdot \vec{r}}$. Also $\vec{k}\cdot \vec{r}=kr\cos \theta$ and $gradf(r)=\frac{df}{dr} grad r$. Then I can write
$$grad e^{ikr\cos \theta}=ik\cos \theta e^{i \vec{k}\cdot \vec{r}} \frac{\vec{r}}{r}=ik\frac{\vec{k}\cdot \vec{r}}{kr}e^{i\vec{k}\cdot \vec{r}} \frac{\vec{r}}{r}$$
Somehow it is the same result only if $\vec{k}=\frac{(\vec{k}\cdot \vec{r})\vec{r}}{r^2}$ and this is not the same. Right?

Last edited by a moderator: Apr 24, 2017
2. Apr 24, 2017

These two are not the same, and I believe the "error" is that in spherical coordinates $\nabla F(r,\theta,\phi)=\frac{\partial{F}}{\partial{r}}\hat{a}_r+\frac{1}{r} \frac{\partial{F}}{\partial{\theta}} \hat{a}_{\theta}+\frac{1}{r sin(\theta)}\frac{\partial{F}}{\partial{\phi}} \hat{a}_{\phi}$, and not simply $(df/dr) \nabla r$. (In this case, the second term $\frac{\partial{F}}{\partial{\theta}}$ is non-zero and needs to be included.) $\\$ Editing... I don't think spherical coordinates is the best approach either.. (what I wrote is not correct because it doesn't properly account for the angle between $\vec{k}$ and $\vec{r}$.) The best/easiest way to see the result that $\nabla e^{i \vec{k} \cdot \vec{r}} =i \vec{k} e^{i \vec{k} \cdot \vec{r} }$ is to simply write out the expression in Cartesian coordinates and evaluate it. $\\$ Additional editing: It may interest you that the gradient symbol can be written in Latex with " \nabla".

Last edited: Apr 24, 2017