Gradient of scalar product

[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
[tex]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}[/tex]
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?

 

[Charles Link]

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".