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I have to calculate: $$\nabla^2 \frac{e^{ikr}}{r}$$ which I know to be $\displaystyle -k^2 \frac{e^{ikr}}{r} $ (from a lecture).

Doing it by hand: $$ \nabla^2 f(r) = \frac{1}{r^2} \frac{\partial}{\partial{r}} \left(r^2 \frac{\partial}{\partial{r}}\right) \nabla^2 \left(\frac{e^{ikr}}{r}\right) = \frac{1}{r^2} \frac{\partial}{\partial{r}} \left(r^2 \frac{\partial(\frac{e^{ikr}}{r})}{\partial{r}}\right) = \frac{1}{r^2} \frac{\partial}{\partial{r}} \left(r^2 \frac{ikre^{ikr} - e^{ikr}}{r^2}\right) = \frac{1}{r^2} \left(i^2k^2re^{ikr} + ike^{ikr} - ike^{ikr}\right) = -k^2 \frac{e^{ikr}}{r} $$ but Wolfram-Alpha gives this.

Am I missing something or is Wolfram-Alpha wrong?

Apurv
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Manuel
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    Apparently there's a free T-shirt in this for you :-) ... http://math.stackexchange.com/questions/279372/a-limit-wrong-using-wolfram-alpha?rq=1 – MGA Jul 02 '15 at 17:11
  • I doubt that the extraordinary Wolphram page has an error of calculations. However, a few months ago, not long, it had a link titled “complex multiplication” in which appeared only the multiplication of complex numbers and nothing about complex multiplication itself as a more advanced concept. Now this “error” is not anymore: does not appears as a link on elliptic curves (i.e. not ordinary endomorphism). – Piquito Jul 02 '15 at 17:26

2 Answers2

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When I work it out I get exactly what Wolfram Alpha gets.

It looks to me like you simply have the formula for the Laplacian in polar coordinates totally wrong! Check out https://en.wikipedia.org/wiki/Laplace_operator#Two_dimensions .

David C. Ullrich
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  • You are right, on the professor's notes is written "polar coordinates", but he is using spherical coordinates... – Manuel Jul 02 '15 at 17:25
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It's because WolframAlpha assumes polar coordinates and not spherical. Using Mathematica I get this:

$$\text{Simplify}\left[\text{Laplacian}\left[\frac{\exp (i k r)}{r},\{r,\theta ,\phi \},\text{Spherical}\right]\right]=-\frac{k^2 e^{i k r}}{r}$$ whereas in polar coordinates I get this:

$$\text{Simplify}\left[\text{Laplacian}\left[\frac{\exp (i k r)}{r},\{r,\theta \},\text{Polar}\right]\right]=-\frac{e^{i k r} \left(k^2 r^2+i k r-1\right)}{r^3}.$$

Unfortunately I don't know how to tell Alpha how to work in specified coordinates.

krvolok
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