Im a undergradute student, and I have little expirience with vector calculus.

Can someone help me to obation equation 2 from equation 1: where $\vec{k}$ is the wavenumber vector and $\vec{r}$ the position vector. The conxtext is electromanetism.

Equation 1: $\nabla \times\left(\frac{1}{\epsilon(\vec{r})} \nabla \times\left(e^{i \vec{k} \cdot \vec{r}} \vec{u}_{\vec{k}}(\vec{r})\right)\right)=\left(\frac{\omega(\vec{k})}{c}\right)^{2} e^{i \vec{k} \cdot \vec{r}} \vec{u}_{\vec{k}}(\vec{r})$

This is a tip a got: Applying the chain rule to take the curl of the exponential term explicitly, we can move it outside of the operators and cancel from both sides:

Equation 2:\begin{aligned} (i \vec{k}+\nabla) \times \frac{1}{\epsilon(\vec{r})}(i \vec{k}+\nabla) \times \vec{u}_{\vec{k}}(\vec{r}) &=\left(\frac{\omega(\vec{k})}{c}\right)^{2} \vec{u}_{\vec{k}} \end{aligned}

I am aplying this entity twice: $\nabla \times(\psi \mathbf{A})=\psi(\nabla \times \mathbf{A})+\nabla \psi \times \mathbf{A}$

but at the end I have two terms that I think somehow by a physics instuition or explaniantion I can justify they are zero.

Here is my try:

I start applying the identity: $\nabla \times(\psi \mathbf{A})=\psi(\nabla \times \mathbf{A})+\nabla \psi \times \mathbf{A}$

And get: $\nabla \times \frac{1}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})}\left[e^{i \vec{k} \cdot \vec{r}}\left(\nabla \times \mathrm{u}_{\mathrm{k}}(\overrightarrow{\mathrm{r}})\right)+\nabla e^{i \vec{k} \cdot \vec{r}} \times \mathrm{u}_{\mathrm{k}}(\overrightarrow{\mathrm{r}})\right]=\left(\frac{\omega(\vec{k})}{c}\right)^{2} \mathrm{e}^{\mathrm{i} \overrightarrow{\mathrm{k}} \cdot \overrightarrow{\mathrm{r}}} \mathrm{u}_{\mathrm{k}}(\overrightarrow{\mathrm{r}})$

Now I tried to compute:$\nabla e^{\vec{i} \vec{k} \cdot \vec{r}}$

$$ \begin{array}{c} \nabla e^{i \vec{k} \cdot \vec{r}}=\frac{\partial e^{i \vec{k} \cdot \vec{r}}}{\partial x} \hat{\imath}+\frac{\partial e^{i \vec{k} \cdot \vec{r}}}{\partial y} \hat{\jmath}+\frac{\partial e^{i \vec{k} \cdot \vec{r}}}{\partial z} \hat{k} \\ =\left(\frac{\partial i \vec{k} \cdot \vec{r}}{\partial x} \hat{\imath}+\frac{\partial i \vec{k} \cdot \vec{r}}{\partial y} \hat{\jmath}+\frac{\partial i \vec{k} \cdot \vec{r}}{\partial z} \hat{k}\right) e^{i \vec{k} \cdot \vec{r}} \\ =\left\{\left(i \frac{\partial \vec{k}}{\partial x} \cdot \vec{r}+i \frac{\partial \vec{r}}{\partial x} \cdot \vec{k}\right) i+\left(i \frac{\partial \vec{k}}{\partial y} \cdot \vec{r}+i \frac{\partial \vec{r}}{\partial y} \cdot \vec{k}\right) \hat{\jmath}+\left(i \frac{\partial \vec{k}}{\partial z} \cdot \vec{r}+i \frac{\partial \vec{r}}{\partial z} \cdot \vec{k}\right) \hat{k}\right\}=i \vec{k} \end{array} $$ I am not sure if the previous result is correct. But assuming is right: $\nabla \times \frac{1}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})} e^{i \vec{k} \cdot \vec{r}}\left[\left(\nabla \times u_{k}(\vec{r})\right)+i \vec{k} \times u_{k}(\vec{r})\right]=\left(\frac{\omega(\vec{k})}{c}\right)^{2} \mathrm{e}^{i \vec{k} \cdot \vec{r}} u_{k}(\vec{r})$

now, I rename the term:

$$ \left[\left(\nabla \times u_{k}(\vec{r})\right)+i \vec{k} \times u_{k}(\vec{r})\right]=[\nabla+i \vec{k}] \times u_{k}(\vec{r}) \equiv \vec{A} $$ SO $$ \nabla \times \frac{1}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})} e^{i \vec{k} \cdot \vec{r}} \vec{A}=\left(\frac{\omega(\vec{k})}{c}\right)^{2} \mathrm{e}^{\mathrm{i} \overrightarrow{\mathrm{k}} \cdot \overrightarrow{\mathrm{r}}} \mathrm{u}_{\mathrm{k}}(\overrightarrow{\mathrm{r}}) $$ applying the identity again: $$ \nabla \times \gamma \overrightarrow{\mathrm{B}}=\gamma(\nabla \times \overrightarrow{\mathrm{B}})+(\nabla \gamma) \times \overrightarrow{\mathrm{B}} $$ but I do not know if in this case $\gamma$ would be: $\gamma=\frac{1}{\varepsilon_{r}(\vec{r})} e^{i \vec{k} \cdot \vec{r}}$ or $\gamma=e^{i \vec{k} \cdot \vec{r}},$ where $\varepsilon_{r}(\overrightarrow{\mathrm{r}})$ is the dielectric configuration

If I take $\gamma=\frac{1}{\varepsilon_{r}(\vec{r})} e^{i \vec{k} \cdot \vec{r}} $ I get

$$ \left[\frac{1}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})} e^{i \vec{k} \cdot \vec{r}}(\nabla \times \overrightarrow{\mathrm{A}})+\nabla \frac{1}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})} e^{i \vec{k} \cdot \vec{r}} \times \overrightarrow{\mathrm{A}}\right]=\left(\frac{\omega(\vec{k})}{c}\right)^{2} \mathrm{e}^{\mathrm{i} \overrightarrow{\mathrm{k}} \cdot \overrightarrow{\mathrm{r}}} \mathrm{u}_{\mathrm{k}}(\overrightarrow{\mathrm{r}}) $$ But $$ \nabla \frac{1}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})} e^{i \vec{k} \cdot \vec{r}}=\frac{\varepsilon_{r}(\overrightarrow{\mathrm{r}}) \nabla e^{i \vec{k} \cdot \vec{r}}-e^{i \vec{k} \cdot \vec{r}} \nabla \varepsilon_{r}(\overrightarrow{\mathrm{r}})}{\left(\varepsilon_{r}(\overrightarrow{\mathrm{r}})\right)^{2}}=\frac{i \vec{k}}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})} e^{i \vec{k} \cdot \vec{r}}-\frac{e^{i \vec{k} \cdot \vec{r}} \nabla \varepsilon_{r}(\overrightarrow{\mathrm{r}})}{\left(\varepsilon_{r}(\overrightarrow{\mathrm{r}})\right)^{2}} $$

any idea what I am doing wrong?


2 Answers 2


I don't know what context you met this equation in, but it is clearly not supposed to be an identity. It is is rather an equation to be solved for $u_{\bf k}({\bf r})$.

They appear to just be using $$ \nabla e^{i{\bf k}\cdot {\bf r}} = e^{i{\bf k}\cdot {\bf r}} (i{\bf k} + \nabla) $$ twice.

Note that $\nabla e^{i(k_xx+k_yy+k_z z)}= (\partial_x,\partial_y, \partial_z) e^{i(k_xx+k_yy+k_z z)}=e^{i(k_xx+k_yy+k_z z)} (ik_x,ik_y,ik_z)$ where in your unnecessarily complicated notation $(k_x,k_y,k_z) \to k_x {\bf i}+k_y {\bf j}+k_z {\bf k}$. Thus $$ \nabla e^{i{\bf k}\cdot {\bf r}} =e^{i{\bf k}\cdot {\bf r}} (i{\bf k} ), $$ and hence $$ \nabla (e^{i{\bf k}\cdot {\bf r}}u({\bf r}))= e^{i{\bf k}\cdot {\bf r}} (i{\bf k} + \nabla)u({\bf r}) $$

  • $\begingroup$ Hello, I think I didnt express myself correctly before I have rewritten the question. I hope now is more clear what I am trying to do. $\endgroup$
    – Who
    Jan 14, 2021 at 20:33
  • $\begingroup$ I'll add to my answer then $\endgroup$
    – mike stone
    Jan 14, 2021 at 21:06
  • $\begingroup$ Sorry I dont se how: $$ \nabla e^{i{\bf k}\cdot {\bf r}} = e^{i{\bf k}\cdot {\bf r}} (i{\bf k} + \nabla) $$ I have added my procedure on the question. $\endgroup$
    – Who
    Jan 14, 2021 at 23:14
  • 1
    $\begingroup$ It's just Liebnitz rule for derivatives $D (fg)= (Df)g+ f(Dg)$ with $f=e^{ikr}$ and $g=u$. $\endgroup$
    – mike stone
    Jan 14, 2021 at 23:25
  • $\begingroup$ Thats what I try to do. But I get the following:$\nabla e^{i \vec{k} \cdot \vec{r}}=\left\{\left(i \frac{\partial \vec{k}}{\partial x} \cdot \vec{r}+i \frac{\partial \vec{r}}{\partial x} \cdot \vec{k}\right) i+\left(i \frac{\partial \vec{k}}{\partial y} \cdot \vec{r}+i \frac{\partial \vec{r}}{\partial y} \cdot \vec{k}\right) \hat{\jmath}+\left(i \frac{\partial \vec{k}}{\partial z} \cdot \vec{r}+i \frac{\partial \vec{r}}{\partial z} \cdot \vec{k}\right) \hat{k}\right\}\nabla e^{i \vec{k} \cdot \vec{r}}=i \vec{k} ? ?$ $\endgroup$
    – Who
    Jan 14, 2021 at 23:28

Your calculation $\nabla e^{i \vec{k} \cdot \vec{r}}=i \vec{k}$ is incorrect. The correct version is $$\nabla e^{i \vec{k} \cdot \vec{r}}=i \vec{k}e^{i \vec{k}\cdot \vec{r}}$$ without dropping the exponential.

The rest has been explained already.

  • $\begingroup$ First thank you very much for your help. Now I understand what I did wrong when computing $$\nabla e^{i{\bf k}\cdot {\bf r}}$$. But as mentioned earlier: they applied $$\nabla e^{i{\bf k}\cdot {\bf r}} = e^{i{\bf k}\cdot {\bf r}} (i{\bf k} + \nabla)$$ But the second time one wants to apply the same result we have and extra escalar function. Therefore this leads to:$$\nabla \frac{1}{\varepsilon_{r}(\overrightarrow{\mathrm{r}})} e^{i \vec{k} \cdot \vec{r}}$$ $\endgroup$
    – Who
    Jan 16, 2021 at 0:08

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