Fourier transform and vector function

[gabbagabbahey]

I think we are okay...

$$\vec{\nabla} \left( e^{-i \vec{k} \cdot \vec{r}} \right) = \frac{d}{dr_x} \left( e^{-i k_x r_x} e^{-i k_y r_y} e^{-i k_z r_z} \right) +\frac{d}{dr_y} \left( e^{-i k_x r_x} e^{-i k_y r_y} e^{-i k_z r_z} \right) +\frac{d}{dr_z} \left( e^{-i k_x r_x} e^{-i k_y r_y} e^{-i k_z r_z} \right)$$

$$= -ik_x e^{-i \vec{k} \cdot \vec{r}} - i k_y e^{-i \vec{k} \cdot \vec{r}} - i k_z e^{-i \vec{k} \cdot \vec{r}}$$

$$= -i \vec{k} e^{-i \vec{k} \cdot \vec{r}}$$

Yes, your right...phew!...Unfortunately there is still a problem in your $$\vec{f_{\perp}}(\vec{k})$$ somewhere and I haven't been able to pinpoint it yet. Your essentially getting

$$\hat{k} \times \hat{k} \times \vec{f}(\vec{k})$$

which would be zero. I think the answer is supposed to be :


$$\hat{k} \times \vec{f}(\vec{k})$$

so that $$\vec{f_{\perp}}(\vec{k})$$ pulls out the component of $$\vec{f}(\vec{k})$$ perpenciular to $$\vec{k}$$.

 

[LocationX]

why are we allowed to do this:

$$\frac{i\vec{k}}{4 \pi} \times \int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \left( \int_{- \infty}^{\infty} \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } \times \vec{f}(\vec{r'}) d^3r' \right) d^3r = \frac{i\vec{k}}{4 \pi} \times \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } d^3r \right) \times \vec{f}(\vec{r'}) d^3r'$$

 

[gabbagabbahey]

why are we allowed to do this:

$$\frac{i\vec{k}}{4 \pi} \times \int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \left( \int_{- \infty}^{\infty} \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } \times \vec{f}(\vec{r'}) d^3r' \right) d^3r = \frac{i\vec{k}}{4 \pi} \times \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } d^3r \right) \times \vec{f}(\vec{r'}) d^3r'$$

I'm pretty sure that this part is fine:
$$e^{i\vec{k} \cdot \vec{r}}$$
is constant over $$r'$$ and so you can pull it inside the inner intgral.

You can then change the order of integration, integrating first over $$r$$ and then $$r'$$

and $$\vec{f}(\vec{r'})$$ is constant over $$r$$ so the $$\times \vec{f}(\vec{r'})$$ can be pulled out of the integral over $$r$$

However, I still can't find the error.

 

[chenel]

There's no problem.

$$\hat{k} \times \hat{k} \times \vec{f}(\vec{k}) = \hat{k}(\hat{k} \cdot \vec{f}) - \vec{f} \left|\hat{k}\right|^2 = \vec{f_{\perp}} - \vec{f}$$

which is the negative of the transverse component.

You're only missing a minus sign somewhere.

 

[gabbagabbahey]

There's no problem.

$$\hat{k} \times \hat{k} \times \vec{f}(\vec{k}) = \hat{k}(\hat{k} \cdot \vec{f}) - \vec{f} \left|\hat{k}\right|^2 = \vec{f_{\perp}} - \vec{f}$$

which is the negative of the transverse component.

You're only missing a minus sign somewhere.

Wow, I can't believe I thought $$\hat{k} \times (\hat{k} \times \vec{f}(\vec{k}))=0$$. I shouldn't help people with there homework so late at night.

 

[chenel]

Whoops, what I meant was

$$\hat{k} \times \hat{k} \times \vec{f}(\vec{k}) = \hat{k}(\hat{k} \cdot \vec{f}) - \vec{f} \left|\hat{k}\right|^2 = \vec{f_{\parallel}} - \vec{f}$$

Note that it's $$\vec{f}$$ minus its LONGITUDINAL component that makes the the transverse component.

@gabbagabbahey: oh well. you did a lot of good stuff -- I guess we can take one mistake :)

 

[LocationX]

I think i found the missing negative... let me try and wrap things up. Thank you again for all the help.

$$u \vec{\nabla} v = \vec{\nabla} (uv) - v \vec{\nabla} u$$

$$\int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \vec{\nabla} \left( \frac{-1}{|\vec{r}- \vec{r'} | } \right) d^3r = \int_{- \infty}^{\infty} \vec{\nabla}\left( e^{i \vec{k} \cdot \vec{r}}} \frac{-1}{|\vec{r}- \vec{r'} | } \right) d^3r - \int_{- \infty}^{\infty} \left( \frac{-1}{|\vec{r}- \vec{r'} | } \right) \vec{\nabla} e^{i \vec{k} \cdot \vec{r}}$$

$$= \int_{- \infty}^{\infty} \frac{ i \vec{k} e^{i \vec{k} \cdot \vec{r}}} {|\vec{r}- \vec{r'} |} d^3r$$

I had $$= \int_{- \infty}^{\infty} \frac{ - i \vec{k} e^{i \vec{k} \cdot \vec{r}}} {|\vec{r}- \vec{r'} |} d^3r$$

Starting from post 31 including the negative sign, we have:

$$= \frac{\vec{k}}{{k^2}} \times \int_{- \infty}^{\infty} (-e^{-i \vec{k} \cdot \vec{r'} }) \vec{k} \times \vec{f}(\vec{r'}) d^3r'$$

$$= \frac{\vec{k}}{k^2} \times \vec{k} \times \int_{- \infty}^{\infty} (-e^{-i \vec{k} \cdot \vec{r'} }) \vec{f}(\vec{r'}) d^3r'$$

$$= \frac{\vec{k}}{k^2} \times \vec{k} \times - \vec{f}(\vec{k})$$

$$=\frac{\vec{k}}{k^2} \times \vec{k} \times - \vec{f}(\vec{k})$$

$$=\frac{- \hat{k}(\hat{k} \cdot \vec{f}(\vec{k}))}{k^2} + \frac{ \vec{f}(\vec{k}) \left|\hat{k}\right|^2}{k^2}$$

$$= \frac{- \vec{f_{\parallel}}(\vec{k}) + \vec{f}(\vec{k})}{k^2}$$


Does the k^2 belong there?

 

[chenel]

Does the k^2 belong there?

Nope. Be careful: you're mixing $$\hat{k}$$'s and $$\vec{k}$$'s with abandon! Remember that
$$\hat{k} = \frac{\vec{k}}{k}$$
... that should straighten you out.

 

[LocationX]

sorry, i still do not see where i have mixed the khat and kvector

 

[LocationX]

actually, i think i got it... thank you.