Fourier transform and vector function

[LocationX]

I used:
u \vec{\nabla} v = \vec{\nabla} (uv) - v \vec{\nabla} u

Skipping some steps, the integral becomes:

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

= - \frac{i 4 \pi \vec{k}}{k^2} e^{-i \vec{k} \cdot \vec{r'}}

thus:

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

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

I am not sure where to go from here, but is everything up to this step valid?

 

[gabbagabbahey]

I used:
u \vec{\nabla} v = \vec{\nabla} (uv) - v \vec{\nabla} u

Skipping some steps, the integral becomes:

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

= - \frac{i 4 \pi \vec{k}}{k^2} e^{-i \vec{k} \cdot \vec{r'}}

thus:

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

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

I am not sure where to go from here, but is everything up to this step valid?

Everything would be fine, except that I've made a huge mistake:

\hat{r} (i \vec{k} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} (\hat{r} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} = - (\hat{r} \times \hat{r}) \times i \vec{k} + i \vec{k} = 0 + i \vec{k} = i \vec{k}

Is clearly not true.

<br /> \vec{\nabla} \left( e^{-i \vec{k} \cdot \vec{r}} \right) = \hat{r}(-i\vec{k} \cdot \hat{r}) e^{-i \vec{k} \cdot \vec{r}} <br />

Must be used throughout the calculations.

 

[LocationX]

how come this is not true:

\hat{r} (i \vec{k} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} (\hat{r} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} = - (\hat{r} \times \hat{r}) \times i \vec{k} + i \vec{k} = 0 + i \vec{k} = i \vec{k}

Doesnt that just mean: \vec{\nabla} \left( e^{-i \vec{k} \cdot \vec{r}} \right) = -i \vec{k} e^{-i \vec{k} \cdot \vec{r}}

which is what has been used

 

[gabbagabbahey]

how come this is not true:

\hat{r} (i \vec{k} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} (\hat{r} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} = - (\hat{r} \times \hat{r}) \times i \vec{k} + i \vec{k} = 0 + i \vec{k} = i \vec{k}

Doesnt that just mean: \vec{\nabla} \left( e^{-i \vec{k} \cdot \vec{r}} \right) = -i \vec{k} e^{-i \vec{k} \cdot \vec{r}}

which is what has been used

It is only true for the VERY special case that \vec{k}=k\hat{r} when \vec{k} points in a different direction, it is clearly false. And since we are integrating over all possible \hat{r} we can't possibly choose a coordinate system where k always points in the same direction as r.

 

[LocationX]

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}}

 

[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&#039;} |^2 } \times \vec{f}(\vec{r&#039;}) d^3r&#039; \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&#039;} |^2 } d^3r \right) \times \vec{f}(\vec{r&#039;}) d^3r&#039;

 

[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&#039;} |^2 } \times \vec{f}(\vec{r&#039;}) d^3r&#039; \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&#039;} |^2 } d^3r \right) \times \vec{f}(\vec{r&#039;}) d^3r&#039;

I'm pretty sure that this part is fine:
e^{i\vec{k} \cdot \vec{r}}
is constant over r&#039; 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&#039;

and \vec{f}(\vec{r&#039;}) is constant over r so the \times \vec{f}(\vec{r&#039;}) 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&#039;} | } \right) d^3r = \int_{- \infty}^{\infty} \vec{\nabla}\left( e^{i \vec{k} \cdot \vec{r}}} \frac{-1}{|\vec{r}- \vec{r&#039;} | } \right) d^3r - \int_{- \infty}^{\infty} \left( \frac{-1}{|\vec{r}- \vec{r&#039;} | } \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&#039;} |} d^3r

I had = \int_{- \infty}^{\infty} \frac{ - i \vec{k} e^{i \vec{k} \cdot \vec{r}}} {|\vec{r}- \vec{r&#039;} |} 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&#039;} }) \vec{k} \times \vec{f}(\vec{r&#039;}) d^3r&#039;

= \frac{\vec{k}}{k^2} \times \vec{k} \times \int_{- \infty}^{\infty} (-e^{-i \vec{k} \cdot \vec{r&#039;} }) \vec{f}(\vec{r&#039;}) d^3r&#039;

= \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.