Fourier transform and vector function

1. Sep 16, 2008

LocationX

A vector function can be decomposed to form a curl free and divergence free parts:

$$\vec{f}(\vec{r})=\vec{f_{\parallel}}(\vec{r'})+\vec{f_{\perp}}(\vec{r'})$$

where

$$\vec{f_{\parallel}}(\vec{r'}) = - \vec{\nabla} \left( \frac{1}{4 \pi} \int d^3 r' \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \right)$$

and

$$\vec{f_{\perp}}(\vec{r'}) = \vec{\nabla} \times \left( \frac{1}{4 \pi} \int d^3 r' \frac{\vec{\nabla'} \times \vec{f}(\vev{r'})}{|\vec{r}-\vec{r'}|}$$

I am trying to take the Fourier transform of $\vec{f_{\parallel}}(\vec{r'})$ and $\vec{f_{\perp}}(\vec{r})$

I am starting at $\vec{f_{\parallel}}(\vec{r'})$. We know that the fourier transform is given by:

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

$$\vec{f}(\vec{r}) = \frac{1}{(2 \pi)^3} \int_{-\infty}^{\infty} d^3k e^{- i \vec{k} \cdot \vec{r}} \vec{f}(\vec{k})$$

I'm not exactly sure where to begin. If I just plug and chug, we'd have:

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

$$\vec{f}(\vec{k}) = \int_{-\infty}^{\infty} e^{- i \vec{k} \cdot \vec{r}} - \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) d^3r$$

I just do not see a simple way of tacking this problem. Any thoughts would be appreciated.

2. Sep 16, 2008

granpa

please stick with one thread. I have no idea which one to respond to.

you can edit your posts by clicking on the 'edit' button below it. you can even delete the whole message and I suppose the whole thread.

3. Sep 16, 2008

LocationX

oh... yikes... i didnt know i posted so many times... i got errors left and right every time i tried to post

4. Sep 16, 2008

granpa

has it occurred to you that even if you remove the electric field due electric charges (divergence) that you still have a curl component due to electric current (as opposed to light which is what I assume you are looking to perform the fourier transform on).

5. Sep 16, 2008

learning_phys

thanks for the info.

any vector function can be composed of its curl free and divergence free parts according to Helmholtz's theorem. I'm trying to take the FT of the curl free and the divergence free parts.

6. Sep 16, 2008

granpa

oh wait. are you just looking at the electric field? youre not looking at the magnetic field at all?

7. Sep 16, 2008

LocationX

who said anything of fields? I'm just taking the FT of a vector function, specifically the one in the first post. any ideas on where to start?

8. Sep 16, 2008

gabbagabbahey

You mean:
$$\vec{f}(\vec{k}) = -\int_{-\infty}^{\infty} e^{- i \vec{k} \cdot \vec{r}} \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) d^3r$$
right?

Well, just looking at this integral I see that you are integrating the product of a function:
$$u(\vec{r}) \equiv e^{- i \vec{k} \cdot \vec{r}}$$
and the derivative of a function:
$$dv(\vec{r}) \equiv \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) d^3r$$

...so I would try using integration by parts (along with the fundamental theorem for gradients) and see what you get.

9. Sep 17, 2008

LocationX

integration by parts will give:

$$u(\vec{r}) \equiv e^{- i \vec{k} \cdot \vec{r}}$$

$$dv(\vec{r}) \equiv \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right)$$

$$-\int_{\infty}^{\infty} u(\vec{r}) dv(\vec{r}) d^3 r = \left( e^{- i \vec{k} \cdot \vec{r}} \int \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) d^3r \right) |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{d e^{- i \vec{k} \cdot \vec{r}} }{dr} \int \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) d^3r \right) d^3r$$

So applying the fundamental theorem for gradients:

$$\int_a^b \nabla f(x) dx = f(b) - f(a)$$

then $$\int \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) d^3r \right) = \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\infty-\vec{r'}|} d^3 r' \right) - \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|-\infty-\vec{r'}|} d^3 r' \right)$$

something like that? I'm not sure if what was done is completely correct. Also, not sure how to take the derivative of $$e^{-i \vec{k} \dot \vec{r} }$$

any thoughts would be appreciated.

10. Sep 17, 2008

gabbagabbahey

Arrg...sorry, the fundamental theorem for gradients doesn't help, but there is a corollary of the divergence theorem that does:
$$\int_{\mathcal{V}} \vec{\nabla} T d^3 r = \oint_{\mathcal{S}} Td \vec{a}$$
Where $$\mathcal{S}$$ is the surface bounding the volume $$\mathcal{V}$$.

Let's define:
$$v( \vec{r} ) \equiv \frac{1}{4 \pi} \int_{- \infty}^{\infty} \frac{\vec{\nabla '} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r'$$
and
$$u(\vec{r}) \equiv e^{- i \vec{k} \cdot \vec{r}}$$

$$u \vec{\nabla} v = \vec{\nabla} (uv) - v \vec{\nabla} u$$
to rewrite our integral as:
$$\vec{f_{\parallel}} (\vec{k})= \int_{-\infty}^{\infty} u(\vec{r}) \vec{\nabla} v(\vec{r}) d^3 r = \int_{-\infty}^{\infty} \vec{\nabla} (u(\vec{r}) v(\vec{r})) d^3 r - \int_{-\infty}^{\infty} v(\vec{r}) \vec{\nabla} u(\vec{r}) d^3 r$$

Using the aforementioned corollary to the divergence theorem, the first term becomes:
$$\int_{-\infty}^{\infty} \vec{\nabla} u(\vec{r}) v(\vec{r}) d^3 r= \oint_{\mathcal{S}} u(\vec{r}) v(\vec{r}) d \vec{a}$$
where $$\mathcal{S}$$ is the surface that bounds all of space. But(!) $$r= \infty$$ for this surface and $$u(\vec{r}) v(\vec{r}) \rightarrow 0$$ as $$r \rightarrow \infty$$ , so the surface integral vanishes and we are left with
$$\vec{f_{\parallel}} (\vec{k}) = - \int_{-\infty}^{\infty} v(\vec{r}) \vec{\nabla} u(\vec{r}) d^3 r$$

Meanwhile,
$$\vec{\nabla} u(\vec{r}) = \hat{r} \left( \frac{\partial}{\partial r} \right) e^{-i \vec{k} \cdot \vec{r}} = \hat{r} (-i \vec{k} \cdot \hat{r}) e^{-i \vec{k} \cdot \vec{r}} = - \hat{r} (i \vec{k} \cdot \hat{r})u(\vec{r})$$

Using the rules for vector triple products, we obtain:
$$\hat{r} (i \vec{k} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} (\hat{r} \cdot \hat{r}) = \hat{r} \times (\hat{r} \times i \vec{k}) + i \vec{k} = - (\hat{r} \times \hat{r}) \times i \vec{k} + i \vec{k} = 0 + i \vec{k} = i \vec{k}$$

$$\Rightarrow \vec{\nabla} u(\vec{r}) = - i \vec{k} u(\vec{r})$$

$$\Rightarrow \vec{f_{\parallel}} (\vec{k}) = \int_{-\infty}^{\infty} i \vec{k} u(\vec{r}) v(\vec{r}) d^3 r = \frac{i \vec{k}}{4 \pi} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{-i \vec{k} \cdot \vec{r}} \frac{\vec{\nabla '} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r d^3 r' = \frac{i \vec{k}}{4 \pi} \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} \frac{\vec{\nabla '} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) e^{-i \vec{k} \cdot \vec{r}} d^3 r$$

We can simplify this integral further with the help of the following product rule:
$$\frac{ \vec{\nabla '} \cdot \vec{A} }{T} = \vec{\nabla '} \cdot \left( \frac{\vec{A}}{T} \right) - \vec{A} \cdot \vec{\nabla '} \left( \frac{1}{T} \right)$$

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

Applying the divergence theorem to the first term gives:

$$\int_{- \infty}^{\infty} \vec{\nabla '} \cdot \left( \frac{\vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \right) d^3 r' = \oint_{\mathcal{S}} \frac{\vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d \vec{a'}$$

Again, $$\mathcal{S}$$ is the surface bounding all space at which $$r' \rightarrow \infty$$. So, assuming $$\vec{f}(\infty)$$ is finite, the surface integral will vanish.

Meanwhile,

$$\vec{\nabla '} \left( \frac{1}{|\vec{r}-\vec{r'}|} \right) = \frac{ \widehat{(\vec{r}-\vec{r'})}}{|\vec{r}-\vec{r'}|^2}$$

$$\Rightarrow \int_{- \infty}^{\infty} \frac{\vec{\nabla '} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' = - \int_{- \infty}^{\infty} \vec{f}(\vec{r'}) \cdot \frac{ \widehat{(\vec{r}-\vec{r'})}}{|\vec{r}-\vec{r'}|^2} d^3 r'$$

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

Can you evaluate the bracketed integral?

Last edited: Sep 18, 2008
11. Sep 18, 2008

LocationX

wow thanks for the help! I followed most of it but would not have come up with the methods that were used. This is really amazing.

yes, i think i know how to integrate it... i'll try it out and let you know what ive done

thanks a lot for the help. I was wondering how I should tackle the $$\vec{f_{\perp}}$$ term? Since there is a cross product, we cannot use the vector identity $$u \vec{\nabla} v = \vec{\nabla} (uv) - v \vec{\nabla} u$$

would you suggest $$f (\vec{\nabla} \times \vec{A} ) = \vec{\nabla} \times (f \vec{A} ) - (\vec{\nabla}f) \times \vec{A}$$

Last edited: Sep 18, 2008
12. Sep 18, 2008

gabbagabbahey

You shouldn't get zero for the integral, you should get:
$$\int_{- \infty}^{\infty} e^{-i \vec{k} \cdot \vec{r}} \left( \frac{ \widehat{(\vec{r}-\vec{r'})}}{|\vec{r}-\vec{r'}|^2} \right) d^3 r = \frac{4 \pi i \vec{k}}{k^2} e^{-i \vec{k} \cdot \vec{r'}}$$

$$\Rightarrow \vec{f_{\parallel}} (\vec{k}) = \hat{k} \left( \hat{k} \cdot \int_{- \infty}^{\infty} d^3 r' \vec{f}(\vec{r'}) e^{-i \vec{k} \cdot \vec{r'}} \right) = \hat{k} \left( \hat{k} \cdot \vec{f} (\vec{k}) \right)$$

So that $$\vec{f_{\parallel}} (\vec{k})$$ picks out the component of $$\vec{f} (\vec{k})$$ parallel to $$\vec{k}$$.

As for $$\vec{f_{\perp}} (\vec{k})$$ , you will need to start with the vector identity you just posted, but you will probably have to use a couple of other identities as well (I haven't worked it out yet).

I recommend you start by evaluating the bracketed integral above before you begin to calculate $$\vec{f_{\perp}} (\vec{k})$$.

13. Sep 18, 2008

LocationX

yup, i changed my post as you were quoting me since i realized the integral doesnt go to zero.

14. Sep 18, 2008

gabbagabbahey

Out of curiosity, is this problem taken from any particular text? What course is this for?

15. Sep 18, 2008

LocationX

this isnt out of any text, just a regular hw assignment for first semester EM course

I am trying to do the integral, here is where I am:

$$\int_{- \infty}^{\infty} e^{-i \vec{k} \cdot \vec{r}} \left( \frac{ \widehat{(\vec{r}-\vec{r'})}}{|\vec{r}-\vec{r'}|^2} \right) d^3 r = \left( e^{-i \vec{k} \cdot \vec{r}} \int \frac{\widehat{(\vec{r}-\vec{r} ' )}}{|\vec{r}- \vec{r} ' |^2 } d^3r \right) |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \left( (-i \vec{k} \cdot \hat{r}) e^{-i \vec{k} \cdot \vec{r} '} \left(\int \frac{\widehat{(\vec{r}-\vec{r} ' )}}{|\vec{r}- \vec{r} ' |^2 } d^3r \right) \right) d^3r$$

$$\int \frac{\widehat{(\vec{r}-\vec{r} ' )}}{|\vec{r}- \vec{r} ' |^2 } d^3r = - \frac{\widehat{(\vec{r}-\vec{r} ' )}}{|\vec{r}- \vec{r} ' | }$$

assuming the integral above is correct, we get:

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

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

First term goes away since the exponential drops faster than 1/r
It seems the second term is going to be integration by parts again, however, that leaves with evaluating log at -inf which enters complex analysis. Did I make a mistake somewhere?

16. Sep 19, 2008

gabbagabbahey

Yes there are a couple of mistakes. (1) When doing integration by parts in vector calculus, you much fist choose which product rule to use. Remember there are 6 to choose from. Looking at the integral, I see a normal product of a function e^{...} and a vector. The only rule that has anything like that is the rule for gradients:
$$u \vec{\nabla} v = \vec{\nabla} (uv) - v \vec{\nabla} u$$
where clearly the vector function would have to be able to be written as the gradient of some scalar for this to be useful. So, can you write
$$\frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 }$$ as the gradient of any scalar? If so, what is the scalar. This should leave you with two terms to integrate now, but the first one can be transformed into a surface integral using that corollary to the divergence theorem I mentioned earlier. Since the surface is a sphere of infinite radius, the integrand (and thus also the integral) will vanish (plug in $$r \rightarrow \infty$$ into the integrand and notice that it is zero). This will leave you with one term to integrate, what is it?

P.S. if your EM class uses Jackson for a text, I would recommend you also borrow a copy of David J Griffith's 'Introduction to Electrodynamics' much of the first chapter is devoted to vector calculus and provides some very good examples and problems you can work through. His style of writing is also much easier to follow IMO than Jackson's. If you work through several of the problems in the text, you will likely get a feel for how to tackle these kinds of problems.

Last edited: Sep 19, 2008
17. Sep 19, 2008

LocationX

is this the relation:

$$\frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } - \nabla \left( -\frac{1}{|\vec{r}- \vec{r'} | } \right)$$

18. Sep 19, 2008

gabbagabbahey

If you mean:
$$\frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } = \nabla \left( \frac{-1}{|\vec{r}- \vec{r'} | } \right)$$

then yes.

19. Sep 19, 2008

LocationX

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

$$\frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } = \vec{\nabla} \left( \frac{-1}{|\vec{r}- \vec{r'} | } \right)$$

$$\int_{- \infty}^{\infty} e^{-i \vec{k} \cdot \vec{r}} \left( \frac{ \widehat{(\vec{r}-\vec{r'})}}{|\vec{r}-\vec{r'}|^2} \right) d^3 r = \int_{- \infty}^{\infty} e^{-i \vec{k} \cdot \vec{r}} \vec{\nabla} \left( \frac{-1}{|\vec{r}- \vec{r'} | } \right) d^3 r$$

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

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

Not sure how to proceed. Were these the correct steps?

20. Sep 19, 2008

gabbagabbahey

You should explicitly show that the first term vanishes by transforming it into a surface integral and then showing that the integrand of that surface integral is zero on the specified surface (i.e. at r=infty)

There is also a slight error in your second term:

$$\vec{\nabla} \left( e^{-i \vec{k} \cdot \vec{r}} \right) = \hat{r}(-i\vec{k} \cdot \hat{r}) e^{-i \vec{k} \cdot \vec{r}} =-i \vec{k} e^{-i \vec{k} \cdot \vec{r}}$$

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