Fourier transform and vector function

[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.

 

[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.

 

[LocationX]

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

 

[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).

 

[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.

 

[granpa]

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

 

[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?

 

[gabbagabbahey]

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


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

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.

 

[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.

 

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

And start with a vector identity,
$$u \vec{\nabla} v = \vec{\nabla} (uv) - v \vec{\nabla} u$$
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?

 

[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 I've 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 $$f (\vec{\nabla} \times \vec{A} ) = \vec{\nabla} \times (f \vec{A} ) - (\vec{\nabla}f) \times \vec{A}$$

 

[gabbagabbahey]

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... use integration by parts and you should get the term goes to zero?

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

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 $$\vec{f_{\perp}} (\vec{k})$$.

 

[LocationX]

yup, i changed my post as you were quoting me since i realized the integral doesn't go to zero.

 

[gabbagabbahey]

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

 

[LocationX]

this isn't out of any text, just a regular homework assignment for first semester EM course

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

I start with integration by parts:

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

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

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?

 

[gabbagabbahey]

Did I make a mistake somewhere?

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.

 

[LocationX]

is this the relation:

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

 

[gabbagabbahey]

is this the relation:

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

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.

 

[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?

 

[gabbagabbahey]

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

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

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

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


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

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

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:

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

and so your integral becomes

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

 

[LocationX]

I think I can change the volume integral into a surface integral and show it goes to 0 since r->infinity. It's should be similar to your previous post.

However, I'm still not sure where to proceed with the given integral. I don't think integration by parts would work since we will be evaluating a negative log term.

 

[gabbagabbahey]

I think I can change the volume integral into a surface integral and show it goes to 0 since r->infinity. It's should be similar to your previous post.

However, I'm still not sure where to proceed with the given integral. I don't think integration by parts would work since we will be evaluating a negative log term.

Hmmm...well, the integral is just the Fourier transform of the function $$1/|\vec{r}-\vec{r'}|$$ does that help you?

 

[LocationX]

oh wow, got it!

I'm wondering why there is an r' in the exponential from the result that was posted previously.

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

If this is the case, then shouldn't we get:

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

f(k') instead of f(k)

 

[gabbagabbahey]

oh wow, got it!

I'm wondering why there is an r' in the exponential from the result that was posted previously.

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

If this is the case, then shouldn't we get:

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

f(k') instead of f(k)

The $$e^{-i\vec{k} \cdot \vec{r'}}$$ comes from the "spatial shift" rule for Fourier transforms:

$$\mathcal{F}[g(\vec{r}-\vec{a})]=e^{-i\vec{k} \cdot \vec{a}} \mathcal{F} [g(\vec{r})]$$

So the Fourier transform of

$$g(\vec{r}-\vec{r'}) \equiv \frac{1}{|\vec{r}-\vec{r'}|}$$

is given by

$$\mathcal{F}[g(\vec{r}-\vec{r'})] = e^{-i\vec{k} \cdot \vec{r'}} \mathcal{F} [g(\vec{r})] = e^{-i\vec{k} \cdot \vec{r'}} \mathcal{F} \left[ \frac{1}{r} \right] =e^{-i\vec{k} \cdot \vec{r'}} \int_{- \infty}^{\infty} \frac{e^{i\vec{k} \cdot \vec{r}}}{|\vec{r}|} d^3 r= e^{-i\vec{k} \cdot \vec{r'}} \int_{0}^{\infty} \int_{0}^{ \pi} \int_{0}^{2 \pi} \frac{e^{i\vec{k} \cdot \vec{r}}}{r} r^2 sin(\theta) dr d \theta d \phi = \frac{4 \pi e^{-i\vec{k} \cdot \vec{r'}}}{k^2}$$

As for your second question, it is still $$f(\vec{k})$$; the $$r'$$ is essentially a dummy variable since it is integrated over; the real variable is $$\vec{k}$$ and it is not primed in the integral, so there is no reason to prime it after the integration.

You should be ready to tackle $$\vec{f_{\perp}} (\vec{k})$$ now. If you want to post each step, I'll go through it with you step by step.

 

[LocationX]

thank you very much for the guidance.

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

$$v( \vec{r} ) \equiv \frac{1}{4 \pi} \int_{- \infty}^{\infty} \frac{\vec{\nabla'} \times \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r'$$

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

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

not sure if this is allowed, but here is what I've done:

$$\int_{- \infty}^{\infty} \nabla \times \left( e^{-i \vec{k} \cdot \vec{r}} \frac{1}{4 \pi} \int d ^3r' \frac{\nabla ' \times \vec{f}(\vec{r'}}{|r - r'|} \right) d^3r = \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} \vec{\nabla} \times \left( e^{-i \vec{k} \cdot \vec{r}} \frac{1}{4 \pi} \int d ^3r' \frac{\nabla ' \times \vec{f}(\vec{r'})}{|r - r'|} \right)d^2r \right) dr$$

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

using Green's Theorem, the surface integral becomes a contour integral around the surface. Since the contour -> inf, u(r)->0

now we are left with:

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

We can continue and use the identity:

$$\vec{A} \times ( \vec{B} \times \vec{C} ) = \vec{B} ( \vec{A} \cdot \vec{C} ) - \vec{C}(\vec{A} \cdot \vec{B})$$

But I'm not completely confident if what I am doing is valid. Comments are appreciated, thanks!

 

[gabbagabbahey]

thank you very much for the guidance.

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

$$v( \vec{r} ) \equiv \frac{1}{4 \pi} \int_{- \infty}^{\infty} \frac{\vec{\nabla'} \times \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r'$$

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

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

(1) Your very welcome

(2) There should be an arrow over v, since it is a vector quantity.

not sure if this is allowed, but here is what I've done:

$$\int_{- \infty}^{\infty} \nabla \times \left( e^{-i \vec{k} \cdot \vec{r}} \frac{1}{4 \pi} \int d ^3r' \frac{\nabla ' \times \vec{f}(\vec{r'}}{|r - r'|} \right) d^3r = \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} \vec{\nabla} \times \left( e^{-i \vec{k} \cdot \vec{r}} \frac{1}{4 \pi} \int d ^3r' \frac{\nabla ' \times \vec{f}(\vec{r'})}{|r - r'|} \right)d^2r \right) dr$$

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

using Green's Theorem, the surface integral becomes a contour integral around the surface. Since the contour -> inf, u(r)->0

But I'm not completely confident if what I am doing is valid. Comments are appreciated, thanks!

(3)You can't transform a volume integral into a surface integral in this manner.

(4) You've actually tried to use Stoke's theorem (curl theorem) here, not Green's theorem. But either way you can't do this because of (3).

(5)There is a corollary to the divergence theorem which you can use to turn a volume integral into a surface integral:

$$\int_{\mathcal{V}} (\vec{\nabla} \times \vec{A})d^3 r = -\oint_{\mathcal{S}} \vec{A} \times d\vec{a}$$

Where $$\mathcal{S}$$ is the surface bounding the volume $$\mathcal{V}$$. This is the relation you should use to show that the first term vanishes.

(6) For the second term, there is no need to use a vector triple product rule. $$\vec{k}$$ is a constant over r and r' so just pull it out of the integral:

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

And you should be able to use the same product rule and corollary to the divergence theorem to simplify the bracketed integral.

 

[LocationX]

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

$$\int_{- \infty}^{\infty} \frac{\vec{\nabla'} \times \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' = \int_{- \infty}^{\infty} \vec{\nabla} \frac{\vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3r' + \int_{- \infty}^{\infty} \vec{\nabla} \frac{1}{|\vec{r} - \vec{r'} | } \times \vec{f}(\vec{r')) d^3 r$$

$$= -\oint_{\mathcal{S}} \frac{\vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \times d\vec{a} - \int_{- \infty}^{\infty} \frac{\widehat{(\vec{r}-\vec{r'} )}}{|\vec{r}- \vec{r'} |^2 } \times \vec{f}(\vec{r'}) d^3r$$

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

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

Is this correct so far?

 

[gabbagabbahey]

You need to keep better track of your (')'s, you should get:

$$\int_{- \infty}^{\infty} \frac{ \vec{ \nabla '} \times \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' = \int_{- \infty}^{\infty} \vec{ \nabla '} \frac{ \vec{f} (\vec{r'} ) }{|\vec{r}-\vec{r'}|} d^3 r' + \vec{ \nabla '} \frac{1}{|\vec{r} - \vec{r'}|} \times \vec{f}(\vec{r'}) d^3 r'$$

$$= -\oint_{\mathcal{S}} \frac{\vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \times d\vec{a'} + \int_{- \infty}^{\infty} \frac{\widehat{(\vec{r}-\vec{r'} )}}{|\vec{r}- \vec{r'} |^2 } \times \vec{f}(\vec{r'}) d^3r'$$

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

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

Do you know where to go from here?

 

[LocationX]

Shouldn't the relation be: $$\vec{\nabla} \left( \frac{1}{|\vec{r}- \vec{r'} | } \right) = - \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 }$$

So that:

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

I am not sure where to go from here, i was thinking about using the relation that the Fourier transform of 1/|r| = 4pi/k^2 but this does not help too much here

 

[gabbagabbahey]

Shouldn't the relation be: $$\vec{\nabla} \left( \frac{1}{|\vec{r}- \vec{r'} | } \right) = - \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 }$$

Yes, but(!)

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

You need to be careful with your primes.

From here, rearrange the integral:

$$\frac{i\vec{k}}{4 \pi} \times \int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \left( \int_{- \infty}^{\infty} \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } \times \vec{f}(\vec{r'}) d^3r' \right) d^3r = \frac{i\vec{k}}{4 \pi} \times \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \frac{\widehat{(\vec{r}-\vec{r} )}}{|\vec{r}- \vec{r'} |^2 } d^3r \right) \times \vec{f}(\vec{r'}) d^3r'$$
$$= \frac{i\vec{k}}{4 \pi} \times \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} e^{i \vec{k} \cdot \vec{r}} \vec{\nabla} \left( \frac{-1}{|\vec{r}- \vec{r'} | } \right) d^3r \right) \times \vec{f}(\vec{r'}) d^3r'$$

Is there a product rule you can use for the bracketed integral?

 

[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.

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

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...

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

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

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