Symmetry of an Integral of a Dot product

[Skaiserollz89]

Homework Statement

Show that $$\int\int_{-\infty}^{\infty}d\vec{r}d\vec{r'}W(\vec{r})W(\vec{r'}) \vec{r} \cdot \vec{r'}=0$$ due to symmetry. Here, $$W(\vec{r})$$ is a circle function with a value of one inside its radius R and 0 beyond the radius R. Note that ##\vec{r} \cdot \vec{r'}=|\vec{r}| |\vec{r'}|cos(\phi)##, where ##\phi## is the angle between vectors ##\vec{r}## and ##\vec{r'}##.

Relevant Equations

$$\int\int_{-\infty}^{\infty}drdr'W(\vec{r})W(\vec{r'}) \vec{r} \cdot \vec{r'}=0 ;$$
$$\vec{r}=r_x\hat{i}+r_y\hat{y};$$
$$\vec{r'}=r'_x\hat{i}+r'_y\hat{y};$$

This homework statement comes from a research paper that was published in SPIE Optical Engineering. The integral $$\int\int_{-\infty}^{\infty}drdr'W(\vec{r})W(\vec{r'}) \vec{r} \cdot \vec{r'}=0$$ is an assumtion that is made via the following statement from the paper : "Since ##\int\int_{-\infty}^{\infty}drdr'W(\vec{r})W(\vec{r'}) \vec{r} \cdot \vec{r'}=0## , terms that are functions of either ##\vec{r}## or ##\vec{r'}##, and not both, can be added without changing the result of the integration. "

I am just trying to justify how the integral equals zero.

My attempt:

$$\int\int_{-\infty}^{\infty}d\vec{r}d\vec{r'}W(\vec{r})W(\vec{r'}) \vec{r} \cdot \vec{r' };$$ where ##W(\vec{r})=0## when ##r>R## and
##W(\vec{r})=1## when ##r\le R.##

Here I change my integration bounds to reflect the constrains of ##W(\vec{r})## and ##W(\vec{r'})##.

$$\int_{0}^{R}\int_{0}^{R}d\vec{r}d\vec{r'} \vec{r} \cdot \vec{r'}$$

$$\int_{0}^{R}\int_{0}^{R}d\vec{r}d\vec{r'} |\vec{r}| |\vec{r'}|cos(\phi)$$

$$cos(\phi) \int_{0}^{R}|\vec{r}|dr\int_{0}^{R} r'dr'$$

$$cos(\phi) \int_{0}^{R}r dr\int_{0}^{R}r' dr'$$

$$cos(\phi)(\frac{1}{2}R^2)(\frac{1}{2}R^2)$$

$$=\frac{R^4}{4}cos(\phi)$$

Im not sure If I am skipping steps or missing the point all together, but I dont see how this will go to zero. Any assistant and guidance would be greatly appreciated so I might be able to understand this paper I am reading a little bit more.

 

[pasmith]

In terms of plane polar coordinates [itex](r,\phi)[/itex] and [itex](r,\phi')[/itex] you have [itex]\theta = \phi - \phi'[/itex]. To integrate over a circle in the [itex](r',\phi')[/itex] you must not only integrate with respect to [itex]r'[/itex] from 0 to [itex]R[/itex], but also with respect to [itex]\phi'[/itex] between 0 and [itex]2\pi[/itex]. The integral of [itex]\cos(\phi - \phi')[/itex] over a complete period of [itex]2\pi[/itex] is zero, so the entire integral is zero.

 

[Skaiserollz89]

In terms of plane polar coordinates [itex](r,\phi)[/itex] and [itex](r,\phi')[/itex] you have [itex]\theta = \phi - \phi'[/itex]. To integrate over a circle in the [itex](r',\phi')[/itex] you must not only integrate with respect to [itex]r'[/itex] from 0 to [itex]R[/itex], but also with respect to [itex]\phi'[/itex] between 0 and [itex]2\pi[/itex]. The integral of [itex]\cos(\phi - \phi')[/itex] over a complete period of [itex]2\pi[/itex] is zero, so the entire integral is zero.

So the statement in the published paper is assuming that integration with respect to ##\phi## must also be performed for it to go to zero, without explicitly saying so?

 

[vela]

So the statement in the published paper is assuming that integration with respect to ##\phi## must also be performed for it to go to zero, without explicitly saying so?

The original integral you wrote has integration variables ##\vec r## and ##\vec r'##. That's not the same as ##r## and ##r'## as you later wrote.

 

[Skaiserollz89]

The original integral you wrote has integration variables ##\vec r## and ##\vec r'##. That's not the same as ##r## and ##r'## as you later wrote.

I agree. After performing the dot product I am working with scalars at that point, correct? So ##d\vec r## and ##d\vec r'## goes to ##dr## and ##dr'##? Do you think I am missing something?

 

[vela]

I agree. After performing the dot product I am working with scalars at that point, correct?

No, that's not correct. The dot product makes the integrand a scalar, but that has no bearing on the region over which you're integrating.

 

[Skaiserollz89]

No, that's not correct. The dot product makes the integrand a scalar, but that has no bearing on the region over which you're integrating.

Thank you for clarifying. But I think I am a little confused. Would you mind elaborating? I think you are saying that my line that reads ##cos(\phi) \int_{0}^{R}r dr\int_{0}^{R}r' dr'## is incorrect, and should be ##cos(\phi) \int_{0}^{R}r d\vec{r}\int_{0}^{R}r'd\vec{r'}## which simplifies to

$$cos(\phi) \int_{0}^{R}r \hat{r}d{r}\int_{0}^{R}r'\hat{r'}d{r'}.$$

If this notation is correct, how do I continue on from here?

 

[vela]

In two dimensions, the area element ##d\vec r## is equal to ##dx\,dy## in Cartesian coordinates or ##r\,dr\,d\phi## in polar coordinates. (You could, of course, swap the order of ##x## and ##y## or ##r## and ##\theta##.) The advantage of ##d\vec r## is that the notation is independent of any choice of coordinate system.

Here's a simple example:
$$\int (\vec r \cdot \hat i)\,d\vec r = \iint (\vec r \cdot \hat i)\,r\,dr\,d\phi= \iint (\vec r \cdot \hat i)\,dx\,dy$$ Then you can evaluate the dot product in whatever way is convenient and appropriate for the problem, e.g.,
$$\int (\vec r \cdot \hat i)\,d\vec r = \iint (r\cos\phi)\,r\,dr\,d\phi= \iint x\,dx\,dy$$

 

[Skaiserollz89]

Great! I think I've got it now. Ill attempt to work it through here...

Given the initial problem
$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}d\vec{r}d\vec{r'}W(\vec{r})W(\vec{r'}) \vec{r} \cdot \vec{r'}$$

It seems like you are referring to Green's Theorem to turn a line integral into a double integral.

From your example, and from the constraints of ##W(r)## and ##W(r')## on the radius I have...

$$\int_{0}^{R}\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{2\pi} rdrd\phi r'dr'd\phi' |\vec{r}| |\vec{r'}| cos(\phi-\phi') ,$$

where ##\phi-\phi'## is the angle between the two vectors ##\vec{r}## and ##\vec{r'}##. Separating terms I have

$$\int_{0}^{R}r^2dr \int_{0}^{R}r'^2dr'\int_{0}^{2\pi}\int_{0}^{2\pi} d\phi d\phi' cos(\phi-\phi') ,$$

Noting the trig identity that ##cos(\phi-\phi')=cos(\phi)cos(\phi')+sin(\phi)sin(\phi')##,
$$\int_{0}^{R}r^2dr \int_{0}^{R}r'^2dr'\left[\int_{0}^{2\pi}\int_{0}^{2\pi} d\phi d\phi' cos(\phi)cos(\phi')+ \int_{0}^{2\pi}\int_{0}^{2\pi} d\phi d\phi'sin(\phi)sin(\phi')\right] ,$$

$$\int_{0}^{R}r^2dr \int_{0}^{R}r'^2dr'\left[\int_{0}^{2\pi}cos(\phi)d\phi\int_{0}^{2\pi} cos(\phi')d\phi'+ \int_{0}^{2\pi}sin(\phi)d\phi\int_{0}^{2\pi} sin(\phi')d\phi'\right] ,$$

where the terms in the square brackets all individually go to zero due to symmetry in the integration over the entire period of the function, namely...
$$\int_{0}^{2\pi}cos(\phi)d\phi=0 \hspace{.5cm}\text{ and }\hspace{.5cm} \int_{0}^{2\pi}sin(\phi)d\phi=0.$$

Therefore;
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}d\vec{r}d\vec{r'}W(\vec{r})W(\vec{r'}) \vec{r} \cdot \vec{r'}=0$$

 

[pasmith]

Easier is [tex]\begin{split}
\int_0^{2\pi}\int_0^{2\pi} \cos(\phi - \phi')\,d\phi\,d\phi' &= \int_0^{2\pi}\left[ \sin(\phi-\phi') \right]_{\phi=0}^{\phi = 2\pi}\,d\phi' \\
&= \int_0^{2\pi}\sin(2\pi - \phi') - \sin (-\phi')\,d\phi' \\
&= \int_0^{2\pi}0\,d\phi' \\ &= 0.\end{split}[/tex]