# Symmetry of an Integral of a Dot product

• Skaiserollz89
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.

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

pasmith said:
In terms of plane polar coordinates $(r,\phi)$ and $(r,\phi')$ you have $\theta = \phi - \phi'$. To integrate over a circle in the $(r',\phi')$ you must not only integrate with respect to $r'$ from 0 to $R$, but also with respect to $\phi'$ between 0 and $2\pi$. The integral of $\cos(\phi - \phi')$ over a complete period of $2\pi$ 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?

Skaiserollz89 said:
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.

vela said:
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?

Skaiserollz89 said:
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.

vela said:
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?

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

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

## What is the symmetry of an integral of a dot product?

The symmetry of an integral of a dot product often refers to the invariance of the integral under certain transformations, such as rotations or reflections. This can simplify the evaluation of the integral by exploiting these symmetrical properties.

## How does symmetry simplify the evaluation of an integral of a dot product?

Symmetry can reduce the complexity of the integral by limiting the domain of integration or by canceling out certain terms. For example, if the integrand is symmetric about an axis, you might only need to integrate over half the domain and then double the result.

## What are common transformations that reveal the symmetry of an integral of a dot product?

Common transformations include rotations, reflections, and translations. These transformations can show that the integral remains unchanged, indicating symmetrical properties that can be used to simplify the computation.

## Can the symmetry of an integral of a dot product be used in physical applications?

Yes, the symmetry of an integral of a dot product is often used in physics to simplify problems, particularly in electromagnetism, fluid dynamics, and quantum mechanics. Symmetrical properties can lead to simpler equations and more straightforward solutions.

## What mathematical tools are used to identify the symmetry of an integral of a dot product?

Mathematical tools such as group theory, tensor analysis, and vector calculus are often used to identify and exploit the symmetry of an integral of a dot product. These tools help in understanding how the integral behaves under various transformations.

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