- #1

Skaiserollz89

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- 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.

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.

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