0619-Gaussian profile

Gaussian Function and Its Integral

A two-dimensional Gaussian function, which is often used to model the intensity distribution of a beam of light or a point source in astronomical data, is defined as:

$$S(x, y) = A \exp\left(-\frac{(x - x_0)^2 + (y - y_0)^2}{2\sigma^2}\right)$$

Here, $A$ is the amplitude, $(x_0, y_0)$ are the coordinates of the center, and $\sigma$ is the standard deviation of the Gaussian distribution, which dictates the width of the bell curve.

Integrating Over All Space

To find the total flux $F$ of this Gaussian, we integrate it over all $x$ and $y$. The integral of a Gaussian function over its entire range is a well-known result in mathematics and is used extensively in physics and engineering:

$$F = \int_{-\infty}^\infty \int_{-\infty}^\infty S(x, y) \, dx \, dy$$

Substituting the Gaussian function:

$$F = \int_{-\infty}^\infty \int_{-\infty}^\infty A \exp\left(-\frac{(x - x_0)^2 + (y - y_0)^2}{2\sigma^2}\right) \, dx \, dy$$

This can be simplified by shifting the coordinates to center the Gaussian at the origin for the purpose of integration (i.e., let $u = x - x_0$ and $v = y - y_0$), which does not change the integral due to the symmetric nature of the Gaussian:

$$F = A \int_{-\infty}^\infty \int_{-\infty}^\infty \exp\left(-\frac{u^2 + v^2}{2\sigma^2}\right) \, du \, dv$$

This integral can be solved using polar coordinates, where $u = r\cos\theta$ and $v = r\sin\theta$, and $du, dv = r, dr, d\theta$:

$$F = A \int_{0}^{2\pi} \int_{0}^\infty \exp\left(-\frac{r^2}{2\sigma^2}\right) r \, dr \, d\theta$$

The $r$ integral is:

$$\int_0^\infty \exp\left(-\frac{r^2}{2\sigma^2}\right) r \, dr = \sigma^2$$

Thus, when you complete the integral over $\theta$:

$$F = A \cdot 2\pi \sigma^2$$

Summary

The integral $\int_0^\infty r e^{-\frac{r^2}{2\sigma^2}} dr = \sigma^2$ follows from a standard Gaussian integral result, and the extra $r$ factor comes from the conversion to polar coordinates, which accounts for all possible directions in two dimensions. This is why the total flux $F$ of a Gaussian beam, when integrated over all space, equals $2\pi A \sigma^2$. This formula is essential in fields like optics and astronomy for calculating the total energy or total number of photons emitted by a source, assuming the distribution of these quantities follows a Gaussian profile.