# The Half Flux Diameter of a perfectly normal distributed star

Calculating the Half Flux Diameter for a perfectly normal distributed star…

… and why the answer is not 42.

The goal of this article is to “manually” calculate the Half Flux Diameter (HFD) for a perfectly normal distributed star. Initially, I decided to add perfectly normal distributed stars with different σ values as additional unit tests to my focus finder software project. A few of those star images with σ=1, 2 and 3 are illustrated in figure 1.

Previously, I examined the Half Flux diameter (HFD) for a plain image. In that article I cover some aspects in greater detail which I am reusing in this article. If something is unclear I recommend to read this article first.

A 2D image can be represented in the 3D space where the $x$- and $y$-axis are used to express the position of the pixel and the $z$-axis to visualize the intensity (pixel value) – see figure 2 below.

Following a similar approach like for the plain image, the Half Flux Diameter (HFD) for such an image will be derived in the following sections.

## Quick summary

For those who just seek for the facts – here is a quick summary:

• The Half Flux Diameter ($HFD$) for a perfectly normal distributed star is $HFD_{norm-dist} = 2 \cdot \Gamma\left(\frac{2}{3}\right) \cdot \sqrt{2} \cdot \sigma = 2.50663 \cdot \sigma$ where $\sigma$ is the variance of the distribution.
• The result does not depend on $R_{out}$ and also not on the normalization factor of the distribution.
• The expression was derived by converting the $HFD$ formula into an integral and inserting the normal distribution function as intensity (pixel value). The integral was solved by using a relation between the normal distribution and the $\Gamma$-function.
• For the volume integration the second Pappus–Guldinus theorem is used.

## Recap – The Half Flux Diameter for a plain image

Before jumping right into the deviation of the Half Flux Diameter ($HFD$) for a perfectly normal distributed star, it makes sense to quickly recap the approach which was taken to deviate the $HFD$ for a plain image. If this summary is hard to understand, more details can be found in the article The Half Flux Diameter of a plain image.

The Half Flux Radius ($HFR$) is defined as

$$HFR = \frac{WDS}{PVS}=\frac{\sum\limits_{i=0}^{N-1} V_i \cdot d_i}{\sum\limits_{i=0}^{N-1} V_i}$$

For the plain image it was possible to pull the $V=V_i$ in front of the sum because all pixels had the same intensity:

$$WDS = V \cdot \sum\limits_{i=0}^{N-1} d_i$$

For one part of the sum it was then possible to rewrite the sum as an integral:

$$V \cdot \sum\limits_{i=k}^{l} d_i \simeq V\cdot \int_{0}^{R_{out}} r dr$$

In case of normal distributed pixel values the values are different, i.e. $V_i \neq V$. Instead, of pulling the $V$ in front of the sum, now the normal distribution $g(r)$ needs to be part of the integral:

$$\sum\limits_{i=k}^{l} V_i \cdot d_i \simeq \int_{0}^{R_{out}} g(r) \cdot r dr$$

In the following section the equation for the normal distribution $g(r)$ will be derived.

## Deriving the normal distribution function $g(r)$

Since the normal distribution can be explicitly expressed as a formula, the image of a normal distributed star will be expressed by a function $z=f(x,y)$ in the following sections. The common bivariate normal distribution (see wikipedia) is a good starting point:

$$f(x,y)=\frac{1}{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}}\exp{\left(-\frac{1}{2(1-\rho^2)^2}\left[\left(\frac{x-\mu_x}{\sigma_x}\right)^2-2\rho\left(\frac{x-\mu_x}{\sigma_x}\right)\left(\frac{y-\mu_y}{\sigma_y}\right)+\left(\frac{y-\mu_y}{\sigma_y}\right)^2\right]\right)}$$

For the purpose of a perfectly normal distributed star this expression can be simplified a lot. First of all, the correlation between the variables $x$ and $y$ can be assumed as $0$:

$$\rho=0$$

Furthermore, the variance for $x$ and $y$ is expected to be the same (an ideal star is perfectly circular):

$$\sigma_x^2 = \sigma_y^2 = \sigma^2$$

Additionally, the mean for $x$ and $y$ is set to $0$ since the center of the star cen be assumed to be at $(0,0)$. This simplifies the equation.

$$\mu_x=\mu_y=0$$

This results in a simplified version of $f(x,y)$:

$$f(x,y)=\frac{1}{2\pi\sigma^2}\exp{\left(-\frac{1}{2}\cdot\frac{x^2+y^2}{\sigma^2}\right)}$$

Figure 3 shows $f(x,y)$ as a 3D plot (red). The blue line illustrates the function $g(x) = f(x, 0)$ which is a vertical cut through the 3D plot. This function $g(x)$ is very similar to the 1D normal distribution function – the only difference is the normalization factor which in this case is $\frac{1}{2\pi\sigma^2}$ instead of $\frac{1}{\sqrt{2\pi}\sigma}$.

For the next step only the “right” part of $g(x)$ is relevant – i.e. $0 \leq x \leq R_{out}$. Since $g$ will later be rotated around the $z$-axis, $x$ will be from now on replaced by $r$. Setting $y=0$ in $f(x,y)$ directly results in $g(r)$:

$$g(r)=\frac{1}{2\pi\sigma^2}\exp{\left(-\frac{1}{2}\left[\frac{r}{\sigma}\right]^2\right)}, r \geq 0$$

Figure 4 illustrates $g(r)$:

## Calculating the Half Flux Diameter for the normal distribution $g(r)$

$g(r)$ expresses the intensity (pixel values) depending on the distance $r$ from the star center. Therefore, $g(r)$ will replace $V_i$ in the original $HFR$ equation. As a recap $HFR$ is again written down here:

$$HFR = \frac{WDS}{PVS} = \frac{\sum\limits_{i=0}^{N-1} V_i \cdot d_i}{\sum\limits_{i=0}^{N-1} V_i}$$

The calculation will take place in two steps:

1. The enumerator – the weighted distance sum ($WDS$)
2. The denominator – the pixel value sum ($PVS$)

### Part 1 – Calculating the Weighted Distance Sum ($WDS$)

In this section the upper part of the $HFR$ equation – the weighted distance sum $WDS$ will be solved for a perfectly normal distributed star. As previously demonstrated for the plain image, the value of $WDS$ corresponds to volume under the surface which is formed when the function $h(r) = r \cdot g(r)$ is rotated around the $z$-axis.

The function $h(r)$ is given as:

\begin{align} h(r) &= r \cdot g(r) \\ &= r \cdot \frac{1}{2\pi\sigma^2} \exp{\left(-\frac{1}{2}\left[\frac{r}{\sigma}\right]^2\right)} \\ \end{align}

A graphical representation of $h(r)$ is shown in figure 5.

In the next step the area $A_{h(r)}$ below the curve $h(r)$ is determined. One part of the sum of $WDS$ can be expressed as

\begin{align} A_{h(r)}&=\sum\limits_{i=k}^{l} V_i \cdot d_i \\ &\simeq \int_{0}^{R_{out}} h(r) dr \\ &= \int_{0}^{R_{out}} r \cdot g(r) dr \\ &= \int_{0}^{R_{out}} r \cdot \frac{1}{2\pi\sigma^2}\exp{\left(-\frac{1}{2}\left[\frac{r}{\sigma}\right]^2\right)} dr \\ \end{align}

At this point I was stuck for a moment since I had no idea how to solve this integral analytically. However, luckily there is a way to rewrite this integral using the $\Gamma$-function:

$$\int_{0}^{\infty}x^n\cdot e^{-ax^2}dx =\frac{\Gamma\left(\frac{n+1}{2}\right)}{2a^{\frac{n+1}{2}}}$$

for $a>0, n>-1$.

The next step is to transform the integral so that it is easier to match the corresponding parameters $n$ and $a$. Furthermore, the upper limit of the integral $R_{out}$ can be replaced by $\infty$ since $h(r)$ converges against $0$.

\begin{align} A_{h(r)} &= \frac{1}{2\pi\sigma^2}\cdot \int_{0}^{R_{out}} r \cdot \exp{\left(-\frac{1}{2}\left[\frac{r}{\sigma}\right]^2\right)} dr \\ &=\frac{1}{2\pi\sigma^2}\cdot \int_{0}^{\infty} r^1 \cdot \exp{\left(-\frac{1}{2\sigma^2}\cdot r^2\right)} dr \end{align}

Comparing the integral against the left part of the expression above, gives

$$n=1$$

$$a=\frac{1}{2\sigma^2}$$

Putting in those values back into the right part of the expression results in

\begin{align} A_{h(r)} &=\frac{1}{2\pi\sigma^2}\cdot \int_{0}^{\infty} r^1 \cdot \exp{\left(-\frac{1}{2\sigma^2}\cdot r^2\right)} dr \\ &= \frac{1}{2\pi\sigma^2}\cdot\frac{\Gamma(1)}{2a^1} \\ &= \frac{1}{2\pi\sigma^2}\cdot\frac{\Gamma(1)}{2\cdot\frac{1}{2\sigma^2}} \\ &= \frac{\Gamma(1)}{2\pi}\\ &=\frac{1}{2\pi} \end{align}

That means, the area below the function $h(r)$ is equal to $\frac{1}{2\pi}$.

The next step is to calculate the volume of $h(r)$ when it is rotated around the $z$-axis (see figure 6).

One way would be to compose an infinitesimal small volume element consisting of $A_{h(r)}$ and a $d\phi$ and then integrated over $d\phi$ for $0$ to $2\pi$. However, there is another way to calculate the volume of the body – the second Pappus–Guldinus theorem:

The second theorem states that the volume V of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area A of F and the distance d traveled by the geometric centroid of F.

Wikipedia / Pappus of Alexandria and Paul Guldin

So what exactly does that mean in this context? In this case it boils down to three steps:

1. Calculation of the geometric centroid of $A_{h(r)}$
2. Calculation of the distance $C$ the geometric centroid “travels” while rotating around the $z$-axis
3. Calculation of the volume $V$ of the body according to the described rule $V=A_{h(r)} \cdot C$

In this case, since it is a rotation around the $z$-axis, the “traveled” distance of the geometric centroid is just the circumference $C$ of a circle with the radius of the $x$ component of the geometric centroid $x_c$:

$$C=2\pi x_c$$

With this the volume $V$ of the body (which corresponds to the $WDS$ term) can be calculated as

\begin{align} WDS&=V\\ &=A_{h(r)}\cdot C \\ &=\frac{1}{2\pi}\cdot2\pi\cdot x_c \\ &=x_c \end{align}

So what remains is the calculation of $x$-component $x_c$ of the geometric centroid of $A_{h(r)}$.

After some research I realized that the calculation of the geometric centroid of a curvilinear trapezoid is what is needed in this case (see figure 7 below).

According to Wikipedia, the geometric centroid of an area which is defined by the boundaries $A$ and $B$, a curvilinear trapezoid (in this case $h(r)$ with $A=0$ and $B=\infty$) and the $x$-axis, can be calculated as

$$x_c=\frac{\int_{A}^{B}xydx}{S}$$
$$y_c=\frac{\frac{1}{2}\int_{A}^{B}y^2dx}{S}$$

where $S$ is the area of the trapezoid. In this case $S$ is already known:

$$S = A_{h(r)} = \frac{1}{2\pi}$$

As already mentioned above, in this context only $x_c$ is needed. Inserting $h(r)$ into the integral results on the following expression for $x_c$:

\begin{align} x_c &= \frac{\int_{0}^{\infty}r\cdot h(r) dr}{\frac{1}{2\pi}} \\ &=2\pi\cdot \frac{1}{2\pi\sigma^2} \cdot \int_{0}^{\infty}r^2 \cdot \exp{\left(-\frac{1}{2\sigma^2}\cdot r^2\right)} dr \\ \end{align}

Again the relation between the normal distribution and $\Gamma$-function comes to the rescue:

$$\int_{0}^{\infty}x^n\cdot e^{-ax^2}dx =\frac{\Gamma\left(\frac{n+1}{2}\right)}{2a^{\frac{n+1}{2}}}$$

for $a>0, n>-1$.

Again, comparing the integral against the left part of the expression below gives

$$n=2$$

$$a=\frac{1}{2\sigma^2}$$

This way the integral can be transformed into

\begin{align} WDS &= x_c \\ &= \frac{1}{\sigma^2} \cdot \frac{\Gamma\left(\frac{3}{2}\right)}{2\cdot \left(\frac{1}{2\sigma^2}\right)^{\frac{3}{2}}} \\ &= \Gamma\left(\frac{3}{2}\right) \cdot \sqrt{2} \cdot \sigma \\ &= 0.88623 \cdot \sqrt{2} \cdot \sigma \\ & = 1.2533 \cdot \sigma \end{align}

That means $WDS$ – the upper part of the $HFR$ equation has been evaluated to

$$WDS=1.2533 \cdot \sigma$$

That means the upper part of the $HFR$ equation only depends on $\sigma$.

### Part 2 – Calculating the Pixel Value Sum ($PVS$)

Luckily this section will be short. Since the 2D normal distribution function $f(x,y)$

$$f(x,y)=\frac{1}{2\pi\sigma^2}\exp{\left(-\frac{1}{2}\cdot\frac{x^2+y^2}{\sigma^2}\right)}$$

is weighted with the correct “normalization” factor $\frac{1}{2\pi\sigma^2}$, the volume covered by the surface is exactly $1$:

$$PVS=1$$

## Putting everything back together

In a last step both expressions $WDS$ and $PVS$ will be put back together into the $HFR$ equation:

\begin{align} HFR_{norm-dist} &= \frac{WDS}{PVS} \\ &= \frac{\Gamma\left(\frac{3}{2}\right) \cdot \sqrt{2} \cdot \sigma}{1} \\ &= \Gamma\left(\frac{3}{2}\right) \cdot \sqrt{2} \cdot \sigma \\ \end{align}

With this result the final Half Flux Diameter for a perfectly normal distributed star can be derived as

\boxed{\begin{align} HFD_{norm-dist} &= 2 \cdot HFR_{norm-dist} \\ &= 2 \cdot \Gamma\left(\frac{2}{3}\right) \cdot \sqrt{2} \cdot \sigma \\ &= 2.50663 \cdot \sigma \end{align}}

## Conclusion

In this article the Half Flux Diameter ($HFD$) for a perfectly normal distributed star was derived as $$HFD_{norm-dist} = 2 \cdot \Gamma\left(\frac{2}{3}\right) \cdot \sqrt{2} \cdot \sigma = 2.50663 \cdot \sigma$$ where $\sigma$ is the variance of the distribution.

As one can see $HFD_{norm-dist}$ does not depend on $R_{out}$ and also not on the normalization factor of the normal distribution. This actually meets the expectation. Furthermore, it is interesting to note that $HFD_{norm-dist}$ does not match the full width at half maximum (FWHM) of a normal distributed star:

$$FWHM=2\sqrt{2\ln{2}}\sigma \approx 2.3584 \cdot \sigma \neq 2.50663 \cdot \sigma$$

With this result it is now easy to compose another unit test for the focus finder software project.

Clear skies!

Last updated: October 13, 2022 at 11:12 am