The Half Flux Diameter of a perfectly normal distributed star

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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.

σ=1
σ=2
σ=3

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.

Figure 2: The $x$- and $y$-axis represent the pixel position on the image plane, the $z$-axis represents the intensity $I$. The blue circle is the “outer circle” with radius $R_{out}$ up to which the pixel values are considered for the $HFD$ calculation. The green circle is the actual $HFD$ circle. The total volume under the surface is 1 (normalization factor is $\frac{1}{2\pi\sigma^2}$.

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.
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The Half Flux Diameter (HFD) of a plain image

/ Carsten / 1 Comment

Why the Half Flux Diameter (HFD) is not exactly what it suggests… and why it probably does not matter in practice.

For people dealing with astronomical images the Half Flux Diameter (HFD) definitely is a well known parameter. In a few words, the HFD measures how well a star is focused. The measure is relatively robust and also works when the star is far out of focus. In another article I cover the basics of the HFD.

Strange things happening here…

HFD and I were best friends until I started to write some unit tests for the HFD calculation routine in the focus finder project. Basically, the idea was to test the calculation routine with some test images for which the expected outcome was well known. This should have proven that the implemented routine was working correctly. One of those test images was a plain image in which all pixel values had the same value (> 0). Given the definition of the HFD:

“The HFD is defined as the diameter of a circle that is centered on the unfocused star image in which half of the total star flux is inside the circle and half is outside.”

it should be easy to predict the expected outcome. Since the flux is equally distributed across the entire image, it seems obvious that – as the name suggests – the resulting Half Flux Diameter is the diameter for which a corresponding “inner circle” contains exactly 50% of the area. The other 50% of the area should be located in between the “inner” – and the “outer” circle. This is what I assumed in my previous article about the HFD. However, it turned out to be wrong and I unfortunately have to invalidate this part.

Instead, the HFD routine gave a different result. Even for big images (to reduce the error of discretization) the result was off by a mysterious factor of 1,060660172 to the expected value. This gave reason to have a closer look to this enigma…

Quick Summary

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

  • The Half Flux Diameter (HFD) of a plain image is $HFD_{plain-img} = \frac{4}{3} \cdot R_{out}$.
  • The corresponding HFD circle does not exactly split the area of the “outer circle” 1:1 even if the definition of the HFD suggests this – in other words $HFD_{plain-img}$ is not exactly $\sqrt{2} \cdot R_{out}$.
  • The root cause lies in the definition of the HFD itself: Pixels far away from the image center are weighted stronger than those which are very close.

For a few more details and reasoning read on…

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Astrophotography – Some objects for beginners

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stargazing_la_palma_setup

Beginning with astrophotography can be a bit frustrating. You bought all the equipment and now you want to see some results. Now you are out there in the field. It is already getting dark and cold and you still have to get all the cables right… You still have to do the mount alignment and then you face some IT problems… When you fixed all that and you focused the camera, it is probably already dark and it is time to move the telescope to the object of desire.

You catch the first frame and… nothing but a couple of dots. You increase the exposure time but all you get are a few more and brighter dots. Your fingers are getting already stiff and you begin to ask yourself what you are actually doing out here. Your wife is at home on the warm and cosy sofa drinking a delicious cup of tea…. welcome to the world of astrophotography!

In this little article I want to present a few sky objects which are relatively easy to locate because they are so bright. This will usually help to locate the object and center it in the telescope. Even with small exposure times you already get astonishing results.

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