This week I’m participating in the Python in Astronomy workshop at the Lorentz Center in Leiden. It is the greatest workshop/conference/unconference I’ve been to, and it’s only day 3 out of 5. I’ve been developing my git and github skills, and submitted my first pull request (i.e., open-source code contribution) yesterday to SunPy! For this afternoon some of us are planning to sit down and try to hammer out the basics of an astroquery HEASARC module for downloading data from NASA’s HEASARC database!
A note on the twitter hashtag feed: I know that abbreviations can be frustrating for those not at the conference or new to the topics. This being said, twitter inherently does not lend itself to verbosity. If you’re trying to follow along and are getting lost in tweets densely packed with abbreviations and acronyms, please reply and request clarification!
Also, you’ll probably find that some people are tweeting with the hashtag more for non-attendees, and some are aiming their tweets at fellow attendees (I admit I tend to be in the latter camp). You may find it less frustrating to follow a few specific people and not everything from the hashtag.
X-ray binaries are so small that we can’t directly image (i.e., spatially resolve) them, due to a combination of being small in size AND very far away.
When updating my Research page I was curious what a good analogy would be for imagining the projected size of an X-ray binary, and I’ve come up with the following. It’s just using algebra and trigonometry, so you can follow along! 😉
Let’s start with some assumptions for this little exercise: we’ll use a 10-solar-mass stellar black hole (i.e., not supermassive) that’s 2.5 kiloparsecs (~ 8000 lightyears) away. A 10-solar-mass black hole is ~2×10^{34} grams (yes, astronomers tend to use grams — cgs, as you’ll see below, means “centimeters, grams, seconds”, referring to the base units).
The radius of the black hole’s event horizon will be
R_{EH} = (2 G M) / c^2 ,
where G is the gravitational constant (6.674×10^{-8} in cgs units), M is the mass of the black hole, and c is the speed of light in a vacuum (~3×10^{10} in cgs units). Plugging these into the equation gives
R_{EH} ~ 3×10^6 cm,
or ~30 km. The distance, 2.5 kiloparsecs, is 7.7×10^{21} cm. Now we do some trigonometry to get the angular size:
A = arctan(3×10^6 cm / 7.7×10^{21} cm) = 2.2×10^{-14} degrees.
As you can see, this is a tiny, tiny number.
So let’s see how big an analogous object would be if it were on the surface of the moon. The closest distance between the surface of the earth to the surface of the moon is, approximately, 376300 km (which is the distance from center of the Earth to center of the moon, subtracted by the radius of the Earth and the radius of the moon), or 3.763×10^{10} cm.
We now want to know the size of an object that would appear to be 2.2×10^{-14} degrees in radius if it were sitting on the moon. This is
where S is the radius in centimeters. Solving this gives
S = 1.4 x 10^{-5} cm, or 0.14 micrometers in radius.
This is 1000 times smaller than the size of a single strand of human hair. Can you imagine trying to take a picture of a piece of hair that’s on the moon, let alone something 1000 times smaller? I can’t.
Let’s try something else — what about something on the surface of Mars? The smallest distance between the surface of Earth from the surface of Mars is 5.57×10^7 km, or 5.57×10^{12} cm. Using the same equation as before, but with this new distance,
arctan(S / 5.57×10^{12} cm) = 2.2 x 10^{-14} degrees,
gives S = 0.0021 cm = 0.021 mm = 21 micrometers in radius.
This is the size of a human hair (~ 30 – 100 micrometers in diameter), or one quarter of the thickness of a piece of paper!
Understandably, we don’t have instrumentation capable of imaging something this small, which is why we rely on spectral and timing measurements of photons emitted from X-ray binaries instead of just taking a picture.