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Just a curious question, do astrophysicist use the SI units, for example in this equation,

enter image description here

$r = 5pc$, will this be converted to meters? And what does this $\nu$ stand for?

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    $\begingroup$ $\nu$ is proabably the frequency of light. $\endgroup$
    – jkej
    Jan 15, 2013 at 18:18
  • $\begingroup$ i thought of that too , and i asked a lot of my friends and all of them said "probably". So i need a sure answer :) $\endgroup$ Jan 15, 2013 at 18:24
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    $\begingroup$ Ok, it's definitely frequency. Happy? :) $\endgroup$
    – jkej
    Jan 15, 2013 at 18:27

2 Answers 2

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Astronomy does not use SI (mks) units1. CGS (Centimeter-Gram-Second)-Gaussian2 units are the standard in astro fields. I've yet to hear a good explanation for why cgs is used, but gaussian units can definitely be convenient.

In your above equation, $\nu$ (pronounced: 'nu') is the standard symbol for frequency, and has units of inverse time (1/s). The standard astronomical procedure would be to use distance ($r$) in cm, planck's constant ($h$) in erg-seconds, and the luminosity in erg/s.


1: Some experimentalists, who are often associated more closely with 'physics' than astronomy or astrophysics, sometimes do use SI (MKS).

2: Gaussian is a sub-unit system of CGS involved in electricity-and-magnetism. In this system, you can write the electromagnetic force simply as $F = \frac{q_1 q_2}{r^2}$, without the usual numerical prefactor that you see in SI $\left( \frac{1}{4\pi \epsilon_0} \right)$.

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  • $\begingroup$ so basically the units i use depend on the units of the plank constant, is that what you're saying? $\endgroup$ Jan 16, 2013 at 5:33
  • $\begingroup$ In the end, the only thing that matters is unit consistency --- so yes, as long as you use consistent units, you will get a consistent result. You're question also asked what the astronomical standards were, so I tried to answer that as-well. $\endgroup$ Jan 16, 2013 at 17:26
  • $\begingroup$ Your emphasis of the word "not" above is most definitely your emphasis. There is no reason at all why you should not use SI units in Astronomy. The 20th General Assembly of the IAU said "The international system (SI) of units, prefixes, and symbols should be used for all physical quantities except for certain special units". iau.org/publications/proceedings_rules/units $\endgroup$
    – ProfRob
    Aug 14, 2014 at 10:43
  • $\begingroup$ Further, the AAS, which publishes ApJ and AJ specifies in their standards for machine-readable tables that "[to] minimize potential confusion, the AAS machine readable (MR) tables use standard SI units in most cases. " - aas.org/authors/… $\endgroup$
    – ProfRob
    Feb 12, 2016 at 16:31
  • $\begingroup$ @RobJeffries I dont know why you're so attached to this. I'm not saying anything about what should happen, or reasons why things are done this way. I'm saying that empirically, and factually, astronomy tends to use cgs units and solar/convenience (solar-masses, parsecs, etc). On any given day, if one chooses 10 random papers on astro-ph at least 8 of them will use cgs units over mks units. That's all I'm saying. I agree that we should all just use SI... and for damn sure shouldn't use magnitudes. $\endgroup$ Feb 12, 2016 at 17:25
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Yeah, convert it to meters. Other units are used for quick-and-dirty conceptual stuff, but when it comes to the math, you need to keep your units consistent.

$\nu$ is frequency. Hence the exchange:

"Hey, what's new?" asked Alan.

"$\nu$ is frequency!" answered Betty.

Alan blinked twice, and began looking for an exit to this conversation.

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  • $\begingroup$ And is that nu 0 in integral lower limit? $\endgroup$ Jan 15, 2013 at 18:38
  • $\begingroup$ Yeah, although we're at the limit of my understanding of astro. We're definitely dealing with frequency, but what does the equation signify? I see luminous flux per unit area times whatever $a_\nu$ is, divided by the energy of each photon, and integrated by frequency, but this is the first time I've seen this particular equation. Clearly you want to look at it from 5 parsecs away (probably from earth, and it's like 15 light years away). $\endgroup$
    – Will Cross
    Jan 15, 2013 at 19:17
  • $\begingroup$ @thatnerd, this equation concerns the number-flux of photons with a frequency above $\nu_0$. $L/(4 \pi r^2)$ is the flux--as you say--and dividing by energy ($h \nu$) converts that to flux of number of photons. I'm not sure what $a_\nu$ is, but maybe an absorption or transmission coefficient (i.e. calculating the number of photons absorbed, transmitted etc.). Apparently the result is about 1 photon per few years, per square-cm. $\endgroup$ Jan 15, 2013 at 19:57
  • $\begingroup$ Oops, it's a period, not a frequency. And I have to defer to zhermes on the Gaussian units. They're common enough in physics, but I didn't question whether or not they were here, and I'm not an astro guy (I'm a gravity guy), so I don't fully know the conventions. $\endgroup$
    – Will Cross
    Jan 17, 2013 at 13:48

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