Do parsecs scale as an inverse proportion with parallax angle?

Question

high school student here. In astronomy my teacher says that if the parsec angle ($\alpha$) of a star is 1/50 arcsecond, then that star is 50 parsecs away. I'm not really convinced - calculation of distances from parsecs requires using the formula $$\text{distance in AU}=\frac{1 AU}{\tan(\alpha)}.$$ This means that distance is inversely proportional to $tan{\alpha}$. Distance is not inversely proportional to $\alpha$.

I know you can use a small angle approximation, but that feels to me as very dodgy - you can get fairly large discrepancies in the calculated distance depending on whether you use $tan(\alpha)$ or $\alpha$ to find the distance.

So, what is the best practice when calculating distances in parsecs? Is my teacher's approximation acceptable in "real life" astronomy?

Answer 1

For small angles the Laurent series of $1/ \tan(x)$ about zero begins:

$$\frac{1}{\tan(x)} = \frac{1}{x} - \frac{x}{3} - ... = \frac{1}{x} \left( 1 - \frac{x^2}{3} -...\right)$$

so for small $\alpha$ the approximation $1/ \alpha$ will be high by about the fraction $\alpha^2/3$.

This means that the error in the distance determined by parallax will be only 1 part per million if $\alpha$ is 1.73 mrad or about 0.1° or 357 arcseconds.

The distance error will be 1 part per thousand with a parallax of 3.1° or 11,300 arcseconds.

The only way for such large parallax values to occur is if the object is so close that it is under the gravitational influence of the Sun and therefore this method of distance determination is no longer valid, it becomes an celestial mechanics question instead.

Your teacher is demonstrating that a feel for the problem and understanding the phenomenon are just as important as strict adherence to rote equations.

When using pencil and paper astronomers will use $1/ \alpha$ and when writing programs meant for general use and scrutinized, astronomers might then use $ 1 / \tan(\alpha)$.

Answer 2

For angles of magnitude 1/50 arcsecond (of the order $10^{-7}$, the difference between $\tan\alpha$ and $\alpha$ is truly negligible (if alpha is in radians) and only a matter of a choice of units if you don't. It is quite correct to say: distance (in AU) = 1/angle (in radians) to get the distance from the angle in radians.

Using parsecs a means that we don't need to use a different unit for the angle, and the point of parsecs is that we can use "arcseconds". So it is correct to use distance (in parsec) = 1/parallax angle (in seconds)