Means of Measuring Stars

We are bounded in a nutshell of Infinite space: Blog Post #4, Worksheet # 2.1, Problem #3: Measuring Stars

3. You observe a star you measure its flux to be $F_\star$. If the luminosity of the star is $L_\star$,

(a) Give an expression for how far away the star is.

(b) What is its parallax?

(c) If the peak wavelength of its emission is at $\lambda_0$, what is the star’s temperature?

(d) What is the star’s radius, $R_\star$?

(a) As might be evident, this problem, in general, assumes you have completed a few previous exercises that explain in further detail the nature of these questions, so I will be recreating the necessary portions to fill in some of the data.

Now then, the apparent distance of a star that is being observed is directly tied to some of the most important astronomical measurement, Flux $F_\star$ and Luminosity $L_\star$, which in turn are directly correlated with one another. Furthermore, some more information on $F_\star$ and $L_\star$, is that, respectively, they have the units $\frac{erg}{cm^2 \cdot s}$ and $\frac{erg}{s}$ (in standard $c \cdot g \cdot s$ notation). If you look closely at these units, you can see the only difference is that $F_\star$ also contains a division by Area, the total area that the sphere of emitted photons have covered over a certain distance $\delta$. $L_\star$’s units also indicate how it is simply energy over a set time, and if you remember physics class, that’s the same as Power, but only in luminous terms.

So if Flux is only Luminosity divided by an area, then the equation is truly that simple, producing: $$F_\star = \frac{L_\star}{4 \pi \delta^2},$$ and in particular, $4 \pi \delta^2$ representing the area of an expanding sphere with radius $\delta$.

Next we take the equation we just had, $$F_\star = \frac{L_\star}{4 \pi \delta^2},$$ and solve it for distance: $$4 \pi \delta^2= \frac{L_\star}{ F_\star },$$ $$\delta^2= \frac{L_\star}{ 4 \pi \cdot F_\star },$$ $$\delta= \left(\frac{L_\star}{ 4 \pi \cdot F_\star}\right)^{\frac{1}{2}}.$$

And there you have an expression for the distance of the star from the observer, knowing the $F_\star$ and $L_\star$.

(b) A parallax is the difference in point of view of the observer which causes the background behind an object focused on to change. This slight change allows astronomers to measure distances of stars whose parallax is noticeable enough that a distance may be calculated from the triangle made by the Earth’s two points of view (vertexes) at two points in its orbit around the Sun, and the third vertex being the object observed. Parallaxes are measured in arc seconds, the unit below arc minutes and degrees, and when an observed object has a parallax of 1 arc second, the established distance is 1 parsec.

From this interpretation, a pattern starts to emerge, for every one arc second of parallax, there is one parsec of distance to the star, and for every 2 arc seconds of parallax (the change was much more noticeable), there is $\frac{1}{2}$ parsecs. This indicates how parallax and distance are inversely related, described by: $$p[''_{(arcsecond)}]=\frac{1}{\delta} [pc_{(parsec)}]$$ which is analogous to the same formula in terms of radians and Astronomical Units (1AU = the distance from the Earth to the Sun): $$\theta [rad] = \frac{1}{\delta} AU$$ where $1 pc = 2 \times 10^5 AU$ since $1    [radian] =2\times10^5 ['']$ .

(c) For this portion of the problem, the question refers to the derivative of the equation to calculate the radiation emitted from a blackbody, a classification for objects that are perfect emitter of thermal radiation. The original equation is:

$$F_{\lambda} (T)= \pi \frac{2 \left(\frac{c}{\lambda}\right)^2}{c^2} \frac{h \frac{c}{\lambda}}{e^{\frac{h\frac{c}{\lambda}}{kT}}} - 1 ,$$

and after fully differentiating it, the equation becomes:

$$5 = {\frac{h c}{\lambda kT}} \cdot {\frac {e^\frac{h\frac{c}{\lambda}}{kT}}{\lambda e^{\frac{h\frac{c}{\lambda}}{kT}}-1}}$$

which reduces to $$5 = {\frac{h c}{\lambda kT}}$$  since the rest of the equation becomes virtually 1 with almost any value plugged in.

Solving for $\lambda$, and plugging in all the constants: $$c_{(Speed of Light)} = 2.98 \times 10^{10} \frac{cm}{s} ,$$ $$k_{(Boltzmann’s Constant)} = 1.4 \times 10^{-16} \frac{erg}{K} ,$$ and $$h_{(Plank’s Constant)} = 6.6 \times 10^{-27} erg \cdot s ,$$ we get: $$\lambda = \frac {0.2857 cm \cdot K}{T}$$ where T is the temperature at the desired wavelength.

And if you were to look for the Maximum temperature that a blackbody would produce as according to the wavelength, then simply plug in the maximum wavelength into: $$T=\frac {0.2857 cm \cdot K}{\lambda_{Max}} .$$

(d) As to the star’s radius, $R_\star$, there have been several equations and constants that can be combined by analysis of their units to give the desired length measurement. These values are:

$$L_\star = \frac{erg}{s} ,$$ $$\sigma_{( Stephen-Boltzmann Constant)} = \frac{erg}{ s cm^2 K^4} ,$$ $$T^4 = K^4 ,$$ (the multiplication of  $\sigma$ and $T^4$ is the result  of the integration of the equation to calculate the radiation emitted from a blackbody), and: $$4 \pi R_\star^2 = cm^2 ,$$ which describes the cross section of the star. So if all these numbers are put together and then solved for $R_\star$, then we have:

$$L_\star = 4 \pi R_\star^2 \sigma T_\star^4$$

$$R_\star = \left( \frac{ L_\star}{ 4 \pi \sigma T_\star^4} \right)^{\frac{1}{2}}.$$