Blog 18: Normal Galaxies and the Tully Fisher Relation

4) Over time, from measurements of the photometric and kinematic properties of normal galaxies, it became apparent that there exist correlations between the amount of motion of objects in the galaxy and the galaxy's luminosity. In this problem, we'll explore one of these relationships.

Spiral galaxies obey the Tully-Fisher Relation:

$$\begin{align} L \sim v_{max}^4 \end{align}$$

where $L$ is total luminosity, and $v_{max}$ is the maximum observed rotational velocity. This relation was initially discovered observationally, but it is not hard to derive in a crude way:

(a) Assume that $v_{max} \sim \sigma$ (is this a good approximation?). Given what you know about the Virial Theorem, how should $v_{max}$ relate to the mass and radius of the Galaxy?

Using the Virial Theorem, we derived in problem 3 that the mass, radius, and stellar velocity can be related with the equation:

$$\begin{align} M \approx \frac{\sigma^2 R}{G} \end{align}$$

We can rearrange this equation to solve for $\sigma^2$ as follows:

$$\begin{align} M \approx \frac{\sigma^2 R}{G}\\ \sigma^2 = \frac{M G}{R} \end{align}$$

Let's assume that $v_{max} \sim \sigma$, which is a good assumption because we can think of $v_{max}$ as the average velocity of stars in the galaxy, and that they don't vary in speed too mych. Using this assumption, we can use the equation above to show the relationship between $v_{max}$ to the mass and radius of the galaxy as follows:

$$\begin{align} v_{max}^2 \sim \frac{M}{R} \end{align}$$

(b and c) To proceed from here, you need some handy observational facts. First, all spiral galaxies have a similar disk surface brightnesses ($\langle I \rangle = \frac{L}{R^2}$) (Freeman's Law). Second, they also have similar total mass-to-light ratios $(\frac{M}{L})$.

Use some squiggly math (drop the constants and use $\sim$ instead of $=$) to find the Tully-Fisher relationship.

Based on the Freeman's Law, we know that $\langle I \rangle = \frac{L}{R^2}$, which can be re-written as:

$$\begin{align} I &= \frac{L}{R^2}\\ IR^2 &= L\\ R &\sim \sqrt{L}\\ \end{align}$$

Since we also know the mass-luminosity ratio as $\frac{M}{L}$, we can say that $M \sim L$.

We can plug these values for the $v_{max}$ derived in part (a) to get the Tully-Fisher Relation:

$$\begin{align} v_{max}^2 &\sim \frac{M}{R} &\\ v_{max}^2 &\sim \frac{M}{\sqrt{L}} \rightarrow  \text{Substituting for Freeman's Law}&\\ v_{max}^2 &\sim \frac{L}{\sqrt{L}} \rightarrow \text{Substituting for mass-ratio relation}&\\ v_{max}^2 &\sim \sqrt{L} &\\ v_{max}^4 &\sim L &\\ \end{align}$$

This is the Tully-Fisher Relation!

(d) It turns out the Tully-Fisher Relation is so well-obeyed that it can be used as a standard candle, just like the Cepheids and Supernova Ia seen previously. In the B-band ($\lambda_{cen} \sim 445$ nm, blue light), this relation is approximately:

$$\begin{align} M_B = -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3 \end{align}$$

Suppose you observe a spiral galaxy with apparent, extinction-corrected magnitude B = 13 mag. You perform longslit optical spectroscopy, obtaining a maximum rotational velocity of 400 km/s for this galaxy. How distant do you infer this spiral galaxy to be?

We can calculate the distance of the spiral galaxy using its apparent and absolute magnitudes with the distance modulus equation:

$$\begin{align} d = 10^{(\frac{m - M_B}{5} + 1)} \end{align}$$

where $d$ is the distance to the spiral galaxy, $M_B$ is the absolute magnitude of the galaxy, and $m$ is the apparent magnitude of the galaxy. We are told that the apparent, extinction corrected magnitude is $m = B = 13$. Therefore, we just need to find the absolute magnitude of the galaxy.

We can solve for the absolute magnitude of the galaxy using the relation:

$$\begin{align} M_B = -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3 \end{align}$$

where $M_B$ is the absolute magnitude of the galaxy, and $v_{max}$ is the maximum rotational velocity of this galaxy, which is told to be $v_{max}= 400$ km/s. We can plug in maximum rotational velocity into this equation to solve for the absolute magnitude, like so:

  $$\begin{align} M_B &= -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3\\ M_B &= -10 \log \left(\frac{400 \text{ km/s}}{\text{km/s}}\right) + 3\\ M_B &= -23 \end{align}$$

Now that we have both the apparent magnitude, $m = 13$ and the absolute magnitude, $M_B = -23$, we can plug these magnitudes into the distance modulus and solve for the distance, $d$ as follows:

$$\begin{align} d &= 10^{(\frac{m - M_B}{5} + 1)}\\ d &= 10^{(\frac{13 + 23}{5} + 1)}\\ d &= 1.6 \times 10^8 \text{ pc} \end{align}$$

Therefore, the spiral galaxy is $1.6 \times 10^8$ pc away!