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I am doing my first steps in spectroscopy (IFS actually) and how we can learn more about galaxies by using it. I came up with a simple question which, unfortunately, I can not answer:

How can we measure the rotational velocity of a spiral galaxy given that the galaxy is perfectly face-on?

I understand how to do so if the galaxy was tilted or edge-on: By measuring the Doppler shift of the spectral lines for each spatial position. One side will be red, the other blue, and the rest somewhere in between. But what about if the galaxy is perfectly face-on? The Doppler shift will be zero right? Most of the resources, papers etc I find online are about tilted galaxies.

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Sorry, no way to directly measure that speed.

However, you may try to guess the direction of the rotation by looking at the spiral structures. Usually the tips of the arms point in the opposite direction from the direction of rotation. But that is not a general rule, in some cases a so-called "leading structure" has been found, where the tips of the arms point in the direction of the rotation.

There is no sure way to know how a face-on galaxy rotates. That is not a big problem, because there are few of them. There are many more edge-on galaxies (can you see why?)


Here is why:

Imagine you throw into the air a bunch of compact discs (your spiral galaxies) with a kick (a homemade Big Bang), and somebody takes a picture while they are still in the air, dispersed all across the room and randomly oriented in all directions.

Which fraction of them in the picture, are seen face-on and edge-on?. To calculate this, you can pick each one up, and move it from its position, frozen in a three-dimensional point of your room, and stick it on the appropiate point on the surface of a sphere, without changing its orientation relative to you.

Since they are randomly oriented, they will cover the the sphere, with an equal number per unit surface, $N$

The sphere surface element can be written, in spherical coordinates, as: $d \phi d \theta sin \theta$. So you have this number of galaxies per element surface:

$$N d \phi d \theta sin \theta$$

Now, look at the thing from a distant point along the vertical of the north pole. Count how many discs are there between co-latitude $\theta_1$ and $\theta_2$:

$$N \int_{0}^{2\pi}d\phi \int_{\theta_1}^{\theta_2} d \phi d \theta sin \theta $$

(Co-latitude meaning polar angle, i.e. $90$ deg for the equator and $0$ deg for the north pole)

This integral yields the result:

$$2\pi N(cos(\theta_1)-cos(\theta_2))$$

Now, let's say you want to know in which proportion edge-ones appear with respect to face-ones. While the face-ones cover the north polar cap (say, between $0$ and $10$ degrees) the edge-ones are around the equator (say, between $80$ and $90$ degrees). So the calculation is:

$$\frac{2\pi N(cos(80 \deg)-cos(90 \deg))}{2\pi N(cos(0 \deg)-cos(10 \deg))} \approx 11.4$$

Thus, when you randomly pick up a spiral galaxy from a survey, finding it edge-on is above one order of magnitude more likely than finding it face-on.

Here is my basket ball and some failed CDs, serving a higher purpose:

Homemade extragalactic physics! Face-one galaxies are harder to find, than edge-ones

You see, only one "galaxy" is face-on, whereas the equator is covered with edge-ones.

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Thank you for your answer Eduardo! I guess there are more edge-on galaxies than face-on galaxies because everything in the universe is moving and the probabilities of finding a perfectly face-on galaxy are just too low? –  AstrOne Apr 20 '13 at 9:46
    
No, it is something simpler. It arises from a geometrical argument. Try to calculate the fraction of galaxies seen between inclination 0 deg and, say, 10 deg as compared with the fraction between 80 deg and 90 deg, assuming they are all perfectly circle-shaped and randomly oriented in space. I post the solution in a few hours, ok? It is an interesting exercise! –  Eduardo Guerras Valera Apr 20 '13 at 10:02
    
Wow, I have to admit, that was a very good explanation! I was thinking the same way in order to explain it but you also did/explained the math and you made it very clear to me. Thank you my friend! –  AstrOne Apr 20 '13 at 12:33
    
I am sorry I can only vote this answer up once, it deserves more upvotes than this! –  Thriveth Sep 15 '13 at 21:39

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