Tully-Fisher relation?

Question

The Tully Fisher relation can Be used to find the Distances to galaxies. 
Tully Fisher relation: L= V^4
L is the luminosity
V is the rotational velocity of the galaxy in km/s

I went online and did a little research and found out that You can express the Luminosity as the Absolute magnitude. I actually found 2 equations: 
M = 4.8 - 2.5 x log(L/Lsun)
or 
M= -9.5 x log(Rotational velocity in km/s)+2
Where M is the Absolute magnitude

I tried to test these equations On the Andromeda Galaxy. Since I couldn't find the rotation curve Of Andromeda, I couldn't use the second equation. 

Luminosity of Andromeda = 2.6 x 10^10 L_sun

M= 4.8-2.5 x log(2.6 x 10^10)
M= -21.23 <---- absolute magnitude of Andromeda. 

I wanted to test If I my answer was right. So used this absolute magnitude and the apparent magnitude of Andromeda, 3.44, to find its distance, which I already know is 2,540,000 Light years. 
Distance= 3.44-(-21.23)+5)/5)
Distance= 10^5.934= 859,013.522 parsecs
859,013.522 parsecs = 2,801,789.65 light years
Is that a close enough answer, or am I just using the Wrong Equation to find the Absolute Magnitude. 

And since L= V^4
(2.6 x 10^10)^(1/4) = 401.55 km/s is the Rotation curve for Andromeda Right? 

I just want to know the real equation and if my answers are valid.

Frst Grade Rocks! Ω · Accepted Answer

Tully and Fischer used 535 km/s for Andromeda for their 1977 paper: http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1977A%26A....54..661T&data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf

L ∝ V^4, not L= V^4

You might want to back track and work with this how to:  http://universe-review.ca/R02-07-candle.htm

Alternatively, you plug it back in to: 
M= -9.5 x log(Rotational velocity in km/s)+2 
(But I can't get a good number out of this formula...did you miss something in the copying)

Or you could do:  (M.Androm/M.Milky)^9.5 *220km/s
but that only gives you 360km/s


**********************Addendum ******************

Got it all figured out.

the ΔV from the Tully-Fisher paper is the differential radial velocity (based on the width the H2 spectrum), not the radial velocity.  Just divide by 2.  Vr = 267 km/s

Further, you need to be sure to use log₁₀, not ln (I hang around too much with people who use log to mean ln).  I made a plug in error when I first attempted this problem.

Answer

If we use the relationship you used:
M = - 9.5 log Vr + 2
we get Vr = 272 km/s
This is approximately what Tully Fisher used 

However, if we use the Sa curve for M31 from this source ( ref:  http://www.phys.unm.edu/~gbtaylor/astr422/09_A422_Galaxies_II.pdf )we get:

M = -9.5 log Vr + 3.15
-21.23 = -9.5 log Vr + 3.15
log Vr = 2.556
Vr = 360 km/s
This is close to the answer I got above for extrapolating based on the Milky Way Vr

Based on my review, the k values for the relationship
L = k V^4
will vary based on the type of galaxy.  You will also get variations because this is an empirical formula so you will get a a gaussian distribution.  Some things to consider would be age of galaxy, rate of new star formation, mass of galactic center, amount of dark matter halo, etc.

Hope this helps.

Anonymous · Answer

This Site Might Help You.

RE:
Tully-Fisher relation?
The Tully Fisher relation can Be used to find the Distances to galaxies. 
Tully Fisher relation: L= V^4
L is the luminosity
V is the rotational velocity of the galaxy in km/s

I went online and did a little research and found out that You can express the Luminosity as the Absolute magnitude. I...

? · Answer

Why use the Tully - Fischer relationship on the Milky Way.
According to Wikipedia the Sun revolves at 220 KM/sec around the Milky Way. So according to the Tully- Fischer relationship  (M = -9.5 x log (rotation velocity in km/s) + 2) the Milky Way has an absolute magnitude of -20.25 magnitude. So it seems reasonable that Andromeda Galaxy has an absolute magnitude of -21.23. Hope this helps

Trending News

Tully-Fisher relation?

3 Answers