1 About galaxy NGC 7083, redshift and dark matter Suzanne FAYE, Lycée Chaptal, Paris, France Michel FAYE, Lycée Louis-le-Grand, Paris, France Measuring Edwin Hubble’s redshift and Vera Rubin’s dark matter Comenius UPMC Paris Juin 2011

2 Preliminary : SalsaJ colors, pixels and slices Ready now to visit galaxy NGC 7083 !!

3 I - About galaxy NGC 7083 Where? in Indus Constellation (Southern hemisphere) Why Southern hemisphere? Because of very performant telescope ESO – VLT (Chili) Internet: Ask NGC 7083 Right Ascension: 21 hours 35 minutes 45 s Declination: -63 degrees, 54 minutes 15s Apparent Magnitude:12 Apparent Diameters:3.5’ long; 2,0’ wide (slide 5)

4 1 - About Indus Constellation southern hemisphere (visible with VLT, Chili) The constellation was one of twelve constellations created by Pieter Dirkszoon Keyser and Frederick de Houtman between 1595 and 1597, and it first appeared in Johann Bayer's Uranometria of Since Indus was introduced in the 17th century, and lies in the south, it was not known to classical or early cultures thus they produced no mythology concerning it. NGC 7083

5 Answer for the angular sizes of the galaxy: 3,5’ long; 2,0’ wide 2 – Angular dimensions of galaxy NGC Open Google Earth 2 - Affichage/ Explorer / Ciel (Sky) 3 – Look for : NGC 7083: we obtain Right Ascension and Declinaison 4 – Zoom to have full galaxy 5 – Outils (Tools) / Regle ( secondes d’arc) 6 – Make measures (in two perpendicular directions)

6 3 – What is the orientation of the galaxy disc plane; angle i ? Towards observation  i i Answer for angle i : cos(i) = width/length = 2,0 / 3, 5 => i = 55°; sin(i) = 0,82 length width. i i We see as an ellipse what is in fact a circle

7 4 – What is redshift of a galaxy? Absorption lines in the optical spectrum of a distant galaxies (right), as compared to absorption lines in the optical spectrum of the Sun (left). Arrows indicate redshift. Wavelength increases up towards the red and beyond (frequency decreases). See Doppler-Fizeau effect  v /c Sun Galaxy

8 5 – Part of NGC 7083 spectrum, by VLT - ESO Continuum emitted by the core of the galaxy Lines emitted by atoms from the disk of the galaxy

9 6- Have a look at Image/ Informations

10 7 – Which lines did VLT astronomers have sent to us? N nitrogen H hydrogen S sulfur Image Information: CRPIX1 = / Reference pixel CRVAL1 = / Coordinate at reference pixel CDELT1 = / Coordinate increment per pixel CTYPE1 = 'Angstrom ' / Units of coordinate  pixel) = a*(pixel-reference) + b = CDELT1 * (pixel+ 1559) ( Å) Core of the galaxy lines  Be careful: 1 Å = 0.1 nm

11 8- How can we get the exact number of pixels? « Plot Profile! » or ZOOM and count pixels Raie N II a : X = 140, So λ (nm) = ( ) x 0, ,7 → λ = 661,39 nm

12 9 – Calculate redshift for each line Line Spectrum on Earth λ 1 (nm) Spectrum of NGC 7083 X (pixel) => λ 2 (nm) Redshift ∆λ/λ = (   -    V galaxie = c. ∆λ/λ (km/s) c = km/s NIIa654.80X=140   Hα656.28X=156   = NIIb658.35X=178   SIIa671.60X=313   SIIb673.10X=328   679, Let us keep V NGC7083 = 3.06*10 3 km/s  pixel) = CDELT1*(pixel-reference) + b = 0,09870 * (pixel+ 1559) + 493,7 ( nm) Good measurement!

13 10 – What is the distance D of galaxy NGC 7083? Let us use Hubble law : V galaxie = H * D, with H ≈ 73 km.s -1.Mpc -1 1pc = 3,26 a.l. et 1a.l. ≈ 9, m D = V NGC7083 /H = 3060/73 = 42 Mpc = 4,2 x10 7 pc D = 1.4 x10 8 a.l. D = 1,3 x10 24 m

Measuring the size d NGC7083 of the galaxy Our Galaxy, Milky Way : d Milky Way = pc NGC 7083: d NGC7083 = 4, pc = 1,7 * d Milky Way d galaxy = α(en radians) * D α NGC 7083 ≈ 3,5’= 1, rad D = 4,23 x10 7 pc

Have sizes of the galaxy with Image/ Informations and apparent diameters  core ≈ 16 pixels = 13’’ Width of the picture ≈ 289 pixels = 237’’ α NGC 7083 ≈ 3,5’ = 210’’= 256 pixels

16 12bis- Another way to measure the size d core of the core of the galaxy : Plot « vertical »profile. Let us evaluate: d core = 16 pixels; d NGC7083 ≈ 256 pixels => d core / d galaxy = 16/256 et d NGC7083 = 4, pc ; so d core ≈ 2, pc= 8, m The core !

17 II – Dancing with a galaxy Redshift Redshift of the core + « Relative » Doppler shift by rotating around the core

18 1 – Why is the shift of the spectrum constant for r > R ? Dark matter bounded? Turning around the core 2 R Dark matter bounded? Vera Rubin (born 1928) is an astronomer who has done pioneering work on galaxy rotation rates. Her discovery of what is known as "flat rotation curves" is the most direct and robust evidence of dark matter.

19 2 – What is a flat rotation curve? Let us watch Doppler shift ! * Doppler shift  is constant for r > R, which means that the relative speed is then constant * Because of the inclination i of the galaxy plane,  = V relative * sin(i) /c ) Let us imagine that the arms of the dancer are blocked by ??? Dark Matter!!! V rotation

20 3 – How can we measure  ?  pixels ≈ 8 Å or 0,8 nm  core = 16 pixels You can either use quotient in pixel, or use CDELT1: 1 pixel ≈ 1 Å or 0,1 nm; remember sin(i); i = 55 degrees V rotation = [  c / sin (55) We use line H , with rotation shift   / core   ≈ 6630Å So: V rotation ≈ (4/6630)* c/0.82 V rotation ≈ 2, m/s Around the core of the galaxy: mV² / r = G m M/ r² so M core = V² R / G G=6, SI R = d core /2 ≈ ( see slide16 ) 4, m M core = kg  galaxy = 256 pixels H   the brightest  line

21 For the core of the galaxy: mV² / R = G m M core / R² so M core = V² R / G G = 6, SI R = d core /2 ≈ 4, m M core = kg For the whole galaxy: mV² / r whole = G m M whole / r whole ² so M whole = V² r whole / G G = 6, SI r whole = d galaxy /2 ≈ 6, m M whole = 4, kg M whole = 16*M core > Brighting mass Here is dark matter, a challenge for researchers !! ::: !! Bright galaxies, dark matters, by Vera Rubin