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In other words, if matter in the accretion disk is continually being depleted by falling into the black hole or being blown out from the galaxy in the form of jets, then a quasar can continue to radiate only as long as new gas is available to replenish the accretion disk.

In fact, there was more gas around to be accreted early in the history of the universe. Back then, most gas had not yet collapsed to form stars, so there was more fuel available for both the feeding of black holes and the forming of new stars. Much of that fuel was subsequently consumed in the formation of stars during the first few billion years after the universe began its expansion. Later in its life, a galaxy would have little left to feed a hungry black hole or to form more new stars. As we see from [link] , both star formation and black hole growth peaked together when the universe was about 2 billion years old. Ever since, both have been in sharp decline. We are late to the party of the galaxies and have missed some of the early excitement.

Observations of nearer galaxies (seen later in time) indicate that there is another source of fuel for the central black holes—the collision of galaxies. If two galaxies of similar mass collide and merge, or if a smaller galaxy is pulled into a larger one, then gas and dust from one may come close enough to the black hole in the other to be devoured by it and so provide the necessary fuel. Astronomers have found that collisions were also much more common early in the history of the universe than they are today. There were more small galaxies in those early times because over time, as we shall see (in The Evolution and Distribution of Galaxies ), small galaxies tend to combine into larger ones. Again, this means that we would expect to see more quasars long ago (far away) than we do today (nearby)—as we in fact do.

Codependence of black holes and galaxies

Once black hole    masses began to be measured reliably in the late 1990s, they posed an enigma. It looked as though the mass of the central black hole depended on the mass of the galaxy. The black holes in galaxies always seem to be just 1/200 the mass of the galaxy they live in. This result is shown schematically in [link] , and some of the observations are plotted in [link] .

Relationship between black hole mass and the mass of the host galaxy.

In this plot the vertical axis is labeled “Black Hole Mass”. The scale goes from “No black hole” at bottom, “One million solar masses” in the middle and “One billion solar masses” at top. The horizontal axis is labeled: “Mass of Central Bulge”. The scale is arbitrary, with an arrow pointing to the right labeled “Increasing”. A straight white line is drawn from lower left to upper right with illustrations of galaxies along its length. At bottom left is a small spiral galaxy. Moving upward along the line, the galaxies increase in size as do the black dots at the center of each representing black holes. The final image at upper right is a very large elliptical with a very large black hole.
Observations show that there is a close correlation between the mass of the black hole at the center of a galaxy and the mass of the spherical distribution of stars that surrounds the black hole. That spherical distribution may be in the form of either an elliptical galaxy or the central bulge of a spiral galaxy. (credit: modification of work by K. Cordes, S. Brown (STScI))

Correlation between the mass of the central black hole and the mass contained within the bulge of stars surrounding the black hole, using data from real galaxies.

In this plot the vertical axis is labeled “Black Hole Mass” in solar masses. It is a logarithmic scale ranging from 106 at bottom to 1010 at top, in increments of 10n+1. The horizontal axis is labeled “Mass of Central Bulge” in solar masses. It is a logarithmic scale ranging from 109 at left to 1012 at right, in increments of 10n+1. The data (and the dashed line fitting the data) runs straight from lower left to upper right. The legend at upper right describes the data points: blue stars for “Stars/Early-type BCG”, orange stars for “Stars/Early-type non-BCG”, green dots for “Gas/Early-type BCG” and red dots for “Gas/Early-type non-BCG”.
The black hole always turns out to be about 1/200 the mass of the stars surrounding it. The horizontal and vertical bars surrounding each point show the uncertainty of the measurement. (credit: modification of work by Nicholas J. McConnell, Chung-Pei Ma, “Revisiting the Scaling Relations of Black Hole Masses and Host Galaxy Properties,” The Astrophysical Journal , 764:184 (14 pp.), February 20, 2013.)

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, Astronomy. OpenStax CNX. Apr 12, 2017 Download for free at http://cnx.org/content/col11992/1.13
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