Astronomers Take “Baby Picture” of an Incredibly Distant Galaxy

by Jason Major on June 1, 2012

Want to stay on top of all the space news? Follow @universetoday on Twitter

False-color image of galaxy LAEJ095950.99+021219.1 (Credit: James Rhoads/ASU)

Astronomers from Arizona State University have grabbed an image of a dim, distant galaxy, seeing it as it looked only 800 million years after the birth of the Universe. Visible above as a green blob in the center of a false-color image acquired with the Magellan Telescopes at the Las Campanas Observatory in Chile, the galaxy is seen in its infancy and, at 13 billion light-years away, is one of the ten most distant objects ever discovered.

The galaxy, designated LAEJ095950.99+021219.1, was detected by light emitted by ionized hydrogen using the Magellan Telescopes’ IMACS (Inamori-Magellan Areal Camera & Spectrograph) instrument, built at the Carnegie Institute in Washington. In order to even find such a remote object — whose existence had already been suspected — the team had to use a special narrow-band filter on the IMACS instrument designed to isolate specific wavelengths of light.

“Young galaxies must be observed at infrared wavelengths and this is not easy to do using ground-based telescopes, since the Earth’s atmosphere itself glows and large detectors are hard to make,” said team leader Sangeeta Malhotra, an associate professor at ASU who helped develop the technique.

“As time goes by, these small blobs which are forming stars, they’ll dance around each other, merge with each other and form bigger and bigger galaxies. Somewhere halfway through the age of the universe they start looking like the galaxies we see today – and not before.”

– Sangeeta Malhotra, ASU professor 

LAEJ095950.99+021219.1 is seen at a redshift of 7, putting it farther away than any other objects previously discovered using the narrow-band technique.

(What is redshift? Watch “How To Measure The Universe” here.)

“We have used this search to find hundreds of objects at somewhat smaller distances. We have found several hundred galaxies at redshift 4.5, several at redshift 6.5, and now at redshift 7 we have found one,” said James Rhoads, associate professor at ASU and research team leader.

“This image is like a baby picture of this galaxy, taken when the universe was only 5 percent of its current age. Studying these very early galaxies is important because it helps us understand how galaxies form and grow.”

So why does LAEJ095950.99+021219.1 not look much like the galaxies we’re used to seeing in images?

Malhotra explains: “Somewhere halfway through the age of the universe they start looking like the galaxies we see today – and not before. Why, how, when, where that happens is a fairly active area of research.”

The team’s NSF-funded research was published in Astrophysical Journal Letters. Read more on Phys.Org News here.

About 

A graphic designer in Rhode Island, Jason writes about space exploration on his blog Lights In The Dark, Discovery News, and, of course, here on Universe Today. Ad astra!

hunterz85 June 1, 2012 at 8:40 PM

I like galaxy name (number) LAEJ095950.99+021219.1

magnus.nyborg June 2, 2012 at 7:13 AM

They trie the name Bob, but the assistant director already had a kid with that name.

Jason Major June 2, 2012 at 9:06 PM

They could have just spelled it backwards — doh!!!

lcrowell June 2, 2012 at 2:56 AM

It is interesting to see how this compares to the Hubble relationship v = Hd. The Hubble constant, or really parameter is H = 72km/s/Mpc, Mpc = megaparsec. A parsec is 3.26lightyears. The redshift factor is z = v/c = 7. The distance is then d = v/H = (7x3x10^5/72)Mpc = 29166Mpc or 9.5 billion light years. This is somewhat close but far enough to show there is a deviation from the quoted result of 13 billion light years.

It is worth noting that the redshift factor indicates this galaxy is moving outwards 7 times faster than the speed of light. However, it is really framed dragged along by the expansion of space itself. There is however no situation where two galaxies in the same local region pass each other at some velocity faster than light.

So what is going on? Why is there this discrepancy? We can go back to Newton’s law of motion with gravity. The motion of an object in a gravity field is given by the kinetic energy of motion T = 1/2mv^2 and its potential energy V = -GMm/d, where M is all the other masses. The sum is then equal to the total energy

E = 1/2mv^2 – GMm/d.

We now consider the motion of a galaxy according to a scale factor a. So a distance d is scale factored in time by ad or d(t) = a(t)d_0. The velocity is then v = (da/dt)d_0, we now write as a’d_0. The mass M is all the matter from all other galaxies or from the vacuum etc in a region with a radius d, which has a volume V = (4?/3)d^3 = (4?/3)a^3d_0^3. The matter in this region has an average density ? and so M = ?V We now put all of this together in our total energy equation to get

E/m = 1/2a’^2 – (4?G?a^2/3).

This is then set to zero in order to model a cosmology with a flat space, and with more work this gives

(a’/a)^2 = 8?G?/3.

This is the Friedmann-Lemaitre-Robertson-Walker (FLRW) energy equation for the motion of a flat spatial universe. It is interesting that basic Newtonian mechanics captures something which is computed more completely in general relativity. The left hand term is written this way because the Hubble parameter H/c = (a’/a).

Assume there is a vacuum energy that fills space as a constant energy density ?. Further assume this dominates the dynamics of the universe. This equation can be represented as a differential equation

a’ = sqrt{8?G?/3}a,

which has the solution a(t) = a_0exp(t sqrt{8?G?/3}). Sometimes this is written with ? = 8?G?/3, which is the cosmological constant, or with the Hubble parameter (H/c)^2 = 8?G?/3. This the most general solution, but for small time t we can expand this by Taylor series as

a(t) =~ a_0(1 + tH/c}.

here the speed of light converts the time to a distance and so we have the Hubble relation

a(d) ~= a_0(1 + Hd).

If we carry the Taylor expansion to second order we have

a(d) ~= a_0(1 + Hd + (Hd)^2/2).

For d large enough this second order term becomes larger. The distance d is solved by a quadratic equation for a more exact result.

LC

MeyerKaty June 2, 2012 at 5:26 PM

my friend’s aunt made $17398 the previous week. she is making income on the internet and bought a $578000 house. All she did was get lucky and try the steps written on this website===>> ?????? http://getitmust.blogspot.com/m

ClarkTommy63 June 3, 2012 at 7:27 PM

my buddy’s mother got paid $21508 the previous week. she is making money on the internet and bought a $386500 house. All she did was get fortunate and put into work the instructions explained on this web site===>> ?????? http://enternet-Job.blogspot.com

DillardCarl92 June 4, 2012 at 5:24 PM

my classmate’s mom made $ 18460 l ast week. s he is ge tting p aid o n the l aptop a nd mo ved in a $ 571 600 house. A ll sh e did w as g et lu cky an d fo llow the t ips ma de cl ear o n th is si te ===>> ?????? http://workoverenternet.blogspot.com

IVAN3MAN_AT_LARGE June 2, 2012 at 3:57 AM

Comments on this entry are closed.

Previous post:

Next post: