# Finding the redshift

by bobie
Tags: redshift
PF Gold
P: 583
We know that z redshift (+1) ir the ratio between the observed and emitted wavelength.
Could you tell me, in practice, how you can you find z for a star? or can you give me a link where a particular case is explained in detail?

Any object (star, (proto-)galaxy,quasar) emits EMR in a wide range of frequencies.Take a quasar:
 Quasars are extremely luminous and were first identified as being high redshift sources of electromagnetic energy, including radio waves and visible light, that appeared to be similar to stars, rather than extended sources similar to galaxies. Their spectra contain very broad emission lines, unlike any known from stars, hence the name "quasi-stellar". Their luminosity can be 100 times greater than the Milky Way... LBQS 1429-008 ... quasar has a red shift of z = 2.076, which is equivalent to 10.5 billion light years
The first problem is to isolate a single frequency, I suppose. But when you've done that,
- how do you determine what was the original emitted frequency? in the quoted example , if the observed frequency is 2*1014 we must ascertain that the emitted frequency was 6.152*1014, then: 0.00015/0.00004877= 3.076, -1 → z = 2.076. Is this correct?
- when you determine the emitted frequency you find z, then, what is the formula to derive from 2.076 the distance (in time) 10.5 Gy or (in space) 10.5 Gly?

Thanks for your help
 Sci Advisor P: 2,851 Usually we use spectroscopy to determine the emitted frequency. Essentially different elements will have different and unique spectra (spectrum of emitted light) and so you can match the spectrum you see to some element you know. Even though the absolute color is shifted, the relative differences wavelength between two different emission lines are not distorted. Which exact lines you look for for which spectra, I'm not sure though.
 Sci Advisor P: 2,851 Usually we use spectroscopy to determine the emitted frequency. Essentially different elements will have different and unique spectra (spectrum of emitted light) and so you can match the spectrum you see to some element you know. Even though the absolute color is shifted, the relative differences wavelength between two different emission lines are not distorted. Which exact lines you look for for which spectra, I'm not sure though.
PF Gold
P: 583
Finding the redshift

 Quote by Matterwave ...Essentially different elements will have different and unique spectra
Thanks, Matterwave, do you know of any article that explains that in detail?
can you answer the second question?
 Sci Advisor P: 2,851 Perhaps you can start here: http://en.wikipedia.org/wiki/Redshif...interpretation And read the linked articles if you don't know what a term means.
 PF Gold P: 583 I know that article, can you tell me how z=2.076 translates into 10.5 Gy?
 Sci Advisor P: 2,851 Oh, that's from use of Hubble's law. I can't recall the exact calculation right now. It's pretty late and my brain isn't working very well. But this article should cover it: http://en.wikipedia.org/wiki/Cosmolo...ional_velocity Basically you have v=HD, and so z gives you v, which then you can use to get D from D=v/H. But it's a little more complicated since the universe is expanding while the light is moving towards us. Here is also a pretty good tutorial on these topics: http://www.astro.ucla.edu/~wright/cosmo_01.htm
 Sci Advisor PF Gold P: 9,454 Hydrogen lines are visible in most galaxy spectra. The Hydrogen alpha line is particularly strong in many galaxies. See http://www.astro.washington.edu/cour...surements.html for discussion.
 P: 1,857 Matterwave has covered some of the essential details, one of the things to realize is redshift requires knowing the properties and original wavelengths involved. This can be a complicated process as there is different types of influence. doppler shift due to motion of the bodies between emitter and observer gravitational redshift due to photons climbing in and out of a gravitational well (Sachs–Wolfe effect is one of the examples )(http://en.wikipedia.org/wiki/Sachs%E2%80%93Wolfe_effect cosmological redshift redshift due to expansion. so distance measures never really rely on one method, no one method works at all distance scales. so they use a variety of procedures called the cosmic distance ladder. http://en.wikipedia.org/wiki/Cosmic_distance_ladder -parallax (trigonometry) -procedures include luminosity vs distance relations -apparent magnitudes -spectrum analysis of the known light reactions to specific elements. -standard candles (type 1A supernova is one example) etc. some of that is covered in minor detail in this article. its fairly lengthy and doesn't go into all the exact details (its written more as a general try to cover too much at once format lol). However I tried to cover some of the main points in distance measures without getting too technical http://cosmology101.wikidot.com/redshift-and-expansion. keep in mind my signature has numerous articles to clarify any of the points in the article ( I know you have already looked at the Universe geometry article from a previous thread.) one factor to remember is there is numerous cross checks on any distance measure, we cannot rely on any one method with high certainty for all distances. So multiple checks are performed however this should get you started.
PF Gold
P: 583
 Quote by Mordred multiple checks are performed
Thanks Chronos, Mordred.
So H lines are the most popular source, on the whole it is a complex process, but the margin of error is very small. In the case of our quasar, the H line is 6.82 cm, and we get z= 2.076 and we check with other lines, right?.

Now, is the procedure to find out when (10.5 Gy ago, the calculator says: 10.6) light was emitted as simple as Matterwave suggested? Can we get that result directly?

I read that z+1 is the stretch of space, the ratio between the distance at present and the distance at the time of emission, but the values in the calculator never coincide with the R(adius) of the Universe now and then. Is the
distance of the quasar meaured directly through the distance ladder D = 17.67?
Everything is determined by z and D?
is it a virtual distance, since it is outside R , the Hubble radius = 14.4?

- Dnow/ S = Dthen: 17.653/3.076 = 5.74 , the distance at the time of emission
- Dnow = / R = Vnow = 17.653/14.4 = 1.23 , the apparent recession speed
Then, what is and how do we get Vthen 1.24?

Thanks again for your invaluable help
Mentor
P: 6,246
 Quote by bobie I know that article, can you tell me how z=2.076 translates into 10.5 Gy?
A distance-redshift relationship is model-dependent.

We directly measure redshift and apparent magnitude (the brightness we see here of an object). Using standard candles, we convert apparent magnitude to actual (aboslute) magnitude. We then find which values of the parameters Friedmann-Lemaitre-Robertson-Walker universe best fit the redshift-magnitude relationship.

The values of these parameters give us a particlular model. Once we have the model, we can calculate a distance-redshift relationship.

For the standard cosmological model,

http://www.physicsforums.com/attachm...4&d=1370171072,

shows how to do the calculation.

Do you want to try it?

Quote by George Jones
 Quote by nenyan How to convert red shift to parsecs?
This depends on which concept of distance is used. See the first paragraph of the attached file.

 Quote by nenyan s there a simple converting equation?
I wrote up an answer for someone else, who specified z = 0.4. I didn't use dimensionless parameters. See the last two graphs in the attached file.
PF Gold
P: 583
 Quote by George Jones Do you want to try it?
Thanks George and congrats for you article.
Before I embark in a difficult task, can you confirm that it is all conventional?
We cannot evaluate a distance without a cosmological model? does it apply also to near stars or even to the Sun?
I take advantage of your expertise to ask you a question that's bugging me:

I have noticed (in the calculator) that all presumed co-moving distances are outside the real R radius of the Universe. This is strange and might be justified by the fact that they are virtual(?) distances. But what struck me is that also all the real distances of the stars from us at the time when light was emitted in the past were outside the the radius.
Isn't that odd? is it possible that there were no star at all inside the radius of the universe?

-even our quasar (which is relatively near, with z ≈ 2) , when emitted the light that we are observing now, is thought to be over 1 Gly more distant from us than the edge of the universe!

Is that plausible?
Astronomy
PF Gold
P: 23,235
 Quote by bobie ... but the values in the calculator never coincide with the R(adius) of the Universe now and then. Is the distance of the quasar meaured directly through the distance ladder D = 17.67? Everything is determined by z and D? is it a virtual distance, since it is outside R , the Hubble radius = 14.4? - Dnow/ S = Dthen: 17.653/3.076 = 5.74 , the distance at the time of emission - Dnow = / R = Vnow = 17.653/14.4 = 1.23 , the apparent recession speed Then, what is and how do we get Vthen 1.24? Thanks again for your invaluable help
R, in the calculator, does not stand for "Radius of the Universe" it is the radius of a relatively small part of the universe called the "Hubble sphere" the region around us where distances are increasing LESS OR EQUAL to speed of light. MOST OF THE GALAXIES WE OBSERVE ARE OUTSIDE THE HUBBLE SPHERE
That is their actual present-day distance, as best we can estimate it, is greater than R.
So it is misleading/confusing to think of R as the radius of the universe, Bobie. It would be better to refer to it as the HUBBLE RADIUS.

So in your example, the present-day distance to the quasar (if we could pause expansion to give us time to measure with radar or a superlong yardstick or some such) is actually reckoned to be 17.67 Gly, like you say. That's not "virtual" either "biggrin: and it's not "light travel time"or anything like that---it is today's proper distance to the quasar.

And you did the arithmetic right, to get the THEN proper distance "17.653/3.076 = 5.74"

Now you ask "how do we get Vthen 1.24?"

Well obviously it is going to be the proper distance then divided by the Hubble radius then. And you already have found 5.74 Gly as the Dthen the proper distance at the time of emission. So you just have to look up and see what the Hubble radius R was at the time of emission. Then if you divide by it you should get 1.24.

That is what the Hubble radius R is good for. It is the size distance that is increasing at speed c, at the given time in history. And all the other distances are increasing at speed proportional to their size. So the distance 2R is growing at speed 2c, and the distance 3R is growing at speed 3c. That is what Hubble law says---recession speed at any given time is proportional to size of proper distance. So to find the recession speed you simply divide by the Hubble radius!

Check it out and see what you get. I'll bet you get Vthen = 1.24

You are asking good questions and it's great you are getting some hands-on experience with the model.

There is a simple answer to your other question about why most of the galaxies we can see with telescope are at present distance which is greater than R, but I don't want to make this post too long so I'll leave that for later (and somebody else may get to it first.)
 PF Gold P: 583 There is something that escapes me and makes it hard for me to comprehend the rationale of the model. Suppose the Hubble radius is R= 10.95 Gly, and a quasar is at 0.38R: 10.95*.3847 = 4.213 Gly S .....T (Gy) .....R (Gly) Dnow (Gly) Dthen (Gly) DHor (Gly) Vnow (c) Vthen (c) 1.500.8.6029 10.9492 6.320 ......4.213.......... 15.204 0.44 0.38 If expansion is isotropic and omogeneous, after any time the ratio between any to points must remain the same. So if : R becomes Rt =14.4 R*1.315, shouldn't Dt be D*1.315 =4.213*1.315=5.54? why is it 6.32 This discrepancy becomes greater and greater with distance. Why so?
Mentor
P: 6,246
 Quote by bobie If expansion is isotropic and omogeneous, after any time the ratio between any to points must remain the same. So if : R becomes Rt =14.4 R*1.315, shouldn't Dt be D*1.315 =4.213*1.315=5.54? why is it 6.32 This discrepancy becomes greater and greater with distance. Why so?
The ratio of the proper distances to any two galaxies that move with the Hubble flow is constant in time. The Hubble radius, as a function of time, does not follow the worldline of a galaxy that moves with the Hubble flow. Consequently, the ratio of the Hubble radius to the proper distance of a particular galaxy changes with time.
 P: 10 I know that this is pretty unrelated to the question, but one time, my professor told me that the redshift can be large enough that we can't observe objects with visible light anymore. For example, a object could be moving away so fast that the wavelength would be low enough to only be able to be observed with infrared sensors. Meaning we wouldn't see anything with telescopes that used visible light. I thought it was pretty cool! Now back on track!
PF Gold
P: 583
 Quote by George Jones The Hubble radius, as a function of time, does not follow the worldline of a galaxy that moves with the Hubble flow. Consequently, the ratio of the Hubble radius to the proper distance of a particular galaxy changes with time.
Thanks, George, I is this is of the non-uniform expansion/recession speed over time?.
is there a formula that gives you c/Ht? do you happen to know the expansion speed H at every moment from the BB?
I have noticed that the ratio Rt/cTt is almost stable ≈1.5-1.4 from the beginning of the table at T = .0004 Gy, down to z=1.15 T = 5.27 Gy, that is for near half the age of the universe, and then it drops rapidly. Why so?
PF Gold
P: 583
 Quote by marcus ...the proper distance then divided by the Hubble radius then. ... There is a simple answer to your other question about why most of the galaxies we can see with telescope are at present distance which is greater than R
- because the are the oldest, marcus? are they the most massive, too?
.. whatever the reason: if most galaxies are concentrated beyond the Hubble radius, how can the total mass be uniformly distributed over the universe? its volume is only 26 times greater than the Hubble volume

as to Vthen I had done a lot of calc to try to figure out by myself , but the figures are too approximate: I tried on the very first line (S=1090) and .042/.0006 gave 70 vs. 66.18

Thanks a lot for your patience, marcus

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