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Distance Measures

by Madster
Tags: distance, measures
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Madster
#1
Dec27-12, 06:41 AM
P: 22
Hi,

I have a question dealing with distances of, lets say galaxies. I researched a little and saw that there are plenty of coordinates. Mostly used are spherical (celestial) ones, RA α and DEC δ. To have a three dimensional distance measure, one uses the redshift and from there the line of sight distance.
According to Hoog arXiv:astro-ph/9905116v4, this computes like:

[tex] D_C = D_H \int_0^z \frac {dz'}{E(z')} [/tex]

where D_H is a constant and E(z) is
[tex] E(z)=\sqrt{\Omega_M (1 + z)^3 +\Omega_k (1 + z)^2 +\Omega_M } [/tex].
So if I got it correctly I some up little line elements along dz, depending on the cosmology i consider ([tex] \Omega_k=0 [/tex]). But this is just the line of sight distance right? What about the position on the skymap α and δ? How can I compute the distances of objects that differ in redshift z and in celestial coordinates? Does it even make sense as the universe expanded in between?
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Naty1
#2
Dec27-12, 08:37 AM
P: 5,632
Hi Madster:

But this is just the line of sight distance right?
Yes, that is what the author calls it.

Do you understand that this is a theoretical construct; in other words, no single observer can make this observation....This relates to the fact that cosmological 'distance' measures are really four dimensional measures, not a 'three dimensional distance measure' as you state. Some convention for TIME must be made.

If you already understand this, then we must await a post from an expert. If not, try
studying this first diagram: http://en.wikipedia.org/wiki/Metric_expansion_of_space

Two objects: Earth moves along the brown line of the left, a quasar [to serve as a distant object] moves along the yellow worldline on the right and they are moving apart as time increase upwards in the illustration.

The red line is the path of a light beam emitted by the quasar about 13 billion years ago and reaching the Earth in the present day. That is all we can SEE, or observe. The orange line shows the present-day distance between the quasar and the Earth, about 28 billion light years. THAT curve is one of instantaneous time....where we calculate the quasar to be NOW. It is not directly 'observable'.
pervect
#3
Dec27-12, 04:53 PM
Emeritus
Sci Advisor
P: 7,594
Quote Quote by Naty1 View Post
Hi Madster:


Do you understand that this is a theoretical construct; in other words, no single observer can make this observation....
More specifically, to make this sort of measurement, you need a chain of observers, all of whom measure the distance to the next observer in the chain "at the same time".

The results in general depend on the details of the chain, the usual chain in cosmology are called "co-moving" observers, ones that see the cosmic microwave background radiation as isotropic.

The results also depend on how the chain defines "at the same time", usually once you have the chain specified you can assume Einstein clock synchorinzation as the method.

Madster
#4
Dec27-12, 05:38 PM
P: 22
Distance Measures

Quote Quote by Naty1 View Post
Hi Madster:




If you already understand this, then we must await a post from an expert. If not, try
studying this first diagram: http://en.wikipedia.org/wiki/Metric_expansion_of_space

Two objects: Earth moves along the brown line of the left, a quasar [to serve as a distant object] moves along the yellow worldline on the right and they are moving apart as time increase upwards in the illustration.

The red line is the path of a light beam emitted by the quasar about 13 billion years ago and reaching the Earth in the present day. That is all we can SEE, or observe. The orange line shows the present-day distance between the quasar and the Earth, about 28 billion light years. THAT curve is one of instantaneous time....where we calculate the quasar to be NOW. It is not directly 'observable'.

Hi. So in comoving coorinates mentioned by pervect this quasar will be 13 billion light years away no matter how much the scale factor evolves, no? What then is the correct calculus of two randomly selected objects in comoving coordinates?
thanks
pervect
#5
Dec27-12, 07:13 PM
Emeritus
Sci Advisor
P: 7,594
Quote Quote by Madster View Post
Hi. So in comoving coorinates mentioned by pervect this quasar will be 13 billion light years away no matter how much the scale factor evolves, no? What then is the correct calculus of two randomly selected objects in comoving coordinates?
thanks
As far as I know, the current labmda-cdm models hold the spatial slices to be flat, so you could use ordinary trig if you knew the two distances (which cosmologists confusingly call proper distance. even though it requires the use of comoving observers).

But I haven't been able to find a good clear reference to confirm my unfortunately fallible memory about this topic.

I'll refrain from working out the trig unless asked, esp. in light of the other uncertanties.
Madster
#6
Dec28-12, 04:13 AM
P: 22
Quote Quote by pervect View Post
As far as I know, the current labmda-cdm models hold the spatial slices to be flat, so you could use ordinary trig if you knew the two distances (which cosmologists confusingly call proper distance. even though it requires the use of comoving observers).

But I haven't been able to find a good clear reference to confirm my unfortunately fallible memory about this topic.

I'll refrain from working out the trig unless asked, esp. in light of the other uncertanties.
So I use the formula above to get the "proper distant" (comoving) between the observer, us, and the distant object because I integrate from 0, that's again us, to the measured redshift z of the object. OK then but how to get this comoving distance between objects that are at z1=0.1 and z2=0.3 e.g.?
Naty1
#7
Dec28-12, 07:44 AM
P: 5,632
Madster: Cosmological distances, [recession] velocities and the conventions used to determine them are about the most the most confusing subject I've come across in these forums....


from my notes: a combination of discussions in these forums and Wikipedia:

the metric or distance function is the core idea in geometry. Different geometries---different metrics---arise as solutions to the GR equation. In order to define distance one must choose a metric, the distance will be defined in that metric. In cosmology, one must also move the metric across some defined
spacetime curvature. We use a curve of constant time to freeze expansion and get a
snapshot.

Proper distance roughly corresponds to where a distant object would be at a specific moment of cosmological time, measured as if a long series of rulers were stretched out from our position to the object's position at that time, and which can change over time due to the expansion of the universe. Comoving distance factors out the expansion of the universe, by moving the endpoints with the CMBR. This gives a distance that does not change in time due to the expansion of space. Comoving distance and proper distance are defined to be equal at the present time; The universe's expansion results in the proper distance changing, while the comoving distance is unchanged by this expansion….
Putting these together with pervect's post 'flat special slices' he recalls makes sense because the FLRW model assumes homogeneous and isotropic space; if you freeze your measurement time than sounds reasonable that space portion would be virtually flat.

here is another piece.....which may help...

Strictly speaking, neither v nor D in the Hubble formula are directly observable, because they are properties now of a galaxy, whereas our observations refer to the galaxy in the past, at the time that the light we currently see left it. For relatively nearby galaxies (redshift z much less than unity), v and D will not have changed much, and v can be estimated using the formula where c is the speed of light. This gives the empirical relation found by Hubble.
For distant galaxies, v (or D) cannot be calculated from z without specifying a detailed model for how H changes with time. The redshift is not even directly related to the recession velocity at the time the light set out, but it does have a simple interpretation: (1+z) is the factor by which the universe has expanded while the photon was travelling towards the observer.
Marcus, an expert of these forums, often refers people to an online cosmological calculator
here:
http://www.astro.ucla.edu/~wright/CosmoCalc.html

and he explains some aspects of the model in this loooooong discussion:

http://www.physicsforums.com/showthr...e%3F%29&page=7

Another calculator he uses is:


Jorrie's calculatorhttp://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm

In the models, you need to utilize standard LCDM parameters, like dark energy fraction of 0.73, and one 'signpost would be that a galaxy with a redshift of 1.8 is at the CEH .

I believe the standard parameters are already plugged in the NedWright calculator.

Marcus used the calculator and answered an interesting question which puts some pieces in perspective:


The projectile sent at .1c towards a target receding at rate .1c actually does make it! For the same reason (linearly scaled down) that we could today send a flash of light to a galaxy at Hubble distance, i.e. receding at rate c, and it would eventually get there.
......
for a long time it kind of hovers at the same proper distance from the target, just barely hanging on. Because its speed .1c just exactly cancels the rate the thing is getting farther way. But the Hubble constant is still decreasing! According to standard LCDM it is now around 70.4 and destined to decline slowly and level out at (asymptotic) value of around 60 km/s per Mpc.

As soon as H gets appreciably below its present value the projectile will begin to slowly gain ground! But for a long time it is just hanging in there, staying approximately 1.4 billion LY from the target. The target receding at .1c and it going towards at .1c, so no change. Then finally because of declining H, the recession speed at that 1.4 billion LY distance will get slightly less, like .95c. And then the projectile will begin to creep towards it at 0.05c. From then on it is assured of getting there.
Hope this helps.....if you do determine an answer, please post the explanation of what you think you found....sooner or later an expert may stumble here and help.


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