Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about the CMB dipole

  1. Jan 26, 2006 #1
    The Local Groups motion relative to the CMB is said to be around 700 km/s. I would have thought there would be no motion relative to the rest frame of the CMB, and instead expected it to be at a constant distance regardless of motion. Do we have to factor in the speed at which the CMB is receding away from us to account for this dipole? For example, if the recession velocity is just under the speed of light, or 0.9999983 c, this motion then becomes observable?
     
  2. jcsd
  3. Jan 26, 2006 #2

    Garth

    User Avatar
    Science Advisor
    Gold Member

    Hi Vast!

    The CMB is radiated from a physical surface, i.e. the Surface of Last Scattering (SLS) when the universe cooled to about 3,0000K so hydrogen palsma associated into hydrogen gas and the universe became transparent.

    The SLS is all around us and is observed much red-shifted (z ~ 1100), not as a gas at 3,0000K but 2.760K.

    The Earth is moving relative to the SLS as it travels around the Sun, the Sun travels around the galaxy, and the galaxy moves in and with the Local Group. This movement is observed as a dipole in the radiation temperature (or wavelength of peak radiation). That is, the sky looks hotter in one direction than in the opposite direction and the temperature varies as the cosine of the angle in between.

    This dipole variation is one part in 103 smaller than the basic temperature. The primordial anisotropies studied by WMAP are one part in 105 that temperature and are therefore two OOM smaller than the dipole.

    The fact that the universe is expanding, and therefore the SLS is receeding from us, is manifested in the CMB continuing to be red shifted, it does not affect the dipole. The dipole is simply recording our peculiar velocity relative to that surface of about 0.1%c.

    This motion of the Local Group relative to the SLS may be caused by us orbiting the Virgo Cluster or by us being attracted towards some Great Attractor, or both!

    Garth
     
    Last edited: Jan 26, 2006
  4. Jan 26, 2006 #3
    Thanks for the explanation Garth.

    I think part of my confusion arises from thinking that the CMB recession velocity to be 0.9999983 the speed of light. (I’m not sure how accurate this figure is) However, take for example the particle horizon which is receding at exactly the speed of light. If it was theoretically possible to observe the particle horizon, then no matter where we are or whatever motion we had, wouldn’t it always appear to be at a constant distance from us?

    So we would then fail to observe any motion relative to it? But the particle horizon isn’t necessarily a physical surface is it? Or we cannot observe anything prior to the moment the universe became transparent.

    I guess what I’m trying to understand is; do we observe our motion relative to the SLS solely because it is a physical surface receding just under the speed of light?

    I hope I’m making sense.
     
  5. Jan 26, 2006 #4

    Garth

    User Avatar
    Science Advisor
    Gold Member

    We cannot see the particle horizon by definition.
    If we are going to talk about how fast objects at large z are receeding from us we need to define some cosmological velocities i.e. clocks and rulers. Davis & Lineweaver have a nice paper Expanding Confusion:
    common misconceptions of cosmological horizons and the superluminal expansion of the universe


    We measure z, how do we translate this into a recessional velocity? What formula do we use?

    If we use the classical formula
    [tex]v = zc[/tex]
    then objects beyond z = 1 are travelling faster than light.

    If we use the relativistic doppler formula
    [tex]1+z = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}. [/tex]
    then only objects with [itex]z=\infty[/itex] are travelling at the speed of light.

    This is: [tex]v_{pec}(z) = c\frac{(1 + z)^2 - 1} {(1 + z)^2 + 1}[/tex].

    If we use a GR cosmological formula:
    [tex]v_{rec}(t, z) = \frac{c}{R_0}\stackrel{.}{R} \int \frac{dz'}{H(z')}[/tex]
    then we can see objects that are travelling faster than light!

    This is because the light catches up with us as the universe decelerates.

    The SLS can be observed and our velocity relative to it can be measured - by the magnitude of the dipole.

    Garth
     
    Last edited: Jan 26, 2006
  6. Jan 26, 2006 #5
    I quickly skimmed over the paper, but I will have to sit down and read it in depth again, but for now some questions:

    So depending on which formula we use, we get different recession velocities? Do I take this to mean there is disagreement among cosmologists with respect to rescission velocities of distant galaxies? The paper mentioned physicists dislike using recession velocities and would rather use redshifts instead. Or depending on which cosmological model one uses, there are different outcomes?

    I read somewhere that the particle horizon corresponds to [itex]z=\infty[/itex] which was also the speed of light. I would then interpret this to mean that everything inside the particle horizon has a velocity of less than c.
     
  7. Jan 26, 2006 #6

    Garth

    User Avatar
    Science Advisor
    Gold Member

    As I said, talking about velocity involves rulers and clocks. How do you measure 'out there' using rulers and clocks 'down here'? You have to make some definitions and stick by them, realising the differences between the different definitions.

    The cosmological problems arise because actually cosmological red shift does not measure recessional velocity but the expansion of space itself, the galaxies/quasars are being carried along with it.

    So long as you specify what formula you are using to convert there is no confusion, although some authors do not realise that such difference conventions exist, so confusion does exist.

    The particle horizon has an infinite red shift, however under most models that means the 'recessional' velocity is much greater than c. And objects just inside the particle horizon, which we can see, with the space in which they are embedded, are also receeding from us much faster than c. We can see them because the light 'catches up' with us as we 'slow down' because of cosmological deceleration.

    Garth
     
    Last edited: Jan 26, 2006
  8. Jan 27, 2006 #7
    The paper states that all galaxies with redshifts greater than z ~ 1.46 are receding superluminally. This would mean that the CMB with a redshift of z ~ 1100 would be receding at quite a considerable speed!

    Oddly enough they say it obeys Hubbles law, so would this mean the galaxies above z ~ 2 are at greater distances than previously thought?

    Also something which I didn’t quite understand was:

    Does this mean that at the time of emission 13.7 billion years ago, their recssion velocity was 58.1c but has now dropped to 3.2c?
     
  9. Nov 10, 2009 #8
    I have to agree with the question: if the universe is homogeneous and isotropic, shouldn't we be effectively at rest, since all coordinate frames are equivalent?
     
  10. Nov 10, 2009 #9
    We are mixing two different principles here.

    There is the principle that the laws of physics must be the same in all coordinate frames, which is different from saying that the universe does/does not have a preferred coordinate frame.
     
  11. Nov 11, 2009 #10
    To clarify: even though we are moving pretty fast relative to the CMB, the light is the same speed coming from all directions so there is no problem.. but I still have a little trouble with the idea that we are moving relative to the CMB- does there exist some universal momentum conservation which states that in the rest frame of the CMB, momentum of all matter is zero?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question about the CMB dipole
  1. CMB dipole direction (Replies: 2)

Loading...