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We say the Earth is in relative motion with respect to the cosmic microwave background (CMB), causing anisotropies in the CMB spectrum. I have four very simple questions about this.

  1. How is it possible to treat the CMB, a bath of blackbody photons in space, as a reference frame? In particular, this frame must be at rest with respect to something. What is that thing?

  2. What do we mean by anisotropies in a blackbody spectrum?

  3. Why should our motion cause anisotropy, and in particular, dipole anisotropy?

  4. How do we quantitatively express the anisotropy?

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  • $\begingroup$ have a look here forbes.com/sites/startswithabang/2017/06/16/… $\endgroup$ – anna v Jun 10 '18 at 17:21
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    $\begingroup$ (2) A single photon does not have a rest frame, but two or more do, as long as they are not all in the same direction. (2) Not just any anisotropy, but specifically the dipole anisotropy. It is a red/blue shift of the BB radiation. (3) Doppler effect. $\endgroup$ – safesphere Jun 10 '18 at 17:53
  • $\begingroup$ (1) "A single photon does not have a rest frame, but two or more do, as long as they are not all in the same direction." This part I fail to understand. Rest w.r.t what? Which velocity w.r.t what is zero? (2) "Not just any anisotropy, but specifically the dipole anisotropy. It is a red/blue shift of the BB radiation." Photons coming from one direction are blue shifted and opposite direction are redshifted. What will be the effect of BB distribution in the direction in which photons are blue/red shifted? @safesphere $\endgroup$ – SRS Jun 11 '18 at 6:20
  • $\begingroup$ The rest frame of two or more photons is the frame, in which the total momentum is zero. Two photons also have an invariant rest mass: arxiv.org/pdf/0708.4289.pdf . The whole BB spectrum is just Doppler shifted, blue in one direction, red in the other. From this the speed is easily calculated: astronomy.swin.edu.au/cosmos/C/… $\endgroup$ – safesphere Jun 11 '18 at 6:45
  • $\begingroup$ "the whole BB spectrum is just Doppler shifted, blue in one direction, red in the other. " What's the effect on BB spectrum in that case? Does it remain a Planckian distribution but with an altered temperature?@safesphere $\endgroup$ – SRS Jun 11 '18 at 6:48
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Individual photons certainly don't have a rest frame. However, there is a rest frame in which the CMB is almost perfectly isotropic (the deviations from a perfect blackbody spectrum are of the order of 1 part in 100,000), and for convenience we call that the rest frame of the CMB.

That frame is essentially the rest frame of the plasma which emitted the CMB, i.e. the surface of last scattering, adjusted for the Hubble flow.

Our motion causes anisotropy through simple Doppler shifting: the CMB photons coming from the direction we're currently heading towards get blueshifted, the photons in the opposite direction get redshifted.

The Earth's velocity with respect to that frame is a little complicated, because we're orbiting the Sun, which is orbiting within the galaxy, which has its own motion in the local group, etc. Of course all of those motions are operating at different time scales, and different speeds. The shortest period effect is of course due to our orbit around the Sun, but our orbit speed is pretty sedate compared to the other motions I mentioned. So there's noticeable annual variation in the exact amount and location of the anisotropy, but the low period high velocity motions are the major factors controlling the anisotropy.


This famous image (from Wikipedia)

CMB

shows the CMB after it's had the anisotropy corrected. The 1 in 100,000 parts variations I mentioned above are amplified enormously, otherwise the image would look totally uniform. This amplification can only be done after the anisotropy compensation, otherwise the anisotropy would totally dominate the image.

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  • $\begingroup$ If there exists a frame in which the CMB is isotropic, wouldn't the CMB also be isotropic in any frame that is rotating (but not translating) with respect to that frame? How do we choose which of these mutually rotating frames is the rest frame? $\endgroup$ – Tom Jun 10 '18 at 20:25
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    $\begingroup$ @Tom The CMB has structure, so you can tell if you're rotating relative to it. According to Is the universe rotating? the CMB as a whole isn't rotating (that would also give rise to Doppler shifts), they constrain the rotation to less than $10^{-9}$ radians per year at the last scattering surface. $\endgroup$ – PM 2Ring Jun 10 '18 at 23:25
  • $\begingroup$ @PM2Ring What is that frame w.r.t which the CMB is at rest? Also, it is unclear to mean what would it mean to say the "CMB is at rest" because as you said individual photons can't be at rest. What do you mean by the CMB, a photon bath, to be at rest? And w.r.t what? Which relative velocity is zero? $\endgroup$ – SRS Jun 11 '18 at 6:27
  • $\begingroup$ @SRS The Individual photons of the CMB have no rest frame, because they're photons, but the whole collection of CMB photons arriving at Earth at any given moment can be assigned a rest frame in which the sum of the momenta of those photons is zero. It may help to instead consider the rest frame of the spherical surface which emitted those photons 13+ billion years ago. That surface has expanded considerably since then, but that has no effect on the photons once they've been emitted, although the photons have been redshifted by their journey through expanding space. $\endgroup$ – PM 2Ring Jun 11 '18 at 6:48
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    $\begingroup$ @SRS I just noticed you've added a 4th question to your question; I've added a little more info to my answer. $\endgroup$ – PM 2Ring Jun 11 '18 at 7:07
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First of all, how is it possible to think CMB, a bath of blackbody (BB) photons in space, as a reference frame? Doesn't quite match my Newtonian notion of reference frames.

But it does. The references frame of $X$ is one where $X$ has zero momentum, and a photon bath has a measurable momentum and by choosing the right frame that can be made to be zero.

What do we mean by anisotropies in a BB spectrum? Tiny departures from BB distribution?

and

Why should our motion cause anisotropy?

Isotropy is a condition of being the same in every direction. From the point of view of observers not in the frame of the CMB, photons from one direction are blue shifted, those from the opposite direction are red shift1, and those not on the line of relative motion are partially Doppler shifted and exhibit angular aberration (search term "Lorentz focusing").

All of these effect make the view different in different directions.


1 Of course different wavelength means different momentum, so these effects are linked to being in frame where the CMB has a different net momentum.

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  • $\begingroup$ "a photon bath has a measurable momentum" By this, you certainly don't mean momentum of individual photons. Then what momentum do you refer here? @dmckee $\endgroup$ – SRS Jun 11 '18 at 6:32
  • $\begingroup$ @SRS: Photons most certainly do have momentum. Just solve E = pc for p. $\endgroup$ – Kevin Jun 11 '18 at 6:46
  • $\begingroup$ @Kevin I did not say that photons don't have momentum. $\endgroup$ – SRS Jun 11 '18 at 6:50
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    $\begingroup$ @SRS: So take the sum... $\endgroup$ – Kevin Jun 11 '18 at 15:13
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If you are in motion with respect to the CMB, you'll see the photons in front of you blueshifted a bit, and the ones behind you redshifted. We do see this, and the Earth appears to be moving ~600 km/s with respect to the CMB.

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  • $\begingroup$ "If you are in motion with respect to the CMB" What do you mean by being in motion with CMB? What do we mean by CMB being at rest? CMB is a bath of photons. Individual photons can't be at rest. So what does it mean for the CMB to be at rest w.r.t an observer and which observer? @HiddenBabel $\endgroup$ – SRS Jun 11 '18 at 6:23
  • $\begingroup$ It means that when the CMB was created, it was isotropic. All directions should give us photons of the same momentum when we are "at rest" with respect to it. $\endgroup$ – HiddenBabel Jun 11 '18 at 16:54

protected by Qmechanic Jun 10 '18 at 19:18

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