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Wikipedia's Cosmic Microwave Background (CMBR) radiation monopole anisotropy (ℓ = 0) says

When ℓ = 0, the ${\displaystyle Y(\theta ,\varphi )}{\displaystyle Y(\theta ,\varphi )}$ term reduced to 1, and what we have left here is just the mean temperature of the CMB. This “mean” is called CMB monopole, and it is observed to have an average temperature of about Tγ = 2.7255 ± 0.0006 K with one standard deviation confidence. The accuracy of this mean temperature may be impaired by the diverse measurements done by different mapping measurements. Such measurements demand absolute temperature devices, such as the FIRAS instrument on the COBE satellite.

Question: So in this case is the use of the word "anisotropy" vestigial? It came along for the ride when higher order terms (especially dipole) were assigned meanings and implications? The monopole term is not really an anisotropy at all, is it?

It might be the case that monopole anisotropy has a well defined or accepted meaning, and that would be an interesting answer. It's hard to prove a negative though.

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  • $\begingroup$ Reading the (German) article by Schönitzer entitled Der Kosmische Mikrowellenhintergrund und seine Anisotropien I stumbled upon "The primary anisotropies are the effects that originated at the time of the CMB's formation, while the secondary anisotropies arose later on the photons' path through the universe." In other words, it sounds that "anisotropy" is not really an anisotropy in the mathematical sense, but just a description of the overall spatial distribution of the CMB. Any thoughts on that? $\endgroup$
    – B--rian
    Apr 1, 2021 at 7:00
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    $\begingroup$ Any example of a non-wikipedia use of the term? $\endgroup$
    – ProfRob
    Apr 1, 2021 at 7:26
  • $\begingroup$ @ProfRob If you asked me and about anisotropy: I do not have an quotable source, but I have seen many scientists from not-so-mathematical research areas using "anisotropy" rather loosely for a non-constant spatial function, e.g. for heat maps with pronounced minima and maxima more than a standard deviation from the mean. $\endgroup$
    – B--rian
    Apr 8, 2021 at 7:28
  • $\begingroup$ @uhoh Any new insights on your question yet? $\endgroup$
    – B--rian
    Apr 8, 2021 at 7:29

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