# Horizon problem: Angular size of the causality patches on the CMB surface

I'm having difficulties in calculating the angular size of the causally connected regions on the cosmic microwave background (CMB), as seen from Earth today. I read in several documents that this angle is of about $$1^{\circ}$$, but most authors are giving only crude hand waving arguments about that number. See for example that page (look at the last two paragraphs):

https://ned.ipac.caltech.edu/level5/Sept02/Kinney/Kinney4_2.html

I'm trying to reproduce that value by explicit calculations from the standard FLRW metric, in the case of a spatially flat geometry ($$k = 0$$): $$$$\tag{1} ds^2 =dt^2 - a^2(t) (dx^2 + dy^2 + dz^2).$$$$ At observation time $$t_{obs}$$ (today: $$t_{obs} \approx 13,8~\mathrm{Gyears}$$), the proper distance from a given source (emitting light at time $$t_{em} \approx 300~000~\mathrm{years}$$) is given by $$ds^2 = 0$$ (light-like spacetime intervall): $$$$\tag{2} \mathcal{D}(t_{obs}, t_{em}) = a(t_{obs}) \int_{t_{em}}^{t_{obs}} \frac{1}{a(t)} \, dt.$$$$ For example, this distance to the CMB surface, in the case of a dust universe, is found with the scale factor $$a(t) \propto t^{2/3}$$: $$$$\tag{3} \mathcal{D} = 3 \, (\, t_{obs} - t_{obs}^{2/3} \, t_{em}^{1/3}).$$$$ This gives $$\mathcal{D} \approx 40,2~\mathrm{Gly}$$ (more accurate models with radiation gives about $$42$$ or $$45~\mathrm{Gly}$$).

Now, the causally correlated regions on the CMB sphere should have a proper radius of (considering the dust only universe): $$$$\tag{4} R_{causal} = a(t_{em}) \int_0^{t_{em}} \frac{1}{a(t)} \, dt = 3 \, t_{em},$$$$ i.e. $$R_{causal} \approx 9 \times 10^5 ~ \mathrm{ly}$$. As seen from Earth, the angular size of a causal patch should have an angular size $$\alpha_{causal}$$ of: $$$$\tag{5} \alpha_{causal} = 2 \arctan{\Big( \frac{R_{causal}}{\mathcal{D}} \Big)} \approx 0.003^{\circ}.$$$$ Of course, this is much too short, and I'm probably doing a naive calculation. I don't know where I'm making a mistake.

How should I fix the angular size (5)?

EDIT: Apparently, the right formula fixing (5) is the following (the factor 2 is to get the full angular diameter, and not just the angular radius of the causal patch): $$$$\tag{6} \alpha_{causal} = 2 \arctan{\Big( \frac{\displaystyle{\int_{0}^{t_{em}} \frac{1}{a(t)} \, dt}}{\displaystyle{\int_{t_{em}}^{t_{ob}} \frac{1}{a(t)} \, dt}} \Big)},$$$$ but I don't understand why the angle is found by the ratio of the comoving lenghts instead of the proper lenghts.

• In the link below, between page 3 and 4 there's the computation, hope it helps. roe.ac.uk/ifa/postgrad/pedagogy/2006_tojeiro.pdf Sep 15 '19 at 19:41
• @Runlikehell, thanks for the link. Unfortunately, that document isn't clear about the variables used. I'm unable to convert its equations to the expressions I prefer to use (time and scale factor, instead of redshift $z$, Hubble's $H_0$ and some ill-dfefined variables in that document).
– Cham
Sep 15 '19 at 20:15
• I could find the right value if I modify (5) as the following, but I don't understand why I should use the comoving distance and size instead of the proper length: $$\alpha_{causal} = 2 \arctan{\Big( \frac{a(t_{obs}) R_{causal}}{a(t_{em}) \mathcal{D}} \Big)}.$$
– Cham
Sep 15 '19 at 20:40

At time of emission $$t_{em}$$, the length traveled by light from the Big Bang is given by (4) above. This length defines the causal patch on the CMB sphere. But then space is expending. At time of observation $$t_{ob}$$, that length becomes dilated: $$R_{causal}(t_{ob}) = \frac{a(t_{ob})}{a(t_{em})} \, R_{causal}(t_{em}) = 3 \, t_{ob}^{2/3} \, t_{em}^{1/3}.$$ Then, at time of observation, the radius angle sustained by a patch of causality is $$\alpha \approx \frac{R_{causal}(t_{ob})}{\mathcal{D}(t_{ob})} = \frac{1}{(t_{ob}/t_{em})^{1/3} - 1} \approx 1.64^{\circ}.$$
Since what we see comes from the past, the angle should be calculated using the past quantities (using (4) above and $$\mathcal{D}(t_{em}) = \frac{a(t_{em})}{a(t_{ob})} \, \mathcal{D}(t_{ob})$$): $$\alpha \approx \frac{R_{causal}(t_{em})}{\mathcal{D}(t_{em})}.$$ The result is the same.