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I believe I've found the solution to my problem. It's actually very simple.

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.

I believe I've found the solution to my problem. It's actually very simple.

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.

EDIT Some of these calculations are shown on pages 27-29 of that document:

https://dspace.mit.edu/bitstream/handle/1721.1/38370/34591655-MIT.pdf?sequence=2

I believe I've found the solution to my problem. It's actually very simple.

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.

I believe I've found the solution to my problem. It's actually very simple.

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.

I believe I've found the solution to my problem. It's actually very simple.

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 $$ \alpha \approx \frac{R_{causal}(t_{ob})}{\mathcal{D}(t_{ob})} = \frac{1}{(t_{ob}/t_{em})^{1/3} - 1} \approx 1,64^{\circ}. $$

I believe I've found the solution to my problem. It's actually very simple.

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.