Undergrad Why is inflation initial patch length ~ 1 Hubble length ?

[DoobleD]

I'm asking many questions on inflation, I hope this is the last one...

If inflation is correct, our observable Universe would have been a tiny homogeneous patch before inflation started (if it started at all). The length of that initial patch is estimated to be of the order of 1 Hubble length at the time before inflation (see for instance http://www.emu.dk/sites/default/files/guth_inflation.pdf, and 1992 paper, or this video lecture). This turns out to be something like 10^-26 cm or 10^-28 cm, that kind of crazy size, depending on the sources.

Why is the initial patch length estimated to be of the order of 1 Hubble length at that time ? I understand that a patch of that size is causally connected, providing homogeneity. But when we talk of the horizon distance problem of the standard Big Bang, we use the particle horizon as the criteria for causal connection, not the Hubble distance.

 

[George Jones]

But when we talk of the horizon distance problem of the standard Big Bang, we use the particle horizon as the criteria for causal connection, not the Hubble distance.

Section 2.2.2 "Hubble Radius vs. Particle Horizon" from Baumann's notes might (or might not) answer your questions about this.

 

[DoobleD]

Section 2.2.2 "Hubble Radius vs. Particle Horizon" from Baumann's notes might (or might not) answer your questions about this.

Thank you, that reading does address the issue. However I have the impression he just basically says that we use the Hubble radius simply because it's easier than using the particle horizon. Am I understanding it correctly ?

Since the Hubble radius is easier to calculate than the particle horizon it is common to use the Hubble radius as a means of judging the horizon problem. If the entire observable universe was within the comoving Hubble radius at the beginning of inflation—i.e. (a_IH_I)⁻¹ was larger than the comoving radius of the observable universe (a₀H₀)⁻¹—then there is no horizon problem. Notice that this is more conservative than using the particle horizon since χ_ph(t) is always bigger than (aH)⁻¹(t). Moreover, using (a_IH_I)⁻¹ as a measure of the horizon problem means that we don’t have to assume anything about earlier times t < t_I .​