Photon outside of Hubble sphere from perspective of comoving frame

Question

I understand that there have been alot of questions on Hubble sphere, but I seek a more direct answer to this confusion that I have because I still can't resolve it after going through the posts.

Suppose a simpler model than our actual universe where the scale factor grows linearly with time. Write $r=ax$ where $r$ is physical distance, $a$ is scale factor, and $x$ is cofactor distance. I can always redefine the unit of $x$ so that $a(t)=ct$. In this case, the Hubble sphere is at $|x|=1$.

Consider what happens for a leftward photon emitted at $x_0>1$ at some time $t_0$. Looking at the physical world frame, I understand that the photon emitted at $r_0>ct_0$ will never by observed by us sitting at $r=0$, qualitatively because the expansion between us and the photon outweighs the photon travelling leftwards.

Thus, in the co-moving frame, the photon should take infinite time to reach $x=0$. However, the calculations I made doesn't seem to support this, as follows:

$$\frac{dx}{dt} = -\frac{c}{a(t)}$$ $$-dx=\frac{1}{t}dt$$ $$x_0-x=\ln t-\ln t_0$$

And therefore $x=0$ is achieved at time $t=t_0e^{x_0}<\infty$. What went wrong with my analysis?

Answer 1

The Hubble Sphere is not a horizon

You assumption that because the point of emission is receding superluminally, the photon can never be observed, is not correct. This is true for Special relativity, in a static Space, but not in General relativity. As far as I can see, your math is correct for GR, while you still think in terms of SR.

The photon is initially receding from us, that is true, but its velocity from our point of view will be lower than the local cosmological recession velocity, meaning that $\dot{x}_{phot} < 0$. As long as the resulting velocity of the photon is lower than that of the Hubble Sphere, (the limit at which the comoving coordinates recede at $v \geq c$), the latter can catch up with it, and the photon will start approaching us, first slowly, then faster as it gets to still closer regions of Space. Thus, for a region in which the cosmological recession velocity minus the velocity of the photon is smaller than the recession velocity of the Hubble sphere ($v - c < v_H$), your photon will still reach us. This is true for all decelerating and even most accelerating cosmologies (as long as the acceleration is less than exponential), it just requires that the Hubble sphere expands faster than the resulting velocity of the photon for long enough to catch it.

However, your maths seems to show that it is always a finite time. This is definitely not the case. I am not exactly sure where your calculations go wrong, but the finite photon travel time should depend on $x$ at time of emission being lower than some value, beyond which it will diverge.

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PS: As always, I recommend the good old Davis & Lineweaver for a quick but thorough overview of expansion, horizons etc.; this answer is also based on this paper.

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PPS: It just occurred to me that when redefining $x$ (the shift from first to second line in your calculations), you may also have redefined $t$ from being proper time to being conformal time, $\tau = t/a(t)$. In a conformal time frame, the scale factor goes $a(\tau) \rightarrow \infty$ for $\tau \rightarrow \tau_{\rm end}$; with $\tau_{\rm end}$ being some finite number. If my hunch here is correct, then your integral will always be finite in conformal time, but only for an interval of $x$'es will it be finite in proper time and scale factor, too.

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PPPS: Just had a conversation with the professor teaching the Cosmology class at our local department, and we concluded that the problem is that for your particular assumption of a Universe, there is no horizon - it would be infinitely old , and the observable bubble will expand indefinitely.