Observed Events Curve - Addressing a Point of Potential Confusion

In Mathematics for Taking Another Look at the Universe, I presented this:

One of the implications of this chart might not be immediately apparent.  Look at the event at t=10,000 million years ago that can be observed today.  Back at that time, there was a proper distance between the event location and the observer location of 2,963 million light years.

To reach the observer, the photon must traverse that proper distance plus the additional distance due to expansion of space.  There are two odd things about this chart.

For the first, note that the dashed purple line that intersects with the event (x_event,t_event) and has a proper distance at t=0 of ct_event.  If an event happened at time t_event ago, then the apparent distance would be that time multiplied by the speed of light, so also ct_event.  Which means the proper distance of an observed event’s location (at time of observation) is equal to its apparent distance.  This might sound intuitively correct to some, but intuitively incorrect to others.

The space that photon traverses will be expanding the whole time and across the whole distance.  Consider when the photon has traversed half of the initial proper distance (that is to say it is at the location that would have been half of the proper distance at the time that the photon was emitted).  The space that it has traversed will have expanded to some extent already, but any further expansion will not be experienced by the photon.  Therefore, it seems reasonable to imagine that the location from which a photon is emitted will end up further away from the observer than the transit time multiplied by the speed of light.

But this is apparently not the case per the logic above.

The other oddity is that the path of a photon from an observed event 10,000 million years ago will end up further away from the observer before closing in on that observer and eventually being observed.  This does not seem correct at all.  Recall that in a FUGE universe, there is no recession greater than the speed of light which means, intuitively, that the location of observed events should be at locations greater than ct, where t is the transit time (as per the above) and the photons should never have a greater proper distance from the observation location than at the time of emission.

We clearly need to dig into these more deeply.

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Imagine a stationary emitter of some sort, sending photons in our direction for all time.  The only motion of that emitter relative to the observer is due to recession (which in turn is due to expansion). 

In this cleaned up version of the chart, we can see the dashed purple line from t=0 to t≈t₀, which plots the proper distance to the emitter from the observer location over time (I am going to refer to the proper distance to the emitter as d(t) in an attempt at clarity).  The approximate symbol is used for the latter time because as t→t₀ granularity becomes more of a problem, you can’t really distinguish between locations when the entire universe has a radius of one unit of Planck length.  When t≈t₀, the proper distance to the proper distance between of the notional locations of the hypothetical emitter and the observer is, effectively, d(t₀)=0.  At the time of observation, t=0, as discussed above, d(0)=ct_event.

Again for clarity, “proper distance” (at least how I am defining it) is the distance given by the time that it would a photon to travel from one location to another times the speed of light, if there were no expansion.  Wikipedia (paraphrasing from here) explains it like this: “The proper distance 𝑑(𝑡)d(t) between two galaxies at time t is just the distance that would be measured by rulers between them at that time.”  There are problems with both definitions, but hopefully the concept is sufficiently clear.  Compare with comoving distances, which don’t change with expansion (imagine a ruler between two locations was elastic and stretched at the same rate at which space expanded between those locations, so the measured distance would never change).

The dashed purple line can be expressed as:

d(t)=m(t-φ)

The t-intercept, φ, is merely the value of t when d(t)=0, so φ=t₀.

The gradient, m, can be found by using any two points on the line.  We will use the two that we have already discussed, in the form (0,t₀).  So, (ct_event,0) and (x₂,t₂) and:

m=(d₂-d₁)/(t₂-t₁)=-ct_event/t₀

Therefore:

d(t_event)=-ct_event/t₀.(t_event-t₀)=ct_event.(t₀-t_event)/t₀

Generalising (by removing the subscript “event”), and noting that x=ct:

d(t)=x.(t₀-t)/t₀

Multiplying through by c/c:

d(t)=x.(ct₀-ct)/ct₀=x.(ct₀-x)/ct₀=x'

So, for any observable event, the proper time to that event (at the time of the event), is as calculated and charted above.

Following the logic that every observable event is equivalent to a photon emitted at that event (ignoring energy/frequency/wavelength because we only care about observability), then it follows that a distantly emitted photon must pass through all intermediate events on the way to an observer.  And in that case, looking at the chart above, sufficiently distant photons must – at first – increase their proper distance from the observer.

The question then is … how can that be?

It’s a matter of interpretation.  We could select any event along the curve, but we will stick with an observable event that emitted a photon 10,000 million years ago.

That photon travels toward us (when the universe is such that t₀=13,800 million years), at a rate of c, from a location with a proper distance (at that time) of x'(10,000)=2,753 million light years.  Consider this to be the length of a racetrack along which the photon must travel to reach us. 

At that time, where æ is the age of the universe (we need another symbol because t in the equations above expresses time since an event), H(æ=3,800)=0.000263Mly/ly/Mly – using unusual, but significantly more convenient units for this purpose.  A location at a distance of 2,753 light years at that rate, will recess by H(3,800).x'(10,000).Δt=36.2 million light years, where Δt=t₁-t₂=50 million years.  So the photon’s position on the racetrack (after Δt=50 million years) will be 2,740 million light years away.  However, the entire racetrack along which the photon is travelling will have expanded by a factor of 1+H(3,800).Δt, which gives us a proper distance of x'(9,950)=2,775 million light years – which is a greater distance than we started with.  The relevant equation, when generalised, is:

x'(t-Δt)=(x'(t)+H(t₀-t).x'(t).Δt-c.Δt).(1+H(t₀-t₁).Δt)

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The equation above does not look at all like the equation for x' that we have discussed recently.  But it is the very same, just expressed differently.  What follows is the proof.

Using x=ct so that x'(t)=ct.(ct₀-ct)/ct₀ and H(æ)=H(t₀-t)=1/(t₀-t):

x'(t-Δt)=(ct.(ct₀-ct)/ct₀+Δt.ct.(ct₀-ct)/ct₀/(t₀-t)-c.Δt).(1+Δt/(t₀-t))

Rationalising:

x'(t-Δt)=(ct.(t₀-t)/t₀+Δt.ct/t₀-c.Δt).(1+Δt/(t₀-t))

Multiplying the blue term by t₀/t₀ and rearranging, noting that the green term above is just the blue term in a different configuration:

x'(t-Δt)=(ct.(t₀-t)/t₀.+Δt.c.(t-t₀)/t₀).(1+Δt/(t₀-t))

Multiplying the second bracket by (t₀-t)/(t₀-t):

x'(t-Δt)=(ct.(t₀-t)/t₀.+Δt.c.(t-t₀)/t₀).((t₀-t)+Δt)/(t₀-t)

Rationalising and rearranging the terms in the second pair of brackets:

x'(t-Δt)=(ct/t₀.+Δt.c/t₀).(t₀-(t-Δt))

Rearranging:

x'(t-Δt)=c.(t-Δt).(t₀-(t-Δt))/t₀

Multiplying through by c/c:

x'(t-Δt)=c(t-Δt).(ct₀-c(t-Δt))/ct₀

If x'(t-Δt)=x'₂, we could by extension say that x₂=c.(t-Δt), and thus:

x'₂=x₂.(ct₀-x₂)/ct₀

Which, while not arrived at by the most intuitive of proofs (I've tried to make it as clear as possible above), is precisely what we should expect.