Distances between observers using the Lightone7 calculator

[Bandersnatch]

For a given observed object (for example, located at z=1), your calculator's outputs are for an observer at the present time, i.e., at t=13.8 Gyr, a=1, and z=0:

D(then) =5.531702 GlyD(now) = 11.06340 Gly
For the same observed object, for an observer at t=12.46 Gyr, a=0.909, and z=0.1, one may want to know what are the corresponding values of D(then) and D(now). Here are the proper distances in that case obtained from Nick Gnedin’s calculator:

D(then) = 4.83567 GlyD(now) = 8.79212 Gly

(you have your nows and thens mixed around)
You can obtain those values from the calculator, I think, but not directly. Since these are still points on the same lightcone, all it takes is to call the outputs for the desired redshifts, like so:

Then deduct the corresponding distances and multiply by the scale factor at reception (Dnow) or emission (Dthen).
E.g.
$D_{now1-0.1}=(D_{now1}-D_{now0.1})*a_{0.1}=(11.063Gly-1.395Gly)*0.90909=8.789Gly$
the remaining discrepancy likely coming from rounding errors.
I'm not sure how one would integrate this into the calculator in its current form, though. An option to choose the observer's redshift, instead of the default 0, maybe? Seems like a lot of work, and potentially confusing to the users.

 

[JimJCW]

(you have your nows and thens mixed around)
You can obtain those values from the calculator, I think, but not directly. Since these are still points on the same lightcone, all it takes is to call the outputs for the desired redshifts, like so:
View attachment 279751
Then deduct the corresponding distances and multiply by the scale factor at reception (Dnow) or emission (Dthen).
E.g.
$D_{now1-0.1}=(D_{now1}-D_{now0.1})*a_{0.1}=(11.063Gly-1.395Gly)*0.90909=8.789Gly$
the remaining discrepancy likely coming from rounding errors.
I'm not sure how one would integrate this into the calculator in its current form, though. An option to choose the observer's redshift, instead of the default 0, maybe? Seems like a lot of work, and potentially confusing to the users.

I don’t think I have my now’s and then’s mixed around; They are identical to the ones in the table you quoted.

I think the agreement between your calculated result and Gnedin’s result for z=0.1 is coincidental:

D(now1−0.1) = [D(now1)−D(now0.1)] ∗ a(0.1) = (11.063Gly−1.395Gly) ∗ 0.90909 = 8.789Gly​

You can try the case of z=0.5 to confirm it:

D(then) = 2.36642 Gly (Gnedin, z=1, 0.5)​

D(now) = 3.15523 Gly (Gnedin, z=1, 0.5)​

The option you mentioned, choosing the observer's redshift, instead of the default 0, should work. It probably will take a lot of work and potentially confuse the users. I think the calculator in the current form is extremely good already. In addition to its tabular form and chart capability, it also includes the radiation term, unlike some other calculators.

JimJCW

 

[Bandersnatch]

I don’t think I have my now’s and then’s mixed around;

You're right, of course. I misread what you wrote earlier in a manner too embarrassing to detail.

I think the agreement between your calculated result and Gnedin’s result for z=0.1 is coincidental:

D(now1−0.1) = [D(now1)−D(now0.1)] ∗ a(0.1) = (11.063Gly−1.395Gly) ∗ 0.90909 = 8.789Gly​

You can try the case of z=0.5 to confirm it:

D(then) = 2.36642 Gly (Gnedin, z=1, 0.5)​

D(now) = 3.15523 Gly (Gnedin, z=1, 0.5)​

No, it works for that case too, as it should. I'm getting the same values.
The conformal diagram shows best what is is being done here:

The observer sits where the z=0.5 line intersects the lightcone. After subtracting the comoving distances between different z's, we need to multiply them by the appropriate scale factor at a given time to get the proper distances.

 

[JimJCW]

No, it works for that case too, as it should. I'm getting the same values.

You are right. It works for z=0.5 too. The D(now)’s are distances determined by space expansion only and are related to a(z)’s only.

How about D(then)’s? We cannot use the same expression, can we? For example,

D(then1−0.1) = [D(then1)−D(then0.1)] ∗ a(0.1)​

JimJCW

 

[Bandersnatch]

How about D(then)’s? We cannot use the same expression, can we?

For D(then) you use the same distance - i.e. Dnow1-Dnow0.1 - because Dnow is the comoving distance. In my earlier post I suggested that one should use difference between Dthen, but that was wrong. You then multiply it by the scale factor at emission, a(then).
In both cases we start off with the comoving distance, i.e. in comoving terms the emitter and observer are equally distant both at emission and at reception. All we need to know to get the proper distance is what was the scale of the universe at each event.

 

[JimJCW]

For D(then) you use the same distance - i.e. Dnow1-Dnow0.1 - because Dnow is the comoving distance. In my earlier post I suggested that one should use difference between Dthen, but that was wrong. You then multiply it by the scale factor at emission, a(then).
In both cases we start off with the comoving distance, i.e. in comoving terms the emitter and observer are equally distant both at emission and at reception. All we need to know to get the proper distance is what was the scale of the universe at each event.

I thought D(now)’s and D(then)’s are both proper distances; their values are time-dependent. Please explain more about how you derive your results.

Your expression works (Thanks!), for example for the case of (z=1) + (z=0.5),

D(now1−0.5) = [D(now1) - D(now0.5)] × a(1).​

For negative redshift, for example, z = -0.1, I have to flip the sign to make it work:

D[now1−(-0.1)] = [D(now1) + D(now-0.1)] × a(1).​

Why is this so? Why D(now) and D(then) in Jorrie’s chart are reflected backward at the horizontal axis at the present time (z=0)?

JimJCW

 

[Bandersnatch]

I thought D(now)’s and D(then)’s are both proper distances; their values are time-dependent. Please explain more about how you derive your results.

The relationship between the proper and comoving distances is defined as $D_{cmv}a=D_{prop}$, where $a$ is the scale factor. At z=0, when the scale factor a=1, the proper distance is numerically identical to the comoving distance. The D(now) in the Lightcone7 calculator is the proper distance at a=1. So it also gives you the comoving distance.

When you ask the calculator to draw the D(now) vs the cosmic time, you may recognize the shape of the light cone as the second one here:

(from Lineweaver&Davis 2003)
I.e. it plots a comoving distance vs time. The first one is the same as D(then) vs time, btw.

For negative redshift, for example, z = -0.1, I have to flip the sign to make it work:

D[now1−(-0.1)] = [D(now1) + D(now-0.1)] × a(1).
Why is this so? Why D(now) and D(then) in Jorrie’s chart are reflected backward at the horizontal axis at the present time (z=0)?

If you call the D(now) graph on the calculator, you get something like this:

The 'reflection' here being just due to how the graph uses only the positive values for distance. There is no negative side of the distance axis. What it actually shows, is this:

That's because the light cone continues past the observer. I.e. we ask what is the distance between the emitter of a signal we observe now, with a given z, and the receiver of the same signal in the future, who will see it with some other z.
When one asks what is the distance between two points(1,2) on the left hand side of the 0 axis, one gets $\lvert{D1-D2}\rvert$. When the second point is on the right hand side of the graph, the total distance ends up being a sum, not a difference.

 

[JimJCW]

The relationship between the proper and comoving distances is defined as $D_{cmv}a=D_{prop}$, where $a$ is the scale factor. At z=0, when the scale factor a=1, the proper distance is numerically identical to the comoving distance. The D(now) in the Lightcone7 calculator is the proper distance at a=1. So it also gives you the comoving distance.

I understand D(proper) = a×D(comoving) and a=1 at the present time. Thanks for explaining the bouncing off of the D vs. t curves at z = 0.

I would like to know more about your two equations. Let’s use (z=1) + (z=0.5) as an example. Jorrie’s calculator gives,

z​	Scale (a)​	T Gy​	Dnow Gly​	Dthen Gly​
1.000000e+0	5.000000e-1	5.863294e+0	1.106340e+1	5.531702e+0
5.000000e-1	6.666667e-1	8.605170e+0	6.332204e+0	4.221469e+0

Gnedin’s output for Distance between two redshifts gives,

• Comoving distance: 3206.03 h-1Mlyr (By the way, what does this mean?)
• This distance at z=1: 2366.42 Mlyr
• This distance at z=0.5: 3155.23 Mlyr

Your equations work like magic:

[D(now1)−D(now0.5)] × a(1) = (11.063Gly−6.332204Gly) × 0.5 = 2.3654Gly​

[D(now1)−D(now0.5)] × a(.5) = (11.063Gly−6.332204Gly) × 0.666667 = 3.1539Gly​

I am confused by the following:

• “D(now1)−D(now0.5)” is the comoving distance between two everts. The question is between what two everts?
• “D(now1)−D(now0.5) × a(1)” is the proper distance between them at z=1.
• “D(now1)−D(now0.5) × a(0.5)” is the proper distance between them at z=0.5.

How can we equate the above two proper distances between the two events to the proper distances of the emitter at emission and at observation for an observer at z=0.5?

JimJCW

 

[Bandersnatch]

Comoving distance: 3206.03 h-1Mlyr (By the way, what does this mean?)

You mean the units?
h is Hubble constant expressed as a unitless fraction. It's $h=\frac{H_0}{100 km/s/Mpc}$. So, if $H_0$ is, say, 68 km/s/Mpc, then h is 0.68 . As far as I understand, it's used to decouple the distance measure from the uncertainty in the value of $H_0$, since all the distances are sensitive to what $H_0$ we assume to be. That way you don't have to commit to one value of $H_0$ in your work.
To get this distance in light years, you need to divide the value given, by h. So, using the value for $H_0$ that Jorrie's calculator has as the default, we get 3.20603 Gly/0.68, which is about 4.7 Gly. Same as what we get from D(now1)-D(now0.5).

• “D(now1)−D(now0.5)” is the comoving distance between two everts. The question is between what two everts?
• “D(now1)−D(now0.5) × a(1)” is the proper distance between them at z=1.
• “D(now1)−D(now0.5) × a(0.5)” is the proper distance between them at z=0.5.

How can we equate the above two proper distances between the two events to the proper distances of the emitter at emission and at observation for an observer at z=0.5?

It's the comoving distance between two world lines of observers stationary w/r to the Hubble flow, not just two events. All the events on these world lines, throughout the history of the universe, will have the same comoving distance between them.
In our case, we have the light cone intersecting the world lines of these two observers, which gives us an emission event and a reception event. We ask: what is the proper distance between the two observers at the moment of emission of a light signal (multiply the comoving distance by the scale factor at emission), and the proper distance between same at reception of this signal (multiply by $a$ at reception).
Same as when we ask what is the proper distance between us (at z=0) and the emitter of CMB (z=1089), at reception, we take the comoving distance and multiply it by the scale factor at reception (a=1) to get the familiar value of ~45 Gly. And when we want the proper distance to CMB at emission, we multiply the proper distance by ~a=0.00`092. I.e., we get the D(now) and D(then) outputs in the calculator.

 

[JimJCW]

It's the comoving distance between two world lines of observers stationary w/r to the Hubble flow, not just two events. All the events on these world lines, throughout the history of the universe, will have the same comoving distance between them.
In our case, we have the light cone intersecting the world lines of these two observers, which gives us an emission event and a reception event. We ask: what is the proper distance between the two observers at the moment of emission of a light signal (multiply the comoving distance by the scale factor at emission), and the proper distance between same at reception of this signal (multiply by a at reception).

Same as when we ask what is the proper distance between us (at z=0) and the emitter of CMB (z=1089), at reception, we take the comoving distance and multiply it by the scale factor at reception (a=1) to get the familiar value of ~45 Gly. And when we want the proper distance to CMB at emission, we multiply the proper distance by ~a=0.00`092. I.e., we get the D(now) and D(then) outputs in the calculator.

Thanks for your explanations. I cannot do reasoning using world lines and light cones yet. I am happy that your two equations work perfectly. Using Jorrie’s output and MS Excel, I can now produce needed results easily, just like adding two additional columns in Jorrie’s output.

Let me summarize what this can do for me:

Suppose photons are emitted at z=1089 (t=0.38 Myr) at various distances from our location. They will arrive at our location at various times (various z values). I need to know the relation, D(then) vs. t(observe). I spent a lot of time to produce the result using Gnedin’s calculator, but reproduced it in minutes using the new tool. The new result is more accurate because the radiation term is included in Jorrie’s calculator.​

I do have another problem and wonder whether you know a solution. Jorrie’s calculator has the following limit of z values: -0.99 < z ≤ 20000. Do you know any approximate formula for calculating event horizon at the small t (high z) limit? Or do you know a formula of relating the event horizon to small a and t, and large H(t) values?

JimJCW

 

[Bandersnatch]

Do you know any approximate formula for calculating event horizon at the small t (high z) limit? Or do you know a formula of relating the event horizon to small a and t, and large H(t) values?

As long as all you need is an approximation (quite a good one, I think) - in comoving terms, between t=0 and the emission of the CMB, the event horizon shrank by only a few percent. Even less so between t=0 and the limit of the calculator. So, what you can do, is take the comoving distance to the event horizon at t=0, which is something like 62 Gly, and multiply it by whatever $a$ you need.

 

[JimJCW]

As long as all you need is an approximation (quite a good one, I think) - in comoving terms, between t=0 and the emission of the CMB, the event horizon shrank by only a few percent. Even less so between t=0 and the limit of the calculator. So, what you can do, is take the comoving distance to the event horizon at t=0, which is something like 62 Gly, and multiply it by whatever $a$ you need.

I don’t want approximation; I need accurate R(event) values for very small a (or very large z) without integration, for example 1 second after the Big Bang, the time cosmic neutrinos became free.

I am not sure whether the following figure is what you suggested:

JimJCW

 

[Bandersnatch]

I don’t want approximation; I need accurate R(event) values for very small a (or very large z) without integration

I think in this case you either take the approximation, or deal with integration.

I am not sure whether the following figure is what you suggested:

Not this one, but similar. Just use the R(t=0) - i.e. the initial comoving extent of the event horizon (or, equivalently, the maximum comoving future extent of the particle horizon) - instead of R(t=13.8), which is the event horizon today. It will be really bad at low z, but at very high z the values should converge.

 

[JimJCW]

I think in this case you either take the approximation, or deal with integration.Not this one, but similar. Just use the R(t=0) - i.e. the initial comoving extent of the event horizon (or, equivalently, the maximum comoving future extent of the particle horizon) - instead of R(t=13.8), which is the event horizon today. It will be really bad at low z, but at very high z the values should converge.

I am lost. Shown below is a plot of particle horizon vs. t and event horizon vs. t. Both horizons have value 0 at t=0.

JimJCW

 

[Bandersnatch]

OK, look - in general, to get proper distance to any event whatsoever, you can always multiply the comoving distance to it by the scale factor at the time you're interested in. That's what we did before with the distances between observers.
Here, we can take the comoving distance to the event horizon, $D_c$ and the scale factor as our variable. So, $D_p=aD_c$. The problem is, of course, that the $D_c$ is time-variable.
But, since you're interested only in the earliest times, with z ranges below 20000, we only need a tiny fragment of the $D_c$ slope, where it's almost constant. I.e. we only need this bit:

Where above z=20000, the $D_c$ varies only between ~61-62 Glyrs (or 63, whatever the actual value is in LCDM), while the scale factor is changing by orders of magnitude, dominating the result. So for a not-bad approximation, you can assume $D_c$ to be constant*, and pick the maximal value it had at t=0. Which is the same value as that of the comoving distance to the particle horizon at infinite time, as the graph above shows.
*(over this range only; of course it will fail spectacularly if you try to graph it over the remainder of the cosmic time)
To be clear, it's no different than taking the graph you posted in #12, and tacking on a straight line from the edge of the R(event) curve to zero. The argument for its use is that for the earliest times, the missing bit of the slope is really not that much different from a straight line.

Anyway, that's the only idea I've got here.

 

[JimJCW]

Anyway, that's the only idea I've got here.

I use Jorrie’s Tutorial Part III and the following method to calculate the event horizon for values beyond z=20000 using MS Excel:

See attachment​

I got the following result:

See attachment​

One time you mentioned,

One has to stand in front of a mirror, draw the ancient sigil of the electric monkey, and repeat his name three times. @Jorrie @Jorrie @Jorrie​

How does it work and where to find such a mirror? I wonder whether Jorrie can verify this result for me. The input parameters are,

See attachment​

JimJCW

 

[Jorrie]

I use Jorrie’s Tutorial Part III and the following method to calculate the event horizon for values beyond z=20000 using MS Excel:

At such high z, the radiation energy density completely dwarfs everything else and if you simplify the equation for the cosmic horizon radius, it will indeed be just a constant times the scale factor. I get the same value as you, but it surely won't hold all the way down to a=0. Where it fails, I have no clue...

BTW, at my age one don't look in the mirror too often anymore

 

[JimJCW]

At such high z, the radiation energy density completely dwarfs everything else and if you simplify the equation for the cosmic horizon radius, it will indeed be just a constant times the scale factor. I get the same value as you, but it surely won't hold all the way down to a=0. Where it fails, I have no clue...

BTW, at my age one don't look in the mirror too often anymore

I appreciate the help received from you and Bandersnatch. Now I am having an easier time to use your calculator. Some of the results are summarized in the following:

When the cosmic neutrinos became free to move at cosmological time, t=1 s,

• t = 1 second
• a = 2.04E-10
• z = 4.89E9
• R(event horizon) = 12.8 light years

When the cosmic microwave background photons became free to move at t=0.38 Myr,

• t = 0.38 Myr
• a = 9.17 E-4
• z = 1090
• R(event horizon) = 5.67 Mly

As you know, at emission time, photons emitted outside of the event horizon will never reach the observer. Using CMB as an example, at t=0.38 Myr, photons freed outside R=5.67 Mly will never reach our location. This can cause a dwindling effect on the supply of the observable CMB photons and may spoil the Big Bang’s prediction that the blackbody nature of the CMB photon gas is persevered under the expansion of space. Your calculator and the two equations derived by Bandersnatch for ‘Distances between observers using the Lightone7 calculator’ make it easier to demonstrate the dwindling effect caused by the event horizon.

JimJCW

Correction:

When the cosmic microwave background photons became free to move at t=0.38 Myr,

• t = 0.38 Mly
• a = 9.17 E-4
• z = 1090
• R(event horizon) = 56.7 Mly

The CMB photons we are receiving now, had a proper distance of D(then) = 41.6 Mly at t = 0.38 Myr. In the future, no matter how much later, no CMB photons freed outside of D(then) = 56.7 Mly will ever reach our location, i.e., there is a dwindling effect on the supply caused by the expansion of space.