B The distance to the nearest identical Hubble volume in light years?

[Cerenkov]

Hello.

I was wondering if I could please have a little help converting a double exponential that is expressed in metres into light years?

My math simply isn't up to the task.

https://en.wikipedia.org/wiki/Multiverse

Level I: An extension of our universe​

A prediction of cosmic inflation is the existence of an infinite ergodic universe, which, being infinite, must contain Hubble volumes realizing all initial conditions.
Accordingly, an infinite universe will contain an infinite number of Hubble volumes, all having the same physical laws and physical constants. In regard to configurations such as the distribution of matter, almost all will differ from our Hubble volume. However, because there are infinitely many, far beyond the cosmological horizon, there will eventually be Hubble volumes with similar, and even identical, configurations. Tegmark estimates that an identical volume to ours should be about 1010115 meters away from us.
Given infinite space, there would, in fact, be an infinite number of Hubble volumes identical to ours in the universe. This follows directly from the cosmological principle, wherein it is assumed that our Hubble volume is not special or unique.The reason I ask this question is that descriptions of distances within our observable universe in the popular science publications that I can read and understand are usual given in light years.

I suspect that the act of my trying to mentally compare distances within our Hubble volume to the immensities that might lie beyond will be (ahem!) difficult.

Therefore, is it possible to make the comparison between these two different scales of distance if it were expressed as a percentage?

Say, the diameter of our observable universe is (insert percentage) of the distance to the nearest identical Hubble volume?

Or another way that might be more appropriate or informative?

My thanks in advance for any help given.

Cerenkov.

 

[Ibix]

$10^{10^{115}}$ is a 1 with $10^{115}$ zeroes after it. One light year is approximately $10^{16}\mathrm{m}$. So the given distance in meters becomes a one with $10^{115}-16$ zeroes after it in light years.

 

[Ibix]

And our universe is on the order of $10^{11}\mathrm{ly}$ across, so in terms of multiples of that, it's a one with $10^{115}-27$ zeroes.

That is, this distance is so far beyond the scale of anything we can ever observe that your choice of units is pretty much a rounding error.

 

[Jaime Rudas]

/

https://en.wikipedia.org/wiki/Multiverse

Level I: An extension of our universe​

[...]
However, because there are infinitely many, far beyond the cosmological horizon, there will eventually be Hubble volumes with similar, and even identical, configurations. Tegmark estimates that an identical volume to ours should be about 1010115 meters away from us.

There it seems that Wikipedia confuses the observable universe with the Hubble volume, however, the error comes from Tegmark himself:

https://arxiv.org/pdf/astro-ph/0302131.pdf

The most distant visible objects are now about 4×10²⁶ meters away, and a sphere of this radius defines our observable universe, also called our Hubble volume, our horizon volume or simply our universe.

Actually, the Hubble sphere has a radius of 1.4×10²⁶ m while the observable universe has a radius of 4.4×10²⁶ m.

https://en.wikipedia.org/wiki/Hubble_volume

https://en.wikipedia.org/wiki/Observable_universe

 

[Cerenkov]

$10^{10^{115}}$ is a 1 with $10^{115}$ zeroes after it. One light year is approximately $10^{16}\mathrm{m}$. So the given distance in meters becomes a one with $10^{115}-16$ zeroes after it in light years.

Thank you Ibix.

So that would be 10,000,000,000,000,000 light years?

 

[Cerenkov]

And our universe is on the order of $10^{11}\mathrm{ly}$ across, so in terms of multiples of that, it's a one with $10^{115}-27$ zeroes.

That is, this distance is so far beyond the scale of anything we can ever observe that your choice of units is pretty much a rounding error.

So the distance to the nearest identical Hubble volume is 1,000,000,000,000,000,000,000,000,000 times the diameter of our observable universe?

 

[Ibix]

Thank you Ibix.

So that would be 10,000,000,000,000,000 light years?

No. It's a 1 with 9,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,984 zeroes after it. That many light years. Assuming I counted the nines right.

 

[Cerenkov]

https://en.wikipedia.org/wiki/Hubble_volume

https://en.wikipedia.org/wiki/Observable_universe

Could you please explain this a little further for someone with a Basic level of understanding, Jaime?


In cosmology, a Hubble volume (named for the astronomer Edwin Hubble) or Hubble sphere, Hubble bubble, subluminal sphere, causal sphere and sphere of causality is a spherical region of the observable universe surrounding an observer beyond which objects recede from that observer at a rate greater than the speed of light due to the expansion of the universe.The observable universe is a ball-shaped region of the universe comprising all matter that can be observed from Earth or its space-based telescopes and exploratory probes at the present time; the electromagnetic radiation from these objects has had time to reach the Solar System and Earth since the beginning of the cosmological expansion.

I realise from the Wiki links you supplied that our Hubble volume must be smaller than the Observable universe, but I need a bit more clarification please.

Up to this point I seem to have been using the two terms interchangeably.

I believe that I understand how the diameter of the observable universe is calculated. This is done by taking the maximum distance light could have travelled since the Big Bang and then factoring in the expansion of space since that event. Is that correct?

But I'll need some help with the Hubble volume.

If objects beyond the limit of the Hubble volume are seen to be receding at a rate greater than the speed of light then how can regions beyond that be said to be within our observable universe?

At first glance it would seem that these objects are unobservable.

No doubt there are subtleties and nuances that are eluding me.

Could you help please?Thank you,

Cerenkov.

 

[Cerenkov]

No. It's a 1 with 9,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,984 zeroes after it. That many light years. Assuming I counted the nines right.

Ok thanks.

But I'm confused. Your second sentence contained two different units of measurement - metres and light years. I can't seem to write standard form in my posts but you wrote that 'the given distance in metres becomes a one with 10 to the power of 115'.

Then you followed that with the distance in light years saying '16 zeroes after it in light years'.

I assumed that you couldn't possibly mean that the distance was a one with 10 to the power of 115 with an additional 16 zeroes after that, because that would be mixing two different types of unit metres and light years.

Therefore, I took what you said about the number of zeroes (16) and placed them after a 1. Yielding my original figure of 10,000,000,000,000,000 light years.

So my confusion is twofold. First, I don't understand how you intended your initial reply to be read and second I don't understand how you derived the figure with all of those nines, ending in 984.

Could you please explain both of these things?

Thank you,Cerenkov.

 

[Cerenkov]

Hmmm...

Looking back through similar threads I see that Davis and Lineweaver's paper is often referred to.

https://arxiv.org/abs/astro-ph/0310808

Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe​

Should I go there to gain a better understanding of the difference between what is meant by the Hubble volume and the Observable universe?

And if I do is it ok to bring questions arising from my reading of that paper to this thread?Thank you,

Cerenkov.

 

[PeterDonis]

Should I go there to gain a better understanding of the difference between what is meant by the Hubble volume and the Observable universe?

That would be a good place to look, yes. That paper is an excellent reference.

if I do is it ok to bring questions arising from my reading of that paper to this thread?

Certainly!

 

[Cerenkov]

Then I do believe that I will do just that, Vanadium50.

As well as continuing to engage with Ibix and Jaime Rudas, here in this thread.

Seeing as they have been kind enough to help me out.Thank you,

Cerenkov.

 

[Jaime Rudas]

But I'll need some help with the Hubble volume.

If objects beyond the limit of the Hubble volume are seen to be receding at a rate greater than the speed of light then how can regions beyond that be said to be within our observable universe?

At first glance it would seem that these objects are unobservable.

No doubt there are subtleties and nuances that are eluding me.

Could you help please?

Maybe it will help you understand if you look at it this way:

We agree that we will be able to observe anything within the Hubble volume, right?

Suppose the Hubble radius is exactly 1.4×10²³ km.
Suppose the speed of light is exactly 300,000 km/s.
Suppose object A is located one kilometer beyond the Hubble radius (that is, at [1.4×10²³ + 1] km). That object, being beyond the Hubble radius, would be receding faster than the speed of light.
Suppose A emits a ray of light toward us at time T.
At time T+1 s, the ray would have already traveled 300,000 km and would therefore be about (1.4×10²³ - 299,999) km from the Earth.
The above means that this ray of light would already be within the Hubble volume and, therefore, we will be able to see it.

 

[Vanadium 50]

I will do just that, Vanadium50.

?

 

[PeterDonis]

Then I do believe that I will do just that, Vanadium50.

Are you responding to my post #11? You might want to double check who wrote it.

 

[Ibix]

Your second sentence contained two different units of measurement - metres and light years.

It doesn't - it's just really hard to write unambiguously about numbers this large without using standard form. I'll try again. Let's call the diameter of the observable universe $U$. Presumably you're happy that $10^N$ is a 1 followed by $N$ zeroes, and with the unit conversions $1\mathrm{ly}=10^{16}\mathrm{m}$ and $U=10^{11}\mathrm{ly}=10^{27}\mathrm{m}$, yes?

Now let's write about a big distance which is still much smaller than Tegmark's distance - $10^{30}\mathrm{m}$. $$\begin{eqnarray*}
10^{30}\mathrm{m}&=&1,000,000,000,000,000,000,000,000,000,000\mathrm{m}\\
&=&\left(1\text{ with 30 zeroes after it}\right)\mathrm{m}
\end{eqnarray*}$$And let's re-express that in light years:$$\begin{eqnarray*}
10^{30}\mathrm{m}&=&\frac{10^{30}\mathrm{m}}{10^{16}\mathrm{m/ly}}\\
&=&10^{(30-16)}\mathrm{ly}\\
&=&10^{14}\mathrm{ly}\\
&=&100,000,000,000,000\mathrm{ly}\\
&=&\left(1\text{ with 14 zeroes after it}\right)\mathrm{ly}
\end{eqnarray*}$$And let's repeat the exercise in terms of $U$:$$\begin{eqnarray*}
10^{30}\mathrm{m}&=&\frac{10^{30}\mathrm{m}}{10^{27}\mathrm{m}/U}\\
&=&10^{(30-27)}U\\
&=&10^{3}U\\
&=&1,000U\\
&=&\left(1\text{ with 3 zeroes after it}\right)U
\end{eqnarray*}$$OK so far?

Now let's try the distance Tegmark referred to, which is $10^{10^{115}}\mathrm{m}$. Spoiler: it's going to be exactly the same as above, except that everywhere I wrote 30 I'm now going to write $10^{115}$. $$\begin{eqnarray*}
10^{10^{115}}\mathrm{m}&=&\left(1\text{ with }10^{115}\text{ zeroes after it}\right)\mathrm{m}
\end{eqnarray*}$$And again in light years:$$\begin{eqnarray*}
10^{10^{115}}\mathrm{m}&=&\frac{10^{10^{115}}\mathrm{m}}{10^{16}\mathrm{m/ly}}\\
&=&10^{\left(10^{115}-16\right)}\mathrm{ly}\\
&=&\left(1\text{ with }\left(10^{115}-16\right)\text{ zeroes after it}\right)\mathrm{ly}
\end{eqnarray*}$$And in terms of $U$:$$\begin{eqnarray*}
10^{10^{115}}\mathrm{m}&=&\frac{10^{10^{115}}\mathrm{m}}{10^{27}\mathrm{m}/U}\\
&=&10^{\left(10^{115}-27\right)}U\\
&=&\left(1\text{ with }\left(10^{115}-27\right)\text{ zeroes after it}\right)U
\end{eqnarray*}$$And of course, $10^{115}-16$ is 9, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 983, and that's the number of zeroes in the distance Tegmark calculated if you express it in light years. And that's so close to $10^{115}$ that you'd just round it off anyway, right?

So the basic answer to your question is that $10^{10^{115}}\mathrm{m}$ is not materially different from $10^{10^{115}}$ in any other unit of measure you care to use, be it Planck lengths ($10^{\left(10^{115}+44\right)}l_P$) or the diameter of the observable universe ($10^{\left(10^{115}-27\right)}U$). It's just so big that the change in the number of zeroes gets lost in the rounding.

 

[Cerenkov]

Are you responding to my post #11? You might want to double check who wrote it.

Oops! Sorry about that Peter and Vanadium50.

I can see what happened.

At the bottom of the page there's the Suggested For... section where similar threads are offered. I looked over some of those and then remembered that the Davis / Lineweaver paper was relevant. The name Vanadium50 must have cropped up in my searching and I unconsciously carried it back to this thread.Cerenkov.

 

[Cerenkov]

Maybe it will help you understand if you look at it this way:

We agree that we will be able to observe anything within the Hubble volume, right?

Hmmm... I can't in all honesty agree with that sentence, Jaime.

Not because I actively disagree with it or with anything else you've written here but because my understanding of what can or cannot be observed within the Hubble volume isn't up to the task.

That is why I'm going to the Davis / Lineweaver paper to rectify my errors and solidify my understanding.

Suppose the Hubble radius is exactly 1.4×10²³ km.
Suppose the speed of light is exactly 300,000 km/s.
Suppose object A is located one kilometer beyond the Hubble radius (that is, at [1.4×10²³ + 1] km). That object, being beyond the Hubble radius, would be receding faster than the speed of light.
Suppose A emits a ray of light toward us at time T.
At time T+1 s, the ray would have already traveled 300,000 km and would therefore be about (1.4×10²³ - 299,999) km from the Earth.
The above means that this ray of light would already be within the Hubble volume and, therefore, we will be able to see it.

Ok, I can see the logic of your argument, Jaime.

As far as I can see this would mean that beyond the cited Hubble radius there's a region containing galaxies that can emit rays of light in our direction which we will be able to observe.

So I envisage a pair of nested spheres centred upon us, the observers. But this would also hold true for any other observers located anywhere else in the entire universe. Because our location is not considered to be in any way special.

The inner sphere is the Hubble volume and beyond the outer sphere is where your argument ceases to hold good. Beyond that outer sphere light emitted from any galaxies there cannot be observed by us. Without consulting Davis / Lineweaver or any other source, I would suppose that this second sphere would constitute what we call our Observable Universe. Is that right?Thank you,

Cerenkov.

 

[PeroK]

If future observations show our universe to be large but finite, then all these calculations and all the speculation about an infinite number of identical Earths across the universe will seem rather farcical.

It's also doubtful that we will ever have data to verify a prediction on this scale. Even if they are backed by a valid mathematical model.

 

[Cerenkov]

It doesn't - it's just really hard to write unambiguously about numbers this large without using standard form. I'll try again. Let's call the diameter of the observable universe $U$. Presumably you're happy that $10^N$ is a 1 followed by $N$ zeroes, and with the unit conversions $1\mathrm{ly}=10^{16}\mathrm{m}$ and $U=10^{11}\mathrm{ly}=10^{27}\mathrm{m}$, yes?

Well, as you can see Ibix, the quotation function hasn't done justice to your efforts.
However, skipping back and forth between your post and this reply I will endeavour to follow things.

Yes, I tentatively agree that 10 ^N is a 1 followed by N zeroes.
Yes, I tentatively agree your value of U in light years and metres.

Now let's write about a big distance which is still much smaller than Tegmark's distance - $10^{30}\mathrm{m}$. $$\begin{eqnarray*}

Yes, I tentatively agree that a 1 with 30 zeroes expresses the standard form of 10³⁰
.
I'm going to break my reply down into manageable chunks Ibix, rather than risk it becoming overlong.

More to follow.

 

[Cerenkov]

If future observations show our universe to be large but finite, then all these calculations and all the speculation about an infinite number of identical Earths across the universe will seem rather farcical.

It's also doubtful that we will ever have data to verify a prediction on this scale. Even if they are backed by a valid mathematical model.

I agree PeroK.

But just look at the fruit my initial question has given rise to.

I was labouring under the misapprehension that the Hubble volume and the Observable Universe were one and the same. Now I see that is not so and want to know how and why. So my question has gifted me with something new to discover and understand. This is an unlooked for development and benefit.

Yes, my question was about highly speculative matters and if future observations do detect a positive curvature in the shape of the universe then all talk about infinities and identical earths will indeed become farcical.

But for the moment, such speculation has been helpful to me and so I'm glad I asked my question.Thank you,

Cerenkov.

 

[Cerenkov]

\end{eqnarray*}$$And let's re-express that in light years:$$\begin{eqnarray*}
10^{30}\mathrm{m}&=&\frac{10^{30}\mathrm{m}}{10^{16}\mathrm{m/ly}}\\
&=&10^{(30-16)}\mathrm{ly}\\
&=&10^{14}\mathrm{ly}\\
&=&100,000,000,000,000\mathrm{ly}\\
&=&\left(1\text{ with 14 zeroes after it}\right)\mathrm{ly}
\end{eqnarray*}

Yes, I can see the workings of your math, Ibix.

$$And let's repeat the exercise in terms of $U$:$$\begin{eqnarray*}
10^{30}\mathrm{m}&=&\frac{10^{30}\mathrm{m}}{10^{27}\mathrm{m}/U}\\
&=&10^{(30-27)}U\\
&=&10^{3}U\\
&=&1,000U\\
&=&\left(1\text{ with 3 zeroes after it}\right)U

I seem to see you concluding this calculation with a figure representing 1,000 times the diameter of U, the observable universe. Which, according to my current understanding would 1,000 times 93 billion light years, the generally accepted diameter of the observable universe.

\end{eqnarray*}$$OK so far?

Tentatively, yes.

 

[Cerenkov]

Now let's try the distance Tegmark referred to, which is $10^{10^{115}}\mathrm{m}$. Spoiler: it's going to be exactly the same as above, except that everywhere I wrote 30 I'm now going to write $10^{115}$. $$\begin{eqnarray*}
10^{10^{115}}\mathrm{m}&=&\left(1\text{ with }10^{115}\text{ zeroes after it}\right)\mathrm{m}
\end{eqnarray*}$$And again in light years:$$\begin{eqnarray*}

Accepted.

10^{10^{115}}\mathrm{m}&=&\frac{10^{10^{115}}\mathrm{m}}{10^{16}\mathrm{m/ly}}\\
&=&10^{\left(10^{115}-16\right)}\mathrm{ly}\\
&=&\left(1\text{ with }\left(10^{115}-16\right)\text{ zeroes after it}\right)\mathrm{ly}

Accepted.

\end{eqnarray*}$$And in terms of $U$:$$\begin{eqnarray*}
10^{10^{115}}\mathrm{m}&=&\frac{10^{10^{115}}\mathrm{m}}{10^{27}\mathrm{m}/U}\\
&=&10^{\left(10^{115}-27\right)}U\\
&=&\left(1\text{ with }\left(10^{115}-27\right)\text{ zeroes after it}\right)U
\end{eqnarray*}$$

Accepted.

 

[Cerenkov]

And of course, $10^{115}-16$ is 9, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 983, and that's the number of zeroes in the distance Tegmark calculated if you express it in light years. And that's so close to $10^{115}$ that you'd just round it off anyway, right?

Accepted and agreed.

So the basic answer to your question is that $10^{10^{115}}\mathrm{m}$ is not materially different from $10^{10^{115}}$ in any other unit of measure you care to use, be it Planck lengths ($10^{\left(10^{115}+44\right)}l_P$) or the diameter of the observable universe ($10^{\left(10^{115}-27\right)}U$). It's just so big that the change in the number of zeroes gets lost in the rounding.

Aha! I had an inkling that comparing Tegmark's cited distance to the nearest identical Earth to any kind of measurement we use inside our observable universe was going to (ahem!) difficult.

That was the phrase I used Ibix... but little did I know (or envisage) just how ridiculous the idea of making this comparison would be. Now I can see why PeroK said what he did.

But I need to thank you personally for putting in the effort here and for helping me understand the process of the calculation more clearly.

I also need to thank Peter Donis and Jaime for their help too.

As I said to PeroK, this thread has spawned an unexpected side-benefit for me. Better understanding and further research into the differences between the Hubble volume and the Observable universe. My next stop will be that Davis / Lineweaver paper.

No doubt I will have questions about it and I look forward to bring them here to continue learning from you guys.Many thanks,

Cerenkov.

 

[Jaime Rudas]

That is why I'm going to the Davis / Lineweaver paper to rectify my errors and solidify my understanding.

Before consulting the Davis & Lineweaver paper, I would recommend that you read this:

https://people.smp.uq.edu.au/TamaraDavis/papers/SciAm_BigBang.pdf

As far as I can see this would mean that beyond the cited Hubble radius there's a region containing galaxies that can emit rays of light in our direction which we will be able to observe.

So I envisage a pair of nested spheres centred upon us, the observers. But this would also hold true for any other observers located anywhere else in the entire universe. Because our location is not considered to be in any way special.

The inner sphere is the Hubble volume and beyond the outer sphere is where your argument ceases to hold good. Beyond that outer sphere light emitted from any galaxies there cannot be observed by us. Without consulting Davis / Lineweaver or any other source, I would suppose that this second sphere would constitute what we call our Observable Universe. Is that right?

Yes, that´s right.

 

[Cerenkov]

Thank you Jaime.

I will check out that link soon.Cerenkov.

 

[syfry]

If future observations show our universe to be large but finite, then all these calculations and all the speculation about an infinite number of identical Earths across the universe will seem rather farcical.

What's funny about such thought experiments, is that even if Earths would repeat... unless I'm mistaken about how the probabilities would work out, we'd be much more likely to encounter many more alternate histories of Earth before we'd encounter an exact match. Simply because the odds of an exact match would be greater.

Heck, I'm not sure if it's even possible for quantum randomness to replicate on such a scale for an identical planet, since the possibilities of each state seems infinite.

And not only would the Earth have to be the same, but its sun too, and its moon, and all the asteroids that impacted, plus the supernova that created the sun's nursery, so then the galaxy too, and everything that went into making it.

All seems way too far fetched.

 

[Jaime Rudas]

What's funny about such thought experiments, is that even if Earths would repeat... unless I'm mistaken about how the probabilities would work out, we'd be much more likely to encounter many more alternate histories of Earth before we'd encounter an exact match. Simply because the odds of an exact match would be greater.

Greater?

 

[syfry]

Yes, greater odds. But after reading my reply again, thought of even more things that might take the odds beyond any chance of possibility:

All of the galaxies and superclusters that we've discovered would have to match up in order for the Earth to be identical in history, and, the filament structure of galaxies, plus, all of the galaxies in the Hubble deep field images.

In order to repeat the Earth exactly, the entire visible universe would have to repeat. And the angle at which James Webb has captured the earliest galaxies.

If we found the identical Earth, then it would have to find us at the same time, by the same person, in order for it to be identical. But, then, that would probably mean we arrived from different directions, therefore breaking the chance for an identical scenario.

 

[PeterDonis]

In order to repeat the Earth exactly, the entire visible universe would have to repeat.

Yes. And if our universe is spatially infinite, this will be the case. In fact, our entire visible universe will repeat an infinite number of times. The very, very, very large number given in post #7 is the best estimate of how far away the closest exact copy of our visible universe is if our universe is spatially infinite.

If we found the identical Earth, then it would have to find us at the same time, by the same person, in order for it to be identical.

Yes, that's all included in the statement that our entire visible universe repeats.

But, then, that would probably mean we arrived from different directions, therefore breaking the chance for an identical scenario.

I don't know what you mean by "arrived from different directions", but the "repeats" claim is that our exact visible universe repeats. Everything in the repeat is identical.

All seems way too far fetched.

I don't think you have fully understood the argument, or fully grasped the implications of our universe being truly spatially infinite. I strongly suggest that you think very carefully before posting any further.