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.

 

[PeterDonis]

take the odds beyond any chance of possibility

If our universe is truly spatially infinite, then no odds are "beyond any chance of possibility", no matter how small. Again, I don't think you have fully grasped the implications of "infinite".

 

[syfry]

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.

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.

Yeah, I did consider that after thinking about it more. I do work a job and am often in a hurry when posting so might catch things only while thinking about it afterward. Thanks for catching that! Can you address the part about quantum repeats? Is the wavefunction of particle positions supposed to be infinite, and, if so, even in an infinite universe, can every possibility truly repeat in that case?

Also, it is true that odds are you'd probably encounter countless numbers of different Earths with alternate histories on the way to finding the identical Earth. If the odds would work out that way. Please correct any inaccuracy in that.

And, what I meant by the directions. Well, to be an identical Earth, it would have to send an explorer out searching for an identical Earth in the same way you went out searching. Does that mean you'd cross paths? And does that mean the path to and from each Earth would be mirror images?

Something should differ, the paths wouldn't be symmetrical is what I was getting at but didn't quite word it right. Fair warning that I'm in a hurry again to sleep for another job early in the morning, so might've worded something unclear in the rush.

 

[Jaime Rudas]

Well, to be an identical Earth, it would have to send an explorer out searching for an identical Earth in the same way you went out searching. Does that mean you'd cross paths? And does that mean the path to and from each Earth would be mirror images?

No, they don't have to be mirror images. Suppose E(0) is Earth and E(1) is the nearest identical Earth in the north direction. If you send an explorer, N(0), north towards E(1), then your identical twin at E(1) will send an explorer, N(1), north towards E(2), and your identical twin at E(-1) will send an explorer, N(-1), north towards E(0).

This scenario is entirely indistinguishable from a closed and finite universe.

 

[PeroK]

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.

The basic idea is that there are only finitely many configurations for a region of the universe of fixed volume. You then have the question of whether identical means instantaneously identical; or, has remained continuously identical since the Big Bang; or, additionally, is destined (in some sense) to remain identical for the lifetime of the universe. If nature is genuinely non-deterministic (in terms of QM, for example), then even two systems that are instantaneously identical will tend to diverge.

This all makes mathematical sense, although whether that mathematics is applicable to our universe is a moot point. There are a number of key assumptions that need to hold.

For one thing, if spacetime is a continuum with infinitely many points, you could argue that there are an (uncountably) infinite number of configurations and an (uncountably) infinite number of ways for a system to evolve over time.

For example: there are only finitely many chess positions, as long as a piece anywhere on a given square counts as the same position. And, the game evolves in a well-defined finite sequence of moves. But, if we are looking for two physically identical systems (down to the quantum state of the N particles that constitute the board and the pieces), then it is questionable whether there are only a finite number of configurations. Moreover, the question of continuous time evolution becomes potentially problematic.

Even if you can maneouvre mathematically here, there's a further question of whether your mathematical model is an exact representation of the physical system. "The map is not the terrain" etc.

Moreover, there is a simple, well-defined process to determine whether two chess positions are identical. But, it's not clear what process you would follow to confirm that two chess boards are identical in terms of their quantum states. Let alone confirm that two Hubble volumes are identical.

IMO, this is a not a mathematical question (or a question of probability theory, per se), but a question of how meaningful the physical conclusions are, given the number of untested (and potentially untestable) underlying assumptions.

 

[Jaime Rudas]

For one thing, if spacetime is a continuum with infinitely many points, you could argue that there are an (uncountably) infinite number of configurations and an (uncountably) infinite number of ways for a system to evolve over time.

It's right. If spacetime is a continuum, the probability that the distance between two particles in two different volumes is exactly the same is zero.

 

[Vanadium 50]

Is there any science to this? Or are we arguing philosophy? Is a world identical to ours except that you asked for a cheesebrurger last week "close enough"? This seems more like a topic for late-night undergraduate musings than something subject to scientific inquiry.

 

[PeterDonis]

Can you address the part about quantum repeats? Is the wavefunction of particle positions supposed to be infinite, and, if so, even in an infinite universe, can every possibility truly repeat in that case?

Remember we are talking about repeats of our observable universe, which is finite in size. Any "wave functions" we use in a quantum model would be restricted to that finite volume. And in that finite volume there are only a finite number of possible quantum states. That is the basis for the "repeat" claim.

it is true that odds are you'd probably encounter countless numbers of different Earths with alternate histories on the way to finding the identical Earth.

I don't know that anyone has actually calculated this, but it seems plausible.

Well, to be an identical Earth, it would have to send an explorer out searching for an identical Earth in the same way you went out searching. Does that mean you'd cross paths?

No, because the identical explorer to you from "repeat Earth" would be going away from you. You are setting off from Earth towards repeat Earth. The repeat Earth explorer would be setting off from repeat Earth towards repeat repeat Earth, in the opposite direction from Earth.

And does that mean the path to and from each Earth would be mirror images?

No, because that would mean the explorer was not from repeat Earth, but from a different almost-copy of Earth where he set out in the opposite direction, towards Earth instead of towards repeat repeat Earth.

 

[syfry]

This seems more like a topic for late-night undergraduate musings than something subject to scientific inquiry.

Was thinking how this sounds like a vsauce episode

 

[syfry]

Doesn't a particle have a chance to be found at any distance toward infinity? So if the universe is infinite, and no matter how infinitely large, any particle on a repeat Earth can instead be found at the boundaries of infinity, so that a certain exact match could be mathematically impossible. Perhaps.

 

[Jaime Rudas]

any particle on a repeat Earth can instead be found at the boundaries of infinity

The boundaries of infinity is an oxymoron.

 

[syfry]

The boundaries of infinity is an oxymoron.

Of course, but all I'm saying is that whatever size area we choose as possible for Earth to repeat, any of its particles can appear beyond that area.

So the problem with any absolute replication down to the particle layer is that any of the components can be found outside of the replication, to the farthest reaches of infinity.

If I understand correctly the probability of where to find any particle.

 

[PeroK]

No, because that would mean the explorer was not from repeat Earth, but from a different almost-copy of Earth where he set out in the opposite direction, towards Earth instead of towards repeat repeat Earth.

This is where, IMO, the repeat Earth theory loses touch with the pure mathematics on which it relies. All you can do is find a planet that is identical to Earth in every respect at a given point in time. But, such a planet will immediately (almost certainly) diverge from Earth as microscopic interactions produce different outcomes from the same initial conditions. And, you don't know which repeat Earth will still be a repeat Earth one second later.

To put in mathematical terms. The following statement is false:$$\exists \ P: \forall t: \Psi_P(t) = \Psi_E(t)$$Where $E$ is planet Earth and $P$ is some planet. But, the following statement could be true:$$\forall t, \exists \ P: \Psi_P(t) = \Psi_E(t)$$Or, to put it another way, there is no time-independent identical Earth:$$\forall P, \exists t: \Psi_P(t) \ne \Psi_E(t)$$In general, if we want to invoke the paradoxes of infinite sequence, we cannot pick and choose the subtlties of infinite sequences that suit some narrative.

To go back to the original calculation. What that says is that for a given $t$ we would expect to find an identical Earth (let's stick to Earth rather than an entire Hubble volume!) within a distance $d$. But, the probability that the "identical" Earth is still identical one second later is practically zero. Instead, if you believe the theory, another planet within that distance that was previously nearly identical to Earth is now identical to Earth. The key point is that the planets identified as identical Earths are changing all the time.

That said, the whole thing is fantasy rather than real physics. All of this is completely untestable.

 

[KobiashiBooBoo]

Maybe it's easier to understand if one stated there are an infinite amount of Earths exactly like our own at all times of history and evolution. (In which case it is conjecture and risks the thread being closed.)

 

[PeterDonis]

All you can do is find a planet that is identical to Earth in every respect at a given point in time.

You will find that, yes, but in a spatially infinite universe, if you look further, you will find, not just an Earth, but an entire observable universe, that matches ours through its entire history. (I'm not clear on whether Tegmark also claims that the entire future will be the same.)

 

[PeterDonis]

All of this is completely untestable.

Agreed.

 

[PeterDonis]

Doesn't a particle have a chance to be found at any distance toward infinity?

No. Here you are using a very simple concept of a wave function in non-relativistic QM. But the cosmological models being used in the argument discussed in this thread are based on relativistic quantum field theory, where the statement you are making can't even be formulated as you state it. Non-relativistic QM doesn't even cover this domain. The key statement in QFT for the argument is that, within a finite volume, there are only a finite number of possible states. AFAIK that statement is valid in QFT.

 

[Jaime Rudas]

You will find that, yes, but in a spatially infinite universe, if you look further, you will find, not just an Earth, but an entire observable universe, that matches ours through its entire history.

And this would be valid for any given moment in the future.

 

[KobiashiBooBoo]

And this would be valid for any given moment in the future.

Everything could be exactly the same except a blade of grass is missing from the Browns/Steelers game (non-scientific, I know, sorry)

 

[256bits]

The Wiki gives a reference to the Tegmark estimate paper,

1. Tegmark, Max (2003). "Parallel Universes". Scientific American. 288 (5): 40–51. arXiv:astro-ph/0302131. Bibcode:2003SciAm.288e..40T. doi:10.1038/scientificamerican0503-40. PMID 12701329.

from the PDF, Tegmark's words are,

Just an estimate using the Pauli exclusion principle.
Low balled, and not repeatable, ie one doesn't find another 'earth ' every set of this distance - could be less, could be more depending again upon chance. [ ie there could be an exact earth in the sphere just outside our own - whose to know ]

 

[PeroK]

The Wiki gives a reference to the Tegmark estimate paper,

1. Tegmark, Max (2003). "Parallel Universes". Scientific American. 288 (5): 40–51. arXiv:astro-ph/0302131. Bibcode:2003SciAm.288e..40T. doi:10.1038/scientificamerican0503-40. PMID 12701329.

from the PDF, Tegmark's words are,
View attachment 332312
View attachment 332313

Just an estimate using the Pauli exclusion principle.
Low balled, and not repeatable, ie one doesn't find another 'earth ' every set of this distance - could be less, could be more depending again upon chance. [ ie there could be an exact earth in the sphere just outside our own - whose to know ]

Here's a key question. We have a statement:

In a region of the universe of radius $10^{10^{115}} \ m$ there is another Hubble volume identical to the Hubble volume centred on the Earth.

Is Tegmark saying:

a) That statement is definitely, 100% guaranteed to be true?

b) That statement may be true, with some well-defined probability?

c) That statement may be true, but there is no good estimate of how likely it is to be true?

In other words, how much confidence should we have in the conclusions of the paper? We can all accept c). It's possible, of course. But is it likely?

He could partially answer the question by saying "if such and such a theory is correct and if these assumptions hold, then the statement is definitely true." But, that only shifts the question to how likely it is that the theory is correct and all the necessary assumptions hold? The distances involved in the paper are so large that the current lower bound for the size of the universe is hardly significant. How can we say with any certainty that the universe has a radius at least $10^{10^{115}}$ times larger than that of the observable universe?