How big is 10^10^123?

How big is 10^10^123? That’s the number represented by 1 followed by 10^123 zeros, and 10^123 is 1 followed by 123 zeros. Obviously it’s a big number. But how big? Much bigger than 10^10? Much bigger than 10^10^122? The answer is yes. Here is my attempt to comprehend it.

Why 10^10^123?

Penrose’s depiction of the Creator at work, on page 444 of his
1999 book The Emperor’s New Mind.

The number 10^10^123 appears in Sir Roger Penrose’s paper “Time-Asymmetry and Quantum Gravity,” published in Quantum Gravity 2 in 1981:

Imagine the Creator, armed with a pin which is to be placed at one spot [among possible universes,] thereby determining the state of our actual universe. [...] We are led to estimate that the accuracy of the Creator’s aim must have been at least on the order of 10^10^88 parts in 10^10^123. [...] I cannot recall seeing anything in physics whose accuracy is known to approach, even remotely, a figure like one part in 10^10^123.

The question under discussion is the smoothness of our universe right after the Big Bang. The universe was remarkably smooth, but it had small irregularities, which eventually led to galaxies and stars and people. Had it been too smooth or too rough, stars would not have formed. By Penrose’s estimate, the universe hit a sweet spot only 1/10^10^123 wide, and thanks to that luck we’re here now to wonder about it.

Penrose has consistently stuck by this number, citing it as recently as in his 2004 book The Road to Reality. He argues that the estimate holds under a wide range of possible assumptions about our universe’s history. As you can imagine, some philosophers have jumped on this number and others like it as evidence of God’s providence at the Big Bang. I won’t get into that discussion now. I just want to try comprehending how big 10^10^123 is.

10^10^123 versus 10^123

Here’s a clue as to how big this number is: Compare what happens if you divide 10^10^123 by 10^10^122 to what happens if you divide 10^123 by 10^122. If you divide 10^123 by 10^122, you get ten. Every time you add 1 to the exponent you’re multiplying by 10. So the number grows quickly, but at least you can picture that growth. Here’s a picture right here:

Each square has ten times the area of the previous square.
The labels give the number of pixels in each sqaure.

Or you can watch the classic “Powers of Ten”:

You can see that 10^n gets very big very fast as n grows. But numbers with a second exponent, as in 10^10^n, get much bigger much faster. Instead of multiplying the number by 10 when you add 1 to the exponent, you are now raising the number to the power of 10. Expressed in math:

So, whereas 10^123 divided by 10^122 gets you all the way down to 10, 10^10^123 divided by 10^10^122 gives you 10^10^122.95424, pretty much what you started with. This means that, at this size, raising the exponent by just 1 blows the magnitude so much higher that the previous number becomes unimaginably negligible in comparison—unimaginably more negligible than the volume of a proton compared to the volume of the observable universe (that ratio is “just” 10^126, by the way, or about 10^10^2).

Clearly, trying to find an analogy for the size of 10^10^123 will be tough.

Building a human by chance

Here’s a thought experiment: What are the odds of taking a bunch of atoms, shaking them randomly in a human-sized box, and seeing them spontaneously assemble into a full human being? Let’s say a specific human being, like Derrick Rose. To have an idea of how unlikely this is, imagine if you were to take a box of Legos, shake it randomly, and see a model spaceship spontaneously assemble. It’s so unlikely that you and everyone you know could spend their entire lifetimes shaking and you’ll never see anything. Now replace the Legos with atoms and the model spaceship with Derrick Rose.

This calculation is going to be very approximate, so for those concerned, we won’t distinguish between the probability of a Derrick Rose with or without a tear in his ACL, for example.

I came up with odds of about 1 in 10^10^29, give or take a couple magnitudes in the top exponent. The rest of this section goes through the details of how I got that, using two different methods. You might want to skim past it.

Let’s start with the volume of Derrick Rose. He’s listed as 6′3″, so we’ll use a 2 m³ box for shaking our atoms.

Next is the atomic composition of Derrick Rose. Using this table as a reference, and Rose’s listed weight of 190 lb, let’s approximate that he comprises 8×10^27 atoms, which are 64 percent hydrogen, 23 percent oxygen, and 12 percent carbon.

8×10^27 atoms, arranged just right.

Now the first method for finding the odds: we calculate the entropies, S₁ and S₂, of the initial and final states respectively. We then use the statistical definition of entropy, S = k ln W, where k is Boltzmann’s constant and W is the density of states. The probability of finding a spontaneous entropy S₂ instead of S₁ is W₂/W₁, or exp(S₂/k) / exp(S₁/k).

Using Wolfram Alpha’s monatomic gas tool we compute S₁ for the randomly shaking atoms that would compose Derrick Rose, and get about 1.1 MJ/K of entropy. We let the atoms have a temperature of 300 K, though it doesn't matter much. The magnitude of W₁ is about 10^10^29.

As for the entropy of Derrick Rose, this article goes through multiple considerations of entropy and negative entropy in the human body, but that doesn't matter much either. No matter how many of these values we include in the entropy of Derrick Rose, W₂ has negligible magnitude compared to W₁. This means that it doesn’t matter much if we care to distinguish between a healthy Derrick Rose, an injured Derrick Rose, a cryogenically frozen Derrick Rose, or a statue of Derrick Rose carved from a block of wood. The difference in probability among those variations isn’t significant compared to the remote odds of getting anything interesting at all out of a random mix of atoms. So we’ll call the odds 1 in 10^10^29. The 29 is approximate, and though an approximation in the top exponent creates a huge range of uncertainty, it’s good enough for me.

Next, the second method for approximating the odds: how many ways are there to distribute 8×10^27 atoms in a volume of 2 m³? This method is much more approximate but should get us a similar magnitude for the probability. The volume of a single bonded hydrogen atom is about 1.2×10^(–31) m³. If we partition Rose’s volume into boxes of that size, we have about 1.7×10^31 boxes. Let’s round it off at 10^31 boxes, give or take a factor of 10.

Next we count the number of ways to arrange 8×10^27 atoms in 10^31 boxes. That’s w(8×10^27, 10^31) in the notation of Bose–Einstein statistics,* or (8×10^27 + 10^31 – 1)! / ((8×10^27)! × (10^31 – 1)!), where the exclamation point represents the factorial. We aren't being careful about the indistinguishability of the atoms or the number that can fit in each box, but it's fine for approximating.

Those big factorials are way too big for a computer to compute, but, mercifully, Wolfram Alpha returns approximate answers. We get about 10^10^30 ways to distribute atoms in a 2 m³ volume. That's similar enough to what we got with the first method.

So...if you skipped all that physics...the odds of randomly shaking a bunch of atoms and seeing them instantly assemble into Derrick Rose is 1 in approximately 10^10^29.

Repeating the experiment

Now let’s imagine we performed this atom-shaking experiment a million times, and Derrick Rose spontaneouly appeared in each experiment a million times in a row. The odds of this happening would be 1/(10^10^29)^1,000,000 = 1/10^10^35. The number 10^10^35 is a lot bigger than 10^10^29, but still negligible compared to 10^10^123.

If we did this every second from the Big Bang until today—meaning we spontaneously produced a Derrick Rose every second for 14 billion years—we’d have beaten odds of about 1 in 10^10^46. That’s still not even close to 10^10^123.

What if we performed the experiment once every second using every available atom in the observable universe? (Let’s ignore the problem that these atoms are 90% hydrogen, and Derrick Rose isn’t.) That would give us about 10^53 new Derrick Roses, every second, for 14 billion years. That's 10^70 total Derrick Roses. The odds of this achievement are about 1 in 10^10^99. Still not even close! Remember how much bigger that number grows with even a small change in that second exponent.

A sample of about 10^2 Derrick Roses. Credit Doug Thonus.

Finally, the punchline: we would arrive at odds of about 1 in 10^10^123 if we wanted to produce 10^94 Derrick Roses in 10^94 tries. That would be 100 trillion Derrick Roses for every single atom in the observable universe.

Conclusion

10^10^123 is really big. So big that in comparison, having people spontaneously pop out of air seems like a reasonable possibility.

* Edit: (6/29/12) I don't know why I didn't just use n choose k for this calculation. That's simpler and just as approximate, and the magnitude is pretty much the same. I even did that calculation before writing the post. A friend called me out on this today, so now I write this edit of shame. But as he pointed out, “Bose-Einstein” sounds a lot more impressive to non-physics people.