Quantum mechanics question (bordering on philosophy)

Factual Questions
21

Two things.

First, say that you do have three basketball-sized chunks of matter appearing somewhere in the universe every second. So what? The sun alone loses 4.5 million tonnes of matter every second. Three basketball-sized chunks in the universe would be too small for anyone to notice.

But I also disagree with the way douglips approached the problem. A better way to think of it is to check all the possible states of the 7.6 * 10[sup]26[/sup] molecules in the room. And that’s 7.6 * 10[sup]26[/sup] factorial, a number so huge that the entire universe would not suffice to write it down. (Even if you divide by a large number of interchangeable equivalent positions, it’s still way larger than anything we can conceive.)

The probability of a visible quantum effect is therefore still much too low to observe - or to occur - in the lifetime of a universe.

22

I was using logarithmic math because the numbers are so huge. You don’t need to wait 1.6 * 10[sup]27[/sup] years for it to happen in one room. You have to wait 10 raised to the power of (1.6 * 10[sup]27[/sup]) years, which is as Exapno says a number so big you can’t write it down in the universe.

Now, I don’t understand Exapno’s model: Where does 7.610[sup]27[/sup]! come from? That is the number of ways I can order 7.610[sup]27[/sup] molecules in a set, but I don’t see what that has to do with the number of states available to the room full of gas unless we are saying that it’s really a room close packed with gas molecules, which would be some kind of a degenerate gas like the conduction band electrons in a metal. In other words, I think this number is only valid if our room still contains the same amount of gas, but is about 1 cubic foot big. (I got that 1 cubic foot by estimating that liquid gas would be about the same molecular density as liquid water. 1260 moles of water will fit in 22 liters, which is less than 1 cubic foot.)

I think the number of states available to a gas molecule in the room is vastly more than that - approximately the volume of the room divided by the volume of the gas molecule, which is a staggeringly hugely mind boggling number, much much much bigger than 7.610[sup]27[/sup]. Whatever that number is, call it q, the number of states available in the room is q!/(q-7.610[sup]27[/sup]). Am I missing something, Exapno?

So, this huge number would be way way bigger than my number. The reason for the difference between Exapno’s number and mine is that there is a HUGE number of states of these particles that put them in my corner box. If you only accepted one of Exapno’s states it would be the same as requiring all particles to be within 1 cubic foot or so, and in a particular order, as if the molecules are numbered or something - whereas I give them a generous 8 cubic feet to roam around in.

Let’s play around with that number 7.610[sup]27[/sup]! though. Note that Stirling’s approximation tells us that n! is approximately sqrt(2pi) n[sup]n+1/2[/sup] e[sup]-n[/sup]. Let’s say n is 7.6 * 10[sup]27[/sup], so n+1/2 is really just still n. Let’s further throw out that sqrt(2pi) for an approximation. Let’s change n[sup]n[/sup] to an exponent of e. log[sub]10/sub is 7.6 * 10[sup]27[/sup] * 28, or about 2.110[sup]29[/sup]. Multiply by ln(10) gives us 4.910[sup]29[/sup], which is ln(n[sup]n[/sup]). So, we can write n[sup]n[/sup] as e[sup]4.910^29[/sup].

So, n! is about e[sup]4.910^29[/sup]e[sup]-7.610^27[/sup], which is about e[sup]4.810^29[/sup] or about 10[sup]210^29[/sup]. My number is 10[sup]1.6 * 10^27[/sup], or about 1% of n!.

23

And I just realized I somehow multiplied Exapno’s number by 10: 7.6 * 10[sup]27[/sup] when it should be 10[sup]26[/sup]. But you get my point, I hope.