Your own equation, “c = sqrt(d^2 + s^2)” becomes sqrt(1.414^2 + 1^2) = sqrt(2+1).
Everything I know about simple linear algebra would collapse if the diagonal of a cube is other than 1.732. The number is easy to remember because George Washington was born in 1732.
(How do I remember Washington’s birth year? It’s the same as √3 !) Or rather, to avert another episode of pedantry, Washington’s birth year is almost the square root of 3 million.
Are you arguing that absolute zero is reachable? Because that is what zero translational motion implies in your classical model.
Even in highschool physics at “absolute zero” you would use the internal energy or:
K[sub]avg[/sub] = 3/2 k[sub]B[/sub]T
Where:
k[sub]B[/sub] = Boltzmann’s constant (1.38·10−23 J·K−1)
T = temperature in Kelvin
For the kinetic energy you would probably use:
v[sub]avg[/sub] = sqrt((8/π)*(k[sub]B[/sub]*T/m))
If you were working from statistical mechanics, I would expect the following eigenvalues for zero potential particle in a box as a starter.
ε[sub]l[sub]x[/sub],l[sub]y[/sub],l[sub]z[/sub][/sub] = h[sup]2[/sup](l[sup]2[/sup][sub]x[/sub] + l[sup]2[/sup][sub]y[/sub] + l[sup]2[/sup][sub]z[/sub]) / 8mL[sup]2[/sup]
HS wouldn’t use that but even in High School kinetic-molecular theory would use:
KE = 1/2 * m * v[sup]2[/sup]
They wouldn’t use basic elastic collisions, and the math and apparent paradoxes that arise from trying to use elastic collisions of two mixing monatomic gases.
A ideal monatomic gas has three degrees of freedom so *f = 3 * thus C[sub]v[/sub] = (3/2)R
For a mixture two monatomic gasses you will run into problems and singularities but setting f = 1 and using basic elastic collision kinematics won’t get you any place close to the concept of entropy or thermodynamic equilibrium.
None of this work at the above levels will disprove the first and second law of thermodynamics or invalidate the arrow of time.
I don’t think this is quite true or may be only true for Earth-like atmospheres. The bottom of that range must be significantly smaller than the size of the Earth. Otherwise, Venus, which is somewhat smaller than Earth and has only 91% of the surface gravity, should not be able to hang on to an atmosphere that’s 90 times the density of Earth’s.
OK, this is mostly due to Venus’ atmosphere being mostly CO[sub]2[/sub], which is a heavier gas than oxygen, nitrogen, and water vapor. But it wouldn’t surprise me at all if most Earth-sized planets have an atmosphere that has much more carbon dioxide than Earth. Also significantly denser. Probably no where near as much as Venus, but more than enough to be toxic to humans.
At any rate, if we ever do find a prospective Earth-like colony world, we may have to spend a couple-three thousand years modifying its atmosphere to be like Earth’s. OK, maybe we can use nanotech to do the job faster, but then we’ll have a world covered in nanobots. That may not be much of an improvement.
Wow. Talk about a consistent, almost pathological, inability to comprehend the simplest of English.
No, I’m not going to push the Button Which Must Not Be Mentioned. But going forward, I will avoid *all *attempts at dialog with you.
I agree that the diagonal of the square is 1.41421 or so. So the diagonal of the rectangle that has one side as 1.41421 and one side as 1 is the square root of (1.41412 squared plus 1) or square root of 3, which is 1.7 or so.
Oops - I see septimus has already given virtually the same explanation as I have. Ah well.
And yet it only needs 38 doublings to colonise all 400 million stars, which allows a generous 26315 years between each doubling. What am I missing?
I think the problem is that exponential growth is impossible in practice. The colonisation wave could only continue to expand in steps that take 26315 years for as long as there are stars within easy reach; eventually every star on the edge of the colonisation wavefront would be already claimed by your neighbor, so you’d need to travel further (and faster) to get to an empty star. What would have started as a leisurely expansion taking tens of thousands of years between hops would turn into a rapid scramble at impossible speeds.
I do think that in practice the colonisation wave would speed up a bit as the colonisers gain more experience and technological know-how, but the ‘scramble’ effect’ would make colonisation subject to diminishing returns as well. Assuming a constant speed of colonisation slows down the colonisation of the entire galaxy to a leisurely ~100 million years - a long slow process, but still a fraction of the age of the galaxy.
I should have asked my daughter. She’s a mathematical biologist. This is a growth curve; the phase of exponential growth only lasts a limited period, then a slow, arithmetical growth sets in.
To help people who may have been confused by my misguided attempts to ask for a cite in GQ, here is a page that will explain, using a one-atom gas how free energy and entropy are related in kinetic theory.
http://umdberg.pbworks.com/w/page/105011406/Example%3A%20The%20one%20atom%20gas
Remember that for something to be a thermodynamic system it does need to be a closed system thus the need to have a constant pressure or constant volume. If you don’t want to do the math above a way that may help to think about it is that the uncertainty which degree of freedom has the energy becomes less absolute and so translational, rotational and vibrational energy and the relation between free energy and entropy can be seen in that domain above.
This page from that same site will help explain why “fluctuations”
http://umdberg.pbworks.com/w/page/104491534/How%20energy%20is%20distributed%3A%20Fluctuations
Statistical mechanics as a tool is only valid if you have a large number of degrees of freedom typically with a large number of particles for ideal gasses.
Just as you can’t tell the odds on a coin flip with a single toss you can’t use the statistical form of the 2nd law for a small number of particles. But that is a limitation of statistics and not the underlying concepts and the one link above will help you understand the single atom implications of entropy as a thought experiment.
Your analysis is missing any sense of the distance between the stars.
Moving another order of magnitude implies that it takes 4000 years for us to reach Alpha Centauri, the nearest star system. A lot of biological evolution and bottleneck can occur in 100 million years. Even a little stellar evolution might happen.
I’m not saying this is impossible though: I’m just saying the numbers are tougher than I thought. Expansion at 186 miles per second is still an order of magnitude faster than our fasted rockets for example. OTOH, we could probably posit multiple civilizations expanding across the galaxy. That could cut down on the necessary speed.
Multiple civilisations expanding towards each other would speed things up, but it is unlikely that two or more civilisations would start expanding at the same time at different locations in the galaxy.
If you consider the stellar evolution and metallicity of the stars and solar systems in the Milky Way, there have been sufficient planetary environments available for the evolution of advanced civilisations for at least the last billion years, possibly two billion (or more). That means any civilisation that emerged a billion years ago would have had ample time to spread throughout our galaxy, and it would be a fairly hefty cosmic coincidence if two or more civilisations of this kind were to emerge at the same time.
That is, unless civilisations emerge with remarkable frequency (in which case we’d be up to our neck in them).
Since this snark is obviously directed at me, kindly point to a claim I made for which I did not comply with a cite request. No, not some claim made by the fantastic strawman you fabricated out of whole cloth — a claim I made.