Can you build a better thermos than empty space?

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lgw
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Can you build a better thermos than empty space?

Postby lgw » Sun Nov 20, 2016 2:36 am UTC

The universe is cooling - slowly now, but quickly in the early days. It is theoretically possible to make a thermos (well, some sort of vast Dewer jar with very reflective surfaces), filled with material at the temperature of the CMBR, such that the contents cool slower than the expanding universe? (Assuming a starting temperature where you could have solid reflective surfaces.)

It seems to me that if you made it large enough, the cube-square law would limit the inevitable heat loss, and the volume of the container would remain the same despite the expansion of space. Am I missing something fundamental?
"In no set of physics laws do you get two cats." - doogly

Tub
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Re: Can you build a better thermos than empty space?

Postby Tub » Sun Nov 20, 2016 11:29 pm UTC

The cube-square law says that a larger thermos will take longer to reach thermal equilibrium. Heat energy is proportional to volume, heat loss is proportional to surface.
But heat transfer is also proportional to the temperature difference, so there is no limit to the heat loss. A big thermos can maintain a larger temperature difference against the cooling CMBR outside, but there's no size at which the thermos won't cool down.


However, if you plan to make the thermos really big, it'll be ripped apart by the metric expansion of space. The space inside expands, but the walls of your thermos don't. You might try to fill it with stuff that's just heavy enough to counteract the expansion without collapsing, but I don't think there's a solution that'll remain stable.

lgw
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Re: Can you build a better thermos than empty space?

Postby lgw » Mon Nov 28, 2016 7:36 pm UTC

Thanks for the reply.

Tub wrote: A big thermos can maintain a larger temperature difference against the cooling CMBR outside, but there's no size at which the thermos won't cool down.


Ah, sure, but the universe keeps cooling down, so as long as the thermos cools slower, it will always be warmer, right? I'm having fun with the idea that this could be a source of power in the distant future.


Tub wrote:However, if you plan to make the thermos really big, it'll be ripped apart by the metric expansion of space. The space inside expands, but the walls of your thermos don't. You might try to fill it with stuff that's just heavy enough to counteract the expansion without collapsing, but I don't think there's a solution that'll remain stable.


Well, that's a point: with materials we understand, there's a maximum size to any structure you can build before it collapses under it's own weight1. But the metric expansion of space is very weak, right? Gravity is expected to hold the galaxy together against it (excepting a "big rip" end to the universe), and a solid object would be so many orders of magnitude stronger.


1Can you spin an arbitrarily large structure such that the net load from gravity and spin is bounded? That is, could you make an arbitrarily large structure from steel girders if you spun it just right? A shape that gives gravity the same curve as the stress from spinning?
"In no set of physics laws do you get two cats." - doogly

Tub
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Re: Can you build a better thermos than empty space?

Postby Tub » Mon Nov 28, 2016 10:06 pm UTC

lgw wrote:Ah, sure, but the universe keeps cooling down, so as long as the thermos cools slower, it will always be warmer, right?

For a long time, but not forever. Eventually, the thermos will decay, rupture, and the whole system will reach the inevitable thermal equilibrium. Entropy is merciless.

lgw wrote:But the metric expansion of space is very weak, right?

Depends on the scale and the time. Right now, at scales below galaxy-size, it's weak. At scales way above galaxy clusters, gravity doesn't seem to do much to stop the expansion. Unless you have the ability to summon additional matter at will (you don't!), redistributing lots of galaxies into the shape of a thermos will not suddenly make them gravitationally bound if they weren't before.
The larger your scale is, the worse it gets. Take it to the extreme: if your thermos is larger than the currently observable universe, how is it supposed to remain bound if the far corners cannot communicate with each other?

It may get worse in the future, depending on whether (and how) the expansion accelerates.

lgw wrote:1Can you spin an arbitrarily large structure such that the net load from gravity and spin is bounded? That is, could you make an arbitrarily large structure from steel girders if you spun it just right? A shape that gives gravity the same curve as the stress from spinning?

A huge, solid ball (or disk) won't work. The outer parts require a different spinning speed than the inner parts, so the structure cannot remain solid.

A simple hollow ring might work. I cannot find the formulas - you'd need to make it light enough that the required rotation does not exceed the speed of light. Of course, a lightweight steel girder ring isn't very stable, and I wouldn't be surprised to see it torn apart by even slight gravitational irregularities or gravitational waves. If the ring has a weight of several galaxies, even a little tug from that mass is enough to break a small steel girder. Make sure the universe is completely empty, non-expanding and free of quantum fluctuations before starting construction.
If the universe expands, you'll be getting the event horizons between observable and unobservable parts again, and no structure can exceed that size.

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Internetmeme
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Re: Can you build a better thermos than empty space?

Postby Internetmeme » Wed Dec 07, 2016 6:28 am UTC

Let's first deal with the equations that will govern our thermos's heat transfer.

Heat can be transfered by conduction, convection, and radiation.

q_tot = q_cond + q_conv + q_rad

Immediately, the conductive term goes away, because that only applies to solids.

To give ourselves the absolute best, most ideal conditions, we'll put this thermos in the middle of some crazy intergalactic void where the density is approximately 0 kg/m^3. This puts our convective heat transfer down to approximately 0.

So all we care about is radiative heat transfer.

Let's mine for some more information to figure out how we're going to build this thermos.

It is theoretically possible to make a thermos (well, some sort of vast Dewer jar with very reflective surfaces), filled with material at the temperature of the CMBR, such that the contents cool slower than the expanding universe? (Assuming a starting temperature where you could have solid reflective surfaces.)

So, starting temperature of the thermos is 2.7k.

And we have a "reflective" surface.

q(1->2) = sigma*E_1 * F_1,2 * A_1 * (T_1 ^4 - T_2 ^4)

In English, the rate of heat transfer from body 1 to body 2 is equal to the stefan-boltzman constant times the emissivity of 1 times the "F" factor (which is how much of object 2 object 1 "sees") times the area of object 1, times the difference of the temperature of both, to the 4th.

So, sigma is a constant.
We'll get back to E in a second.
Because the thermos is surrounded on all sides by the universe, we'll say it "can't miss" the universe, so F = 1
From what I can tell, a normal 16oz thermos is 9.7" tall by 3.2" wide, or in units that don't suck, 0.25 meters tall with a radius of 0.08m. The surface area, assuming a cylindrical thermos, is then 0.17 m^2.

Well that's not good, so far. We *need* something, anything at all, on the righthand side to equal zero if we're going to have zero heat transfer. Our last hope is emissivity. If this were a perfectly reflecting thermos, emissivity would be 0, and we might be able to save some small portion of heat from the inevitable heat death of the universe. Why? Because that would make the entire righthand side of the equation equal to zero, and thus no heat would be transferred. But, there is no known material that perfectly reflects, so it must be greater than zero.

Let's wrap all of that up into something we'll call resistance, defining the resistance to be equal to 1/(sigma*E*F*A).

We get the equation q = (T1^4 - T2^4)/resistance

T1 is the temperature of the stuff inside the thermos, and T2 is the temperature of the stuff inside the

Because we know resistance is always greater than zero (sigma is a constant, area must be greater than zero, E cannot be zero, and F is equal to 1), the only way our rate of heat transfer can be zero is if T1 and T2 are equal... which would mean the stuff inside the thermos is the same temperature as the stuff outside the thermos, which defeats the purpose of having a thermos.

So, our first conclusion is that it is impossible to build a thermos that can maintain heat forever.

The good news, however, is that if we can build an ultra-uber-super-duper-reflective surface, we might be able to temporarily hold onto some heat. The problem is that if there is any sort of emissivity at all, heat is going to be transfered, so it's not going to stay at the initial temperature indefinitely. As long as the rate of heat transfer (governed by our emissivity) is lower than the rate the universe is cooling, we're succeeding in the goals outlined in the OP. But that would require a magical material with an insanely low emissivity, which may not even be physically possible.

But there is some silver lining to all of this. It turns out that it takes an infinite amount of time for two objects to reach a perfect thermal equlibrium. So, even though there is no practical chance of our thermos cooling slower than the universe, there is still the bright side that the thermos will always be slightly hotter (even though it the difference will be so small that we can't even hope to measure it) than the rest of the universe, assuming it doesn't decay.
Spoiler:


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