Actually I think this fails for ordinary reasons. Key question: how are you getting energy out of lowering the helium?
If you mean the helium is chemically bound to the sheets (through adsorption) then you’ll need to use energy to release it
If you mean the helium is trapped in balloons, then it will be neutrally buoyant in the ambient helium atmosphere unless you expend energy to compress it.
Big thank you! I made an edit to the post, clarifying this point (the one in my previous reply). Do you think I need to address the point of enclosing the gas directly?
Ah! Good question. Perhaps I should have given more details.
I think enclosing the helium (without compression) is the way to go. And the density will spike close to the shell (gas comming from the asteroid will accumulate there). You will this have:
*The atmosphere, with the greatest density.
*The vicinity of the shell wall, with a spike in density.
*Space in between, very close to a vacuum.
You will be able to lower the enclosed helium through a lot of space without any bouyancy, turning potential energy into work.
Why do you think gas will accumulate close to the shell? This is not how gases work, the gas will form an equilibrium density gradient with zero free energy to exploit.
Here’s why I believe a slight density increase near the shell is not only possible but statistically inevetable:
The shell can be placed sufficiently far from the asteroid such that escaping helium atoms travel nearly radially from the surface. Most will have very low kinetic energy, having just barely escaped the gravitational potential well.
Sufficiently far from the asteroid, these atoms feel almost no gravity. They’re essentially coasting in near-inertial trajectories.
If the shell were absent, these atoms would simply escape into space. But with the shell in place, their outward motion is halted. They bounce off the shell instead of escaping.
Since their approach is slow and nearly radial, many will strike the shell with low momentum. After bouncing, some will scatter at non-radial angles and may linger in the vicinity. Given enough atoms and time, this creates a diffuse accumulation zone close to the shell. A mild, geometry-induced density spike.
This isn’t a violation of thermodynamic equilibrium. It’s a boundary condition effect arising from system geometry and the kinematics of slow-moving atoms arriving at a barrier. The system as a whole trends toward equilibrium, but that equilibrium includes local features shaped by containment. Also, we do not have to wait for equilibrium before collecting atoms. It is enough that we calculate that at some point in time there will be a density spike close to the shell.
So yes, under these assumptions, some degree of helium accumulation near the shell is to be expected.
No, when you carry through the calculations you will find that in equilibrium the density is monotonic with distance from the asteroid.
One easy way to see this: if there were increased density near the shell without any counterbalancing force attracting them to the shell, then there would be a net flow of particles away from the shell reducing the density. So this cannot be an equilibrium.
There may be transient microscopic density variations, but no macroscopic ones (absent some sort of Maxwell’s Demon).
It is also an incorrect assumption that the motion is nearly radial. At all heights the direction distribution is still uniformly random.
OK, so the issue here is that you’ve switched from a thermodynamic model of the gas atoms near the asteroid to one which ignores temperature at the shell. I’m not going to spend any more time on this because while it is fun, it’s not a good use of time.
One of the properties of the second law is that if you can’t find a single step in your mechanism which violates it, then the mechanism overall cannot violate it. Since you claim that every step in the process obeys the second law, the entire process must obey the second law. Even if I can’t find the error I can say with near-certainty that there is one.
For anyone reading: Please note that I do not claim this is a perpetual motion machine. I do not claim the setup breaks the second law. In fact, I claim the opposite, and I even think I have found a mechanism where entropy does increase with the required scaling. I think the answer to why the second law doesn’t break may be important and interesting.
Actually I think this fails for ordinary reasons. Key question: how are you getting energy out of lowering the helium?
If you mean the helium is chemically bound to the sheets (through adsorption) then you’ll need to use energy to release it
If you mean the helium is trapped in balloons, then it will be neutrally buoyant in the ambient helium atmosphere unless you expend energy to compress it.
Big thank you! I made an edit to the post, clarifying this point (the one in my previous reply). Do you think I need to address the point of enclosing the gas directly?
Ah! Good question. Perhaps I should have given more details.
I think enclosing the helium (without compression) is the way to go. And the density will spike close to the shell (gas comming from the asteroid will accumulate there). You will this have:
*The atmosphere, with the greatest density.
*The vicinity of the shell wall, with a spike in density.
*Space in between, very close to a vacuum.
You will be able to lower the enclosed helium through a lot of space without any bouyancy, turning potential energy into work.
Why do you think gas will accumulate close to the shell? This is not how gases work, the gas will form an equilibrium density gradient with zero free energy to exploit.
Here’s why I believe a slight density increase near the shell is not only possible but statistically inevetable:
The shell can be placed sufficiently far from the asteroid such that escaping helium atoms travel nearly radially from the surface. Most will have very low kinetic energy, having just barely escaped the gravitational potential well.
Sufficiently far from the asteroid, these atoms feel almost no gravity. They’re essentially coasting in near-inertial trajectories.
If the shell were absent, these atoms would simply escape into space. But with the shell in place, their outward motion is halted. They bounce off the shell instead of escaping.
Since their approach is slow and nearly radial, many will strike the shell with low momentum. After bouncing, some will scatter at non-radial angles and may linger in the vicinity. Given enough atoms and time, this creates a diffuse accumulation zone close to the shell. A mild, geometry-induced density spike.
This isn’t a violation of thermodynamic equilibrium. It’s a boundary condition effect arising from system geometry and the kinematics of slow-moving atoms arriving at a barrier. The system as a whole trends toward equilibrium, but that equilibrium includes local features shaped by containment. Also, we do not have to wait for equilibrium before collecting atoms. It is enough that we calculate that at some point in time there will be a density spike close to the shell.
So yes, under these assumptions, some degree of helium accumulation near the shell is to be expected.
No, when you carry through the calculations you will find that in equilibrium the density is monotonic with distance from the asteroid.
One easy way to see this: if there were increased density near the shell without any counterbalancing force attracting them to the shell, then there would be a net flow of particles away from the shell reducing the density. So this cannot be an equilibrium.
There may be transient microscopic density variations, but no macroscopic ones (absent some sort of Maxwell’s Demon).
It is also an incorrect assumption that the motion is nearly radial. At all heights the direction distribution is still uniformly random.
This sort of issue is what people invented numbers and equations for.
OK, so the issue here is that you’ve switched from a thermodynamic model of the gas atoms near the asteroid to one which ignores temperature at the shell. I’m not going to spend any more time on this because while it is fun, it’s not a good use of time.
One of the properties of the second law is that if you can’t find a single step in your mechanism which violates it, then the mechanism overall cannot violate it. Since you claim that every step in the process obeys the second law, the entire process must obey the second law. Even if I can’t find the error I can say with near-certainty that there is one.
For anyone reading: Please note that I do not claim this is a perpetual motion machine. I do not claim the setup breaks the second law. In fact, I claim the opposite, and I even think I have found a mechanism where entropy does increase with the required scaling. I think the answer to why the second law doesn’t break may be important and interesting.