4 ms·

I've often read that absolute zero (0 K) is impossible to obtain, but never really understood whether it's just for practical reasons or if there's actually a t

I've often read that absolute zero (0 K) is impossible to obtain, but never really understood whether it's just for practical reasons or if there's actually a theoretical background. As far as I understand, there's no reason why a particle having 0 kinetic energy can't exist. Also, achieving it seems theoretically possible when a particle has kinetic energy equal to one photon, then emits a photon. Obviously it's hard to experimentally produce such a situation, but not theoretically impossible. So, would 0 K theoretically be possible or am I missing something?

ben_w 10y ago
Well, Heisenburg uncertainty becomes more relevant the colder you get. In this case, [uncertainty in position] * [uncertainty in momentum] >= ħ/2, so something with mass but no kinetic energy would be everywhere in an infinite universe with equal probability.
- blauditore 10y ago
  Ah, this definitely makes sense, so it's at the least not observable. Another thing crossing my mind is whether Heisenberg uncertainity becomes relevant for these kind of experiments. I think we're talking about ~10^-7 K, thus ~10^-30 J mean energy. For a hydrogen atom (1 u), that would be roughly p = ~10^28 Ns. Hence, uncertainity of location would be h/p ~= 10^-6 m, i.e. in micrometer range, which seems quite large. Does this make sense? Edit: It's aluminium in the experiment (27 u), but the numbers should be in the same ballpark.
  - ben_w 10y ago
    Sounds about right to me. Wikipedia's list of Orders of magnitude (temperature)* shows atomic waves being coherent over centimetres at 10^-15 K, and a 10^8 difference in temperature is a 10^4 difference in momentum — which fits μm to cm. * https://en.wikipedia.org/wiki/Orders_of_magnitude_(temperature https://en.wikipedia.org/wiki/Orders_of_magnitude_(temperatu...
contravariant 10y ago
Temperature doesn't really make sense for one particle. In fact any one particle will have 0 kinetic energy in some frame of reference. The third law of thermodynamics actually states that there's no systematic way that you can reach 0K, although I haven't found a particularly thorough explanation why. In general it seems to boil down to the fact that no matter which parameter you try to tune there will always be some entropy remaining, so you can't reach T=0 (zero entropy) in a finite number of steps. And then you have quantum mechanical effects that prevent a particle from actually having exactly 0 kinetic energy.

ben_w 10y ago

Well, Heisenburg uncertainty becomes more relevant the colder you get. In this case, [uncertainty in position] * [uncertainty in momentum] >= ħ/2, so something with mass but no kinetic energy would be everywhere in an infinite universe with equal probability.

blauditore 10y ago
Ah, this definitely makes sense, so it's at the least not observable. Another thing crossing my mind is whether Heisenberg uncertainity becomes relevant for these kind of experiments. I think we're talking about ~10^-7 K, thus ~10^-30 J mean energy. For a hydrogen atom (1 u), that would be roughly p = ~10^28 Ns. Hence, uncertainity of location would be h/p ~= 10^-6 m, i.e. in micrometer range, which seems quite large. Does this make sense? Edit: It's aluminium in the experiment (27 u), but the numbers should be in the same ballpark.
- ben_w 10y ago
  Sounds about right to me. Wikipedia's list of Orders of magnitude (temperature)* shows atomic waves being coherent over centimetres at 10^-15 K, and a 10^8 difference in temperature is a 10^4 difference in momentum — which fits μm to cm. * https://en.wikipedia.org/wiki/Orders_of_magnitude_(temperature https://en.wikipedia.org/wiki/Orders_of_magnitude_(temperatu...

contravariant 10y ago

Temperature doesn't really make sense for one particle. In fact any one particle will have 0 kinetic energy in some frame of reference. The third law of thermodynamics actually states that there's no systematic way that you can reach 0K, although I haven't found a particularly thorough explanation why. In general it seems to boil down to the fact that no matter which parameter you try to tune there will always be some entropy remaining, so you can't reach T=0 (zero entropy) in a finite number of steps. And then you have quantum mechanical effects that prevent a particle from actually having exactly 0 kinetic energy.