# -273 degree Celcius

Discussion in 'Physics & Math' started by Saint, Mar 3, 2017.

1. ### Q-reeusValued Senior Member

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https://futurism.com/science-explained-hottest-possible-temperature/
Of course you have to trust particle physicists have a good handle on the reasoning behind it. A mere trillion degrees - believed to be easily exceeded in the core of a newly formed neutron star, is outright frigid by comparison to 'absolute hot' i.e Planck temperature.

3. ### exchemistValued Senior Member

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So far as I know there is no upper limit. Thermal motion can always be increased by adding yet more energy, though matter dissociates more and more as various energy thresholds are crossed, i.e. molecules break down into fragments and eventually atoms, then atoms start to get electrons knocked out of them, forming plasmas of ions, then eventually I imagine even atomic nuclei would start to come apart. But all the particles thereby liberated would carry a lot of kinetic energy in random motion and would thus contribute to the temperature.

Oops, just corrected by Q-reeus. I had not heard of Planck temperature before. So I've learned something: thanks for asking the question.

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5. ### Q-reeusValued Senior Member

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At least SF is chugging along this way

. The basic reasoning behind Planck temperature as an 'absolute limit' is that centre-of-energy collisions at such a temperature are expected to frequently generate micro-black holes. Which then immediately 'evaporate' via Hawking radiation to form more particles than the typically two involved in the initial collision. Thus more degrees of freedom are generated for the same overall energy. Which thereby quenches any rise in temp above that level. Nature may or may not agree with that scenario, but no real prospect of finding out by experiment!

7. ### DaveC426913Valued Senior Member

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The highest temp would be limited by relativistic velocities of atoms.

8. ### exchemistValued Senior Member

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Hmm, is that right? Is temperature defined in terms of the velocity of particles or their kinetic energy? I must admit I thought it was the latter (E mean = 1/2kT, for each degree of freedom), in which case I'm not at all sure that relativistic limits on velocity would prevent the temperature from increasing.

But I admit to being rusty on all this.....

9. ### DaveC426913Valued Senior Member

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How is kinetic energy not limited by c?

10. ### exchemistValued Senior Member

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Well I'm rusty and no expert on relativity, but I had thought that, in a particle collider for example, one used an expression for KE like this:
KE = {1/√(1-v²/c²) -1}. mc² and that this allowed the the KE to be increased indefinitely without being limited by c.

Is that a misunderstanding on my part and, if so, what is the correct way to describe the increase in energy of a particle accelerated close to c?

11. ### DaveC426913Valued Senior Member

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You mean because m is not limited? Sure. You could always use more massive particles, or more of them. I was sorta talking about KE per unit particle.

12. ### rpennerFully WiredStaff Member

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Well then you are using the wrong physics.

The expression for kinetic energy of a free massive particle in special relativity is $E - E_0 = mc^2 \left( \left(1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}} - 1 \right) = mc^2 \frac{c - \sqrt{c^2 - v^2}}{\sqrt{c^2 - v^2}}$
Let $x = \sqrt{c^2 - v^2}$ then $\lim_{v\to c} E(v) - E_0 = mc^2 \lim_{x\to 0} \frac{c - x}{x} = \infty$

So there is no velocity-limited upper limit to kinetic energy in special relativity.

But since atoms break apart at energies of a few eV, in the limit of high energies, everything is (normal) plasma and then quark-gluon plasma and these electrically charged (and then chromodynamically charged) particles give rise to strong fields which give rise to pair-production of particles. So at high temperatures, thermal equilibrium means normal matter and antimatter should be at near equal levels and the rate of pair production should equal the rate of annihilation. The problem is the universe is not 50% antimatter so dense normal matter is a barrier to getting very, very hot.
https://en.wikipedia.org/wiki/R136a1
https://en.wikipedia.org/wiki/Pair-instability_supernova

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13. ### DaveC426913Valued Senior Member

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Yeah. Wut he sed. ^

14. ### Q-reeusValued Senior Member

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Connection between them is? How is dense ordinary matter a barrier to getting 'very very hot'? Given it's acknowledged ordinary matter naturally transforms into a particle/antiparticle plasma at sufficiently high temperatures. How hot is 'very very hot'?

15. ### rpennerFully WiredStaff Member

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Electrons have a rest-energy of 0.511 MeV, so when particles have an energy per degree of freedom near that, you can expect pair-production to be statistically likely.
1 GK is probably too low, so I imagine it to become significant between 3 GK and 6GK and certainly by 10GK. It's not an absolute barrier, but likely a turning point on the graph of heat capacity of matter.

https://www-users.cs.york.ac.uk/~susan/bib/ss/ast/model.pdf (I would like to see an update of this paper)

In the 1960's another turning point was found near 2 TK.
https://en.wikipedia.org/wiki/Hagedorn_temperature

16. ### Q-reeusValued Senior Member

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Hmm...this article has a passage on Hagedorn temperature(s) and it's assumptions and limited applicability: https://en.wikipedia.org/wiki/Absolute_hot

In short, at various temperatures there are marked changes in specific heat owing to new species being generated. With a hypothesized ultimate hot that may be a few orders of magnitude lower than the Planck temperature ~ 10^32 K, which itself may not be an ultimate limit. Still much unknown physics to say anything more concrete.

17. ### DinosaurRational SkepticValued Senior Member

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The properties of a Bose-Einstein Condensate seem to be experimental evidence strongly supporting the Uncertainty Principle.

Atoms or particles near absolute zero (circa -273 Celsius) are almost motionless. Note that temperature is a measurement of particle motion.

Since they are almost motionless, their positions becomes indeterminate. The particles become almost indistinguishable, losing their individual identity.

18. ### OysteinRegistered Senior Member

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Seems like it would be much easier to determine position. Explain what you mean.
Explain that too.

19. ### DaveC426913Valued Senior Member

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It would be more efficient if you just read up on Bose-Einstein Condensates.

20. ### OysteinRegistered Senior Member

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I know where the old guy got the statements he made. I just wanted him to explain, not parrot. Can you explain without quoting?

21. ### exchemistValued Senior Member

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I'm with Oystein: I don't follow this either. Surely a Bose-Einstein condensate is a state in which a large number of bosons all collectively inhabit the ground state. I do not recall any mention of particles losing their identity, or of the uncertainty principle needing to be invoked in order to explain it.

Can you refer me to a source that describes the phenomenon the way you are doing?

22. ### SchmelzerValued Senior Member

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No. Absolute zero temperature is simply the state of minimal energy. The ground state. This would not violate any uncertainty principles.

23. ### DaveC426913Valued Senior Member

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Well this is a site for kids, but it describes it succinctly:
"So, it's cold. A cold ice cube is still a solid. When you get to a temperature near absolute zero, something special happens. Atoms begin to clump. The whole process happens at temperatures within a few billionths of a degree, so you won't see this at home. When the temperature becomes that low, the atomic parts can't move at all. They lose almost all of their energy.

Since there is no more energy to transfer (as in solids or liquids), all of the atoms have exactly the same levels, like twins. The result of this clumping is the BEC. The group of rubidium atoms sits in the same place, creating a "super atom." There are no longer thousands of separate atoms. They all take on the same qualities and, for our purposes, become one blob. "

http://www.chem4kids.com/files/matter_becondensate.html

"Bose-Einstein condensate (BEC), a state of matter in which separate atoms or subatomic particles, cooled to near absolute zero (0 K, − 273.15 °C, or − 459.67 °F; K = kelvin), coalesce into a single quantum mechanical entity—that is, one that can be described by a wave function—on a near-macroscopic scale."
https://www.britannica.com/science/Bose-Einstein-condensate