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Is the Planck temperature ($1.416×10^{32}$ Kelvin), the hottest possible temperature that can ever be reached, with absolute zero as it's opposite analog?

All I know is a particle with that temperature would have energy high enough, so that its gravitational force would be comparable to other forces and the behavior is not well understood without a theory of quantum gravity. But is it possible to have a state with a higher temperature ? If not, are there any thermodynamic reasons ?

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The Planck temperature isn't the hottest possible temperature in the same sense that zero Kelvin is a theoretical minimum. It is simply the temperature at which it's black body radiation is of the order of the Planck length.

The Planck length is the length scale at which it is theorised that quantum-gravitational effects become significant. Quantum-gravity is a hot topic of research and we simply do not have sufficient understanding of it to know the significance of the Planck length at this point.

So in summary, the Planck temperature is just the temperature at which we need to refer to quantum gravity to have a proper handle on what is going on. Since we don't have a proper grasp of quantum-gravity, anything that hot is just too hot to understand.

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  • $\begingroup$ That's what I suspected, just wanted to verify because wiki calls such states to be absolute hot. $\endgroup$
    – smiley06
    Commented Jan 25, 2016 at 10:39
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We need to think about what it means for one system to be "hotter" than another, or to have a temperature $T$. Thermodynamics defines the temperature of systems in thermal equilibrium. If an energy level $i$ of energy $U_i$ has degeneracy $g_i$, its occupation level $\propto g_i \exp \left( -\beta U_i \right) $ for some constant $\beta$ at thermal equilibrium. We then define $T := \tfrac{1}{k_B \beta}$. When all energy is at the lowest level, $T = 0$. When $T \in \left( 0,\,\infty\right)$ high-energy levels are occupied, but their occupation-to-degeneracy ratio is lower than for low-energy levels. When $T=\infty$ , this low-more-occupied-than-high disparity is replaced with equality, i.e. occupation $\propto g_i$ . Clearly, we are gradually moving energy up to higher energy levels, thereby increasing the mean energy. But if you obtain a state of even higher mean energy by obtaining a high-more-occupied-than-low disparity, mathematically $T < 0$ . These "negative-temperature" systems (which can be experimentally prepared but are short-lived) are "hotter" than infinitely hot ones (or indeed systems at the Planck temperature), rather than being below absolute zero (which is impossible).

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  • $\begingroup$ $ T<0$ are "hotter than infinitely hot ones'' ?? !! I knew there was this asymptotic rate of cooling at 0 K, so going below that is impossible. $\endgroup$
    – smiley06
    Commented Jan 25, 2016 at 10:36
  • $\begingroup$ @smiley06 You can't go "below" (i.e. colder than) absolute zero, but "negative temperatures" are actually "above" (i.e. hotter than) absolute zero, or $10^3 \text{K}$, or $+\infty \text{K}$. This isn't the usual ordering of $\mathbb{R}\cup\left\{\infty\right\}$. $\endgroup$
    – J.G.
    Commented Jan 25, 2016 at 10:39
  • $\begingroup$ Can you provide me with a reference for this principle ? $\endgroup$
    – smiley06
    Commented Jan 25, 2016 at 10:41
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    $\begingroup$ The textbook example is population inversion in a three- or four-level laser. Sorry; I'm not sure how one hyperlinks in comments. en.wikipedia.org/wiki/Negative_temperature#Lasers en.wikipedia.org/wiki/Population_inversion#Three-level_lasers $\endgroup$
    – J.G.
    Commented Jan 25, 2016 at 11:06

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