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Space Science

Is There Such a Thing As Absolute Hot? 388

AlpineR writes "Is there an opposite to absolute zero? An article from PBS's NOVA online explains several theories of the maximum possible temperature. Maybe it's the Planck temperature, 10^32 K, beyond which the known laws of physics break down. Or maybe just 10^30 K, the limit of some versions of string theory. If space is actually 11-dimensional then the maximum temperature could even be as low as 10^17 K, attainable by the Large Hadron Collider. Or maybe infinite temperature wraps around to negative temperature and absolute hot is the same as absolute cold."
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Is There Such a Thing As Absolute Hot?

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  • Re:Speed (Score:5, Informative)

    by Cyberax ( 705495 ) on Monday December 24, 2007 @12:11PM (#21806432)
    Nope.

    Temperature depends on particles _energy_. At low temperature particle energy is calculated as E=m*v^2/2, but if you start to get closer to the light speed then the _MASS_ of a particle will grow. So you can get arbitrarily large energy as you approach the "c" limit.
  • Re:Speed (Score:5, Informative)

    by BlueParrot ( 965239 ) on Monday December 24, 2007 @12:20PM (#21806510)
    Temperature is defined in terms of the energy, not speed. At high velocities the mass of particles grow with their speed as per Einstein's theory, so even thou the top speed is limited, energy is in fact not. As a particle's speed aproaches the speed of light its energy diverges. This is in fact WHY you can't accelerate particles to the speed of light. As you get closer to C the particle mass starts growing rapidly so larger and larger amounts of energy is needed for smaller and smaller increments in speed. Thus you can't accelerate a particle to C using only a finite amount of energy. This effectively means that realitivity doesn't limit temperature. There may of course be other limits involved.
  • by explosivejared ( 1186049 ) <hagan@jared.gmail@com> on Monday December 24, 2007 @12:22PM (#21806534)
    Maxwell-Boltzman probably wouldn't apply anyways, because at >10^32 K it would be pretty hard to be in thermal equilibrium. As for your question... maybe I just don't understand the physics enough, but wouldn't temperature still be defined as the average of atomic vibration, in this case the very large atomic vibrations.
  • by sbaker ( 47485 ) * on Monday December 24, 2007 @12:22PM (#21806538) Homepage
    Temperature is basically the average kinetic energy of the particles, and kinetic energy is half the mass times velocity squared, when things start to get very hot, the particles would eventually start getting up to relativistic speeds.

    This has lead some people to suggest that the cosmic speed limit (the speed of light) imposes a cosmic temperature limit - but that's NOT the case.

    As things start to move closer and closer to the speed of light, relativity says that their mass increases (as seen from the perspective of an outside observer). Whilst there is a cosmic speed limit - as you approach it, your mass increases without limit. Since unlimited mass and finite velocity means unlimited kinetic energy, relativity does not impose a cosmic temperature limit.

    If there is a cosmic temperature limit, it's caused by something else.
  • by fermion ( 181285 ) on Monday December 24, 2007 @01:02PM (#21807034) Homepage Journal
    Most of us have difficulty differentiating heat and temperature [uh.edu]. I am not even going to try to come up with a simple definition here. But, as the referred transcript states, if you have a very thin gas, temperature does, in some way, relate directly to motion. Therefore, absolute zero is approximately defined as the point where the atom in gas, where the atoms do not hit each other often, would stop moving. At present, I know of now peer review paper reporting 0 K reached, though some groups have come very close.

    So the question of maximum temperature is not so silly. There are a number of ways to approach it from various definitions. If we have a few atoms in a large space, then perhaps we can drive those atoms to the speed of light, but no further. If we think of it thermodynamically, as Dr. Lienhard suggests, then we can ask is there an limit to the heat that can be driven between two systems. Such a limit would suggest a maximum temperature if we assume newtons law of cooling, which is itself is approximate, can be applied a large temperature differences, which it probably cannot.

    In any case, nature, at least we way that science approaches it, appears to abhor vacuums [marinelayer.com] and black holes, both of which seem to exist, but don't seem to make sense. The question is apt as we have seem that assuming infinities do us little good.

  • by Roger W Moore ( 538166 ) on Monday December 24, 2007 @01:18PM (#21807220) Journal
    But no one has tied relativity to quantum mechanics yet. Therefore those speed limits only apply to a narrow vision of the universe.

    Sorry but Special Relativity and Quantum Mechanics are very well integrated: it was first done by Dirac in ~1932 and led to the prediction of anti-matter which was discovered a few years later with the positron (anti-electron). The Dirac (along with the Klein-Gordon and Proca) equations form the underpinnings of Quantum Field Theory which is what we use in particle physics to describe all the fundamental particles of nature (that we know of) and how they interact (except via gravity). This has Lorentz invariance built into it and is a complete union of QM and SR.

    What is harder is to unify QM and GR. This has not been successfully done yet. You can create a quantized gravitational field relatively easily but the problem is that you have to specify a maximum energy scale in order to normalize it (in 3+1D at least). This is bad because there is no justification for a maximum energy scale once you include gravity where the physics will change. Hence either the theory is wrong or there is something else at some really high energy. In either case you cannot use it to make meaningful predictions and so we say we have no valid way, yet, to unify QM nd GR.
  • by Anonymous Coward on Monday December 24, 2007 @01:20PM (#21807234)
    You just discovered the Planck Temperature :)
    http://en.wikipedia.org/wiki/Planck_temperature [wikipedia.org]

    Temperature can be expressed in electron volts:
    http://en.wikipedia.org/wiki/Temperature [wikipedia.org]

    electron volts can be used as a unit of mass:
    http://en.wikipedia.org/wiki/Electronvolt [wikipedia.org]

    Based on the Planck Units, all mass above the Planck Temperature turns to energy.

    So if you compress a gas so that one Planck Mass of the gas fits inside a sphere one Planck Length in diameter, it would reach the Planck Temperature and instantly becomes a black hole and turn completely into energy via hawking radiation.

    The plank mass is 2.176 × 10-8 kg
    Convert mass to electron volts, you get:
    (speed of light ^2) * (plank mass) = 1.22064661 × 10^28 electron volts

    Convert electron volts to kelvin, you get:
    (speed of light ^2) * (plank mass) * 11604 K = 1.41643833 × 10^3 Kelvin

    Which is plank's temperature.

    (feel free to correct me, I'm basically parroting Wikipedia)
  • by Mr. Underbridge ( 666784 ) on Monday December 24, 2007 @01:23PM (#21807268)

    I have to wonder about the definition of temperature at such high energies. I would think it would be difficult to envisage a situation where you have anything resembling a Maxwell-Boltsman distribution at 10^33 K, so just what is meant with temperature in this case?

    If you're referring to exp(-B/kT), then the high temperature will swamp the B (activation energy), meaning that all states will be effectively uniformly populated. So at infinity, I believe a Botzmann distribution ends up as pretty much a uniform distribution.

  • by Roger W Moore ( 538166 ) on Monday December 24, 2007 @01:35PM (#21807418) Journal
    Temperature is directly related to the velocity of the atoms in a gas or plasma.

    No - it is directly related to the kinetic energy of the atoms in a gas and the electrons and ions/nuclei in a plasma (there are no atoms in a plasma). In classical physics this is 0.5mv^2 but this is just the low energy approximation of the true KE which is "ymc^2-mc^2" where y=gamma=1/sqrt(1-v^2/c^2). As you can see this has no upper bound.

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