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A What is the formula 1/(dS/dE)>>0 and how does it apply?

  1. Feb 19, 2017 #1
    I am only aware that the formula has to do with entropy/thermodynamics. I could really use the help on how it applies in physics and what the formula is really about.
     
  2. jcsd
  3. Feb 20, 2017 #2

    stevendaryl

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    In that equation, [itex]S[/itex] is the entropy and [itex]E[/itex] is the energy. In thermodynamics, temperature can be defined as:

    [itex]\frac{1}{T} = \frac{dS}{dE}[/itex]

    So your inequality just says [itex]T \gg 0[/itex]. So the temperature is well above absolute zero.
     
  4. Feb 20, 2017 #3

    stevendaryl

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    That definition of temperature assumes that entropy increases with energy (so [itex]T[/itex] is always positive), which is true for classical thermodynamics, but for systems with a discrete number of states, it's possible for [itex]S[/itex] to decrease with [itex]E[/itex], which leads to the bizarre notion of a negative absolute temperature.
     
  5. Feb 20, 2017 #4
    So the inequality is saying that the temperature is above zero? If so, you talk of the "bizarre notion" of a negative absolute temperature that some people infer, but that is not possible correct?
     
  6. Feb 20, 2017 #5

    stevendaryl

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    The symbol [itex]\gg[/itex] means "much greater than". So the temperature isn't just positive, it's pretty high.

    Negative temperatures are not possible in classical thermodynamics, but there are quantum systems where a negative temperature is possible. A negative temperature means that the entropy goes down instead of up when the system gets more energy.
     
  7. Feb 20, 2017 #6
    Alright! Thank you so much you have helped me better understand this and even went deeper into the meaning without making it hard to understand. I really couldn't thank you enough :)
     
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