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Regarding heat

  1. Oct 24, 2007 #1
    My first post!

    Bear with me here - I'm a weekend warrior of particle physics and am just learning because it amazes me.

    In my journey through nuclear fusion and quantum theory, I first learned that our sun isn't big enough to overcome the Coulomb barrier between two protons necessary to allow them to fuse, that the basic force of gravity isn't enough, (Insert need for quantum fields here etc). Then I later read that the sun also isn't hot enough at the core, (the first article never mentioned heat...), and that both heat and gravity are necessary for fusion. My question is, what is heat then? If temperature can slow down or speed up particle movement, then why isn't it a basic force?

    I learned that something as simple as the strong nuclear force is caused by the transfer of gluons, what sub-atomic particle causes temperature change then?

    Go easy on me, but not too easy. My IQ hovers around 120, but when I want to learn something I'm relentless.
  2. jcsd
  3. Oct 24, 2007 #2
    Temperature and heat are statistical quantities that are needed to describe many particle systems using only a few variables. They are not fundamental quantities. Suppose that you know everything there is to know about atoms, nuclei, the forces between them etc. Then you can calculate how a certain intial state will evolve in time.

    If you try to use your knowledge to the Sun and try to compute the energy output of the Sun, then you'll face a formidable problem. The Sun consists of a huge number of particles and it is not possible in practice to consider all the reactions between all the particles in your computations. Instead, what you should do is use statistical methods.

    Now, you know that to do statistical computations, you need to know about the probabilities to find the system in certain states. E.g. if you throw dice and add up all the numbers the dice are showing, you can estimate this number because you know that the numbes 1 to 6 are equally likely.

    Suppose you have a box containing particles. Then given all the information you have, e.g. the volume and the total energy thast is inside the box, there is some finite number of physical states, X, the particls can be in (this is a consequence of quantum mechanics). The laws of physics say that all these X states are equally likely (actually this is a postulate that is consistent with the laws of physics, it hasn't been rigorously proven). So, any one particular state has probability 1/X.

    Now, suppose you have two such boxes and you bring them into contact so that energy can flow between them. The number of states for the particles in box 1 is X but the value of X depends on the energy contained in this box, so let's write it as X(E1) where E1 is the energy in box 1. Similarly box 2 can be in Y(E2) possible states if the energy in that box is E2. The combined system of box 1 and 2 together can thus be in X(E1)*Y(E2) possible states.

    If energy can flow between box 1 and box 2 then we have to consider box 1 and box 2 together as a single isolated box. All possible states are equally likely for this box. This means that an energy distribution over box 1 and box 2 that is consistent with the most number of states for the combined box is the most likely energy distribution.

    So, the energy E1 and E2 in boxes 1 and 2 will change until
    X(E1)*Y(E2) becomes maximal. E.g. E1 can decrease and E2 will then increase by the same amount. This energy flow is what we call "heat". It is thus a purely statistical effect. Now, if you know some calculus you know that you can compute the maximum by taking setting a derivative equal to zero. In this case we can do this as follows. If E1 cnages E2 will also change because the total energy in the box must remain constant, say E. So E2 = E - E1

    The total number of states is thus always:

    X(E1)*Y(E - E1)

    The logarithm of the total number od states is:

    Log[X(E1] + Log[Y(E-E1)]

    Thate the derivative w.r.t. E1 and set it equal to zero:

    d Log[X(E1]/dE1 + d Log[Y(E-E1)]/dE1 = 0 --->

    d Log[X(E1]/dE1 = - d Log[Y(E-E1)]/dE1 ---->

    d Log[X(E1]/dE1 = d Log[Y(E2]/dE2

    In the last line we have introduced E2 again and used that the derivative w.r.t. E2 is minus the derivaive w.r.t. Now this equations says that the energies in the boxes in equilibrium will become such that some quantity that only depends on the property of box 1 will be come equal to the same kind of property of box 2. This quantity is used to define temperature.
  4. Oct 25, 2007 #3
    Thanks muchly Count. Another tangent to explore. :)
  5. Oct 26, 2007 #4
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