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Thermal equillibrium

  1. Jan 13, 2013 #1
    as we know that atomic collisions are perfectly ellastic
    perfectly ellastic collisions are given by
    v1=u1(m1-m2)+2(m2)u2 / m1+m2
    and v2=u2(m2-m1)+2(m1)u1 / m1+m2
    when m1=m2
    v1=u2
    v2=u1

    we know heat is stored in the form of vibrational energy(kinetic energy)
    so to exchange energy they should collide
    according to ellasticity the should exchange their energies
    T1° will become T2° and T1° will become T2°
    the how will they come to equillibrium
     
  2. jcsd
  3. Jan 14, 2013 #2

    Avodyne

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    Your formulas apply only in one dimension. In one dimension, it is true that colliding particles will not thermalize.
     
  4. Jan 16, 2013 #3
    this is just a doubt in ellasticity and not thermodynamics
    in 2d collisions
    when m1=m2
    v1=(u2+u1)cos(x/2)
    v2=(u2-u1)sin(x/2) [scource : http://en.wikipedia.org/wiki/Elastic_collision ]
    initial total energy(T.E)=1/2(m^2)(u1^2 +u2^2)
    final T.E=1/2(m^2)((u2-u1)^2)

    according to physics laws initial T.E should be equal to final T.E

    hence , 1/2(m^2)(u1^2 +u2^2)=1/2(m^2)((u2-u1)^2)

    (u1^2 +u2^2)=((u2-u1)^2)

    which is only possible when u1 or u2 is equal to zero
    but when u1 & u2 are not equal to zero they violate the law of conservation of energy

    why is that so????
     
    Last edited: Jan 16, 2013
  5. Jan 16, 2013 #4

    Bill_K

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    Please move this to the Classical Physics forum. This has nothing to do with Quantum Mechanics.
     
  6. Jan 17, 2013 #5


    could you tell me how to do it
    i'm new to PF
     
  7. Jan 17, 2013 #6

    Avodyne

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    Your formulas do not appear in the wikipedia article that you cite, and they are not correct.
     
  8. Jan 18, 2013 #7

    look for 2d collisions i just modified the relative initial velocity as u1+u2

    and before that plz substitute m1=m2
     
  9. Jan 18, 2013 #8

    Avodyne

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    Your formulas are from the 1d section.
     
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