Doubts T06

Good Evening
I have a doubt in the T06.1
When you ask us to calculate the average potential energy you mean, the time average?, or the average like

\int d\Gamma V(x) \exp{-\beta H}/ \Omega ?

Comentarios

  1. Thomas22 de marzo de 2020 a las 2:48 p.m.

    The average with respect to the canonical ensemble -- not the time average. In fact you should use the equipartition theorem as presented in the lecture notes.

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    Respuestas
    1. BrandonCamberos23 de marzo de 2020 a las 6:00 p.m.

      Ok, I was reviewing and I think it would be like this:

      = -\frac{1}{\beta}\ \frac{\partial [Ln(\Omega)]}{\partial \beta} - \frac{k_B T}{2}

      Am I right?

      PD: I tried to put the formula as image but I couldn't

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    2. BrandonCamberos23 de marzo de 2020 a las 6:01 p.m.

      Este comentario ha sido eliminado por el autor.

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    3. BrandonCamberos23 de marzo de 2020 a las 6:02 p.m.

      Average(H_pot) =*

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  2. Bianca Glez23 de marzo de 2020 a las 8:09 a.m.

    Good morning. I have another questions: in the part 2, is eta a constant? And in the third part, (V; p) is a vector, isn't it? Is not a binomial coeficient. Thank you.

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  3. Thomas23 de marzo de 2020 a las 10:41 a.m.

    Hi Bianca, I am sorry, but I don't understand to which part you refer to.
    Or do you refer to problem 3. There, "p" is pressure and "V" is volume.
    The problem there, is about classical thermodynamics -- not statistical physics.

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  4. Bianca Glez23 de marzo de 2020 a las 11:17 a.m.

    Hi, is next to (i)An expanssion.... (V; p)= l (V_b; p_b)+(1-l)(V_a; p;a).
    My question was that if it is matrix notation, I mean, V= l *V_b+(1-l)*(V_a); p=l*p_b+(1-l)*p_a.

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