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N bead Gaussian chain

  1. Oct 6, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider a system made up of joining together ##N## beads and ##N-1## springs. The positions of the beads is indicated by ##N## real numbers ##\left\{x_i\right\}_{i=1,...N}.## The Hamiltonian which characterises it is $$\mathcal H = \frac{k}{2}\sum_{i=1}^{N-1}(x_{i+1}-x_i)^2,$$ where k is a real and positive constant. Assume that the first bead is fixed at the origin (##x_1=0)##.

    Find the partition function of the system. Change variables to ##\left\{\sigma\right\}_{i=1...N}## where ##\sigma_i = x_{i+1}-x_i##.

    2. Relevant equations
    Assume the positions of the beads follow Boltzmann distribution ##\text{exp}(-\beta \mathcal H)##

    3. The attempt at a solution
    N beads and N-1 springs, so two springs connected to each bead. The position of the beads (except the one fixed at the origin) may extend to ##\pm \infty##. Then the partition function is the sum over all possible positions, so can write $$Z_N = \int_{-\infty}^{\infty} \text{d}x_1 \text{d}x_2 ... \text{d}x_N \text{exp}\left(-\frac{\beta k}{2} \sum_{i=1}^{N-1} \sigma_i^2\right) \delta(x_1),$$ where the delta function enforces the fixed position of the first bead. I can write this as $$Z_N = \int_{-\infty}^{\infty} \text{d}x_1 \text{d}x_2 ... \text{d}x_N \prod_{i=1}^{N-1} \text{exp}\left(-\frac{\beta k}{2} \sigma_i^2\right) \delta(x_1),$$ but how to continue?

    ##d \sigma_1 = dx_2## because ##x_1## is fixed but ##d \sigma_2 = d(x_3-x_2)##.

    Thanks!
     
  2. jcsd
  3. Oct 6, 2015 #2

    TSny

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    Are you sure you can allow the bead's x coordinates to vary independently between -∞ and ∞? That would mean the beads can pass through each other and springs would overlap each other.
     
  4. Oct 7, 2015 #3

    CAF123

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    Hi TSny,
    Ah I was imagining the problem in more than one dimension but I see that there is only one coordinate (x) labeling the positions of the beads. I suppose I could write the integrals as $$\int_{0}^{\infty} dx_2 \int_{x_2}^{\infty} dx_3....\int_{x_{N-1}}^{\infty} dx_{N}$$ if assuming they lie in positive x or replace infinity with -infinity if in negative x?
     
  5. Oct 7, 2015 #4

    TSny

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    Yes, that looks right. What is the Jacobian determinant associated with the change of variables from x's to σ's?
     
  6. Oct 8, 2015 #5

    CAF123

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    I think that is just one so can write ##dx_{i} \rightarrow d\sigma_{i-1}, i=2...N##

    I actually asked a TA about this question while in workshop and he said that indeed the limits would extend to +/- infinity. He said the problem is to viewed as a single point fixed but that the springs do not all need to be in a line. However, what then is the justification for the form of the Hamiltonian?
     
  7. Oct 8, 2015 #6

    TSny

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    I agree, the form of the Hamiltonian would not make sense if the beads are not restricted to the x axis.

    I also agree that the Jacobian determinant is 1.
     
  8. Oct 8, 2015 #7

    CAF123

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    Ok, thanks I'll raise this point with my TA next time I'm in class. If we use the limits for the integrals as in #3, how would I begin to evaluate such a Gaussian integral?
     
  9. Oct 8, 2015 #8

    TSny

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    What are the limits of integration for each ##\sigma##?
     
  10. Oct 8, 2015 #9

    CAF123

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    Ah, they're just from 0 to infinity. Ok so we have one gaussian integral to the (N-1)th power.
     
  11. Oct 8, 2015 #10

    TSny

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    Yes.
     
  12. Oct 9, 2015 #11

    CAF123

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    Many thanks. If I use the other interpretation of the problem and assume that the beads do not have to lie on a line then the result is that the limits on the integral are from ##-\infty## to ##+\infty## (i.e computing the ones in the OP) and so each Gaussian integral differs only by a half from the case we considered. Is this in any way to be expected?

    Would you agree that if the beads were not in a line, then there should be a y (and z) component of extension within the Hamiltonian too? Even if we suppose that they are not in a line and still restrict the motion of each bead to be in 1D, there will still be elongations of each spring in the y direction so I think what the TA said is maybe incorrect.

    The next part of the question asks about the average energy of the system. I know that for a discrete system ##\langle E \rangle = \sum_i E_i P(E_i)## but can you help me set up the form for this question? Since the variables are continuous there will be integrals now.

    Thanks!
     
  13. Oct 9, 2015 #12

    TSny

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    Yes, I think there would be y and z variables as well as x variables if the chain is allowed to flop around in 3D, and the variables could vary from minus infinity to plus infinity.

    It seems odd that your Hamiltonian does not include any kinetic energy.

    For <E>, you can use the second formula here https://en.wikipedia.org/wiki/Parti...s)#Calculating_the_thermodynamic_total_energy
    It works for partition functions that are integrals as well as sums, as you can easily verify.
     
  14. Oct 9, 2015 #13

    CAF123

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    Thanks, I will speak to the TA about this. I got (N-1)/2*kT for the average energy. The equipartition theorem allows 1/2 kT per degree of freedom in each energy. Each bead can only move in one 1D and the Hamiltonian was modelled with only potential energy so 1/2 kT per particle. One bead is fixed so therefore we have (N-1)/2 kT, as obtained. That seem fine? I guess this means the 1D interpretation of the problem makes more sense.

    The final part of the problem asks for the probability that the bead is at position y. It gives the formula $$P_N(y) = \frac{\prod_{i=1}^{N-1} \int d\sigma_i \delta(x_N-y) e^{-\beta \mathcal H}}{\prod_{i=1}^{N-1}\int d\sigma_i e^{-\beta \mathcal H}}$$ I can rewrite the delta as $$\delta(x_N - y) = \int_{-\infty}^{\infty} \frac{dq}{2\pi} e^{iq(x_N-y)}$$ and then to separate the ##\sigma_i## integrals try to complete the square.

    ##x_N = \sum_{i=1}^{N-1} \sigma_i##, so need to rewrite, after inserting everything into the boltzmann distribution, $$-\frac{\beta K}{2}\left(\sigma_i^2 - \frac{2i}{\beta K} \sigma_iq\right).$$ Is it possible to complete the square here such that the sigma and q terms decouple?
     
  15. Oct 9, 2015 #14

    TSny

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    Yes, you can complete the square. However, I think the resulting integral only comes out nice if you are integrating ##\sigma## from ##-\infty## to ##+\infty##, rather than from 0 to ##+\infty##. I hope I haven't led you astray by interpreting the problem in such a way that the beads must stay on the x axis without passing through each other.
     
  16. Oct 10, 2015 #15

    CAF123

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    Ok I can write $$-\frac{\beta K}{2}\left(\sigma_i^2 - \frac{2i}{\beta K}\sigma_i q\right) = -\frac{\beta K}{2} \left(\sigma_i - \frac{iq}{\beta K}\right)^2 - \frac{q^2}{2\beta K}$$ so I will need to evaluate, putting everything together $$\int_{-\infty}^{\infty} e^{-iqy - iq^2/2\beta K} \prod_{i=1}^{N-1} \int d_{\sigma_i} e^{-\frac{\beta K}{2}(\sigma_i - iq/\beta K)^2}$$ Setting ##z= \sigma_i - iq/\beta K##, if I take a lower limit of ##\sigma_i## to be ##-\infty## then the lower limit in z is also this, whereas if I take the lower limit to be 0 then the integral has finite lower boundary so becomes more difficult to calculate. Is this what you mean?

    Using -infinity as the lower boundary I get the probability to be $$P_N(y) = \sqrt{\frac{\beta K}{2\pi}} e^{-\beta K y^2/2}$$ which looks right since if I integrate this over all positions y, I get one.

    So I think either way the problem is to be interpreted as beads constrained to move along the x axis, either moving through each other, in which case the limits on the sigmas extend from -infinity to infinity, or not. Otherwise there should be added terms to the Hamiltonian and more integrals over the y and z degrees of freedom.
     
    Last edited: Oct 10, 2015
  17. Oct 10, 2015 #16

    TSny

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    That all looks good to me.
     
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