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Statistical physics - average length of polymer

  1. Dec 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A polymer consits of ##10^{20}## monomers, each 2 nm long. One end of the polymer is hanged at the ceiling on the other end, we have a load of ##4\cdot 10^{-10}g##.
    Calculate the average potential energy!


    2. Relevant equations



    3. The attempt at a solution

    Well... I don't know...

    Each monomer adds ##acos\vartheta ## to the total length. So... all monomers together ##\sum_{i=0}^{N}acos\vartheta _i ## Now how do i get the average value from that? That i don't know...

    I also tried using phase integral ##e^{-\beta F}=\int_{0}^{2\pi }exp(\beta mg\sum_{i=0}^{N}acos\vartheta _i )d\varphi ## from where the potential energy would be equal to ##\frac{\mathrm{d} \beta F}{\mathrm{d} \beta }## ....

    Can anybody please help me here? Thanks!

    So the key here is to somehow calculate the average ##cos\vartheta ## where ##\vartheta ## can be anything between 0 and ##\pi##
     
    Last edited: Dec 28, 2013
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  3. Dec 28, 2013 #2

    TSny

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    The integration should be over all of phase space; i.e., over all orientations of each of the monomers.

    Think about how to specify the orientation of a monomer in 3D space.

    Are you sure you have 1026 monomers? :uhh: (How long would the polymer be in light years if all of the monomers were in the same orientation?)

    What are the units for the load?
     
  4. Dec 28, 2013 #3
    I've edited the first post: It's ##10^{20}## monomers and the units of the load are grams.

    About integration over all of phase space...:
    Well, yes, ##\varphi ## is the angle between in the xy plane, so ##\varphi \in \left [ 0,2\pi \right ]## and ##\vartheta ## is angle between the z (vertical) axis and is ##\vartheta \in \left [ 0,\pi \right ]##

    I somehow managed to make calculate the average ##\cos \vartheta## but I don't know how to continue:

    ##<\cos \vartheta >=\frac{\int_{0}^{2\pi }\int_{0}^{\pi }\cos\vartheta exp(\beta mgacos\vartheta )d\vartheta d\varphi }{\int_{0}^{2\pi }\int_{0}^{\pi }exp(\beta mgacos\vartheta )d\vartheta d\varphi }##

    if ##z=\cos \vartheta## and ##u=z## than ##du=dz##, therfore

    ##<\cos \vartheta >=\frac{\int_{-1}^{1 }uexp(\beta mgau )du}{\int_{-1}^{1}exp(\beta mgau )du }##

    ##<\cos \vartheta >=\frac{1}{\beta mga}\frac{e^{\beta mga}(mga-1)+e^{-\beta mga}(mga+1)}{e^{\beta mga}-e^{-\beta mga}}##

    ##<\cos \vartheta >=coth(\beta mga)-\frac{1}{\beta mga}=\doteq 0.5186##

    So the average potential energy of load is ##W_p=-mgNa<\cos \vartheta >=-0,495J##

    That is IF this is not completely wrong?

    Now I have to calculate the free energy (because the second part of the problems is to calculate the change of heat if we hang load twice as heavy!)

    Using equation ##F=<E>-TS=W_p-TS## where S stands for entropy and T for temperature, I could get the heat but i would need F... ??? :/
     
  5. Dec 28, 2013 #4

    TSny

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    The area of a patch on a unit sphere is ##\sin\theta \; d\theta d\varphi##. It appears that you left out the ##\sin\theta## but somehow made up for it when you did your substitution. I think your final result looks ok, but I can't check your numbers since I don't know what temperature you're working at.

    To find ##F## I would suggest finding the partition function ##Z## which will entail an integration similar to what you have already done. You can use ##Z## to get the average energy as a check on the way you already found the average energy.
     
  6. Dec 28, 2013 #5
    Oh, I completely forgot about Jacobian (that ##\sin \vartheta##). Lucky substitution I guess, but thank you for noticing!

    ##10^{20}## monomers, each 2 nm long at load of ##4\cdot 10^{-10} g## and at constant temperature of 300 K.

    Hmm... about F... Will this work:

    ##e^{-\beta F}=\int_{0 }^{2\pi}\int_{0}^{\pi }\sin \vartheta e^{\beta mga\cos \vartheta}d\vartheta d\varphi##

    ##e^{-\beta F}=2\pi\int_{-1}^{1}e^{\beta mgau}du##

    so

    ##-\beta F=ln(\frac{2\pi}{\beta mga})+ln(e^{\beta mga}-e^{-\beta mga})## ?
     
  7. Dec 28, 2013 #6

    TSny

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    Does your first equation take into account that you have N monomers? Shouldn't ##F## end up being extensive (proportional to N)?
     
  8. Dec 28, 2013 #7
    No no, that's for one monomer only..

    For N monomers should be something like...

    ##e^{-\beta F}=2\pi N\int_{-1}^{1}e^{\beta mgau}du##

    and

    ##-\beta F=ln(\frac{2\pi N}{\beta mga})+ln(e^{\beta mga}-e^{-\beta mga})##

    If i am not mistaken...
     
  9. Dec 28, 2013 #8

    TSny

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    No. Notice how your result for F is not proportional to N.

    Note $$e^{\sum{x_i}} = \Pi e^{x_i}$$ and refer to your expression for ##e^{-\beta F}## in your first post.
     
  10. Dec 28, 2013 #9
    hmm, does this mean that ##e^{-\beta F}=\int e^{-\beta mga< \cos >N} d \Gamma ## ?

    That would mean that ##-\beta F=Nln(\frac{2\pi}{\beta mga})+Nln(e^{\beta mga}-e^{-\beta mga})##
     
  11. Dec 28, 2013 #10

    TSny

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    What does ##d\Gamma## denote? Why does the average, <cos>, appear in the integral?

    The partition function is an integral over all the angle variables of all the monomers (2N variables in all).

    The integrand should be ##e^{-\beta E}=e^{ \beta mga \sum{\cos \theta_i}}##. You can write the integration as a product of identical integrals such that each integral is over just the angles of one monomer. (Use ##e^{a+b} = e^ae^b##.)

    Somehow, this looks correct. You can write the argument of the last log in terms of sinh(βmga) if you want.
     
  12. Dec 29, 2013 #11
    We used notation ##d\Gamma ## for integrating over all variables in phase space, so ##d\Gamma \equiv d\vartheta d\varphi##.
    So, 2N variables, because each monomer is described by two angles for N monomers that's 2N variables. That's something I should probably know in the middle of the night...
    Ok, now let me put it this way... I know exactly what you are trying to say here but since you already started it, let me show you where it all stops for me...

    So partition function for one monomer should be something like ##e^{-\beta F}=\int_{0}^{2\pi }e^{\beta mga\sum_{i=0}^{N}\cos \vartheta _i}d\varphi##

    which is, like you already mentioned it, equal to:

    ##e^{-\beta F}=2\pi e^{\beta mga\sum_{i=0}^{N}\cos \vartheta _i}##

    ##e^{-\beta F}=2\pi \prod_{i=0}^{N}e^{\beta mga\cos \vartheta _i}##

    For N monomers,, that is 2N variables, I'm guessing that's ##e^{-\beta F}=(2\pi )^{2N}\prod_{i=0}^{N}e^{\beta mga\cos \vartheta _i 2N}##

    Which probably means that i was wrong by factor 2 here: ##-\beta F=Nln(\frac{2\pi}{\beta mga})+Nln(e^{\beta mga}-e^{-\beta mga})##

    But, my point is... that, i have got absolutely no idea what to do with this product now ##e^{-\beta F}=(2\pi )^{2N}\prod_{i=0}^{N}e^{\beta mga\cos \vartheta _i 2N}##

    That is why i've put ##<\cos >## in integral - to avoid this product.
     
  13. Dec 29, 2013 #12

    TSny

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    The sum occurring on the right side is over all monomers. So this can't be the partition function of just one monomer.

    Suppose N = 2 so there are only 2 monomers. Then $$Z = \int e^{b\sum_{i=0}^{2}\cos \vartheta _i}d\Gamma= \int e^{b(\cos \vartheta_1+\cos \vartheta_2)}\sin \vartheta_1 d\varphi_1d\vartheta_1 \sin\vartheta_2d\varphi_2d\vartheta_2$$
    where ##b \equiv \beta mga##. [Above expression edited to correct for leaving out the ##\sin## functions.]

    Can you factor this expression into an integral over ##\varphi_1## and ##\vartheta_1## times another identical integral over ##\varphi_2## and ##\vartheta_2##?
     
    Last edited: Dec 29, 2013
  14. Dec 29, 2013 #13
    that's factor ##2\pi ## for each integration by ##d\varphi _i## and factor ##\frac{1}{b}e^{b\cos \vartheta _i}## for each integration by ##d\vartheta _i##.

    for two monomers: ##Z=\frac{(2\pi )^2}{b^2}e^{b(\cos \vartheta _1 + cos \vartheta _2)}##
     
  15. Dec 29, 2013 #14

    TSny

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    What about the integration over ## \vartheta _1## and ##\vartheta _2##?


    The key point is that ##Z## for two monomers may be written as

    $$Z = \int e^{b(\cos \vartheta_1+\cos \vartheta_2)}\sin \vartheta_1 d\varphi_1d\vartheta_1 \sin\vartheta_2d\varphi_2d\vartheta_2 = \int e^{b\cos \vartheta_1}e^{b\cos \vartheta_2}\sin \vartheta_1 d\varphi_1d\vartheta_1 \sin\vartheta_2d\varphi_2d\vartheta_2 $$
    $$=\int e^{b\cos \vartheta_1}\sin \vartheta_1 d\varphi_1d\vartheta_1 \int e^{b\cos \vartheta_2}\sin\vartheta_2d\varphi_2d\vartheta_2$$

    The two double-integrals are identical (just a renaming of the dummy integration variables). So,
    $$Z =\left[ \int e^{b\cos \vartheta_1}\sin \vartheta_1 d\varphi_1d\vartheta_1 \right]^2$$

    How would this extend to N monomers?
     
  16. Dec 29, 2013 #15
    Well ##Z=(\int e^{b \cos \vartheta _1}\sin \vartheta _1d\vartheta _1 d\varphi)^N=\frac{(2\pi)^N}{b^N}e^{b\cos \vartheta _1 N}## or...?

    If yes... Stilll... What to do with ##\cos \vartheta _1 ## in exponent?
     
  17. Dec 29, 2013 #16

    TSny

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    OK. Good.

    You didn't do the integration over ##\vartheta_1##.

    ##\int e^{b \cos \vartheta _1}\sin \vartheta _1d\vartheta _1 d\varphi = 2 \pi \int e^{b \cos \vartheta _1}\sin \vartheta _1 d\vartheta _1##

    You need to evaluate this integral.
     
  18. Dec 30, 2013 #17
    I fell so stupid right now... :D
    Is it not
    ##2\pi \int e^{b\cos \vartheta _1}\sin \vartheta _1 d\vartheta _1=2\pi \int e^{bu}du=\frac{2\pi }{b}e^{b\cos \vartheta _1}##
     
  19. Dec 30, 2013 #18

    TSny

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    OK (except for the overall sign of your result). But you need to evaluate at the limits. (I've been too lazy to type the limits of integration.)
     
    Last edited: Dec 30, 2013
  20. Dec 30, 2013 #19
    Aha...

    ##Z=(\frac{2\pi}{b})^N(e^{b}-e^{-b})^N##

    That is all, am I right?
     
  21. Dec 30, 2013 #20

    TSny

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    Yes, that's right.
     
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