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Vibrations of Triatomic Molecules, Landau-Lifshitz

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data

    https://ia701205.us.archive.org/11/items/Mechanics_541/LandauLifshitz-Mechanics.pdf [Broken]

    Page 72 of the book itself, but Page 81 of the PDF. Problem 1.

    2. Relevant equations

    See Section 24 of the book.

    3. The attempt at a solution

    So far I have completed and understand the kinetic energy part of both the transverse and longitudinal modes of oscillation of the molecules for Problem 1, as well as the potential energy for the longitudinal mode. For some reason I cannot figure out, after many hours of tedious thinking why they have the potential energy they do for the transverse case.

    If anyone could explain this, it would be very helpful.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 23, 2014 #2

    TSny

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    Hello, electricspit.

    Consider arbitrary vertical displacements of the atoms as shown in the attached figure. Please show us your attempt at expressing the angle δ in terms of y1, y2, y3, S12, and S23.

    EDIT: I was assuming your question has to do with the expression given by LL for δ. If not, are you asking why they take the potential energy to be of the form (1/2)k2l2δ2?
     

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    Last edited: Mar 23, 2014
  4. Mar 23, 2014 #3
    Hello TSny,

    Thanks for the reply, I was asking the latter rather than the former, but if that is the correct diagram for this case I am completely wrong in my initial assumptions.

    They express delta in terms of y1, y2, y3, and l alone, so I'm wondering where the S12, S13 come into play.

    My attempts at finding delta were obviously wrong because I wasn't using a diagram like this because they mentioned delta being the angle ABA with y1=y3, so I assumed it was the open angle at y2.

    Does this make sense?

    EDIT: Well I reread the LL question definition of delta and it seems that I missed the word "deviation". How unfortunate. Now I have:

    [itex]\delta=\frac{y_1-y_2}{\ell}+\frac{y_3-y_2}{\ell}[/itex]

    Which is correct.

    Now for the potential energies I had started with:

    [itex]V=\frac{1}{2}k(y_1-y_2)^2+\frac{1}{2}k(y_3-y_2)^2[/itex]

    Which, resting upon the fact that:

    [itex]y_1-y_2=\frac{\ell\delta}{2}[/itex]
    [itex]y_1=y_3[/itex]

    Would give me:

    [itex]V=\frac{1}{4}k\ell^2\delta^2[/itex]

    Which is half of the answer they're getting.
     
    Last edited: Mar 23, 2014
  5. Mar 23, 2014 #4

    TSny

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    OK. For arbitrary displacements, you should be able to show that ##\delta = \theta_1- \theta_2 ## where ##\theta_1 = \sin^{-1}\frac{y_2-y_1}{S_{12}}##, etc. See attached figure. For small displacements and keeping terms only to first order, you get the expression for ##\delta##.

    Note that they use a different force constant k2 for the transverse displacement compared to the linear displacement constant k1. The force constant k2 for the bending mode will depend on the peculiar stiffness properties of the molecular bonds and does not necessarily have a simple relationship to k1.

    It seems to me that the form they take for the potential energy just comes from the initial assumption that the potential energy depends only on ##\delta##: ##V(\delta)##. Then, expanding: [tex]V(\delta) \approx V(0)+V’(0)\delta +\frac{1}{2} V’’(0) \delta^2 + ….[/tex]
    You can take ##V(0) = 0## and argue that ##V’(0) = 0##.

    So, ##V(\delta) \approx \frac{1}{2} V’’(0) \delta^2## = (const.) ##\delta ^2##

    They choose to write the constant in the last expression as ##\frac{1}{2}k_2l^2##, which essentially defines ##k_2##.
     

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    Last edited: Mar 23, 2014
  6. Mar 23, 2014 #5
    Ah yes I was just lazy and wrote [itex]k[/itex] when I really meant [itex]k_2[/itex]. I was thinking that [itex]k_2[/itex] must have had some other definition, that they probably omitted.

    I definitely understand that [itex]V(0)=0[/itex] and [itex]V'(0)=0[/itex] because it is an energy minimum. Taking the constant to be this value seems too "hand-wavy" for me at this moment in time, as I could have taken it to be anything to define some arbitrary force constant [itex]k_2[/itex]. The way I'm taking it right now is that at [itex]\delta=0[/itex] the curvature of the potential function will depend on the distance between the two molecules which is given by [itex]\ell[/itex] therefore [itex]V''(0)\propto\ell^2[/itex].

    Would this be a correct assumption?

    EDIT: Also thank you for your help!

    of course I was just think and maybe he expanded at [itex]\ell\delta[/itex]:

    [itex]V(\ell\delta)=\frac{1}{2}V''(0)\ell^2\delta^2[/itex]

    and assuming:

    [itex]V(\ell\delta)\propto \delta^2[/itex]

    we have:

    [itex]V(\ell\delta)=\frac{1}{2}k_2\ell^2\delta^2[/itex]
     
    Last edited: Mar 23, 2014
  7. Mar 23, 2014 #6

    TSny

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    I suppose your argument is OK for the presence of ##l^2## in ##V##. I tend to think of ##l^2## as being there so that ##k_2## will have the same units as ##k_1##. ##l## is the only natural length in the problem.

    But your argument might be fine.
     
  8. Mar 23, 2014 #7
    Ah I see where you're coming from that actually makes more sense than my expansion which might be more arbitrary of a choice than LL.

    I appreciate your insight. If you have the patience and time, I was wondering if you knew how they use 24.2 to obtain the expression in Problem 2:

    [itex](y_1-y_3)\sin{\alpha}-(x_1+x_3)\cos{\alpha}=0[/itex]

    I'm a bit confused about the displacement vectors [itex]u_1[/itex] and [itex]u_2[/itex] they are using.

    Sorry for all the questions but this is my first real mechanics course (my university doesn't have a second or first year course) and my professor is literally copying his notes from LL without going into any detail of what anything means, so I'm basically following this book by myself in my spare time.
     
  9. Mar 23, 2014 #8

    TSny

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    In applying 24.2, note that ##\vec{r}_{10} = l\sin\alpha\hat{i} + l\cos\alpha\hat{j}## and ##\vec{u}_{10} = x_1\hat{i} + y_1\hat{j}##, etc.
     

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  10. Mar 23, 2014 #9
    Thank you so much for your help!
     
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