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I Error in Landau Lifshitz Mechanics?

  1. Jan 24, 2017 #1

    Demystifier

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    The Landau Lifshitz book "Mechanics" has a good reputation of one of the best books, not only on classical mechanics, but on theoretical physics in general. Yet, I have found a serious conceptual error (or at least sloppyness) in it.

    Sec. 23 - Oscillations of systems with more than one degree of freedom:

    In the paragraph after Eq. (28) they say that ##\omega^2## must be positive because otherwise energy would not be conserved. That's wrong. One can take negative ##k## and positive ##m##, which leads to negative ##\omega^2##, solve the equations explicitly, and check out that energy is conserved.

    In the next paragraph they present a mathematical proof that ##\omega^2## is positive, but for that purpose they seem to implicitly assume that ##k## is positive, without explicitly saying it. A priori, there is no reason why ##k## should be positive (except that otherwise the solutions are not oscillations, but it has nothing to do with energy conservation).
     
    Last edited: Jan 24, 2017
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  3. Jan 24, 2017 #2

    TeethWhitener

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    In eqs 23.2 and 23.3 they mention that potential and kinetic energy are both given by positive definite quadratic forms.

    Edit: I agree that the energy conservation argument is strange. ##\omega^2<0## just implies an unstable equilibrium, but as far as I can tell, energy is still conserved.
     
  4. Jan 24, 2017 #3

    Demystifier

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    Exactly!
     
  5. Jan 24, 2017 #4

    Andy Resnick

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    This is all on page 67, right?

    I'm not sure I follow. If we set x(t) = Ae(iω+β)t, the equation of motion will have an imaginary component. Imaginary components are associated with dissipative processes, so energy is not conserved.
     
    Last edited: Jan 24, 2017
  6. Jan 24, 2017 #5

    TeethWhitener

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    As far as I can tell, there are 3 cases for ##x(t) = A\exp(i\omega t)##. First is ##\omega \in \mathbb{R}##, in which case, ##\omega^2 \geq 0## and we get boring old oscillatory motion. The second case is when ##\omega \in \mathbb{C}##. As you've pointed out, if ##\omega## has both a real and an imaginary component, there will be an imaginary component in the EOM. The third case is when ##\omega## is strictly imaginary, which is the case that Demystifier and I referred to above. In this case, there's no imaginary component in the EOM and energy is still conserved, so it's unclear from the argument given in the book that ##\omega^2## must be positive. The only thing I can think is that L&L previously stipulated that the potential and kinetic energies are both positive definite.
     
  7. Jan 24, 2017 #6

    dextercioby

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    People, just read the book CAREFULLY. The k (spring constant) is assumed real and positive on page 58, just at the beginning of section 21. This assumption never changes in the sequel, thus by equation 21.6 omega is real and positive (his square, too!), being the square root of a real and positive number. The comment made by L&L on page 67 (which is the same as the comment by Andy Resnick above) implies the existence of the so-called ;damped oscillations; which L&L had not discussed in the text and which are, indeed, dissipative (hence with non-conserving energy) dynamical systems. The damped oscillations are subject of section 25 and, guess what, that k (spring constant) is also assumed to be positive...
     
  8. Jan 24, 2017 #7

    TeethWhitener

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    That's fine for the 1d case, but what about section 23? Is the generalization that simple? L&L state that the potential and kinetic energies are positive definite quadratic forms, but that just implies that the matrices ##k_{ij}## and ##m_{ij}## have positive eigenvalues. But the crux of their proof that ##\omega^2 >0## relies on those matrices being real. Is a symmetric matrix with positive eigenvalues always real?

    I guess my question is: to obtain ##\omega^2>0##, is it enough to assume that the potential and kinetic energies are positive definite, or must we also enforce that the mass and spring constant matrices are real?
     
  9. Jan 24, 2017 #8

    Andy Resnick

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    If I understand what you are saying, my response is that for a purely imaginary frequency, x(t) will either go to zero or infinity- one is boring and the other unphysical.

    However, there are instances when purely imaginary frequencies are used in electromagnetics- Hamaker functions and constants used to analyze van der Walls interactions, for example. The dielectric constant (which can be complex-valued) is taken to depend on a complex-valued frequency. The purely imaginary (frequency) component is used to construct the Hamaker function. For some reason, I think this generates thermophysical properties (as compared to electrophysical properties).
     
  10. Jan 25, 2017 #9

    Demystifier

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    When wave function has imaginary frequency, then it violates unitarity, implying that there is dissipation into the environment. But when ##x(t)## has imaginary frequency, it does not need to have anything to do with dissipation. Indeed, for negative ##k## the Hamiltonian does not have an explicit dependence on time, so energy is conserved and there is no dissipation.
     
  11. Jan 25, 2017 #10

    Demystifier

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    Positivity implies reality.
     
  12. Jan 25, 2017 #11

    Demystifier

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    That's all fine. ##\omega^2## is positive because ##k## is positive, period. But the point is that the argument that ##\omega^2## must be positive because otherwise energy would not be conserved is simply not a good argument.
     
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