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Clarify remark from Landau's Mechanics

  1. Apr 17, 2010 #1
    While discussing the small oscillations of particles about a stable equilibrium, Landau writes

    Where q is the generalized co-ordinate.

    Section 21, Volume 1

    1. How do you know such a polynomial expansion for q is allowed? How do you know it exists? After all, this is any old U with its first derivative 0 at q0.
    2. Why does the co-efficient of [tex]\dot{q}^{2}[/tex] have to be a function of q? I thought it'd be a constant.
     
    Last edited: Apr 17, 2010
  2. jcsd
  3. Apr 17, 2010 #2

    dx

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    A generalized coordinate q will be some function of the ordinary cartesian coordinate x. For concreteness, lets say q = x³, or x = q1/3.

    The kinetic energy is then

    [tex] T = \frac{1}{2} m \dot{x}^2 = \frac{m}{18q^{4/3}} \dot{q}^2 = \frac{1}{2} a(q) \dot{q}^2 [/tex]

    with a(q) = m/9q4/3
     
  4. Apr 17, 2010 #3

    dx

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    Expand U(q) as a taylor expansion around q0. The first order term will be zero since q0 is an equilibrium point.
     
  5. Apr 17, 2010 #4
    Oh! I didn't think of that. Thanks. What about question 1?
     
  6. Apr 17, 2010 #5

    dx

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    [tex] U(q) = U(q_0) + \frac{dU}{dq}_{q_0} (q - q_0) + \frac{d^2U}{dq^2}_{q_0} \frac{(q - q_0)^2}{2} + ... [/tex]

    [tex] U(q) - U(q_0) = \frac{d^2U}{dq^2}_{q_0} \frac{(q - q_0)^2}{2} + ... [/tex]
     
    Last edited: Apr 17, 2010
  7. Apr 17, 2010 #6
    I know how to Taylor expand that function. My question was how do you know such a thing exists for any potential.
     
  8. Apr 17, 2010 #7

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    You can just think of this formula as something that applies for any potential which is twice differentiable at q0. Most physically meaningful potentials will of course be infinitely differentiable.
     
  9. Apr 17, 2010 #8
    Okay! Thank you!
     
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