Clarify remark from Landau's Mechanics

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While discussing the small oscillations of particles about a stable equilibrium, Landau writes

...The potential U(q) for small deviations can be expressed as a polynomial
[tex] U(q) - U(q_{0}) = \frac{1}{2}k(q - q_{0})^{2} [/tex]

...


The kinetic energy of a free particle in one dimension is generally of the form
[tex]\frac{1}{2}a(q)\dot{q}^{2}[/tex]

...

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.
 
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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.
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
 
  • #3
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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.

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

I know how to Taylor expand that function. My question was how do you know such a thing exists for any potential.
 
  • #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.
 
  • #8
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Okay! Thank you!
 

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