Clarify remark from Landau's Mechanics

[WiFO215]

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
U(q) - U(q_{0}) = \frac{1}{2}k(q - q_{0})^{2}

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

...

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 \dot{q}^{2} have to be a function of q? I thought it'd be a constant.

 

[dx]

2. Why does the co-efficient of \dot{q}^{2} 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, let's say q = x³, or x = q^1/3.

The kinetic energy is then

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

with a(q) = m/9q^4/3

 

[dx]

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.

 

[WiFO215]

Oh! I didn't think of that. Thanks. What about question 1?

 

[dx]

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} + ...

U(q) - U(q_0) = \frac{d^2U}{dq^2}_{q_0} \frac{(q - q_0)^2}{2} + ...

 

[WiFO215]

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} + ...

U(q) - U(q_0) = \frac{d^2U}{dq^2}_{q_0} \frac{(q - q_0)^2}{2} + ...

I know how to Taylor expand that function. My question was how do you know such a thing exists for any potential.

 

[dx]

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.

 

[WiFO215]

Okay! Thank you!