# Clarify remark from Landau's Mechanics

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.

Last edited:

dx
Homework Helper
Gold Member
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, lets say q = x³, or x = q1/3.

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/9q4/3

dx
Homework Helper
Gold Member
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.

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

dx
Homework Helper
Gold Member
$$U(q) = U(q_0) + \frac{dU}{dq}_{q_0} (q - q_0) + \frac{d^2U}{dq^2}_{q_0} \frac{(q - q_0)^2}{2} + ...$$

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

Last edited:
$$U(q) = U(q_0) + \frac{dU}{dq}_{q_0} (q - q_0) + \frac{d^2U}{dq^2}_{q_0} \frac{(q - q_0)^2}{2} + ...$$

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

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

dx
Homework Helper
Gold Member
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.

Okay! Thank you!