I Taylor Series Expansion of Quadratic Derivatives: Goldstein Ch. 6, Pg. 240

[Ben Geoffrey]

Can anyone tell me how if the derivative of n(n') is quadratic the second term in the taylor series expansion given below vanishes. This doubt is from the book Classical Mechanics by Goldstein Chapter 6 page 240 3rd edition. I have attached a screenshot below

 

[Orodruin]

It does not vanish and it is not what is being said. What is being said is that the expression for the kinetic energy already contains a quadratic term through the derivatives. Anything in the expansion of m that is of linear order or higher will therefore not contribute to the kinetic energy at quadratic order (eg, the linear term will lead to an overall cubic term) and can be neglected for small oscillations.

 

[Ben Geoffrey]

Can you tell me how the linear term in n becomes cubic when substituted in equation 6.5 ?

Appreciate your patience in dealing with amateurs

 

[Orodruin]

Something quadratic times something linear is something cubic.
$$
\mathscr O(x^2) \mathscr O(x) = \mathscr O(x^3)
$$

 

[Ben Geoffrey]

No but derivative of n is quadratic and n is linear, if we multiply both we we would get derivative of n square times n right ?

 

[Orodruin]

No but derivative of n is quadratic and n is linear, if we multiply both we we would get derivative of n square times n right ?

Which makes the term cubic in $\eta$. Both $\eta$ and $\dot\eta$ count.

Also, it is not clear what you mean by "derivative of $\eta$ is quadratic". It is not, the point is that the expression for the kinetic energy is already quadratic in the small oscillations due to the appearance of $\dot\eta_i \dot \eta_j$. Multiplying it by something that is linear therefore results in a cubic term.

 

[Ben Geoffrey]

No my doubt is x times derivative of x, will become x² ?

 

[Orodruin]

No my doubt is x times derivative of x, will become x² ?

It will be of second order in the small oscillations. When you assume that the oscillations are small, imagine writing $x = \epsilon y$, where $\epsilon$ is a small fixed number. Then expand everything in $\epsilon$. You should find that $x$ and $\dot x$ are of first order in $\epsilon$.

 

[Ben Geoffrey]

Can you see I've understood it right, if x is displacement, in small oscillation problem we take x = x_o + η where x_o initial position is made to coincide with origin and hence becomes zero and there we have only the small displacement η . Now when substituting the first order Taylor expansion of m_ij into 6.5 in the second term we have to multiply the derivative of n_i and the derivative of n_j times n. That part is what I don't understand ?

 

[Ben Geoffrey]

could you explain that part a little more ? what is the derivative of n ? and how does multiplying with n make it quadratic in n ? I understand if you multiply two same linear terms like x and x they become quadratic. But how come in this case ?

 

[Orodruin]

When you insert the expansion of m into 6.5 you get a term proportional to $\eta_k \dot \eta_i \dot\eta_j$. This term is obviously cubic in the small oscillations. Why do you doubt this?

 

[Ben Geoffrey]

So its like the order of a differential equation ? y multiplied by y' . will have an order 2 ?