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

In summary: 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?The derivative of ##\eta## is quadratic, but the second term in the Taylor expansion 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.In summary, the expression for the kinetic energy already contains a
  • #1
Ben Geoffrey
16
0
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
osc.JPG
 

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  • #2
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.
 
  • #3
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
 
  • #4
Something quadratic times something linear is something cubic.
$$
\mathscr O(x^2) \mathscr O(x) = \mathscr O(x^3)
$$
 
  • #5
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 ?
 
  • #6
Ben Geoffrey said:
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.
 
  • #7
No my doubt is x times derivative of x, will become x2 ?
 
  • #8
Ben Geoffrey said:
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##.
 
  • #9
Can you see I've understood it right, if x is displacement, in small oscillation problem we take x = xo + η where xo 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 mij into 6.5 in the second term we have to multiply the derivative of ni and the derivative of nj times n. That part is what I don't understand ?
 
  • #10
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 ?
 
  • #11
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?
 
  • #12
So its like the order of a differential equation ? y multiplied by y' . will have an order 2 ?
 

1. What is the Taylor series expansion of quadratic derivatives?

The Taylor series expansion of quadratic derivatives is a mathematical method used to approximate a function at a specific point by using its derivatives. It involves creating a polynomial expression using the function's derivatives evaluated at the point of interest.

2. Why is the Taylor series expansion important in Goldstein's Ch. 6?

In Goldstein's Ch. 6, the Taylor series expansion is used to approximate the motion of particles in a potential field. This is important because it allows for the prediction of the behavior of particles in a given system and helps in understanding the dynamics of the system.

3. How is the Taylor series expansion related to quadratic derivatives?

The Taylor series expansion uses the derivatives of a function, including the quadratic derivatives, to create a polynomial expression. This allows for a more accurate approximation of the function at a specific point.

4. What is the significance of the quadratic term in the Taylor series expansion?

The quadratic term in the Taylor series expansion represents the second derivative of the function. It is important because it determines the curvature of the function at the point of interest and can greatly affect the accuracy of the approximation.

5. Can the Taylor series expansion be used for functions other than quadratic?

Yes, the Taylor series expansion can be used for any type of function with derivatives. However, the accuracy of the approximation may decrease if the function is not well-behaved or has a high degree of curvature at the point of interest.

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