mij is a diagonal matrix, the coeficients are dependent of the coordinates, so, in the example he does he uses polar coordinates, we can see that in the first term of mij=m11 which is given by mr(dot)^2 the mass depends of the coordinates not by its velocity or acceleration, so m11= m then in the second term mij=mr^2 is easy to see that the mass depends of the coordinate r^2 because is its position. If there was in the first term for example mr(dot) (theta) mij=m11 would be m(theta) instead of m because is depending of the coordinate (theta).
We don't discard any lower order terms. A Taylor expansion is a polynomial approximation of a function at a given point in R^n. Since this is the theory of small oscillations, higher order terms in eta are not going to improve the accuracy of our results much. So we drop them. The first two are dropped by using the equilibrium condition.
First term is discarded by shifting of zero of the potential. ie, coz we can. For second term, first derivative of potential is zero at equilibrium point.
There was a glitch in the matrix :)
Lecture 1 (video 1 & 2) is on its way to youtube now.
We do upload all our videos into playlists.
very smooth voice, great to listen to. thanks
Satisfied.!! videos abreast my taste and hunger.
sir, i am not able to understand the concept of mij in kinetic energy! please explain it!
Bro, did you get that mij kinetic energy ? I am also struggling with it for some days now.
mij is a diagonal matrix, the coeficients are dependent of the coordinates, so, in the example he does he uses polar coordinates, we can see that in the first term of mij=m11 which is given by mr(dot)^2 the mass depends of the coordinates not by its velocity or acceleration, so m11= m then in the second term mij=mr^2 is easy to see that the mass depends of the coordinate r^2 because is its position. If there was in the first term for example mr(dot) (theta) mij=m11 would be m(theta) instead of m because is depending of the coordinate (theta).
How can we just discard the constants in taylor expansion of potential
We don't discard any lower order terms. A Taylor expansion is a polynomial approximation of a function at a given point in R^n. Since this is the theory of small oscillations, higher order terms in eta are not going to improve the accuracy of our results much. So we drop them. The first two are dropped by using the equilibrium condition.
First term is discarded by shifting of zero of the potential. ie, coz we can. For second term, first derivative of potential is zero at equilibrium point.