Monday, 27 May 2019

Lagrangian formulation of the problem: small oscillations around an equilibrium


I'm having trouble understanding how some conclusions are made in my book. I'm studying from a coursebook based on Goldstein's "Classical Mechanics", here's what's written in my book, with my problems under each section:



Assume a system with conservative forces and holonomic, time independent constraints, described by n generalized coordinates qk.
The Lagrangian is L=TV, with the kinetic energy:


T=kl˙qk˙ql(i12miriqk.riql)=k,l12mkl(q1,...,qn)˙qk˙ql
...





  • I understand how to get the first part, but I don't understand the second equality. What is meant by "mkl"?



...
a polynomial of the second degree in the generalized velocities. The potential energy V(q1,...,qn) only depends on the generalized coordinates.


The system has a point of equilibrium (q0k=q01,...q0n) when all generalized forces are 0 in this point:


Qk=(Vqk)0=0


If the system is in a point of equilibrium at a certain starting time, with the starting velocities 0, then the system will remain in that point of equilibrium. In other words: qk(t)=q0k is the solution to the Lagrange equations


lmkl¨ql+lmmklqm˙ql˙qm=lm12mlmqk˙ql˙qmVqk


with these initial conditions.





  • This last equation I don't understand at all, I know what the Lagrange equation is but when filling it in I don't see how you would get this result.



Answer



The mkl is a "mass matrix". In Cartesian coordinates the kinetic energy is T=imi2v2i with v2i=v2i,x+v2i,y+v2i,z. For simplicity let us consider the case of one particle. We can write this in matrix form as T=12(vx,vy,vz)(mi000mi000mi)(vxvyvz)=12klmklvkvl. The latter expression generalizes to the case of several particles, just imagine a bigger matrix.


But typically when using the Lagrangian formalism Cartesian coordinates are not the most convenient. Now, since vk=˙xk, if we change coordinates to q=q(x), we have ˙qk=kqkxj˙xj and since the coordinate change is invertible, ˙xk=jxkqk˙qj. Therefore in the new coordinates the expression for the kinetic energy is T=12ijklmklxkqi˙qixlqj˙qj=12˜mkl˙qk˙ql where ˜mkl=12ijmijxiqkxjql. (This, of course, is the transformation law for a rank 2 tensor, because that is what the mass matrix is.)


Since we are allowed to do any coordinate transformations, the entries in the matrix ˜mkl may be non-constant functions of the coordinates. Then we see that (1) is the first equation in your question.


For a concrete example where the mass matrix has non-constant entries you can try to find the mass matrix in spherical coordinates, then compare with the expression for the kinetic energy in spherical coordinates. (The latter should be in Goldstein if you don't remember it.)


For the second part of your question, it is really just Newton's first law. If the system is at rest in a position where V/qj=0, it is at rest with no forces acting on it. The more rigorous way to to understand it is to write down the Euler-Lagrange equations and check that qk(t)=q0k is a solution, then appeal to a uniqueness theorem for solutions to differential equations.



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