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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 $q_k$.
The Lagrangian is $L=T-V$, with the kinetic energy:

$T=\sum_{kl}\dot q_k\dot q_l (\sum_i \frac{1}{2}m_i\frac{\partial\vec r_i}{\partial q_k}.\frac{\partial\vec r_i}{\partial q_l}) = \sum_{k,l}\frac{1}{2}m_{kl}(q1,...,q_n)\dot q_k \dot q_l$
...

  • I understand how to get the first part, but I don't understand the second equality. What is meant by "$m_{kl}$"?

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a polynomial of the second degree in the generalized velocities. The potential energy $V(q_1, ..., q_n)$ only depends on the generalized coordinates.

The system has a point of equilibrium ($q_k^0=q_1^0, ... q_n^0$) when all generalized forces are $0$ in this point:

$Q_k=-(\frac{\partial V}{\partial q_k})_0=0$

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: $q_k(t)=q_k^0$ is the solution to the Lagrange equations

$\sum_l m_{kl}\ddot q_l+\sum_l\sum_m\frac{\partial m_{kl}}{\partial{q_m}}\dot q_l\dot q_m=\sum_{lm}\frac{1}{2}\frac{\partial m_{lm}}{\partial q_k}\dot q_l \dot q_m -\frac{\partial V}{\partial q_k}$

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.
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The $m_{kl}$ is a "mass matrix". In Cartesian coordinates the kinetic energy is $$T = \sum_i \frac{m_i}{2} v_i^2$$ with $v_i^2 = v_{i,x}^2 + v_{i,y}^2 + v_{i,z}^2$. For simplicity let us consider the case of one particle. We can write this in matrix form as $$T = \frac{1}{2} (v_x, v_y, v_z) \begin{pmatrix} m_i & 0 & 0 \\ 0 & m_i & 0 \\ 0 & 0 & m_i \end{pmatrix} \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \frac{1}{2} \sum_{kl} m_{kl} v_k v_l.$$ 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 $v_k = \dot{x_k}$, if we change coordinates to $\mathbf{q} = \mathbf{q}(\mathbf x)$, we have $$\dot{q}_k = \sum_k \frac{\partial q_k}{\partial x_j} \dot{x_j}$$ and since the coordinate change is invertible, $$\dot{x_k} = \sum_j \frac{\partial x_k}{\partial q_k} \dot{q}_j. $$ Therefore in the new coordinates the expression for the kinetic energy is $$T = \frac{1}{2} \sum_{ijkl} m_{kl} \frac{\partial x_k}{\partial q_i} \dot{q}_i \frac{\partial x_l}{\partial q_j}\dot{q}_j = \frac{1}{2} \tilde{m}_{kl} \dot{q}_k\dot{q}_l \tag{1} $$ where $\tilde{m}_{kl} = \frac{1}{2} \sum_{ij} m_{ij} \frac{\partial x_i}{\partial q_k} \frac{\partial x_j}{\partial q_l}.$ (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 $\tilde{m}_{kl}$ 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 $\partial V /\partial q_j = 0$, 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 $q_k(t) = q_k^0$ is a solution, then appeal to a uniqueness theorem for solutions to differential equations.

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  • $\begingroup$ I think I understand it now, although I'm not sure if your conversion of coordinates is correct, as well as the expression for the second $m$ $\endgroup$ – Joshua Jun 10 '14 at 16:35

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