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For a general natural system, the kinetic energy part of the lagrangian may be written as $$T = \frac{1}{2}\sum_{ij} a_{ij}(q_{1}, q_{2}, ..., q_{n})\dot{q}_{i}\dot{q}_{j}.$$ For $n = 2$, $$T = \frac{1}{2} a_{11}\dot{q}^{2}_{1} + a_{12}\dot{q}_{1}\dot{q}_{2} +\frac{1}{2} a_{22}\dot{q}^{2}_{2}$$ since the kinetic matrix $a_{ij}$ is symmetric. To generate new coordinates that are orthogonal such that $$T = \frac{1}{2} a'_{11}\dot{q}'^{2}_{1} + \frac{1}{2} a'_{22}\dot{q}'^{2}_{2},$$ my textbook says to simply set $q'_{2} = q_{2}$ and $q_{1}' = q_{1} + \frac{a_{12}}{a_{11}}q'_{2}$. However how would this be done for $n > 2$? (I have tried $q_{3}' = q_{3}, \, q_{2}' = q_{2} + \frac{a_{23}}{a_{22}}q'_{3}, \, q_{1}' = q_{1} + \frac{a_{12}}{a_{11}}q'_{2} + \frac{a_{13}}{a_{11}}q'_{3}$ with no success.)

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  • $\begingroup$ Are you assuming that the coefficients $a_{ij}$ depend on $q$ and $t$? $\endgroup$ – Qmechanic May 4 at 17:39
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It's not necessary to find eigenvectors when diagonalizing a quadratic form such as $T=\frac 12 M_{ij} \dot q_i \dot q^j$. This form is what you have in a "small oscillation" problem where the coeffeicients $M_{ij}$ are $q$-independent numbers. Instead you just keep completing squares. Suppose you have an expression like $$ Q=x^2-y^2 -z^2+ 2xy-4xz+6yz = \left(\matrix{x,y,z}\right)\left(\matrix{1& 1&-2\cr1& -1&3\cr -2&3&-1}\right)\left(\matrix{x\cr y\cr z}\right). $$ Complete the square using $x$: $$ Q= (x+y-2z)^2 -2y^2+10yz-5x^2, $$ so the terms outside the parenetheses no longer contain $x$. Now complete the square in $y$: $$ Q= (x+y-2z)^2- (\sqrt 2y-\frac 5{\sqrt 2} z)^2 +\frac{15}{2} z^2. $$ Now set $$ \xi= x+y-2z, \\ \eta= \sqrt 2y-\frac {5}{\sqrt 2} z ,\\ \zeta = \sqrt {\frac {15}{2}} z, $$ so that
$$ Q=\xi^2-\eta^2+\zeta^2 $$ is diagonal. Because kinetic energy terms are positive definite, when you apply this method you will always be able to reduce the kinetic energy to $$ T= \dot Q_1^2+\dot Q_2^2+\ldots+ \dot Q_n^2. $$ You don't need eigenvectors because you are trying to find matrices $A$ such that $$ M= A^T{\rm diag}(1,1,\ldots,1) A. $$ This task is much easier than the eigenvalue problem where you seek $A$ such that $$ M= A^{-1} {\rm diag}(m_1,m_2,\ldots,m_N) A. $$

This method is also why you can simultaneously diaginalize the $M$ and $V$ matrices in the small oscillation Lagrangian $$ L= \frac 12 M_{ij} \dot q_i \dot q^j- \frac 12 V_{ij}q_iq_j $$ even though $M$ and $V$ do not necessarily commute. First you diagonalize the $M$ to get a matrix with all 1's. The $V$ matrix now becomes a new matrix $\tilde V=A^TVA$ --- but it is still symmetric so you can diagonalize the new $\tilde V$ with an orthogonal matrix $\tilde A$ which does not disturb the already diagonalized KE and end up with $$ L= \frac 12 \sum_i \left\{\dot Q_i^2 - \Omega^2_i Q_i^2\right\}. $$

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  • $\begingroup$ Thank you very much for the detail response! $\endgroup$ – Hello May 4 at 19:34
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Instead of attempting to generalize the transformations that have been described in your textbook for a 2-dimensional system ($q_1,q_2$ are the coordinates), you might want to consider diagonalizing your general kinetic matrix. This can be done by computing the eigenvectors of your matrix $a_{ij}$ and using them to find the symmetry transformation that takes you to the diagonal basis. That will work for arbitrary dimensions.

Edit: I am assuming that $a_{ij}$ are independent of the $q_i$s and $t$, as is the case in your example. I made this assumption because it seemed to be the case in your examples. Moreover, if that is not the case then I am not sure if I will really call it a 'Kinetic term' because then you have some non-linear theory. I come to associate a standalone kinetic term with linear theory.

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  • $\begingroup$ and then just say that $\bold{q}' = D\bold{q}$ where $D$ is the matrix with the eigenvectors of $a_{ij}$ which under similarity transform diagonalise it? $\endgroup$ – Hello May 4 at 15:03
  • $\begingroup$ Yes! That's correct. $\endgroup$ – nGlacTOwnS May 4 at 16:08

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