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In this question the term kinetic matrix of a Lagrangian is used, but I cannot find a general definition anywhere in wikipedia or elsewhere.

Can anyone tell me the precise definition? What is it used for primarily, what is its raison d'être?

Qmechanic
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Roland Salz
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2 Answers2

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A broad class of Lagrangians is of the form $$L(\mathbf q,\dot{\mathbf q})= \frac{1}{2}\sum_{i,j}g_{ij} \dot q^i \dot q^j - V(\mathbf q)$$

where $g_{ij}$ are the components of a (generally $\mathbf q$-dependent) symmetric bilinear form; this is the "kinetic matrix" to which the other question refers. In this class of Lagrangians, the canonical momentum components are given by

$$p_i = \frac{\partial L}{\partial \dot q^i} = g_{ij} \dot q^j$$

In order for the Legendre transform to be well-defined, $g_{ij}$ must be non-degenerate (and hence invertible), in which case the Hamiltonian is given by

$$H(\mathbf q,\mathbf p) = \frac{1}{2} (g^{-1})^{ij}p_i p_j + V(\mathbf q)$$ In the case that $g_{ij}$ is degenerate, the naive Legendre transform fails to be well-defined (as is the case in your linked question) and one must be more careful.

J. Murray
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    $g_{ij}$ is also a function of $\mathbf q$ – Eli Feb 13 '21 at 21:16
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    @Eli Yep, edited to clarify that point. – J. Murray Feb 13 '21 at 21:20
  • Thanks a lot! I assume your definition also holds in the more general case where T and L explicitly depend on t and where we have three terms for the kinetic energy, one of them being the one you described, homogeneous of order 2, which contains the kinetic matrix. – Roland Salz Feb 14 '21 at 09:56
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    @RolandSalz That's such a radical generalization that I would be cautious about using the same terminology without very explicitly making clear what you mean. In particular, the well-definedness of the Legendre transform requires non-degeneracy of the Hessian matrix $H_{ij} = \frac{\partial^2 L}{\partial \dot q^i \partial \dot q^j}$. In the class of Lagrangians in my answer, $H_{ij}=g_{ij}$, but in more general cases the invertibility of $g$ does not imply invertibility of $H$. – J. Murray Feb 14 '21 at 21:00
  • Can't we generalize to the case $L(\mathbf q,\dot{\mathbf q},t)= \frac{1}{2} g_{ij} \dot q^i \dot q^j + b_k \dot q^k +a_0 - V(\mathbf q)$, where g, b, and $a_0$ are all independent of $\dot q$, but may depend on q and t? In this case, described by Goldstein, chapter 8, the two additional terms add nothing to the Hessian of L, it remains identical with $g_{ij}$. This is actually the case I meant, somewhat imprecisely. I didn't intend to carry the generalization any further than that. – Roland Salz Feb 15 '21 at 19:27
  • @RolandSalz In that case, sure. I’d still be careful to define my terms, though. – J. Murray Feb 15 '21 at 19:31
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I totally concur with J. Murray's explanation, which is excellent, IMO. But because you seem to be somewhat in the darkness about where the need for a matrix comes from in the Lagrangian formalism, let me add an approach which is to be understood as complementary of the previous one. Let's say your mechanical system has $n$ Cartesian coordinates. The Lagrangian is $T-V$, $T$ being the kinetic energy, and $V$ the potential energy. You can convince yourself that the Lagrangian formalism gives you Newton's equations quite easily. Now, suppose you want to describe your system's motion in terms of more convenient variables, say $q_i$. These are generally called "generalised coordinates". If you change variables and use basic results of calculus, you get, $$ \frac{1}{2}\sum_{i}m_{i}\dot{x}_{i}^{2}= $$ $$\frac{1}{2}\sum_{i}m_{i}\dot{x}_{i}\dot{x}_{i}=\frac{1}{2}\sum_{i=1}^{n}m_{i}\sum_{j=1}^{r}\frac{\partial x_{i}}{\partial q_{j}}\dot{q}_{j}\sum_{k=1}^{r}\frac{\partial x_{i}}{\partial q_{k}}\dot{q}_{k}= $$ $$ =\frac{1}{2}\sum_{j,k=1}^{r}\left(\sum_{i=1}^{n}m_{i}\frac{\partial x_{i}}{\partial q_{j}}\frac{\partial x_{i}}{\partial q_{k}}\right)\dot{q}_{j}\dot{q}_{k} $$ By calling, $$ g_{jk}=\sum_{i=1}^{n}m_{i}\frac{\partial x_{i}}{\partial q_{j}}\frac{\partial x_{i}}{\partial q_{k}} $$ you can see that a matrix naturally arises in the kinetic part when you try to free Newton's formalism from the chains of having to express everything in terms of Cartesian coordinates. I hope that helps explain the raison d'être you're looking for.

PS: In general $r\leq n$ (the number of generalised coordinates is less or equal than $n$, the number of Cartesian coordinates, because there may be constraints.)

joigus
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