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Given the Lagrangian,

$$L = \frac{1}{2}\dot{q}_i M_{ij} \dot{q}_j + f(q,\dot(q),t)$$

where $M_{ij}$ is a non-degenerate matrix and $q,\dot{q}$ are generalised coordinates & velocities and summation over repeated indices is assumed. I came across a question which asks to show only the symmetric part of $M$ enters the Euler-Lagrange equations. My attempt was to simply calculate the E.O.M as being:
$$ \frac{1}{2}\frac{d}{dt}(M_{ij}\dot{q}_j+\dot{q}_jM_{ji}) +g(q,\dot{q},t)=0 $$ where $g$ depends on the derivatives of $f$.

So I was able to show this, however I am confused as to why only the symmetric part stays. If anyone could clarify my misunderstanding that would be great.

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    $\begingroup$ Is the antisymmetric part present in the Lagrangian? $\endgroup$ – Herr_Mitesch Apr 10 '18 at 16:57
  • $\begingroup$ Technically, the coordinates could also be fermionic, so this question is then pointless. $\endgroup$ – DanielC Apr 10 '18 at 20:14
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The symmetric part of $M$ is defined to be $M_S = (M + M^T)/2$.

Note that $\dot{q}_j M_{ji} = M_{ji}\dot{q}_j = M^T_{ij}\dot{q}_j$, so you can write your EOM as $$\frac{d}{dt}((M_{ij}+M^T_{ij})/2 \,\dot{q}_j +\dots = \frac{d}{dt}(M_{S,ij} \dot{q}_j)$$

More fundamentally,you can split $M$ into its symmetric and antisymmetric part, $M = M_S + M_A$, where $M_S^T = M_S$ and $M_A^T = -M_A$. Since $\dot{q}_i\dot{q}_j =\dot{q}_j\dot{q}_i$ we can see that $$\dot{q}_i M_{A,ij}\dot{q}_j = \dot{q}_j M_{A,ji}^T \dot{q}_i = -\dot{q}_i M_{A,ij} \dot{q}_j = 0$$

So $\dot{q}^T M \dot{q} = \dot{q}^T (M_S + M_A) \dot{q} = \dot{q}^T M_S \dot{q} + 0 $. Only the symmetric part of $M$ actually contributes to the Lagrangian.

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