# Question about units in Lagrangian dynamics (inertia matrix)

I have a 3 degree of freedom system and my equation of motion is like this:

$$M(q)q_{dd} + C(q,q_d)q_d+G(q)~=~0$$

$M(q)$: inertia matrix

$C(q,q_d)$: Coriolis-centrifugal matrix

$G(q)$: potential matrix.

where $q_d$ is the first derivative of q etc and q is a vector of my variables.

$q=[θ,γ,a]^T$

"θ" and "γ" are angles in (rad) and "a" is length in (m).

Now my question is this: I have 3 equations with 3 unknowns but 2 variables are in terms of angles and 1 in terms of distance, so the elements of my inertia matrix don't have the same units. Is that wrong? I mean each equation is consistent in units. The first 2 have units of kg(m/s)^2 while the 3rd has units of kg(m/s^2). I can make the whole thing dimensionless but I read somewhere that the inertia matrix should be symmetric(which it is symmetric at the moment). If I make my equation of motion dimensionless then inertia matrix won't be symmetric anymore. What shall I do?

Any suggestions are welcome

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That is both an advantage and a disadvantage of the Lagrangian formulation. – ja72 Aug 26 '12 at 13:07

If you do it right, the three equations of motion have different units, so that the coefficients M_{ij} are symmetric. The reason they are symmetric is because the Lagrangian

$$L = {1\over 2} M_{ij} \dot{q_i}\dot{q_j} - V(q)$$

is manifestly symmetric in exchanging index i and j. This is true even when the $q_i$ have different units--- the M's then have different units.

When you find the eqauation of motion:

$${d\over dt}( M_{ij} \dot{q}_j) + {\partial V \over \partial q_i} = 0$$

The $M_{ij}$ are still symmetric. But once you expand it out to second order form:

$$M_{ij} \ddot{q}_j + ... = 0$$

There is a temptation to divide the i-th equation by $M_{ii}$. This ruins the symmetry of M, so don't do that. The equality of $M_{ij}$ and $M_{ji}$ when $q_j$ and $q_i$ have different units comes from the fact that in this case $M_{ii}$ and $M_{jj}$ have different units.

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