# Can Lagrange multipliers be tensorial quantities?

Let's say I have a many-rigid-bodies-system subject to the constraint that some of the distances must be constant over time. The usual Lagrangian for the system is $$\mathcal L=T(\{\dot q_n\})-V(\{q_n\})+\sum_k\lambda_k\left(\lvert\vec r_k(\{q_n\})\rvert^2-r_0^2\right).$$

Is re-writing the Lagrangian as:

$$\mathcal L=T(\{\dot q_n\})-V(\{q_n\})+\sum_k\left[\vec r_k(\{q_n\})-\vec r_0\right]^T\mathbf \Lambda_k\left[\vec r_k(\{q_n\})+\vec r_0\right],$$

where ${\bf\Lambda}_k$ is a symmetric tensor (not necessarily $\lambda_k\bf I$), and $\vec r_0$ is a constant vector with the desired length, for instance $$\vec r_0=\frac{r_0}{\sqrt d}\sum_{i=1}^d\hat e_i,$$ a valid formulation to increase the size of the parameter space in case my integration scheme has an unbalanced number of equations?

• In your 1st equation you have each $\lambda_k$ multiplied by its corresponding expression for the constraint, which is the way it should be. But for your tensor analogue (the 2nd eqn.), you have each $\Lambda_k$ somehow embedded within each constraint expression rather than standing outside the constraint expression and multiplying it as was done in the first equation. Not even sure what these tensor constraint terms with embedded $\Lambda_k$ are supposed to mean. So I would say that if there is some tensor analogue to the normal Lagrange multiplier approach, then equation 2 isn't it. – Samuel Weir Sep 6 '17 at 2:07
• Thank you. Your observation makes total sense. So, I've edited the question to make sure that $\Lambda_k$ acts upon all the entities. – GeoArt Sep 6 '17 at 2:49
• I know virtually nothing about this example, but from a quick search I found Lanczos tensors, which the page describes as originally being introduced as Lagrange multipliers. – Tyberius Sep 6 '17 at 3:23
• Seems that in the first equation the constraints are that all the $r_k$ vectors are required to all have the length $r_o$. Not sure what you mean by a generalized (and "non-isotropic"?) form of these constraints. You may want to explain in your question exactly what you're trying to achieve. – Samuel Weir Sep 6 '17 at 4:23
• OP asks (v5): Does it make sense? Well, the two Lagrangians make mathematical sense, but they represent inequivalent physical systems. – Qmechanic Sep 6 '17 at 9:34

Yes, they definitely can. Indeed, you're already doing pretty much just that: in essence, if you have $N$ particles, you've already defined a vectorial function $\mathbf f$, taking values in $\mathbb R^N$ and given by $$\mathbf f(\{q_n\})= \left(\left(\lvert\vec r_1(\{q_n\})\rvert^2-r_0^2\right), \ldots , \left(\lvert\vec r_N(\{q_n\})\rvert^2-r_0^2\right)\right),$$ and you mostly have a single vectorial Lagrange multiplier $\boldsymbol \lambda\in\mathbb R^N$, so that your lagrangian is given by $$\mathcal L=T(\{\dot q_n\})-V(\{q_n\})+\boldsymbol\lambda\cdot\mathbf f(\{q_n\}).$$ This generalizes transparently to higher-order tensors if you have a multi-dimensional array of constraint functions.