# In the context of field-theoretic constrained dynamics, do we have the freedom to choose the Lagrange multipliers to be time-independent?

Let us work in a box $$(t,\overrightarrow{x}) \in [0,1] \times [0,1]^3$$. For any function on this box, we impose some Dirichlet boundary condition on the temporal direction and periodic boundary conditions on the spatial directions.

Let $$i,j,k,l \in \{1,2,3\}$$, which we use as spatial indices.

Now, think of some action functional $$$$S[f_l,T_{ij}]=\int_0^1 dt \int_{[0,1]^3}d^3\overrightarrow{x}\mathcal{L}(f_l(\overrightarrow{x}, t),T_{ij}(\overrightarrow{x}, t),\frac{\partial f_l}{\partial t}, \partial_k f_l)$$$$ subject to the constraints $$$$T_{ij}-\frac{\partial_i f_j + \partial_j f_i}{2}=0.$$$$

(All field components are set to be real-valued.)

I am using the field-theoretic language now, so the Lagrange multipliers introduced to enforce these constraints must form some (symmetric) tensor field $$\Psi_{ij}(\overrightarrow{x}, t)$$, so that the modfied action is

$$$$S_{mod}[f_l,T_{ij},\Psi_{ij}]=\int_0^1 dt \int_{[0,1]^3}d^3\overrightarrow{x}\Bigl[\mathcal{L}(f_l,T_{ij},\frac{\partial f_l}{\partial t}, \partial_k f_l)-\Psi_{ij} \cdot \Bigl(T_{ij}-\frac{\partial_i f_j + \partial_j f_i}{2}\Bigr)\Bigr].$$$$

where, the constraints are realized as the Euler-Lagrange equations of $$S_{mod}$$ with respect to $$\Psi_{ij}$$'s.

Now, my question is:

Can we assume that $$\Psi$$ does NOT depend on time explicitly, that is, $$\frac{\partial \Psi_{ij}}{\partial t}=0$$?

I think since the constraint does NOT contain any time derivative, it is valid on each time slice. Thus, there is no impediment for assuming $$\frac{\partial \Psi_{ij}}{\partial t}=0$$.

• Are the constraints time dependent? Seems like they are...
– hft
Aug 30, 2022 at 15:36
• Yes, the constraints themselves are functions of time also, since $T_{ij}$ and $f_l$ are. But I want to know if I can set the "Lagrange multiplier" to be time-independent in this specific case. Aug 30, 2022 at 15:38
• Since it is the action functional that you are minimizing, you have to take the Lagrange multipliers to be dependent on both space and time in general.
– hft
Aug 30, 2022 at 16:45
• See, for example, Arfkin and Weber "Mathematical Methods for Physicists" at section 17.7 "Variation subject to constraint"
– hft
Aug 30, 2022 at 16:46

You'll need to assume it to be time dependent in general for the variational principle to give you the correct constraint at all time. This is done by varying $$\Psi$$. Intuitively, the constraint is at each instant event in space-time, so you'll need a multiplier for each event, i.e. a $$x,t$$ dependence of $$\Psi$$. If you don't assume time dependence, the variational method would only give you: $$\int_0^1dt \, T_{ij} = \int_0^1 dt \, \frac{\partial_if_j+\partial_j f_j}{2}$$ which is weaker.
1. Since OP wants to impose the constraint at any time (and space), then the Lagrange multiplier $$\Psi$$ should depend on time (and space), cf. lpz's answer.
2. Another question is if $$\Psi(x,t)$$ is a constant of motion, i.e. if its solution is independent of time? This depends on OP's Lagrangian density $${\cal L}$$ and it's symmetries.