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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 \begin{equation} 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) \end{equation} subject to the constraints \begin{equation} T_{ij}-\frac{\partial_i f_j + \partial_j f_i}{2}=0. \end{equation}

(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

\begin{equation} 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]. \end{equation}

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 $\frac{\partial \Psi_{ij}}{\partial t}=0$.

However, I cannot give a more rigorous justification for this assumption.. Could anyone please help me?

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  • $\begingroup$ Are the constraints time dependent? Seems like they are... $\endgroup$
    – hft
    Aug 30, 2022 at 15:36
  • $\begingroup$ 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. $\endgroup$
    – Keith
    Aug 30, 2022 at 15:38
  • $\begingroup$ 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. $\endgroup$
    – hft
    Aug 30, 2022 at 16:45
  • $\begingroup$ See, for example, Arfkin and Weber "Mathematical Methods for Physicists" at section 17.7 "Variation subject to constraint" $\endgroup$
    – hft
    Aug 30, 2022 at 16:46

2 Answers 2

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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.

Note that when actually solving the equations for a stationary point, it may be useful to assume time independence to easily find solutions.

Hope this helps.

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  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.

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