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?

  • $\begingroup$ 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


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.

  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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.