# Why is energy not conserved when the "surface is moving in time"?

Consider a particle moving in the field of a conservative potential $$V(\bf{r})$$ which is constrained to move on a surface $$\sigma(\bf{r},t)=0$$. Here $$\bf{r}$$ denotes the radius vector of the particle. Solving the Euler-Lagrange equation by Lagrange multipliers, it is found that the total mechanical energy changes at a rate, $$\left( \bf{\dot p} -F\right)\cdot {\bf{\dot r}} = \dot W= \lambda \frac {\partial \sigma}{\partial t}$$ Where does the energy go? My belief is that the surface $$\sigma$$ "moves in time" through continuous elastic transformations given by the explicit time variation, and the change in the elastic P.E. is compensated by the mechanical power. However, I am not sure about this and need some authentic resources telling about this phenomenon. Thanks!

I haven't checked your formula, but it is trivial that a time dependent constraint can do work. Just think of the floor of an elevator as an example for $$\sigma(x,t)$$. Do you think the elevator raises the energy of you moving in the cabin? I would say yes. Is the energy state of the elevator/environment part of the described system? No, it is not, the surface is just a constraint of the action minimization and as such not part of the Lagrangian. Hence the energy that you gain cannot be reflected by an equal but opposite energy change of your surroundings within the Lagrange multiplier approach.