A particle constrained to always move on a surface whose equation is $\sigma (\textbf{r},t)=0$. Show that the particle energy is not conserved

Question

In Goldstein's Classical mechanics question 2.22

Suppose a particle moves in space subject to a conservative potential $V(\textbf{r})$ but is constrained to always move on a surface whose equation is $\sigma(\textbf{r},t)=0$. (The explicit dependence on $t$ indicates the surface is moving.) The instantaneous force of constraint is taken as always perpendicular to the surface. Show analytically that the energy of the particle is not conserved if the surface moves in time. What physically is the reason for non-conservation of the energy under this circumstance?

So the way I see it we can use the surface equation as the constraint, hence the generalised force is given by $\sum_{i}\lambda\frac{\partial\sigma}{\partial q_i}$ with $\lambda$ being a lagrange multiplier.

The energy is given by $E=T+V$, and the energy is conservative if $\frac{dE}{dt}=0$. From the question we know $\frac{dV}{dt}=0$. so we have to show that $\frac{dT}{dt} \ne 0$.

Now since the surface changes over time I imagined the kinetic energy would look something like this

$$T=\frac{1}{2}m(v^2+\dot{\sigma}^2)$$

Where $v$ is the particle velocity and $\dot \sigma=\frac{\partial \sigma}{\partial t}$ is the surface velocity. However this doesn't explicitly depend on time (t) . so $\frac{dE}{dt}=0$.

I then thought to decompose $v$ as a function of the surface tangent vectors i.e. $ v = \frac{\partial \sigma}{\partial \textbf{r}}\frac{\partial \textbf{r}}{\partial t}$. So that the kinetic energy becomes

$$T=\frac{1}{2}m(\frac{\partial \sigma}{\partial \textbf{r}}\cdot \frac{\partial \sigma}{\partial \textbf{r}}\dot{\textbf{r}}^2 + \dot{\sigma}^2) $$

However, again I dont see any explicit dependence on time (t) . Unless it is assumed that $\frac{\partial \sigma}{\partial \textbf{r}}$ is explicitly dependent on t? I am not sure where I am going wrong.

Answer 1

I believe a proof can be given closer to the spirit of classical mechanics, at least if assuming the constraint is holonomic in nature. The unconstrained system is a particle moving in a central potential having the Lagrangian $L(q,\dot{q},t)$, obeying the usual Euler-Lagrange equations and having an energy function $h(q,\dot{q},t)=\sum_{j}\dot{q}_{j}(\frac{\partial L}{\partial \dot{q}_{j}})-L(q,\dot{q},t)$. This energy function is conserved and also equal to the system's total energy, as follows from the structure of the Lagrangian and Euler's theorem for homogeneous functions.

The constrained problem involves the Lagrangian $L^{\prime}$

$L^{\prime}(q,\dot{q},t)=L(q,\dot{q},t)+\mu(t)\sigma(q,t)$

I've allowed the Lagrange multiplier to depend on time due to the requirement for the force of constraint to always be perpendicular to the moving surface, and as such it needs to move along with it. The Euler-Lagrange equations for $L^{\prime}$ are

$\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{j}})-\frac{\partial L}{\partial q_{j}}+\mu(t)\frac{\partial \sigma}{\partial q_{j}}=0$

You then take the total time derivative of L' and attempt to reconstruct $h$, the energy function of the unconstrained Lagrangian $L$. Using $\frac{\partial L}{\partial t}=0$ and $\dot{\sigma}=\sum_{j}\frac{\partial \sigma}{\partial q_{j}}\dot{q}_{j}+\frac{\partial \sigma}{\partial t}=0$, you should ultimately get

$\dot{h}=\mu\frac{\partial \sigma}{\partial t}$

So, the reason why energy is not conserved is due to the constraining surface moving in time, which physically becomes equivalent with a time dependent potential. Note that otherwise the energy would be conserved, much like the unconstrained case conserves energy due to the central potential not varying with time.