Consider a system with generalized coordinates $u_1, u_2$ and $u_3$ such that $u_1$ and $u_2$ are dependent through the following holonomic constraint \begin{equation} G(u_1, u_2)=0. \end{equation} It is also given that generalised force corresponding to each coordinate is zero.

Kinetic energy of the system is given by \begin{equation} T(u_1, u_2, u_3, \dot{u}_1,\dot{u}_2, \dot{u}_3)=\frac{1}{2}\dot{\bf{u}}^TD(\textbf{u})\dot{\textbf{u}} \end{equation} where $\textbf{u}=[u_1, u_2, u_3]^T$ and $D(\textbf{u})$ is positive definite for all $\textbf{u}$.

The potential energy of the system is given by a function $U(\textbf{u})$. Will the total energy $T+U$ be constant?

  1. Assume

    • (i) that the kinetic term $T$ is quadratic in generalized velocities $\dot{\bf u}$;

    • (ii) that the potential term $U$ is independent of the generalized velocities $\dot{\bf u}$; and

    • (iii) that the Lagrangian $L=T-U$ does not depend explicitly on time.

  2. Case without holonomic constraints. The energy $h=\dot{\bf u}\cdot \frac{\partial L}{\partial \dot{\bf u}}-L=T+U$ is conserved because the Lagrangian $L=T-U$ does not depend explicitly on time, cf. e.g this Phys.SE post.

  3. Case with holonomic constraints $G({\bf u})\approx 0$ without explicit time dependence.

    • (i) Either we can formally eliminate variables such that there are no holonomic constraints left; or

    • (ii) alternatively, we can introduce Lagrange multipliers, which we add to the list of variables ${\bf u}$, and add terms of the form 'Lagrange multiplier times $G$' to the potential term $U$. (The notion of potential energy will be unaltered on-shell.)

    In both cases, the form 1 is maintained, and we can apply the conclusion from section 2: Energy is still conserved.

| cite | improve this answer | |
  • $\begingroup$ What if the constraint is such that one variable can’t be written in terms of other explicitly? In that case we can use a Lagrange multiplier and construct a new Lagrangian. Would the argument in section 2 still work? $\endgroup$ – user602132 Oct 8 '19 at 22:46
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Oct 8 '19 at 23:21

your example

you have three 3 degree of freedom $u_1,u_2,u_3$ (not generalized coordinate) and 1 constraint equation $g(u_1,u_2)=0$ so you have 2 generalized coordinate .

I see two cases:

I) form the constraint equation $g(u_1,u_2)=0$ you can obtain explicit for example $u_2=u_2(u_1)$ so your position vector (mechanical system) is:

$$\vec{r}=\vec{r}(u_1,u_3)$$ $$\vec{v}=\vec{\dot{r}}=\frac{\partial \vec{r}}{\partial u_1}\dot{u}_1+\frac{\partial \vec{r}}{\partial u_3}\dot{u}_3$$






The kinetic energy plus potential energy is conserved

II) If you can't eliminate one of the degree of freedom from the constraint equation then:

from: $$g(u_1,u_2)=0\quad \Rightarrow\quad \frac{\partial g}{\partial u_1}\dot{u}_1+\frac{\partial g}{\partial u}_2\dot{u}_2=0$$

so $$\dot{u}_2=\dot{u}_2(u_1,u_2,\dot{u}_1)\quad u_2=\int \dot{u}_2\,dt$$

so again like case I the kinetic energy plus potential energy is conserved

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.