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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?

Qmechanic
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user602132
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2 Answers2

2

  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.

Qmechanic
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  • 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? – user602132 Oct 08 '19 at 22:46
  • I updated the answer. – Qmechanic Oct 08 '19 at 23:21
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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:

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

$\Rightarrow$

$$T=T(u_1,\dot{u}_1,u_2\,\dot{u}_2)=m\,\frac{1}{2}\,\vec{v}^T\,\vec{v}$$

$$U=U(u_1,u_3)$$

so:

$$\frac{d}{dt}(T+U)=0$$

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

Eli
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