# Conservation of total energy for a system with holonomic constraints

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 $$$$G(u_1, u_2)=0.$$$$ It is also given that generalised force corresponding to each coordinate is zero.

Kinetic energy of the system is given by $$$$T(u_1, u_2, u_3, \dot{u}_1,\dot{u}_2, \dot{u}_3)=\frac{1}{2}\dot{\bf{u}}^TD(\textbf{u})\dot{\textbf{u}}$$$$ where $$\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.

• 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 8 at 22:46
• I updated the answer. – Qmechanic Oct 8 at 23:21

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