Question

Prove that a particle constrained to stay on a surface $\mathbf{f}\left(x,y,z\right)\mathbf{=}\mathbf{0}$, but subject to no other forces, moves along a geodesic of the surface. Hint: The potential energy$\mathbf{V}$is constant, since constraint forces are normal to the surface and so dono work on the particle. Use Hamilton’s principle and show that the problem offinding a geodesic and the problem of finding the path of the particle are identicalmathematics problems.

Short answer

It has been proved that a particle constrained to stay on a surface $\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0$, but subject to no other forces, moves along a geodesic of the surface.

The potential energy $\mathrm{V}$is constant, since constraint forces are normal to the surface and so do no work on the particle, the geodesics acceleration vector $\ddot{\mathrm{r}}$is:

$\ddot{\mathrm{r}}‐\mathrm{dr}=0$

Where$\mathrm{dr}$ is the small displacement tangent of the surface$\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0$

Step-by-step solution

State the Hamilton principle

According to the Hamilton principle, a variational problem for a functional based on a single function, the Lagrangian, determines the dynamics of a physical system. The Lagrangian may contain all physical information about the system and the forces operating on it.

Given parameters

Given a surface $\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0$and subject to no other forces, moves along a geodesic of the surface.

Also given that the potential energy is constant.

Find the geodesics acceleration

Since the geodesics acceleration vector$\ddot{\mathrm{r}}$is normal to the surface.

Thus,

$\ddot{\mathrm{r}}‐\mathrm{dr}=0$,

Where denotes the small displacement tangent to surface $\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0$.

As we know that $\mathrm{L}=\mathrm{T}-\mathrm{V}$.

Thus, from the Hamilton’s principle:

And by the Newton’s second law $\mathrm{m}\ddot{\mathrm{r}}=-\nabla \mathrm{V}$

Let assume that potential energy is due to constraint $\mathrm{f}=0$.

Therefore, the Gradient of $\mathrm{V}$is normal to the surface.

And$\nabla \mathrm{V}‐\mathrm{dr}=0$if $\mathrm{dr}$is a displacement tangent to surface.

Then,

$$\begin{gathered} m\ddot{r}‐dr=‐\nabla V·dr;\nabla V‐dr=0 \\ \ddot{r‐dr=0} \end{gathered}$$

So, the geodesics acceleration vector is normal to the surface and gives localid="1664354192539" $\ddot{\mathrm{r}}‐\mathrm{dr}=0$.