Does the gradient of potential energy exist independent of coordinates?

Question

Potential energy $U(\vec{r})$ of a conservative force field $\vec{F}$ is defined as a function whose variation between positions $\vec{r}_A$ and $\vec{r}_B$ is the opposite of the work done by the force field to move a pointlike body between the positions considered.

Gradient $U(\vec{r})$ is usually introduced as the vector field: $$ \vec{\nabla} U = \frac{\partial U}{\partial x}\hat u_x+\frac{\partial U}{\partial y}\hat u_y+\frac{\partial U}{\partial z}\hat u_z.$$

It is then shown that $\vec{F}=-\vec{\nabla} U$.

So it seems that to define the gradient, cartesian coordinates are necessary. But is this true? What happens if the potential energy is not expressed as a function of $x,y,z$?

Answer 1

So it seems that to define the gradient, cartesian coordinates are necessary. But is this true?

No, it is not true. See, for example, this website, which explains that the gradient of a scalar function can be described in a coordinate-free manner as: $$ df = \vec \nabla f \cdot \vec {dr}\;, $$ where the vector-looking thing $\vec \nabla f$ is indeed a vector.

What happens if the potential energy is not expressed as a function of x,y,z?

Then if you still want to work with components, you can perform a change of variables (and invoke, say, the chain rule of differentiation). This is also explained at the previously-cited website, where it is written, that a general coordinate form of the gradient looks like: $$ \vec\nabla f = \frac{\partial f}{\partial q^i} g^{ij}\vec e_j\;, $$ where the $q^i$ are generalized coordinates, $\vec e_j = \frac{\partial \vec r}{\partial q^j}$, and $g^{ij}$ is the inverse metric tensor.

Answer 2

How to obtain The $~\vec\nabla~$

starting with the position vector $~\vec R=\vec R(q_i)~,i=1~..3~$ where $~q_i~$ are the generalized coordinates

from here the Jacobian Matrix $~\mathbf J_{i,j}=\frac{\partial R_i}{\partial q_j}~,\mathbf J=3\times 3~$

$$\vec\nabla=\hat{\mathbf{J}}\begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \\ \frac{\partial}{\partial q_3} \\ \end{bmatrix}$$

where $~\hat{\mathbf{J}}~$ is the column norm of the Jacobian matrix