# Does the gradient of potential energy exist independent of coordinates?

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 $$x,y,z$$?

• Look here physics.stackexchange.com/q/749428
– Eli
Commented Jun 13, 2023 at 17:19
• Del in cylindrical and spherical coordinates Commented Jun 13, 2023 at 17:28
• You said it yourself: nature expresses the gradient of a potential as a force, which itself can be understood as that which would accelerate a body. Coordinates don't exist in nature at all. They are merely house numbers that we bolt on to do numerical calculations because it is too tedious to build an experimental setup to let nature calculate the results for us in "analog" form. At least the latter is the view on physics from the seat of the experimental physicist. Commented Jun 13, 2023 at 19:30
• @FlatterMann "It is too tedious to build an experimental setup to let nature calculate the results for us in "analog" form." - Sorry, but I don't understand what you mean. Could you please elaborate? Commented Jun 13, 2023 at 21:34
• @Filippo: Theory is a description of what we observe. Everything in our theory either corresponds to a physical observable (in one form or another) or it has gauge degrees of freedom that have to be dealt with. A scalar potential is only defined up to a constant, but the force is an immediate observable (it's proportional to the acceleration of a test mass). Nature "calculates" the theory for us because we made the theory to match nature's "output". Does nature use coordinates for that? No. Those are abstractions we use to make our lives easier. Commented Jun 13, 2023 at 23:12

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.

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