# I Partial derivative of potential energy and work

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1. Apr 3, 2016

### Soren4

For a conservative force $\vec{F}=-\vec{\nabla} U \implies dW=-\vec{\nabla}U \cdot d\vec{s}$

Where $d\vec{s}$ is the infinitesimal vector displacement.

Does the following hold?

$-\frac{\partial U}{\partial \vec{s}}=-\vec{\nabla} U \cdot d\vec{s}=d W$, i.e. the infinitesimal work is minus the directional derivative of $U$ in the direction of $\vec{s}$.

2. Apr 3, 2016

### jamie.j1989

No this doesn't make sense, you've written what the spatial derivative of the potential is, its $\vec{F}$ not dW.

3. Apr 3, 2016

### Staff: Mentor

Your equation is wrong. It should read $dU=\vec{\nabla} U \cdot \vec{ds}=-dW$
The interpretation is the infinitesimal work in the direction of $\vec{ds}$ is equal to minus the directional derivative of U in the direction of $\vec{ds}$.

4. Apr 3, 2016

### Soren4

@jamie.j1989 Thanks for the answer, i just applied the rule that uses nabla for directional derivatives, so how to correctly write the directional derivative?

@Chestermiller Thanks a lot for the reply, where is the exactly the mistake? Is it the fact of not having specified that the infinitesimal work is in the direction of $\vec{ds}$ ? Moreover how is the directional derivative of $U$ written explicitly, if $\frac{\partial U}{\partial \vec{s}}$ is wrong?

5. Apr 3, 2016

### jamie.j1989

The directional derivative of $U$ is $\vec{\nabla}U = \hat{i}\frac{\partial U}{\partial x}+\hat{j}\frac{\partial U}{\partial y}+\hat{k}\frac{\partial U}{\partial z}= -\left(\hat{i}F_x +\hat{j}F_y+\hat{k}F_z\right)$

6. Apr 3, 2016

### Staff: Mentor

No, $\vec \nabla U$ is the gradient of U (a vector). The directional derivative of U along a direction specified by a unit vector $\hat n$ is $\vec \nabla U \cdot \hat n$ (a scalar). If you go a distance $ds$ in the direction of $\hat n$, then the change in U is $dU = (\vec \nabla U \cdot \hat n) ds = \vec \nabla U \cdot (\hat n ds) = \vec \nabla U \cdot \vec{ds}$.

7. Apr 3, 2016

### Staff: Mentor

Thanks a lot for the reply, where is the exactly the mistake? Is it the fact of not having specified that the infinitesimal work is in the direction of $\vec{ds}$ ?
No. It's that you set dU/ds equal to a differential, which, of course, it is not.

Moreover how is the directional derivative of $U$ written explicitly, if $\frac{\partial U}{\partial \vec{s}}$ is wrong?
If $\vec{s}$ is position vector from an arbitrary origin to a position in space, and $\vec{s}+\vec{ds}$ is a position vector from the origin to an adjacent position in space, then the change in potential U between $\vec{s}$ to $\vec{s}+\vec{ds}$ (in the direction of $\vec{ds}$ is given by:
$$dU=\vec{\nabla} U\cdot \vec{ds}$$The derivative of U with respect to distance in the direction of $\vec{ds}$ is obtained by dividing by the magnitude of the differential postion vector $\vec{ds}$:
$$\frac{dU}{|\vec{ds}|}=\vec{\nabla} U\cdot \left(\frac{\vec{ds}}{|\vec{ds}|}\right)=\vec{\nabla} U\cdot \vec{n}_s$$
where $\vec{n}_s$ is a unit vector in the direction of $\vec{ds}$.