0
$\begingroup$

If we have a force whose component along $\vec{r}$ is $F_r$. where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$. then the force is = derivative of $U$ (potential energy) wrt $r$.

So my question is : if $F_r = \frac{dU}{dr}$
then: $F_r = \frac{\partial{U}}{\partial{x}} + \frac{\partial{U}}{\partial{y}}+ \frac{\partial{U}}{\partial{z}}$ ...is this correct? where $\partial$ is partial differential

$\endgroup$
2
  • $\begingroup$ $F=-\nabla U$. This formula encodes force being equal to the negative of the gradient of the potential and is enough to calculate what you are asking for. This is basically equal to what you have written but with proper unit vectors added which are missing in the equation given above. $\endgroup$
    – Lost
    Oct 14 at 8:34
  • $\begingroup$ @Lost ok , thanks a lot. $\endgroup$ Oct 14 at 8:59
0
$\begingroup$

the force is \begin{equation} \vec{F} = - \frac{\partial U} {\partial{\vec{r}}} =-\left(\frac{\partial U} {\partial{x}}; \frac{\partial U} {\partial{y}}; \frac{\partial U} {\partial{z}}\right) \end{equation}

$\endgroup$
2
  • $\begingroup$ so force is not derivative of U wrt r but partial derivative of U wrt r?? (with -ve sign) $\endgroup$ Oct 14 at 8:53
  • $\begingroup$ what is wrt? = with respect to? Yes, partial derivative $\endgroup$ Oct 14 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.