0
$\begingroup$

The formula for change in potential energy of a system made up of object A and B is:

change in potential energy of system = - (dot product of [conservative force vector of A on B] and the [displacement vector of B])

But what if we instead consider:

change in potential energy of system = - (dot product of [conservative force vector of B on A] and the [displacement vector of A])

Will the result be the same either way we do it?

An example would be appreciated.

$\endgroup$
3
  • $\begingroup$ Why not try it yourself with Newton’s inverse-square law tor gravity? Or Coulomb’s Law for electrostatics? $\endgroup$
    – G. Smith
    Aug 10, 2020 at 22:33
  • $\begingroup$ I did in this question: physics.stackexchange.com/questions/572463/… , but I did not get the same result. Is the displacement vector used in this formula an absolute displacement vector or just relative to the other object? $\endgroup$ Aug 10, 2020 at 22:35
  • $\begingroup$ Hi and welcome to physics SE. Please, use laTex notation for formulae. It's about writing them in between of dollar symbols like this $ E=\vec{F}\cdot \vec{r}$ , and laTex commands inside. See here: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – FGSUZ
    Aug 10, 2020 at 22:40

1 Answer 1

2
$\begingroup$

If you'd physically think about such a situation, where we have A and B as two point masses at rest, being attracted by the gravitational force, nature doesn't care what you think is the relative position of A with respect to B or B with respect to A.

But even if you think about this problem mathematically, the expression for the forces on particles A and B would be -

$$\vec{F_B}=\frac{-G m_A m_B}{r^2} \hat{r} \qquad \qquad \vec{F_A}=\frac{+G m_A m_B}{r^2} \hat{r}$$

Here, $\vec{r_B}=r \hat{r}$ (displacement vector of B) and $\vec{r_A}=-r \hat{r}$ (displacement vector of A) and $\hat{r}$ is the unit vector pointing from A to B.

Now, if you try to calculate the potential energy difference using either A or B as the origin, it's $$P.E.= \int_{\infty}^{r}\vec{F_A} \cdot d\vec{r_A}= \int_{\infty}^{r}\vec{F_B} \cdot d\vec{r_B}=\frac{-Gm_A m_B}{r}$$

Therefore, this again proves that nature doesn't care what you choose as the displacement vector direction or the origin. The potential energy expression would be consistent anyway.

$\endgroup$
6
  • $\begingroup$ Thanks for your response. What if for example, you chose your point of reference, you chose the earth. Now consider lifting a ball on earth to some height. In this reference frame, the displacement of earth and the ball are not opposite, right? $\endgroup$ Aug 11, 2020 at 0:16
  • $\begingroup$ You can equally well argue that the ball is lifted from the earth at some H units of height or the earth moved away from the ball at some H units of height. The only difference would be the direction of the displacement vector $\endgroup$
    – Tachyon209
    Aug 11, 2020 at 0:19
  • $\begingroup$ Also, I forgot that we were dealing with a position variable force so I edited the PE equation now with the appropriate integrals. Sorry for that error. $\endgroup$
    – Tachyon209
    Aug 11, 2020 at 0:20
  • 1
    $\begingroup$ So when considering the displacement for potential energy, you have to consider one of the objects as your reference frame, right? $\endgroup$ Aug 11, 2020 at 0:25
  • 1
    $\begingroup$ Ok thanks for you help. $\endgroup$ Aug 11, 2020 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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