# Question about choice of objects when calculating potential energy

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.

• Why not try it yourself with Newton’s inverse-square law tor gravity? Or Coulomb’s Law for electrostatics? Aug 10, 2020 at 22:33
• 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? Aug 10, 2020 at 22:35
• 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/… Aug 10, 2020 at 22:40

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.

• 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? Aug 11, 2020 at 0:16
• 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 Aug 11, 2020 at 0:19
• 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. Aug 11, 2020 at 0:20
• So when considering the displacement for potential energy, you have to consider one of the objects as your reference frame, right? Aug 11, 2020 at 0:25
• Ok thanks for you help. Aug 11, 2020 at 15:08