Is potential energy defined only for internal conservative forces?

[zenterix]

Homework Statement

While re-reading a chapter on potential energy and conservation of energy, I noticed that there are some conceptual questions I still have about potential energy.

Relevant Equations

The chapter I am reading has the following quote

When only internal conservative forces act in a closed system the sum of the changes of the kinetic and potential energies of the system is zero.

Is gravitational potential energy defined only for internal conservative forces?

For reference, this is the chapter.

Suppose we have two objects: object 1 and also object 2.

If we consider the system to be both objects, then we can compute the work done by the pair of internal conservative forces

$$\int_A^B \vec{F}_{21}\cdot d\vec{r}_1+\int_A^B\vec{F}_{12}\cdot d\vec{r}_2\tag{1}$$

$$=\ldots$$

$$=\int_A^B \vec{F}_{21}\cdot d\vec{r}_{21}\tag{2}$$

where $A$ and $B$ represent states A and B of the system.

(2) is be definition of the integral itself equal to $\Delta K_{sys}$.

For a gravitational field, we end up with

$$W=Gm_1m_2\left ( \frac{1}{r_f}-\frac{1}{r_i} \right )$$

which shows that work depends only on the difference in initial distance $r_i$ and final distance $r_f$ between the two masses.

Change in potential energy of the system is defined as

$$\Delta U=-W=-\Delta K_{sys}$$

and by defining the potential energy at some reference point we obtain a potential energy function.

Now suppose our system is just object 1.

Object 2 creates a gravitational field and object 1 is in it. The force on object 1 is now an external force.

If object 2's position is fixed, then its gravitational field is also fixed at every point in space.

We can calculate the work done by this field on object 1 from an initial state to a final state and this will be path independent.

Thus, it seems we could define a potential energy function for object 1.

Of course, and here is what I am not really grasping very well, object 1 has a gravitational field of its own, and this affects object 2, which affects the latter's gravitational field if object 2's position changes due to acceleration.

 

[PeroK]

Who knows what the author means? We use GPE in three cases:

1) For objects near the surface of the Earth we have a GPE of $mgh$, based on a uniform force.

2) For planetary orbits we have a GPE of $-\frac{GMm}{r}$, based on an inverse square force and a stationary central mass $M$.

3) For two-body problems we have a GPE of $-\frac{Gm_1m_2}{r}$, where the GPE applies to the two-body system and is shared between the two bodies.

You can try to sum that up in a single sentence if you want. Why bother?

 

[haruspex]

Is potential energy defined for this system of just object 1?

Potential energy is a property of a system. Thinking of it as a property of just object 1 would be rather Zen, like one hand clapping.

 

[zenterix]

Who knows what the author means? We use GPE in three cases:

1) For objects near the surface of the Earth we have a GPE of $mgh$, based on a uniform force.

2) For planetary orbits we have a GPE of $-\frac{GMm}{r}$, based on an inverse square force and a stationary central mass $M$.

3) For two-body problems we have a GPE of $-\frac{Gm_1m_2}{r}$, where the GPE applies to the two-body system and is shared between the two bodies.

You can try to sum that up in a single sentence if you want. Why bother?

This is all well and fine.

But in each case, there was a step where you had to define $\Delta U$ as $-W$.

In 1), we assume a constant gravitational field in which the object is situated. We ignore the field of the object's effect on the Earth.

This seems to be the case I described in my OP in which object 2's position is fixed and thus its gravitational field is fixed.

Not sure what the distinction is between your cases 2) and 3).

My question is about what happens if instead of analyzing a pair of conservative forces in a system, we split the system and consider just one of the forces.

 

[Orodruin]

like one hand clapping

 

[PeroK]

My question is about what happens if instead of analyzing a pair of conservative forces in a system, we split the system and consider just one of the forces.

That's what we do in cases 1 and 2. The force on the mass or planet is external, in the sense that the mathematics takes nothing else into consideration. These basic models consider only the one force and the motion of one body.

 

[PeroK]

Not sure what the distinction is between your cases 2) and 3).

In 3) we consider the motion of both bodies around their common centre of mass. This is called clapping with both hands.

 

[zenterix]

Potential energy is a property of a system. Thinking of it as a property of just object 1 would be rather Zen, like one hand clapping.

But isn't this kinda what is done when we make the approximation for an object near the surface of the Earth?

We are disregarding the gravitational force on the Earth and only considering the gravitational force of the Earth on the object. This field is considered fixed and the object is in the field. The fact that the field is fixed is what seems to make such an approximation possible.

When we analyze a system of two objects,

we start with a pair of forces. Each force is a function of a position vector. When we calculate the work done by the pair in moving the system from a state A to a state B, we compute the work done by each of the forces in the pair.

However, after a few steps we end up with

$$\int_A^B\vec{F}_{21}\cdot d\vec{r}_{21}\tag{1}$$

which is the work integral of a single force.

$\vec{F}_{21}$ obeys the following relationship

$$\vec{F}_{21}=\frac{d^2\vec{r}_{21}}{dt^2}\frac{m_1m_2}{m_1+m_2}\tag{2}$$

which shows that (1) is calculating the work done on a "reduced mass" system (with mass $\mu=\frac{m_1m_2}{m_1+m_2}$) with position $\vec{r}_{21}$ inside a force field $\vec{F}_{21}$ that is fixed in time for each $\vec{r}_{21}$.

Thus, to compute the work (and potential energy function) for a system of two masses it seems we've reduced the problem to a single mass $\mu$ inside of a fixed force field (thus, fixed gravitational field).

We know that $\vec{F}_{21}(\vec{r}_{21})=\frac{Gm_1m_2}{r_{21}^2}\hat{r}_{21}$.

The gravitational field is

$$\frac{\vec{F}_{21}}{\mu}=\frac{G(m_1+m_2)}{r_{21}^2}\hat{r}_{21}$$

So we can solve (1) and we end up with a potential energy function of $U(r_{21})=\frac{Gm_1m_2}{r_{21}}$

 

[zenterix]

That's what we do in cases 1 and 2. The force on the mass or planet is external, in the sense that the mathematics takes nothing else into consideration. These basic models consider only the one force and the motion of one body.

So then you are confirming we can define a potential energy function for a single mass under the influence of an external conservative force?

 

[PeroK]

So then you are confirming we can define a potential energy function for a single mass under the influence of an external conservative force?

Why not?

 

[zenterix]

Why not?

That is precisely the main question in the OP. The textbook I am reading specifically says "internal conservative forces". I was wondering why they explicitly used the adjective "internal".

 

[PeroK]

That is precisely the main question in the OP. The textbook I am reading specifically says "internal conservative forces". I was wondering why they explicitly used the adjective "internal".

Well, you'll have to ask the author. Technically, that statement is about the specific case of an internal force. It says nothing about the case of an external force.

If he'd said "men wear shoes". That does not imply that women don't wear shoes.

 

[zenterix]

My next question would be: does the external conservative field have to be constant in time?

Suppose we have a single mass in a gravitational field created by a mass outside the system.

We move the internal mass from position A to position B. During transit from A to B, the internal mass exerts a force on the external mass, thus accelerating the latter and changing its gravitational field.

Is the gravitational field in which the internal mass is in still conservative?

 

[PeroK]

My next question would be: does the external conservative field have to be constant in time?

In general, a time-dependent field cannot be conservative. If you remain at rest, your PE could change. You should be able to see a counterexample like that yourself.

 

[zenterix]

This seems me to be the reason for considering pairs of third law gravitational forces.

A fixed conservative external field allows us to define a potential energy function for an object in the system.

If the external field is that of an external mass then it seems we have an issue.

If we use the approximation that the external mass isn't affected by the internal mass's field (perhaps because the external mass is very large), then we fall into the case of a fixed conservative external field.

But if we don't use this approximation, then because the external field is changing it isn't path independent any more and we can't define a potential energy function for the internal mass.

Finally, if we add the external mass into the system, then as far as I can tell, it is the sum of the pair of gravitational forces that is conservative.

Now, all of what I will say next is speculation.

It seems that when we say "conservative" or "path independent" there is an implicit inertial reference frame, let's call it $S$.

From $S$'s perspective, the gravitational field of a each mass individually is time-dependent. After all, each mass is accelerating.

The total gravitational field created by the two masses seems, however, to be conservative (from some inertial reference frame).

Work done in such a gravitational field to take $m_1$ from $\vec{r}_{1,A}$ to $\vec{r}_{1,B}$ and $m_2$ from $\vec{r}_{2,A}$ to $\vec{r}_{2,B}$ is the same as the work done in another (fictitious) gravitational field, $\frac{G(m_1+m_2)}{r_{21}^2}\hat{r}_{21}$, created by a single point mass $m_1+m_2$, to take mass $\frac{m_1m_2}{m_1+m_2}$ from position $\vec{r}_{21,A}$ to $\vec{r}_{21,B}$.

The latter is conservative, and so I think I can conclude from this that the total gravitational field created by the two masses separately is also conservative.

 

[jbriggs444]

A fixed conservative external field allows us to define a potential energy function for an object in the system.

You can also define a potential for a fictitious force. In that case there is no external interaction force at all.

Take for example the centrifugal force that arises in a rotating frame of reference. This is a force that is proportional to $r$. It has a potential that is proportional to $r^2$. Since the centrifugal force acts away from the center of rotation, the potential would normally be taken as zero at the center and as increasingly negative elsewhere.

If you do mechanics in the uniformly rotating frame, energy conservation works just fine. An object travelling away from the center seems to spiral away at an ever increasing speed. It gains kinetic energy. Exactly as much kinetic energy as it is losing in increasingly negative potential energy.

Obviously, you can only define a potential for fictitious forces that are conservative.