Are all conservative forces internal?

[walking]

Work done by conservative forces adds potential energy (by definition of conservative force), and potential energy can only be internal to a system. Thus, conservative forces must be internal to a system.

Is this reasoning correct?

 

[Herman Trivilino]

Work done by conservative forces adds potential energy (by definition of conservative force), and potential energy can only be internal to a system. Thus, conservative forces must be internal to a system.

Is this reasoning correct?

No. Take, for example, a baseball and planet Earth. They exert gravitational forces on each other that are internal to the ball-Earth system. When work is done by gravity to move the two objects closer together the potential energy of the system decreases ($W=-\Delta U$).

On the other hand, if you consider the ball alone to be your system the gravitational force exerted on it by Earth is external to the system.

 

[walking]

If we define a system to be a free-falling ball then gravity is external to the system. Now say the ball is 5kg and falls 5m under gravity. Then the work done by gravity is 25g. Now assume the ball is lifted 5m, then gravity does -25g on the ball. So the total work done by gravity is 0. Also, this would apply for any closed path traveled by the ball.

Thus, gravity is a conservative force with respect to the ball. Why can't we then say that gravity gives the ball potential energy? Is it because it lies outside out system and thus the ball cannot be said to have energy in itself (it needs the Earth in order to fall back down, and since the Earth is not there the system cannot be said to have the potential energy associated with the earth)?

 

[walking]

No. Take, for example, a baseball and planet Earth. They exert gravitational forces on each other that are internal to the ball-Earth system. When work is done by gravity to move the two objects closer together the potential energy of the system decreases ($W=-\Delta U$).

On the other hand, if you consider the ball alone to be your system the gravitational force exerted on it by Earth is external to the system.

Thanks, I had another similar question here:

https://www.physicsforums.com/threads/potential-energy-and-conservative-forces.972929/

 

[walking]

I know that potential energy only exists for conservative forces, but is it true that these forces also have to be internal to the system for potential energy to make sense?

In other words, is it true that external conservative forces cannot add potential energy to a system?

If this is the case, then why do Kleppner and Kolenkow define potential energy without making the assumption that the conservative forces are internal? Their definition is roughly as follows:

The work done by a conservative force depends only on the endpoints of the path. Hence for conservative forces it is possible to define a function such that the work done satisfies $\int_{\mathbf{r}_a}^{\mathbf{r}_b}\mathbf{F}\cdot d\mathbf{r}=\text{function of}\ (\mathbf{r}_b)\ -\ \text{function of}\ (\mathbf{r}_a)$

or

$\int_{\mathbf{r}_a}^{\mathbf{r}_b}\mathbf{F}\cdot d\mathbf{r}=-U(\mathbf{r}_b)+U(\mathbf{r}_a)$

where $U(\mathbf{r})$ is defined by the above expression and is called the potential energy function.

 

[vanhees71]

I don't know what you mean by "internal". I only know "external fields", which are (electric or gravitational) fields assumed to be generated by some matter not taken into account in the equations of motion. That's of coarse and approximation which makes only sense if you can consider the particles in the equation of motion as "test particles" in the sense that they don't change too much the external fields.

For the (Newtonian) theory of gravity, e.g., you can give the external field as $\vec{g}(\vec{r})$. The equation of motion for a test particle in this field is then given as
$$m \ddot{\vec{r}}=m \vec{g} \; \Rightarrow \; \ddot{\vec{r}}=\vec{g}.$$
The gravitational field has a potential,
$$\vec{g}(\vec{r})=-\vec{\nabla} \phi(\vec{r}).$$
You can use this picture, e.g., to describe the motion of a planet around the sun in our solar system. Since the mass of the sun is so large, you can simplify this somewhat by using the gravitational field of the Sun as an external field. For a spherical sun, you'd give it in terms of the potential as
$$\phi(\vec{r})=-\frac{G M_{\text{sun}}}{r}.$$

Now of course there are also interaction forces, e.g., if you take into account that the Sun has a finite muss. Then you describe the force by an interaction potential
$$V(\vec{r}_1,\vec{r}_2)=-\frac{G m_1 m_2}{|\vec{r}_1-\vec{r}_2|},$$
and the equations of motion for the two bodies, now forming a closed system, are
$$m_1 \dot{\vec{x}}_1=-\vec{\nabla}_1 V=-\frac{G m_1 m_2 (\vec{r}_1-\vec{r}_2)}{|\vec{r}_1-\vec{r}_2|^3} = \vec{F}_{12}$$
and
$$m_2\dot{\vec{x}}_2=-\vec{\nabla}_2 V=-\frac{G m_1 m_2 (\vec{r}_1-\vec{r}_2)}{|\vec{r}_2-\vec{r}_1|^3} = \vec{F}_{21}.$$
Note that automatically Newton's 3rd Law is fullfilled here,
$$\vec{F}_{21}=-\vec{F}_{12}.$$

 

[walking]

I don't know what you mean by "internal". I only know "external fields", which are (electric or gravitational) fields assumed to be generated by some matter not taken into account in the equations of motion.

What do Tipler and Mosca mean when they say the following:

"Since the ball alone is our system, its mechanical energy is entirely kinetic."

They are saying that when a system consists only of a ball, it cannot have potential energy such as GPE. This is what lead me to believe that potential energy can only come from internal conservative forces.

Any explanation would be appreciated.

 

[Dale]

That is a little bit of an odd comment. I am not sure that most authors would say the same thing.

Do they give some justification about why external forces are not considered?

 

[Dale]

Note, there were three almost identical threads, which I have combined.

 

[Herman Trivilino]

I know that potential energy only exists for conservative forces, but is it true that these forces also have to be internal to the system for potential energy to make sense?

Yes.

In other words, is it true that external conservative forces cannot add potential energy to a system?

Huhh? This is a separate issue. External forces of any kind can change the internal energy of a system, this includes the potential energy of that system.

They are saying that when a system consists only of a ball, it cannot have potential energy such as GPE. This is what lead me to believe that potential energy can only come from internal conservative forces.

I agree. It's common for textbooks to refer to the gravitational potential energy $mgh$ as the potential energy of a ball. But it's really the potential energy of the ball-Earth system.

 

[vanhees71]

What do Tipler and Mosca mean when they say the following:

"Since the ball alone is our system, its mechanical energy is entirely kinetic."

They are saying that when a system consists only of a ball, it cannot have potential energy such as GPE. This is what lead me to believe that potential energy can only come from internal conservative forces.

Any explanation would be appreciated.

Sure, they consider a lonely ball in an empty (I guess Newtonian) universe. Then there's nothing it could interact with, and thus there are neither external fields nor other particles it can interact with. There cannot act any forces on it. Then it's total energy is just kinetic energy, and it moves uniformly, i.e., with constant velocity (Newton's Axiom I).

 

[vanhees71]

I agree. It's common for textbooks to refer to the gravitational potential energy $mgh$ as the potential energy of a ball. But it's really the potential energy of the ball-Earth system.

No, it's the potential energy of the ball with the Earth taken into account only as providing an external gravitational field. The back reaction of the motion of the ball with the Earth is neglected (which of course is very well justified given that the mass of the ball is tiny compared that of the Earth). See also my posting #6.

 

[jbriggs444]

which of course is very well justified given that the mass of the ball is tiny compared that of the Earth

and given that the chosen frame of reference is one in which the gravitational field is static -- i.e. given that the Earth is at rest.

 

[walking]

No, it's the potential energy of the ball with the Earth taken into account only as providing an external gravitational field. The back reaction of the motion of the ball with the Earth is neglected (which of course is very well justified given that the mass of the ball is tiny compared that of the Earth). See also my posting #6.

I think I understand what you are saying now. Basically, a ball can still have PE even if our system only consists of the ball, as long as we take into account the Earth as an external field. If we don't do this however, then the ball can only have KE. So if I am correct, the upshot is that PE can come from external conservative forces as well?

(But then what did Mister T mean when he said "yes" to my question:
"I know that potential energy only exists for conservative forces, but is it true that these forces also have to be internal to the system for potential energy to make sense? ")

I am not saying that Tipler and Mosca are wrong, but would I be justified in saying that the way they worded it was very confusing? I mean they didn't mention that by letting the ball be our system, we are excluding everything else in the universe (like you said) and simply considering it as a ball in an empty universe.

Mabye I am still misunderstanding.

 

[walking]

That is a little bit of an odd comment. I am not sure that most authors would say the same thing.

Do they give some justification about why external forces are not considered?

No, they simply state that if the ball alone is our system then it can only have KE.

 

[walking]

I probably should have done this at the start but I have decided to just post a picture of the section in Tipler and Mosca as I may be distorting things slightly due to my misunderstanding.

 

[walking]

x

x

x

x

I have attached a picture of the problem.

 

[walking]

No, it's the potential energy of the ball with the Earth taken into account only as providing an external gravitational field. The back reaction of the motion of the ball with the Earth is neglected (which of course is very well justified given that the mass of the ball is tiny compared that of the Earth). See also my posting #6.

I posted this question on physics SE as well and this was one of the replies:

"I know that conservative forces can exist outside a system, but I thought that potential energy only exists for internal conservative forces.
It does. But for an object to have gravitational potential energy the "system" has to be the object together with the earth. That makes the force of gravity an internal (to the system) conservative force.
Regarding your comment to @AlephZero about the ball in free fall, the ball alone does not have gravitational potential energy. It is the ball/earth system that possesses gravitational potential energy. For this reason you can't say the ball alone has potential energy. If it were not in a gravitational field, its acceleration could just as well been due to any external force such that F=mg=. This, as I understand it, is what led Einstein to the equivalency principal and eventually to the General Theory of Relativituy.
Hope this helps."

He seems to be saying that the Earth has to be part of the system. You seem to be saying that the Earth can be outside of the system and the ball can still have PE. I am very confused.

He also says "For this reason you can't say the ball alone has potential energy" which seems to agree with Mister T.

 

[Herman Trivilino]

He also says "For this reason you can't say the ball alone has potential energy" which seems to agree with Mister T.

Right. That is the way many college-level introductory physics textbooks approach the topic. The authors seek a definition of work that is consistent in both mechanics and thermodynamics.

 

[walking]

Right. That is the way many college-level introductory physics textbooks approach the topic. The authors seek a definition of work that is consistent in both mechanics and thermodynamics.

Are you saying that this definition doesn't apply to more advanced physics? If so I am happy to accept this definition for the time being but I am still confused as to why Kleppner and kolenkow's definition of PE (see post #5) doesn't assume the conservative force is internal.

 

[jbriggs444]

Are you saying that this definition doesn't apply to more advanced physics? If so I am happy to accept this definition for the time being but I am still confused as to why Kleppner and kolenkow's definition of PE (see post #5) doesn't assume the conservative force is internal.

For me, classifying a conservative force as internal complicates things.

If the force is internal then we have to contemplate the third law partner force and the internal object on which that third law partner force acts. If that other object moves then the result is that we do not have a static vector field and it becomes incorrect to talk about the path integral of work done around a curve being an invariant property of the endpoint positions. [If the partner object does not move as a result of the forces applied on it then we might as well consider it to be external]

It is [arguably] better to keep things simple and accept a static vector field as background and then be able to talk about path independence and potential energy as a function of position.

One alternative, I suppose, would be to treat "potential energy" as a pairwise property of interacting particles. We could require the attractive or repulsive force to be an unchanging function of separation alone and prove that regardless of speed, timing or how the particle pair is twirled and twisted through space, the path integral of work done by the two sides of the force pair add to zero as long as starting and ending separations are the same. For a suitably well-behaved interaction force, the result would be that one could ascribe a "potential energy" to the particle pair based on their separation alone.

 

[walking]

x

I don't understand most of your post unfortunately as I am quite new to physics (Tipler and Mosca chapters 1-8 = all the physics I know). I think I understand the gist of what everyone is saying though. Basically PE only exists when the object exerting the force is internal to the system, and in more advanced physics, this is always assumed(?) So in advanced physics, there is no internal/external distinction. However, even in advanced physics, if we explicitly defined a system as not consisting of the conservative force, then PE doesn't exist, eg defining a system to exclude the Earth would mean GPE doesn't exist for objects in that system.

Is this correct?

 

[Herman Trivilino]

Are you saying that this definition doesn't apply to more advanced physics?

No. I'm saying that when an author writes a college-level introductory physics textbook a choice must be made as to how to define work. My comments in this thread appear to be consistent with the way the authors of your textbook are presenting it.

 

[walking]

No. I'm saying that when an author writes a college-level introductory physics textbook a choice must be made as to how to define work. My comments in this thread appear to be consistent with the way the authors of your textbook are presenting it.

Thanks. I do appreciate the help I am getting but I'm just going to give up on understanding this for the time being as my questions haven't been directly addressed eg post #5, or if they have I am just too beginner to understand the level of physics that is being used in the explanations.

 

[vanhees71]

I think I understand what you are saying now. Basically, a ball can still have PE even if our system only consists of the ball, as long as we take into account the Earth as an external field. If we don't do this however, then the ball can only have KE. So if I am correct, the upshot is that PE can come from external conservative forces as well?

(But then what did Mister T mean when he said "yes" to my question:
"I know that potential energy only exists for conservative forces, but is it true that these forces also have to be internal to the system for potential energy to make sense? ")

I am not saying that Tipler and Mosca are wrong, but would I be justified in saying that the way they worded it was very confusing? I mean they didn't mention that by letting the ball be our system, we are excluding everything else in the universe (like you said) and simply considering it as a ball in an empty universe.

Mabye I am still misunderstanding.

If you take the ball and the Earth as your system. There's not simply a potential, but an interaction potential, i.e., a function containing both the position vector of the ball and the position vector of the earth. The fundamental symmetries of Galilean space-time (homogeneity of space, homogeneity of time, isotropy of space, Galilei-boost invariance) dictate that the total energy (or even more importantly the Hamiltonian) of the two-body system must be of the form
$$H(\vec{p}_1,\vec{p}_2,\vec{x}_1,\vec{x}_2)=\frac{\vec{p}_1^2}{2m_1} + \frac{\vec{p}_2^2}{2m_2} + V(|\vec{x}_1-\vec{x}_2|).$$
You cannot say that "there's only kinetic energy", but you also cannot say, there's "kinetic energy" and "potential energy" of each particle. You can only say, there's total energy consisting of the sum of the kinetic energies of the particles (each are indeed "single-particle observables") and the interaction potential (which is a "two-particle observable").

For an $N$-particle system there are, in principle, a whole hierachy of "$k$-particle contributions" (with $k \in \{1,2,\ldots,N \}$) contributions to the Hamiltonian. Fortunately in many cases you can truncate the hierarchy to 1- and 2-particle contributions, i.e., the single-particle kinetic energies + the sum over pair-potentials over all possible particle-pairs.

To really understand particle dynamics, you have to learn analytical mechanics, i.e., mechanics in terms of the Hamilton principle of least action. That's well worth the effort since all fundamental physics rests on this principle, and a good understanding of the most simple case, i.e., Newtonian classical mechanics, is a very good fundament for all further studies of physics.