I Conservative forces and internal and external forces

[fog37]

Hello Everyone,
This is not a completely new dilemma but I have been discussing it with several people and teachers and read different physics textbooks and continue to get different perspectives, sometimes overlapping sometimes not.

Let me explain and summarize:

• Forces are interactions between two agents. Forces be contact or noncontact forces. Noncontact forces are what we describe as force fields.
• The system is what we define, i.e. it includes the entities we choose. Everything else represents the environment.
• Forces between entities inside the system are internal forces. Forces between internal agents and external agents in the environment are external forces.
• For internal forces, the net internal force $F_{net_{internal}}=0$ always. However, the work done each of this internal forces may be different, hence $W_{net_{internal"}} \neq 0$ in general. If this work is nonzero, it means that different forms of energies inside the system are transformed into different types but the total system's energy remains constant.
• Forces, beside being categorized as internal or external to the system, can also be classified as either "conservative" or "nonconservative". Almost all textbooks do that. However, new authors claim that all forces, deep down, are conservative and the conservative/nonconservative discussion confuses students.
• Let's focus on conservative forces. A conservative force is an interaction between two agents mediate by a force with the following characteristics:

a) the force, when it does nonzero work, keeps the the total mechanical energy $ME = KE_s + PE_s$ of the system constant.
b) the force path integral (i.e. its work) is path-independent and only depends on the initial and final points
c) the force is the negative of the gradient of a scalar function $U(x,y,z)$. Same as saying that the curl of F is zero everywhere.
d) Nonconservative forces are not only of the dissipative type. Propulsive forces are also nonconservative
e) conservative forces only depend on position variables $x,y,z$ and not time or speed or else (if ME conservation is a requirement)

• Potential energy $U$, all textbooks state, is the energy of the system: energy shared by the internal components of the system only. For example, if entities $A$ and $B$ are internal system's components and $C$ is external to the system, potential energy $U$ is only for/between $A$ and $B$. Entity $C$ does not participate in $U$.
• b) Assume the force between A and B is conservative. The external force between A and C and the external force B and C cannot be conservative because a external conservative force would change $ME$ of the system contradicting a). I argue that internal forces can be either conservative or nonconservative but external forces cannot be conservative.

Trivial example: a rock and planet Earth.

Case 1: rock R and earth E=system. The force $F_g$ between R and E is internal and conservative.

Case 2: rock=system and Earth=environment. The force $F_g$ between R and E is external and does not conserve the system's $ME$. However, such an external force still satisfies b) and c)... So $F_g$ is "conservative" in some sense but does not conserve $ME$. Also, I don't think we can attribute potential energy $U$ to the system since R and E are not both components inside in the system...

THANK YOU!

 

[hmmm27]

On the one hand, thank you very much for being able to spell.

On the other hand... it's a bit confusing.

eg:

Trivial example: a rock and planet Earth.

Case 1: rock R and earth E=system. The force between R and E is internal and conservative.

There is no singular "force between". Each is independent, exerted by one on the other.

Case 2: rock=system and Earth=environment. The force between R and E is external and does not conserve the system's .

In the context of R, the normal force exerted by E is external.

I'm not sure how "conservation" enters into it.

"conservation" is self-defining : two steel balls bouncing off each other within in a sphere is "conservative" if you want to ignore the noise and heat generated, and dissipative if you do not (etc.) .

If you lose something to the outside of the system or outside the chosen context (eg: heat which stays in the system) it's non-conservative

However, such an external force still satisfies b) and c)... So is "conservative" in some sense but does not conserve . Also, I don't think we can attribute potential energy to the system since R and E are not both components inside in the system...

Within the system of R, PE has been added : the rock is compressed (very slightly) from Fg.

 

[fog37]

Hello,

Sorry, my scenario is a rock (R) that is at a certain height above the Earth's ground (E).

Sure, every force has a paired force from Newton's 3rd law. The conservation of ME, which is the sum of the system's KE and U, is ensured if the force involved is conservative. That works fine if the conservative force is internal to the system.

But what if the force is external to the system? Can it still be called conservative (i.e. it conserves ME) if it is now external?

 

[hmmm27]

The distance between the rock and the Earth does not create a PE that is internal to a system that consists only of the rock. What mechanical energy do you think exists and is being conserved ?

 

[fog37]

PE is energy of configuration, positional energy. The rock and Earth, being at a mutual distance, have PE together...PE is a form of mechanical energy along with KE. Conservative forces keep ME=PE+KE constant while conservative forces don't

 

[curious_mind]

I think the definition of conservative forces should be (b), (c) and that the force $\vec{F}$ should satisfy $\nabla \times \vec{F} = 0$ .

Notice that the path of integration in (b), or gradient in(c) and curl in (d) can be performed with arbitrary space points in 3D euclidean space whereas when you separate system and environment, you are dividing the euclidean 3D space I think. So, when you talk Mechanical energy of system remain conserved for subpart of euclidean space , then it gets into conflict with (b), (c) and $\nabla \times \vec{F} = 0$ .

Therefore, it is a bit incomprehensible perception of conservative forces to regard that it does not alter mechanical energy of the system, we are to separate entire space into system and the environment.

However, new authors claim that all forces, deep down, are conservative and the conservative/nonconservative discussion confuses students.

There is some uncontrollable factors due to which the heat is dissipated in a thermodynamic system. We have deterministic global equation of gravity - in terms of Newton's universal law of Gravitation. Heat is not something deterministic. So I think even if gravity is outside the system, it should be regarded as conservative but Friction should regarded as non-conservative.As the last note, I feel this is more about matter of perspective, if the concept is understood, then it is not worth effort to standardize the consensus about the issue.

 

[vanhees71]

One has to distinguish the two treatments of the problem and write down equations of motion to make things clear. Let's take the example of the Kepler problem, i.e., the motion of a planet around the Sun. Since in our solar system the mass of any planet is much smaller than the mass of the Sun, we can approximate the problem such that we consider the Sun as "infinitely heavy". Then one can choose an inertial reference frame, where the Sun is at rest at the origin, and the planet moves in the gravitational field of the Sun. Assuming a spherical Sun, we get the equation of motion
$$m \ddot{\vec{x}}=-\frac{G m M}{r^3} \vec{x},$$
where $G$ is Newton's gravitational constant, $m$ the mass of the planet and $M$ the mass of the Sun. Further $r=|\vec{x}|$ with $\vec{x}$ being the position vector of the planet.

This "external force" has obviously a potential,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x})\; \quad V=-\frac{G m M}{r}.$$
So we have an equation of motion for the planet in an external conservative force field, and energy conservation,
$$E=\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=\text{const}$$
as well as angular-momentum conservation,
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}=\text{const}$$
hold true. That's due to the symmetries of the problem, i.e., time-translation invariance and rotation invariance for rotations around the origin (i.e., the center of mass of the "infinitely heavy" Sun).

You can also treat the problem as a closed system of two interacting "particles". Here the gravitational force is treated as a true interaction between the two bodies. The equations of motion now are
$$m \ddot{\vec{x}}_P=-\frac{G m M}{|\vec{x}_P-\vec{x}_S|^3}(\vec{x}_P-\vec{x}_S),$$
$$M \ddot{\vec{x}}_S=-\frac{G m M}{|\vec{x}_S-\vec{x}_P|^3}(\vec{x}_S-\vec{x}_P).$$
The interaction force has a potential,
$$V(\vec{r}_P,\vec{r}_S)=-\frac{G m M}{|\vec{x}_P-\vec{x}_S|}$$
and the equations of motion get
$$m \ddot{\vec{x}}_P=-\nabla_P V, \quad M \ddot{\vec{x}}_S=-\nabla_S V.$$
The energy
$$E=\frac{m}{2} \dot{\vec{x}}^2 + \frac{M}{2} \dot{\vec{x}}_S^2 + V(\vec{x}_P,\vec{x}_S)=\text{const},$$
because
$$\dot{E}=m \dot{\vec{x}}_P \cdot \ddot{\vec{x}}_P + M \dot{\vec{x}}_S \cdot \ddot{\vec{x}}_S+ \dot{\vec{x}}_P \cdot \vec{\nabla}_P V + \dot{\vec{x}}_S \cdot \vec{\nabla}_S V=0.$$
Obviously Newton's 3rd Law is fulfilled and this immediately leads to
$$\ddot{\vec{R}}=0, \quad \vec{F}=\frac{1}{m+M} (m\vec{x}_P+M \vec{x}_S),$$
which is the center-of-mass theorem, i.e., the center of mass moves like a free particle,
$$\vec{R}=\vec{R}_0 + \vec{V} t, \quad \vec{V}=\text{const}.$$
It's now convenient to introduce the relative vector $\vec{r}=\vec{x}_P-\vec{x}_S$. After some algebra you get
$$\vec{x}_P= \vec{R} + \frac{M}{m+M} \vec{r}, \quad \vec{x}_S=\vec{F}-\frac{m}{m+M} \vec{r}.$$
Plugging this in either of the equations of motion and using $\ddot{\vec{R}}=0$ you get the equation of motion
$$\mu \ddot{\vec{r}}=-\frac{G m M}{r^3} \vec{r}=-\frac{G \mu (M+m)}{r^3} \vec{r}, \quad \mu=\frac{m M}{M+M}=-\vec{\nabla}_{\vec{r}}V,$$
where
$$V=V(\vec{x}_P,\vec{x}_S) \equiv V(\vec{r})=-\frac{G m M}{r}.$$
That's the equation of motion for a "quasi-particle" with mass $\mu$ in the external force field given by the potential $V(\vec{r})$.

It's easy to see that also the total angular momentum is conserved
$$\vec{J}=\vec{L}_P + \vec{L}_S=m \vec{x}_P \times \dot{\vec{x}}_P + M \vec{x}_S \times \dot{\vec{x}}_S=\text{const}.$$
In terms of the center of mass and relative coordinates this reads
$$\vec{J}=M \vec{R}_0 \times \vec{V} + \vec{L}_{\text{rel}}, \quad \vec{L}_{\text{rel}}=\mu \vec{r} \times \dot{\vec{r}}.$$
Of course also $\vec{L}_{\text{rel}}=\text{const}$.

 

[fog37]

One has to distinguish the two treatments of the problem and write down equations of motion to make things clear. Let's take the example of the Kepler problem, i.e., the motion of a planet around the Sun. Since in our solar system the mass of any planet is much smaller than the mass of the Sun, we can approximate the problem such that we consider the Sun as "infinitely heavy". Then one can choose an inertial reference frame, where the Sun is at rest at the origin, and the planet moves in the gravitational field of the Sun. Assuming a spherical Sun, we get the equation of motion
$$m \ddot{\vec{x}}=-\frac{G m M}{r^3} \vec{x},$$
where $G$ is Newton's gravitational constant, $m$ the mass of the planet and $M$ the mass of the Sun. Further $r=|\vec{x}|$ with $\vec{x}$ being the position vector of the planet.

This "external force" has obviously a potential,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x})\; \quad V=-\frac{G m M}{r}.$$
So we have an equation of motion for the planet in an external conservative force field, and energy conservation,
$$E=\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=\text{const}$$
as well as angular-momentum conservation,
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}=\text{const}$$
hold true. That's due to the symmetries of the problem, i.e., time-translation invariance and rotation invariance for rotations around the origin (i.e., the center of mass of the "infinitely heavy" Sun).

You can also treat the problem as a closed system of two interacting "particles". Here the gravitational force is treated as a true interaction between the two bodies. The equations of motion now are
$$m \ddot{\vec{x}}_P=-\frac{G m M}{|\vec{x}_P-\vec{x}_S|^3}(\vec{x}_P-\vec{x}_S),$$
$$M \ddot{\vec{x}}_S=-\frac{G m M}{|\vec{x}_S-\vec{x}_P|^3}(\vec{x}_S-\vec{x}_P).$$
The interaction force has a potential,
$$V(\vec{r}_P,\vec{r}_S)=-\frac{G m M}{|\vec{x}_P-\vec{x}_S|}$$
and the equations of motion get
$$m \ddot{\vec{x}}_P=-\nabla_P V, \quad M \ddot{\vec{x}}_S=-\nabla_S V.$$
The energy
$$E=\frac{m}{2} \dot{\vec{x}}^2 + \frac{M}{2} \dot{\vec{x}}_S^2 + V(\vec{x}_P,\vec{x}_S)=\text{const},$$
because
$$\dot{E}=m \dot{\vec{x}}_P \cdot \ddot{\vec{x}}_P + M \dot{\vec{x}}_S \cdot \ddot{\vec{x}}_S+ \dot{\vec{x}}_P \cdot \vec{\nabla}_P V + \dot{\vec{x}}_S \cdot \vec{\nabla}_S V=0.$$
Obviously Newton's 3rd Law is fulfilled and this immediately leads to
$$\ddot{\vec{R}}=0, \quad \vec{F}=\frac{1}{m+M} (m\vec{x}_P+M \vec{x}_S),$$
which is the center-of-mass theorem, i.e., the center of mass moves like a free particle,
$$\vec{R}=\vec{R}_0 + \vec{V} t, \quad \vec{V}=\text{const}.$$
It's now convenient to introduce the relative vector $\vec{r}=\vec{x}_P-\vec{x}_S$. After some algebra you get
$$\vec{x}_P= \vec{R} + \frac{M}{m+M} \vec{r}, \quad \vec{x}_S=\vec{F}-\frac{m}{m+M} \vec{r}.$$
Plugging this in either of the equations of motion and using $\ddot{\vec{R}}=0$ you get the equation of motion
$$\mu \ddot{\vec{r}}=-\frac{G m M}{r^3} \vec{r}=-\frac{G \mu (M+m)}{r^3} \vec{r}, \quad \mu=\frac{m M}{M+M}=-\vec{\nabla}_{\vec{r}}V,$$
where
$$V=V(\vec{x}_P,\vec{x}_S) \equiv V(\vec{r})=-\frac{G m M}{r}.$$
That's the equation of motion for a "quasi-particle" with mass $\mu$ in the external force field given by the potential $V(\vec{r})$.

It's easy to see that also the total angular momentum is conserved
$$\vec{J}=\vec{L}_P + \vec{L}_S=m \vec{x}_P \times \dot{\vec{x}}_P + M \vec{x}_S \times \dot{\vec{x}}_S=\text{const}.$$
In terms of the center of mass and relative coordinates this reads
$$\vec{J}=M \vec{R}_0 \times \vec{V} + \vec{L}_{\text{rel}}, \quad \vec{L}_{\text{rel}}=\mu \vec{r} \times \dot{\vec{r}}.$$
Of course also $\vec{L}_{\text{rel}}=\text{const}$.

WOW :) Much appreciated the detailed explanation vanhees71!

 

[fog37]

One has to distinguish the two treatments of the problem and write down equations of motion to make things clear. Let's take the example of the Kepler problem, i.e., the motion of a planet around the Sun. Since in our solar system the mass of any planet is much smaller than the mass of the Sun, we can approximate the problem such that we consider the Sun as "infinitely heavy". Then one can choose an inertial reference frame, where the Sun is at rest at the origin, and the planet moves in the gravitational field of the Sun. Assuming a spherical Sun, we get the equation of motion
$$m \ddot{\vec{x}}=-\frac{G m M}{r^3} \vec{x},$$
where $G$ is Newton's gravitational constant, $m$ the mass of the planet and $M$ the mass of the Sun. Further $r=|\vec{x}|$ with $\vec{x}$ being the position vector of the planet.

This "external force" has obviously a potential,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x})\; \quad V=-\frac{G m M}{r}.$$
So we have an equation of motion for the planet in an external conservative force field, and energy conservation,
$$E=\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=\text{const}$$
as well as angular-momentum conservation,
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}=\text{const}$$
hold true. That's due to the symmetries of the problem, i.e., time-translation invariance and rotation invariance for rotations around the origin (i.e., the center of mass of the "infinitely heavy" Sun).

You can also treat the problem as a closed system of two interacting "particles". Here the gravitational force is treated as a true interaction between the two bodies. The equations of motion now are
$$m \ddot{\vec{x}}_P=-\frac{G m M}{|\vec{x}_P-\vec{x}_S|^3}(\vec{x}_P-\vec{x}_S),$$
$$M \ddot{\vec{x}}_S=-\frac{G m M}{|\vec{x}_S-\vec{x}_P|^3}(\vec{x}_S-\vec{x}_P).$$
The interaction force has a potential,
$$V(\vec{r}_P,\vec{r}_S)=-\frac{G m M}{|\vec{x}_P-\vec{x}_S|}$$
and the equations of motion get
$$m \ddot{\vec{x}}_P=-\nabla_P V, \quad M \ddot{\vec{x}}_S=-\nabla_S V.$$
The energy
$$E=\frac{m}{2} \dot{\vec{x}}^2 + \frac{M}{2} \dot{\vec{x}}_S^2 + V(\vec{x}_P,\vec{x}_S)=\text{const},$$
because
$$\dot{E}=m \dot{\vec{x}}_P \cdot \ddot{\vec{x}}_P + M \dot{\vec{x}}_S \cdot \ddot{\vec{x}}_S+ \dot{\vec{x}}_P \cdot \vec{\nabla}_P V + \dot{\vec{x}}_S \cdot \vec{\nabla}_S V=0.$$
Obviously Newton's 3rd Law is fulfilled and this immediately leads to
$$\ddot{\vec{R}}=0, \quad \vec{F}=\frac{1}{m+M} (m\vec{x}_P+M \vec{x}_S),$$
which is the center-of-mass theorem, i.e., the center of mass moves like a free particle,
$$\vec{R}=\vec{R}_0 + \vec{V} t, \quad \vec{V}=\text{const}.$$
It's now convenient to introduce the relative vector $\vec{r}=\vec{x}_P-\vec{x}_S$. After some algebra you get
$$\vec{x}_P= \vec{R} + \frac{M}{m+M} \vec{r}, \quad \vec{x}_S=\vec{F}-\frac{m}{m+M} \vec{r}.$$
Plugging this in either of the equations of motion and using $\ddot{\vec{R}}=0$ you get the equation of motion
$$\mu \ddot{\vec{r}}=-\frac{G m M}{r^3} \vec{r}=-\frac{G \mu (M+m)}{r^3} \vec{r}, \quad \mu=\frac{m M}{M+M}=-\vec{\nabla}_{\vec{r}}V,$$
where
$$V=V(\vec{x}_P,\vec{x}_S) \equiv V(\vec{r})=-\frac{G m M}{r}.$$
That's the equation of motion for a "quasi-particle" with mass $\mu$ in the external force field given by the potential $V(\vec{r})$.

It's easy to see that also the total angular momentum is conserved
$$\vec{J}=\vec{L}_P + \vec{L}_S=m \vec{x}_P \times \dot{\vec{x}}_P + M \vec{x}_S \times \dot{\vec{x}}_S=\text{const}.$$
In terms of the center of mass and relative coordinates this reads
$$\vec{J}=M \vec{R}_0 \times \vec{V} + \vec{L}_{\text{rel}}, \quad \vec{L}_{\text{rel}}=\mu \vec{r} \times \dot{\vec{r}}.$$
Of course also $\vec{L}_{\text{rel}}=\text{const}$.

Hello again Vanhees71.

You detailed reply confirms that:

a) internal forces can be conservative or nonconservative
b) external forces can also be conservative or nonconservative. In your example, Earth is first considered outside of the system and then included into the system
c) the physics is the physics regardless of how what we define the system

Just to be clear:
does potential energy $U$ belong only to the system components, in virtue of their reciprocal distance, or can potential energy be shared between system components and entities outside the system?

Thanks!

 

[vanhees71]

As this example shows, when making approximations it depends on which parts of the bigger (closed) system to include into what you call "system" and the "rest" or "environment".