Potential Energy and Potential -- Systems versus Particles

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When I first learned about these subjects, I did what was intuitive to me and treated particles as if they carried potential energy. I would do this similarly for rigid bodies where I would also treat them as a particles with their body's mass at the center of mass. This wasn't helped by numerous textbooks and other sources supporting this type of thinking. Life was simpler back then.

My interpretation is now as follows. We first define our system. Then we have these equations: Worktotal= ΔKsystem, Workexternal = ΔEsystem, -1*Workinternal, conservative = ΔUsystem, and Workexternal + Workinternal, non-conservative = ΔEmechanical. Conservative forces have to be internal to our system to define potential energy. Also, the potential energy we define is of our total system, not divided among the components of our system in some manner. Does anything seem wrong with my interpretation so far?

I don't want this initial post to get too long, so I'll cut to some questions I have:

1) We construct potential energy functions from the equation ΔUsystem = -1*Workinternal, conservative. Let's talk about gravitational potential energy in a system where m1 << m2, such as a ball-earth system. We have two internal, conservative forces in this system, the equal and opposite gravity forces. However, as the earth is much more massive than the ball, the gravity force by the ball on the earth essentially does no work on the earth; the earth isn't going to experience much of a ΔK ∴ the resulting -ΔU contribution from the earth's ΔK will also be negligible. Is this what leads to our approximation that "the potential energy of the ball-earth system" = "the potential energy of the ball"?

2) What about in other systems where we shouldn't ignore the second, third, etc. internal conservative forces? Take a horizontal spring-mass system. Why is it that we ignore the equal and opposite force that the mass exerts on the end of the spring? In our potential energy function calculation, we only consider minus the work of the spring force onto the mass.

3) Plenty of sources state that electric potential is the potential energy per unit charge that a test charge will possess at some point in space (relative to an arbitrary origin of course). However, doesn't this immediately contradict the notion of potential energy belonging to a system and not one particle (in this case the test charge)? I'm thinking that something similar to 1) is going on, perhaps because we're dealing with an electrostatic field.
 

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  • #2
Nugatory
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1) The potential energy is ##mgh##, but it is sloppy language to call it the "potential energy of the ball" instead of the "potential energy of the system". This sloppiness is somewhat excusable because if we release the ball and convert the potential energy to kinetic energy, the kinetic energy all ends up in the ball - but it's still sloppy.
2) the only time we consider only the work on the mass is when the other end of the spring is attached to the earth or other immobile object. The problem reduces to the first case.
3) Any source that uses the exact wording you've used ("the potential energy that a test charge will possess") is being a bit sloppy. The potential energy is possessed by the system as a whole. In this case the sloppiness is excusable because you're supposed to be treating the natural-language words as commentary; the path integral that defines the potential is authoritative.
 
  • #3
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Should the potential energy of a system be thought of (maybe defined?) as the work needed to assemble the system in a manner that doesn't change the system's kinetic energy? Maybe there's a better way of describing what I'm trying to say. (Edit: Apparently, this leads into the topic of something called self-energy.)

This seems to be the thinking used in, for example, how one of my textbooks discusses the electric potential energy of a system of charges. All of the charges are assembled through an applied force that is equal and opposite to the Coulomb force the charge being acted on is experiencing from the other charges in the system. The superposition of the Coulomb force is relied upon in obtaining the result.

Taking this same charge example again to ask my next question, we are also obviously, after we have moved a charge from ∞ to its desired position, continuing to apply a force to this charge to keep it stationary, correct? (My book doesn't discuss this, perhaps because it is trivial or maybe because I'm supposed to assume that the charges will stay in place already due to this being electrostatics.) This applied force will be equal and opposite to the net Coulomb force that charge experiences from the other charges in the system. The key here would be that because the charge is held stationary, no external work is done on this charge and our calculation for Usystem is not disrupted.
 
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  • #4
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So I guess in addition to my last post, I want to ask some more questions for now:
1) So once again talking about the "potential energy of a test charge" (according to many sources), relative to some reference point, being defined as the product of q_0 * ∅(x,y,z), is the "sloppiness" here similar to my ball-earth system question? Is the key here the fact that it's an electrostatic field, so once we release this test charge, what we have called its potential energy will be turned into only the test charge's kinetic energy; all of the other charges are static.

2) How would this be defined in an electrodynamic field (if it is)?

3) But still in our first example, our static arrangement of charges have potential energy. (According to one of my textbooks, it is the work necessary to bring in all the charges in the system into their arrangement from them first being infinitely separated.) Even with all we've discussed, what do you think of this interpretation of things that someone introduced to me?: We define our system and we consider its self-energy, due to internal interactions, and its potential energy, due to external interactions. We can choose a rigid object as our system and can treat it as if it has gravitational, elastic, electric, etc. potential energy.

Ex: Let's take two (isolated) planets in space of mass m1 and m2. Taking m1 as our system, we would say that its gravitational PE is -1*Gm1m2/r. Similarly, if we were to take m2 as our system, we would say that the grav. PE of m2 is the same quantity. However, if we were to take both m1 and m2 as our system, -1*Gm1m2/r would be treated as the system's self-energy. The system's potential energy would arise from its external interactions.

How does this all sound to you?
 
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  • #5
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1) The potential energy is ##mgh##, but it is sloppy language to call it the "potential energy of the ball" instead of the "potential energy of the system". This sloppiness is somewhat excusable because if we release the ball and convert the potential energy to kinetic energy, the kinetic energy all ends up in the ball - but it's still sloppy.
Just to check what I understood:
  • the increase of potential energy of the Earth-ball system (compared to the condition when the ball is at the Earth surface) is ##mgh##: we are assuming external Work done against the "gravitational" force on the Earth due to the ball is negligible (basically we are neglecting Earth displacement measured in the inertial system of reference of the Earth-ball center of mass); thus is reasonable to compute total external Work as done actually all against the "gravitational" force acting on the ball due to the Earth
  • similar topic about kinetic energy: upon releasing the ball is reasonable to assume overall potential energy of the Earth-ball system all ends up in the kinetic energy of the ball alone - again, in the inertial system of reference of the Earth-ball center of mass, we can neglect the kinetic energy of the Earth.
Does it make sense ? Thanks
 
  • #6
anorlunda
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No. Although the velocity of earth is small, its mass is large.

##\frac{mv^2}{2}##

If you want to learn this stuff, begin with two objects having the same mass. Find the equations. Then as step two, ask yourself what happens is one is more massive than the other.

Also get used to calculating things instead of characterizing them as small or big.
 
  • #7
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No. Although the velocity of earth is small, its mass is large.

##\frac{mv^2}{2}##

If you want to learn this stuff, begin with two objects having the same mass. Find the equations. Then as step two, ask yourself what happens is one is more massive than the other.
I believe it is a two-body problem: take ##M## as the mass of the Earth and ##m## of the ball.
We can use the reduced mass ##\mu = \frac {Mm} {M + m}## to work out the displacement vector ## \mathbf r(t)## from the differential equation ##\mu \ddot {\mathbf r} = \mathbf F_{Mm}##.

Consider now the inertial system of reference with origin in the 'Earth-ball system' center of mass: starting from it we can define ##\mathbf r_M## and ##\mathbf r_m## as the vector positions of the Earth and the ball respectively. They are related to ## \mathbf r(t)## as follows:

##\mathbf r_M(t) = \frac {m} {M + m} \mathbf r(t)##
##\mathbf r_m(t) = - \frac {M} {M + m} \mathbf r(t)##

and thus for kinetic energies in that system of reference we have:

##K_M = \frac {M \dot {\mathbf r_M}^2} {2}##
##K_m = \frac {m \dot {\mathbf r_m}^2} {2}##

## \frac {K_M} {K_m} = \frac {m} {M}##

When ##M\gg m##, as in our case we, can neglect the kinetic energy of the Earth and assume the potential energy ##mgh## of the 'Earth-ball' system all ends up in the kinetic energy of the ball alone when it reaches the ground from height ##h##.
 
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