How to interpret this definition of potential energy?

[anon90]

Hello everyone, I'd like to share a doubt I am currently struggling with.
So we know that ΔU=−W, where ΔU is the difference of potential energy and Wthe work done by the force to move the body from point A to point B.

When analyzing this for the gravitational force, since we have U=−GmM/R, with the minus due to the fact that we're dealing with an attractive force, if we pick U(∞)=0, we have that ΔU<0 when a particle is moved from infinity to another point.

That makes sense to me because the work done by the gravitational force is positive, and that means that the kinetic energy is increasing ( ΔK=W); since the total energy must be conserved, when starting at infinity with a null total energy, if K is increasing then U
must decreasing accordingly.

On the other hand, if we move a body from a point A to a point B "outwards", the work done by gravity is negative and that means that gravity will "slow down" the body (aka, the kinetic energy will decrease and the potential one will increase).

Supposing that what I've said so far is actually correct, what confuses me is the definition that some books give when dealing with the potential energy.

More precisely, a possible definition is that "the potential energy of a system is the work done by an external agent against the force to create such a configuration, with the particles of the system starting from a picked reference point".

I'm having a hard time understanding the correct interpretation of the work done by the "external agent", especially in this specific case: here this external agent would do negative work to bring the particles from infinity to a certain configuration. What does someone having to do negative work mean?

That such a configuration would happen even without them doing anything, since the force will do the job in their stead? I think I can sort of tell that's because gravity would pull these particles close to each other without me having to work against it to build the desired configuration, and on the other hand if I start with n particles close to each other and I want to move them apart I have to do work against gravity hindering me, but I cannot say I'm totally comfortable with this explanation.

How should I interpret it?

 

[nasu]

I would rather forget about these "definitions" and stick with the actual definition of potential energy, in terms of the work done by the field. As your questions show once more, the alternative creates nothing but confusion. Look in any serios mechanics book and you will find the PE defined in terms of the work of the field.

 

[wrobel]

Assume that a particle experiences a force ##\boldsymbol F=\boldsymbol F(\boldsymbol r).## Then by definition a potential energy of this force is a function ##V=V(\boldsymbol r)## such that ##-\nabla V=\boldsymbol F##.
Here ##\boldsymbol r## is a position vector of the particle.

Such a function ##V## is not obliged to exist.

If there are a number of particles with position vectors ##\boldsymbol r_1,\ldots,\boldsymbol r_n## and ##\boldsymbol F_i## is applied to i-th particle then a potential energy ##V=V(\boldsymbol r_1,\ldots, \boldsymbol r_n)## of the system is defined as follows
$$\boldsymbol F_i=-\nabla_{\boldsymbol r_i}V$$

 

[PeroK]

What does someone having to do negative work mean?

When you raise a weight, your muscles do positive work by increasing the PE of the weight. Your muscles expend energy in this case. When you lower a weight, your muscles do negative work by absorbing the lost PE of the weight. Some of that energy is transformed to heat in your muscles. In any case, that's negative work.

 

[sophiecentaur]

what confuses me is the definition that some books give when dealing with the potential energy.

You would need to quote one of your books if you want an opinion of what they are trying to say but I suspect your confusion is to do with the reference for displacement and force.

In an elementary problem, as an object is raised from the ground, the distance (height) is conventionally positive and the force to be overcome is negative (down). In the case of an object in space, the reference is at infinity so the displacement is positive, towards a star and so is the force.
If potential is defined in terms of the work done on an object when moving it through a field (the work you need to do) you will get the negative of the work done by the field.
To resolve any confusion, look at a conventional image of a Potential Well (the word "well" is the clue). The PE is always negative, relative to the reference at infinity. That implies that the Work Done On a unit mass as it falls in Negative. Reason from that basic viewpoint and you can't go wrong. (Wanna bet?)

 

[cianfa72]

I would say that the potential energy is a physical quantity associated to the configuration of the overall system we are looking at. So in case of gravity (classical mechanics) when we lift a body we are actually transferring energy to the potential energy (PE) of the overall system 'body + Earth'. Technically speaking there is not PE associated to the body alone.

 

[vanhees71]

The main problem is that in introductory textbooks they confuse potential energy and work, which are related but not the same.

The most general idea is work, and it is introduced because of the so-called "work-energy theorem". Let's consider the most simple case of one particle moving in some external field (like the particle in the gravitational field of the Earth or a charged particle in an electromagnetic field, etc.). Then the equation of motion reads
$$m \ddot{\vec{x}}=\vec{F}.$$
If you have a solution ##\vec{x}(t)## of this equation of motion and you multiply this equation of motion with ##\dot{vec{x}}## you get
$$m \dot{\vec{x}} \cdot \ddot{\vec{x}} = \dot{\vec{x}} \cdot \vec{F}.$$
Now the left-hand side can be written as a time-derivative,
$$\frac{\mathrm{d}}{\mathrm{d} t} E_{\text{kin}}=\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right)=\dot{\vec{x}} \cdot \vec{F}.$$
Integrating this wrt. to time over the interval ##(t_1,t_2)## you get
$$ E_{\text{kin}}(t_2)-E_{\text{kin}}(t_1)= \int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}} \cdot \vec{F}=W.$$
This means that the change of the kinetic energy of the particle is given by the work of the force on the particle along the trajectory (i.e., the solution of the equation of motion) of the particle.

Now there is a class of forces, for which there exists a potential. These are forces, which only depend on the position of the particle (like the gravitational field of the Earth or the force on a charged particlen in an electrostatic field of some fixed charge distribution). By definition this means that the force on the particle is given by a scalar field, ##V(\vec{x})##, the potential of the force, such that
$$\vec{F}=-\vec{\nabla} V.$$
Then the work defined above is
$$W=\int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}} \cdot \vec{F} = -\int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}} \cdot \vec{\nabla} V[\vec{x}(t)]=-\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V[\vec{x}(t)]=-[V(\vec{x}_2)-V(\vec{x}_1)],$$
where ##\vec{x}_1=\vec{x}(t_1)## and ##\vec{x}_2=\vec{x}(t_2)##.
Then you get
$$E_{\text{kin}}(t_2)-E_{\text{kin}}(t_1)=-[V(\vec{x}_2)-V(\vec{x}_1)] \; \Rightarrow \; E_{\text{kin}}(t_2) + V(\vec{x}_2) = E_{\text{kin}}(t_1)+V(\vec{x}_2)=E,$$
i.e., the total energy
$$E=\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=\text{const}.$$
This is the law of energy conservation, and as just shown, it's a special case of the work-energy theorem. It applies only to forces which have a potential in the above defined sense, but then you have the advantage that the energy-conservation law gives you a first integral of the equations of motion, which tells you something about the solution of the equation of motion without the need to know this solution to write this first integral down.

 

[anon90]

-cut-

I don't think I have big issues understanding the conservation of energy, it's more about the concept of negative work and what that implies.

I would rather forget about these "definitions" and stick with the actual definition of potential energy, in terms of the work done by the field. As your questions show once more, the alternative creates nothing but confusion. Look in any serios mechanics book and you will find the PE defined in terms of the work of the field.

You are probably right, still I wanted to try understanding the physical meaning behind it.
Like, I think I'm fine when I'm supposed to understand the potential energy as the work done by the field: for a particle moving from inf towards something the work done by the field is positive, and that makes sense, since the source will attract the particle, the particle's kinetic energy will increase because of that and in order to keep the total energy constant, the potential energy will decrease.
It's the negative work of the agent (or rather, their role in the analogy) that puzzles me.

Assume that a particle experiences a force ##\boldsymbol F=\boldsymbol F(\boldsymbol r).## Then by definition a potential energy of this force is a function ##V=V(\boldsymbol r)## such that ##-\nabla V=\boldsymbol F##.
Here ##\boldsymbol r## is a position vector of the particle.

Such a function ##V## is not obliged to exist.

If there are a number of particles with position vectors ##\boldsymbol r_1,\ldots,\boldsymbol r_n## and ##\boldsymbol F_i## is applied to i-th particle then a potential energy ##V=V(\boldsymbol r_1,\ldots, \boldsymbol r_n)## of the system is defined as follows
$$\boldsymbol F_i=-\nabla_{\boldsymbol r_i}V$$

Yup, I think that's clear.

When you raise a weight, your muscles do positive work by increasing the PE of the weight. Your muscles expend energy in this case. When you lower a weight, your muscles do negative work by absorbing the lost PE of the weight. Some of that energy is transformed to heat in your muscles. In any case, that's negative work.

So in my first example I would consider the system of me, the particle and whatever is generating the field, if I am right.
So if the external agent's work is negative, it means they have to do nothing at all to create that situation, since the rest of the system's work is positive? As, if I want to bring a particle from infinity to a closer point the gravitational field will actually do that for me, since it will attract the particle itself; on the other hand, if I want to move a particle from a certain point to a further one I have to do positive work (field doing negative work, that is trying to stop the particle from moving away)
Although heat is unrelated in this case since we're dealing with conservative forces only.

You would need to quote one of your books if you want an opinion of what they are trying to say but I suspect your confusion is to do with the reference for displacement and force.

In an elementary problem, as an object is raised from the ground, the distance (height) is conventionally positive and the force to be overcome is negative (down). In the case of an object in space, the reference is at infinity so the displacement is positive, towards a star and so is the force.
If potential is defined in terms of the work done on an object when moving it through a field (the work you need to do) you will get the negative of the work done by the field.
To resolve any confusion, look at a conventional image of a Potential Well (the word "well" is the clue). The PE is always negative, relative to the reference at infinity. That implies that the Work Done On a unit mass as it falls in Negative. Reason from that basic viewpoint and you can't go wrong. (Wanna bet?)

Unfortunately they're not in English so I don't think it would be useful, but I translated the definition in my first message.

I would say that the potential energy is a physical quantity associated to the configuration of the overall system we are looking at. So in case of gravity (classical mechanics) when we lift a body we are actually transferring energy to the potential energy (PE) of the overall system 'body + Earth'. Technically speaking there is not PE associated to the body alone.

Yes, you're right we talk about the system's energy, rather than the single body's.The main problem is that in introductory textbooks they confuse potential energy and work, which are related but not the same.

 

[wrobel]

Yup, I think that's clear.

then go step by step from definitions to theorems. 99% of confusions disappear after careful reading of precise formulations.

 

[wrobel]

The main problem is that in introductory textbooks they confuse potential energy and work, which are related but not the same.

exactly

 

[sophiecentaur]

I would say that the potential energy is a physical quantity associated to the configuration of the overall system we are looking at.

I guess that implies that we should always declare what we mean every time. Electric Potential would usually be used to described the work that can be done 'on' charges / currents, rather than how much energy would be taken from a battery.
Seldom a problem, usually, when the meter in your head puts the appropriate sign on the answers you get.