I know that potential energy exists for a system of objects. But why? Is there no potential energy for a single object system? I assume that there is.

Consider the earth-object system. Suppose an external force lifts the object above the earth. Then the work done by it gets stored in the earth-object system as potential energy.Right?

Now, my question is that is the potential energy always zero for a single object system.


If we consider the object only as the system and lift it above the Earth, we are doing positive work on it and filling it with energy and the gravitational field of the Earth is doing negative work on it and draining it of energy. Thus whenever the object is moving it has kinetic energy which is fine. But when it is not moving,i.e. when its weight is equal to the force applied by us, the net work done is zero and so the object should not have any kinetic energy. Does it have potential energy?

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    $\begingroup$ Single object system? Potential energy arises from interaction, which needs two bodies. $\endgroup$ – user36790 Oct 25 '16 at 10:45
  • $\begingroup$ Can't we consider the interaction between the bodies as external force and consider the system as the object only? $\endgroup$ – MrAP Oct 25 '16 at 10:48
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    $\begingroup$ It is not correct to say that the potential energy is zero in this case. It is not defined. $\endgroup$ – garyp Oct 25 '16 at 11:59
  • $\begingroup$ @MAFIA36790, you said "potential energy arises from interaction, which needs two bodies." If we consider the system as earth+object then its fine.Is it mandatory that the object(Earth in this case) exerting force has to be a part of the system? What if we consider the object as the system and the earth as an external object and hence the gravitational force an external force. Then will the object not have potential energy? $\endgroup$ – MrAP Oct 27 '16 at 7:34

To understand your situation, you need to rethink your understanding of the definition of the potential energy. The best definition for the potential energy is: Potential energy is a measure of work done by a potential force upon the object moving while it is experiencing that force. The change in the potential energy of a body between two positions is equal to work of the potential force upon movement of the body between those two positions. So, when you specify Potential energy of the body, it is always the energy of a specific [conservative!] force (or a combination of conservative forces), and with respect to some point of reference that you choose as a zero level of that potential energy. (Note that the position is also with respect to the other body that produces the force.)

When you have a single solid body, i.e. there are no external forces acting on it, you do not have any potential energy defined. So, the answer to your question: in case of a single solid body the potential energy is undefined. Saying that something that is undefined is zero is not strictly correct.

The case is rather different when the body is not solid, i.e. when its shape is not constant. E.g. if the shape of an elastic body can be changed (e.g. the body can be compressed), - then you have a potential energy related to that deformation. In that case, you can consider the body as a combination of different parts ("points" of very small mass) of it that are acting on each other with forces. If the object is fully elastic, each of those forces is potential, and then you can associate potential energy with each of them. Accounting for all of them individually is rather difficult, so, an average potential energy for the entire ensemble of these portions (points) is introduced. In this, fully elastic case, this potential energy is a unique function of the shape of the body (i.e. of the relative positions of all the points within that body).

Some other typical examples where the confusion similar to the OP's occurs is the potential energy of a charged particle (e.g. electron) in an electric field. It looks like there is no other "body". In reality, there is a field (electric), which effectively is the representation of another body(ies) acting on this charged particle. Note that in case of gravity, one can define a gravitational field, thus "hiding" the object that creates it.


Alright, let us dive into some easy stuff first.

  • What actually is potential energy?

Potential energy is defined as the energy possessed by a system by the virtue of it's configuration. This is the hard way. The easy way to understand is of course, building intuition, with some real life examples. Visualize this - you are an archer. Your bow string is at it's natural position, free of any strain. Suddenly, you decide to fire an arrow. You pull the string, and the Kinetic Energy of the string during this pulling gets totally converted into potential energy. When you've stretched it to the maximum, that is the situation where the energy of this system is only potential. But still, what is this energy?

Imagine the string to be at natural length. Now also imagine a 'dot' at this point. When you stretch this string, it goes farther and farther from the reference point, or the 'dot'. So basically, potential energy is the energy the string gains when it is pushed away from a reference point.

The mathematics now. Potential energy is mathematically defined as the negative of work done on a system.

$$ dU = -dW = -F.dr$$

Integrating this,

$$ U_f - U_i = -W = -\int F.dr$$

Now, if you define $ U_i = 0 $ , that leaves you the potential energy of the system at any point away from the reference point, or the point where the energy was $U_i$. So, you see, you could define the surface of earth as the reference point, or the moon etc. Hope you get the point.

To your question now. The potential energy of a single object system depends on some things. Firstly, what kind of potential energy? Are you talking about the energy it has because of some charge lying next to it ( electrostatic energy)? Or due to a huge ball of rocks ( gravitational ) ? Are you saying that the object is lying around in vacuum, with nothing except the object there? Firstly, such a place is almost impossible, but still, if that be the case, that isolated piece doesn't actually have potential energy, because there is no force acting on it, which makes the integral above equal to zero. Hope this helps.

  • $\begingroup$ Can we consider the object on the earth as the system and measure the gravitational potential energy it has due to the earths pull, which is an external force? $\endgroup$ – MrAP Oct 25 '16 at 11:17
  • $\begingroup$ Does considering the object as the system mean you're going in the reference frame of the object? I'm asking because energy is a frame dependent quantity. $\endgroup$ – 111 Oct 25 '16 at 11:20
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    $\begingroup$ You left out two important words. Potential energy is defined as the negative of the internal work done against conservative forces. The word internal implies that there is more than one object. Potential energy is not defined for a system comprising only one object. $\endgroup$ – garyp Oct 25 '16 at 11:57
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    $\begingroup$ @MrAP If you take the system as the object, then the earth does external work on it as you say, and the force of gravity can change the object's kinetic energy. There is no internal work, so no notion of potential energy. What is the volume of a point? You can't answer, because it's not defined. A point has no length, depth or breadth. $\endgroup$ – garyp Oct 25 '16 at 12:19
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    $\begingroup$ See page 131 of "Five Easy Lessons" by Randall Knight. I suspect his textbook has the same discussion, but my copy was stolen years ago, $\endgroup$ – garyp Dec 6 '16 at 13:21

Let us examine your statement:

If we consider the object only as the system and lift it above the Earth, we are doing positive work on it and filling it with energy and the gravitational field of the Earth is doing negative work on it and draining it of energy.

So far, so good.

Thus whenever the object is moving it has kinetic energy which is fine.

And in order to get it moving, we had to add more energy via the lifting force than the gravitational force removed. In other words, the net work is now positive. There is still no change in the potential energy, because the gravitational force is not part of your system.

But when it is not moving,i.e. when its weight is equal to the force applied by us, the net work done is zero and so the object should not have any kinetic energy.

Right. When it's not moving within some defined reference frame, it has no kinetic energy in that reference frame. And the reason the object slows and stops is that, during the slowing process, gravity does more negative work than your force does positive work. It doesn't matter if it has changed position compared to some earlier time. It now has the same total energy as when it started because the net work done on your single-object system is zero. You might want to ask "but what about mgh?" My response is "That tells me how much work gravity did while the object moved."

Does it have potential energy?

No, because you can't count the work done by gravity and gravitational potential energy. It is an either/or situation.


Now, my question is that is the potential energy always zero for a single object system.

Not just zero, but undefined.

How would you measure or describe the potential energy, if gravity is not present (or not a part of your system)? Potential energy is a way to describe how much work some force will do if released (if the body is released from rest). If there is no such force, there is nothing to do this work, so there is no potential energy stored anywhere in the first place.

You can say it like this: Lifting an object causes potential energy to be stored. But what if Earth now suddenly vanishes? What happens to the object when you let go? Will it fall? Of course not - it has nowhere to fall when nothing pulls in it. Stored energy being converted into work when released only happens if there is a force to do that work. If not, no energy is stored in the first place. Removing the Earth also removes this stored potential energy.

  • $\begingroup$ But what about us? We too are also not in the system. What will happen to the energy which we "inject" into the body by doing positive work on it. $\endgroup$ – MrAP Oct 25 '16 at 17:20
  • $\begingroup$ @MrAP Well, when you did that work from your body, that work was indeed stored as potential energy, because gravity was present to pull downwards. If we remove Earth suddenly, we also remove that stored energy. If you try to move the object now, you don't store any energy as potential energy (only as kinetic energy). Potential energy was present for the two-body-system but not in the one-body system. $\endgroup$ – Steeven Oct 25 '16 at 17:23
  • $\begingroup$ What is your point when you say to remove the Earth?When the earth is present, i can consider the object as the system and by applying a displacing force to it i can store potential energy in the body.Right? $\endgroup$ – MrAP Oct 25 '16 at 17:27
  • $\begingroup$ @MrAP What do you mean by considering the object as the system? As should be clear by now, potential energy exists because of two bodies pulling in each other. What is a "system" by your definition? We can consider the object as a system, yes ok - but that doesn't really change anything about the fact that two objects are needed for potential energy to be present. $\endgroup$ – Steeven Oct 25 '16 at 17:45
  • $\begingroup$ One thing i cannot understand is that why cant the object have potential energy if we do not include the Earth into the system. Cannot we consider the gravitational force as external forces and can't external forces affect potential energy? $\endgroup$ – MrAP Oct 25 '16 at 17:52

Yes, it exists, because in physics there is always an observer (you, for example) and you interact with this "single object". In physics it is meant to be an interaction energy. Often this potential energy is negligible with respect to the kinetic energy of yours, but it exists anyway. Physics is not mathematics where objects "exist" in our rich/poor imagination and without any other thing.


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