# I Physical interpretation of the "total potential energy"

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1. Dec 21, 2016

### vco

The principle of minimum total potential energy is frequently used in solid mechanics as an elegant way of deriving the equilibrium equations for an elastic body under conservative forces. This method states that out of all the possible displacement fields that fulfill the boundary conditions, the one which minimizes the total potential energy satisfies equilibrium.

In textbooks, the total potential energy is defined as the sum of the elastic energy of the body and the potential energy of the external forces. The elastic energy is a well-defined concept, but I have been wondering for years what exactly is the physical interpretation of the "potential energy of the external forces", and consequently, the physical interpretation of the total potential energy itself.

A simple example of the application of the method: for a linear spring with stiffness $k$ loaded with a force $F$, the elastic energy is $$U=\frac{1}{2}ku^2,$$ and the potential energy of the external force is defined as $$V=-Fu,$$ where $u$ is the displacement of the spring. The total potential energy of the system is $$\Pi=U+V=\frac{1}{2}ku^2-Fu,$$and by minimizing $\Pi$ with respect to $u$ we end up with the obvious equilibrium equation $$ku-F=0.$$ In every textbook I have seen, the total potential energy and the associated principle are introduced as if they were some fundamental physical concepts that need no explanation or derivation. Even the name "total potential energy" sounds confusing since it cannot be the energy stored in the system; the stored energy is the elastic energy, is it not?

Also, the Wikipedia article says something about the lost potential energy being turned into heat, how does this fit in? Is this related to the forces being applied instantaneously vs quasi-statically?

So, could someone please explain what is the physical meaning of the
1. potential energy of the external forces?
2. total potential energy?

2. Dec 21, 2016

### ShayanJ

Newton's second law, $\vec F=m \ddot{\vec x}$ gives you the acceleration of a particle at any moment in time. The particle is in Equilibrium if there is zero net force acting on it. Now if we define $\vec F=-\vec \nabla V$, the equilibrium point is given by $\vec \nabla V=0$ which gives the extremums of the potential. And also because its force that is observable and two potentials differing only in an additive constant give the same force, the total potential is not a physical quantity and only difference in potential is physical.

3. Dec 21, 2016

### Staff: Mentor

Not all of the system's energy is elastic. Some of the system's energy is stored in F.

4. Dec 22, 2016

### vco

But if we were to suddenly make $F=0$, would not the spring start to vibrate with a total energy equal to the elastic energy of the original scenario?

5. Dec 22, 2016

### Staff: Mentor

Then the potential for F would be time dependent and therefore energy would not be conserved.

6. Dec 22, 2016

### vco

I am still confused.

Another thing to increase my confusion:
Since the potential energy of the external forces is by definition the negative of the work they perform, there seems to be a "hidden" assumption that the external forces are applied instantaneously and not quasi-statically.

In a quasi-static case, the work performed by the external forces is equal to the elastic energy. In the instantaneous case which the principle suggests, the work performed by the external forces is greater than the elastic energy.

In the example, the spring begins to vibrate due to the external force attaining its final value immediately. The external force must attain its final value immediately since the work it performs (the negative of its potential energy) is defined to be $Fu$. The vibration seems to be assumed to dissipate due to damping present in the system, and finally the equilibrium position is attained.

Where does this reasoning go wrong, or does it?

7. Dec 22, 2016

### Staff: Mentor

I have no idea what you mean by this.

This is incorrect. The vibration will happen with or without the external force, depending on the initial conditions.

Oscillations happen when you have a potential with a minimum. Adding a linear potential may change the location of the minimum, but doesn't determine whether or not there is a minimum

8. Dec 22, 2016

### Staff: Mentor

Sorry, I had missed this earlier and had forgotten the original context regarding he principle of minimum potential energy.

Yes, you are correct. Vibrations are assumed to have already dissipated. Alternatively, you can consider that you are looking for equilibrium solution which is, by definition, the solution which is not a function of time.

9. Dec 22, 2016

### vco

Imagine the spring is from the suspension of a car, and you compress it in a press. Or that the spring is one of those demo springs found in physics classrooms, and you compress it with your hand. You do not apply the force instantly as $F$, but you increase it from zero to $F$. The work the force performs is not $Fu$ since the force is not constant during the deformation. If you apply the force infinitely slowly from zero to $F$, i.e., quasi-statically, then the work performed by the force is equal to the elastic energy $\frac{1}{2}ku^2$, which is half of $Fu$. In the instantaneous case, the difference between the work performed by the external force and the elastic energy of the spring in equilibrium manifests itself as vibration, which then dissipates.

The principle of minimum total potential energy implies that the force is applied instantaneously: the potential energy of the external force is stated as $-Fu$ and hence the work it performs is $Fu$, which in turn implies that the force is constant during the deformation.

Someone else has also wondered this issue: http://physics.stackexchange.com/qu...rce-on-particle-attached-to-spring-is-equal-t

Now, I know that the principle of minimum total potential energy applies only to statics, not dynamics. And I also know that the spring example introduced in my first post is static by definition. But when you move from the idealized world to the real world, the example is no longer strictly static. But you still get the correct solution (the equilibrium equation) by applying the principle, even if the work performed by the external force is different than in the idealized world.

I must admit that I am an engineer, so I might be looking at things from a different perspective.

Last edited: Dec 23, 2016
10. Dec 23, 2016

### vco

11. Dec 23, 2016

### Staff: Mentor

You are way overthinking this. We are looking at static solutions, so there is no "apply". The force simply is, it has always been there and will always be there. That is what it means to be static, it doesn't change over time. There is no distinction between quasi-static or instantaneous, since either one is a force changing over time, and we are looking only for static solutions so both the instantaneous and the gradual application are explicitly excluded by our assumption.

Yes, this is unrealistic, but that is not the point. In science and physics and especially in engineering we make assumptions that simplify things. We know that our assumptions are wrong at some level, but they can often serve as a useful approximation. Here the assumption that we are making is the static assumption, so nothing will depend on time.

Yes, this is why we teach and use this particular approximation. The reason it works is that damping forces are a function of the velocity rather than the coordinate, so they don't affect the asymptotic position. So the asymptotic position of a damped system is the same as the static position of the undamped system.

I am an engineer too. You should get used to making assumptions and approximations. You need to have your assumptions in mind when you analyze a problem for two reasons, first they are important because they make analysis easier, second they are important because they can lead to big errors when they are violated.

12. Dec 26, 2016

### vco

Thank you. That was an excellent post.

I am used to making assumptions and approximations. It is just that if there is even the slightest sense of doubt in my mind regarding the mathematical and logical consistency of the "foundations", I feel uncomfortable. It is not satisfactory to know perfectly how to use something and know when it can and cannot be applied, but not knowing why it works at the fundamental level.

I gave things some thought and came to the conclusion that the reason for the confusion was that I did not fully understand the concept of a potential energy function in general.

Ever since from high school I had always associated potential energy functions of forces to either internal forces (elasticity) or external force fields (gravitation). But of course also "regular" external forces not arising from force fields (contact forces) have potential energy as long as they are conservative, even though this is less intuitive.

In the idealized world, a conservative external force would continue to perform work to the body from here to eternity should all other forces dissappear, just like the gravitational force in an infinite gravitational field would. We have defined the external force to be there, and there it shall be no matter what happens. A constant gravitational force is no different from a constant conservative external force in this context.

I think the following misleading convention has contributed to my confusion: in textbooks, the gravitational potential energy is always included in the total potential energy, but the potential energy of the external force is usually excluded from the total potential energy in textbooks not specifically dealing with applied/analytical/structural mechanics.

So unless you disagree, I consider myself enlightened :)

Last edited: Dec 26, 2016
13. Dec 27, 2016

### lychette

What do you mean by 'stored n F' f it is not potential energy

14. Dec 29, 2016

### vco

I think he means that if we wanted to bring the system back from the loaded position to the original position while keeping $F$ as it is, then the work required would be equal to the change in the total potential energy. In this process, the value of the total potential energy would increase from $\frac{1}{2}ku^2 - Fu = -\frac{1}{2}ku^2$ back to zero.

It is a bit tricky to understand if you consider it from a too practical aspect. Of course, the energy that would be released by the spring if $F$ was suddenly removed would be the elastic energy (which in equilibrium appears to be, at least in this example, the negative of the total potential energy). But as long as $F$ is there, we have to include its potential energy in the total potential energy of the system.

Imagine that $F$ was caused by gravity. In this case, it is quite clear that part of the energy of the system is stored in $F$. Think about it, and see that it makes no difference whether $F$ is an externally applied conservative force or a gravitational force. We cannot turn off gravity, and we cannot turn off an external force since we have defined it to be there.

This is how I see it. However, note that I figured this out quite recently, so I am not that sure whether this is correct.

Last edited: Dec 29, 2016
15. Dec 29, 2016

### Staff: Mentor

It is potential energy, just not elastic potential energy. The OP didn't specify the nature of F, but whatever its nature it clearly has an associated potential energy.

That potential energy is the same as the gravitational potential energy of an object with mass $F/g$, so barring any other information that is what I would assume.

16. Dec 29, 2016

### lychette

Many apologies for my typing error....
'.....stored in F' if it is not potential energy.

17. Dec 29, 2016

### vco

The nature of the force in the example is simply that it is constant in magnitude and direction. Therefore, it is conservative and has a potential energy function, like you said. No other restrictions.

18. Jun 11, 2017

### bcl

I'm confused on this also. Lets say you impose work (energy) with magnitude of F*u, but the strain energy magnitude is less than the work. Where did the extra energy go? I expected that the imposed work (F*u) would all go into strain energy.

19. Jun 11, 2017

### Staff: Mentor

Either to another potential energy or to kinetic energy.

20. Jun 12, 2017