Conceptual derivation of (classical mechanical) energy?

1. May 8, 2010

RoyLB

I have a pedagogical question, and a philosophical question, both involving energy:

1) All the derivations I’ve ever seen of kinetic energy in both elementary and advanced texts (or in Physics Forum searches) either simply define T=1/2mv^2 or start by taking the dot product of force and distance. Once either of these steps are done, it’s just a matter of mathematics to get to the work energy theorem. However, pedagogically speaking, both approaches seem forced (if you’ll pardon the pun). Do you know of a more natural way to introduce and motivate the idea of (mechanical) energy? I assume it is natural to start from F=ma, since the non-expert can grasp the concepts of forces, masses, and accelerations pretty easily, especially after some simple experiments.

2) What exactly is energy, anyway? Since work involves a dot product, can it be considered the component of something? If so, what? I suspect that one might need to resort to relativity and/or thermodynamics to get the true picture of what energy is, not unlike how Least Action in classical physics can be “explained” by resorting to
quantum physical considerations
(see http://www.eftaylor.com/pub/QMtoNewtonsLaws.pdf)

Direct answers or pointers to relevant references would be appreciated.

Thanks!
Roy

2. May 8, 2010

Studiot

Energy (and work) is a scalar so can't be a component of anything.
They are both physical concepts.
The dot product is a mathematical concept that results in a scalar.

In traditional mechanical education energy is defined as the capacity to do work. The concept is introduced after work has been defined in terms of forces and distances. Both these entities are subject to relativistic transformations.

So where are you coming from?

3. May 8, 2010

Gerenuk

Here is a way I've discovered. The outline is:
All of our assumptions are
1. conservation of momentum p(v)
2. all forces are inverse square forces (or more generally $\oint \vec{F}\cdot d\vec{r}=0$ or $F=-\nabla V(\vec{r})$)
I make absolutely no assumption about the velocity dependence of the momentum. Momentum is just "something" that is conserved as a total.
I also never use "mass". It is not needed, but whoever wishes can define m=p/v
In the end I derive purely mathematically from vector momentum conservation that a scalar quantity $E=\int v\,dp$ exists, which is conserved provided you take care of another new quantity that we can call potential energy.

$$\frac{d\vec{p}_i}{dt}=\sum_{\stackrel{j}{j\neq i}}\vec{F}_{ij}$$
$$\frac{d\vec{p}_i}{dt}\cdot d\vec{r}_i&=\sum_{\stackrel{j}{j\neq i}}\vec{F}_{ij}\cdot d\vec{r}_i$$
$$d\vec{p}_i\cdot\vec{v}_i&=\sum_{\stackrel{j}{j\neq i}}\vec{F}_{ij}\cdot d\vec{r}_i$$
$$\sum_i \int \vec{v}_i\cdot d\vec{p}_i&=\sum_{\stackrel{i,j}{j\leq i}}\int\left(\vec{F}_{ij}\cdo d\vec{r}_i+\vec{F}_{ji}\cdot d\vec{r}_j\right)$$
If we assume
$$\vec{F}_{ij}&=-\nabla_i V(\vec{r}_i,\vec{r}_j)+\vec{A}\times\frac{d\vec{r}_i}{dt}$$
(conservative force plus any perpendicular part)
and define a new unassigned quantity
$$E_i= \int\vec{v}_i\cdot d\vec{p}_i$$
then the RHS is integrable
$$\sum_i E_i+\sum_{\stackrel{i,j}{j\leq i}}V(\vec{r}_i,\vec{r}_j)&=\sum_i E_i(0)+\sum_{\stackrel{i,j}{j\leq i}}V_{ij}(0)=\text{const}$$

You see that whenever we have only inverse square laws, then conservation of momentum (which is completely undefined yet), implys conservation of a scalar we will call kinetic energy. Note that we need to add a correction factor which we will call potential energy.

If I now combine $E=\int v\, dp$ with E=m (natural units) i.e. E=p/v, then I get all of the expressions for relativistic mechanics.

If on the other hand I unnaturally assume p/v=const=m, then I get classical mechanics.

Last edited: May 8, 2010
4. May 8, 2010

Gerenuk

So you see in my view energy is just an artificial number that happens to be conserved in addition to momentum conservation, since we only have inverse square law forces.

Without inverse square laws, maybe our world wouldn't conserve energy.

(I haven't generalized my proof to non-instantaneous forces yet :( )

5. May 8, 2010

Acut

Conservation of energy holds even with forces that do not follow an inverse square law!
(as in springs, which follow Hooke's law)

6. May 9, 2010

Gerenuk

@Acut: Think about it! Even springs are based in the inverse square law of electrostatics.

7. May 9, 2010

RoyLB

Thanks eveyone for responding

@Studiot - I am an Aeronautical engineer specializing in control systems, so I am familiar with the points you made. I am trying to write a treatise that takes a reader from some exposure to physics and perhaps calculus to where he/she could understand some elementary automatic control theory. I think it would be useful to bring some energy methods to this treatment. However, I realized that I was unable to bridge the gap in a satisfactory way between F=ma and W=Fd or T=1/2mv^2.

@ Gerenuk - I think you've just showed that if all the forces are derivable from a potential function that conservation of energy holds. But that is just the condition for the forces to be conservative (although I'm not sure how A x dr/dt comes into the discussion). I think that 1/r^2 forces are just a subset of conservative forces. To be even more general, Noether's theorem indicates that as long as there's no explicit dependence on time that energy will be conserved.

But, having said all that, your second step is to take the dot product between the forces and the displacement. What motivated this step - except of course that you knew how the answer needed to come out in the end! As a student, I always felt I understood concepts better when all the steps in a derivation made sense and followed naturally from the step before.

- Roy

8. May 9, 2010

jack action

The best definition I've seen for energy is the capacity to create an effect.

As for teaching it, I can't see a simpler image to grasp than force times distance. If you imagine a person pulling a sliding block by a cord, anyone can grasp the concept that the bigger the force, the more energy required and the longer the distance traveled, the more energy required. In fact, I think it's more obvious (intuitively) than the bigger is an acceleration, the bigger is the force in F=ma.

9. May 9, 2010

Gerenuk

Well, I actually derive that something like energy exists at all. Why do you think that $F=-\nabla V$ means anything? Don't forget that you have to imagine to be the first scientist on earth and no-one has ever discovered something like energy.
I suppose if you were to start a proof, you would not only restrict yourself to static *external* potentials, but also you would define $dE=Fds$ for no apparent reason? I just made this proof more general and logical.

The "A x" term is merely to add an admissible degree of freedom without breaking the proof. And this term is needed to represent magnetism.

I vaguely recall, that inverse square forces and zero gradient is equivalent in 3D (I guess provided forces should vanish at infinity).

That energy conservation follows from Noether, is a myth. The Lagrangian formalism guesses an formalism that is already tailored to yield energy results. Noether for energy in a way states that if energy is time independent, then energy is conserved.

With mathematical experience, one sees that one can exploit $\int \nabla V\cdot d\vec{s}=\Delta V$. So I really could have been lead by mathematical simplification ideas only.

10. May 9, 2010

RoyLB

Jack,
You may have a point in that I certainly don't recall having this difficulty during my own undergraduate or even high school physics experience. Perhaps I am making too much of it.

But, as for my other philosophical question, why should force times distance be meaningful? Why is energy a scalar at all? Is energy just a mathematical fiction / convenience, like imaginary numbers (OK - perhaps I opened a can of worms with that statement :-) or is there some physical "reality" to it? I think it is real, in that the energy concept exists in thermodynamics apart from mechanics. Also, I think relativistic physics shows that energy is as real as mass and momentum - in fact they are all aspects of the same mass energy momentum 4-vector. Unfortunately, my understanding of physics is not yet sufficiently sophisticated to make sense of all these ideas.

- Roy

11. May 9, 2010

fluidistic

Here is what I know or believe to be true: in some closed systems (where there's no dissipative force), there's a quantity that is invariant with respect to time translations, i.e. a constant. This number is called the total energy and we can as you know separate it as kinetic and potential energy which aren't necessarily constant. The derivation of the kinetic energy comes from the Lagrangian of the system. And if the Lagrangian doesn't depend explicitly on time, then the energy is conserved.

12. May 9, 2010

Gerenuk

The point is: How do you write down the Lagrangian for a system? You put in your knowledge about kinetic energy or anything equivalent! So you already set up the stage to make it yield energy conservation at some point.

If you just say that there is some vector quantity momentum which is conserved, then your system could be any crazy world.

But as soon as you say, OK, and now I take it as a *new* law that the Lagrangian formalism holds and I will put in mv^2/2 in there, then you are already presupposing that eventually there will be something like a conserved energy.

13. May 9, 2010

fluidistic

I agree that I also found the expression of the Lagrangian of a system with 1 particle somehow artificial.
I don't know if you had a look at a thread of mine I started about a week ago: https://www.physicsforums.com/showthread.php?t=399993. It might be of interest for this part of the discussion.

This requires a lot of hours for me to think in order to fully agree with this.

14. May 9, 2010

RoyLB

@ Gerenuk

I originally asked 2 questions. The first is how to show the derivation of energy from simpler concepts, the second is about what energy actually is. In the first case, I don't need a rigorous proof, I just want each step to be understandable, even if it is purely heuristic. In my mind, taking F dot ds comes out of nowhere, but perhaps that's just me. I feel there ought to be a more elegant way to approach this argument. As for my second question, I don't think your proof really tells me what energy actually is - anymore than any other textbook argument, unless you think that energy really is a sort of convenient mathematical fiction. As I've said earlier, I don't think that this is the case.

I think your proof is probably fine from a mathematics standpoint (although I think you'd need to show where A x dr comes from more explicitly and I don't think you've shown that energy is a constant in this case - i.e. that dE/dt = 0). I think from a physics standpoint you'd have to show that forces are conservative (I'd accept F = dp/dt from experimental reasoning, or even just as a definition). I believe that physicists think that this is indeed the case for the fundamental forces. However, the mathematics and the physics of the conservation of energy still work for dissipative forces, as long as you consider heat to be a form of energy.

In fact I think that the (mass-) energy momentum 4 vector from relativity is the probably the more fundamental physical entity, and that "forces" being the rate of change of momentum can be likely derived from that principle. Its just that we humans can appreciate forces, masses, and accelerations from our everyday experience while energy is more abstract.

- Roy

Last edited: May 9, 2010
15. May 9, 2010

RoyLB

@fluidistic, Gerenuk

I tend to agree with you that the Lagrangian formulation seems contrived. If you check here: http://www.eftaylor.com/pub/QMtoNewtonsLaws.pdf you'll see that the "least action" formulation is better understood from quantum mechanical principles (although I think they've left out some key portions of the argument in that paper)

- Roy

16. May 9, 2010

Studiot

Thank you for your response Roy, I should have perhaps said, where are you going to, rather than where are you coming from, but you anticipated even that!

From what you say I can't see the point of mentioning dot products at all. You may not be aware that physicists and mathematicians work to a slightly different definition of both vectors and scalars. The physicists' one being more restrictive.

If you want to introduce physical concepts in a staged manner I wonder how you introduce the definition of a force?
I was appalled to learn that modern teaching in the UK introduces friction as the first force discussed.

I have found in my experience of gaining understanding of a subject that following the historic route usually suits the best. Historically new concepts mostly arose when the necessary ground had been laid and could be seen in the concept of that work and understanding.
So some development of mechanics preceeded development of what we now call classical thermodynamics and some development of electrical phenomenon was also stirring. So it became obvious that some equivalence of the mechanical work, heat, fluid and electrical energies was called for.

I think those whose educational development followed this route have less trouble with what I might call engineering physics than with say particle or sub particle physics.

17. May 9, 2010

Gerenuk

Yes, and that's why I explained to you why someone who has never heard of F dot ds would still perform this operation. It's purely mathematical logic and an obvious mathematical simplification step.

I get the impression that you are not reading what I write. If you are really interested in answers please read carefully. With the A x dr terms I'm trying to express a more general function that still allows for my proof. I'm not assuming any knowledge about real world physics and the choice F=\nabla V+A x dr is motivated only by the mathematical form of the expression.
And dE/dt=0 follows if you look at the step carefully. The A x dr term just cancels.

Note that at no point I'm using force. Force is just flow of momentum. So it's not a question of experimental reasoning or definitions.

What is heat? It's a macroscopical concept. On the microscopic level there are only bouncing billard balls, so you cannot speak about heat.

The problem with the 4 vector philosophy is, that it is overdetermined and it's a miracle why it is consistent. Given that 3 components are conserved and our world has inverse square laws, my derivation shows that the 4th component is conserved.
It's much more ugly to postulate two laws which could potentially be conflicting. It's nicer if you show how to derivate one of the law from the other.

18. May 9, 2010

RoyLB

I need the dot product anyway, perhaps in both the mathematical and the physical senses. Since you can see I have trouble bringing in the F ds concept, you can imagine that I really have a problem with introductory Controls texts that simply introduce Laplace and Fourier transforms. I plan to use the idea of infinite dimensional "dot products" to motivate these transforms.

I was planning just to use an archetypical spring-mass-damper system to introduce the "intuitive" concept of force. My main concern is not rigor, but elegance. Having said that, I don't want to sacrifice accuracy for simplicity.

I'm really focusing on control systems, so I'll only bring in as much physics and math as I need. I think the historical approach - while fascinating - will take too much time. I'll probably focus the treatise on the engineering physics path and bring in mathematics via hypertext links as appropriate. Perhaps if I am ambitious enough I will add historical sidebars to the text.

I probably don't even really need energy that all that much. After I explain the spring mass damper EOM via the force picture, I want to just mention that it can also be explained by energy transfer, from potential to kinetic to dissipative. This opens the door to more sophisticated methods (I prefer Kane's method to Lagrange, but that's what I studied in grad school)

- Roy

19. May 9, 2010

Acut

Well, I can think of a perhaps more natural way of introducing energy.

Begin with Torricelli:

v^2 = vo^2 + 2*a*delta s

Apply it to free fall

v^2 = vo^2 + 2g(ho - h)

Rearrange:

v^2 + 2gh = vo ^2 + 2gho

You can see that it's a beginning: you have isolated on each side of the equation terms that only relate to the beginning and terms that only relate to the end.
Dividing both sides by 2 is just a mathematical operation. Multiplying it by m and getting the "right" expressions for kinetic and gravitational potential energy is a bit more tricky: use the spring-mass system and show you can do something similar (one side having only information about the past and one side having only information about future) with it. Then it becomes quite a natural thing to multiply the one-half-v^2 term by the mass of the object.

Now, this derivation follows from F=ma (so that you know the object's acceleration, which Torricelli demands) and only some very basic knowledge of kinematics. I think it's a nice way of showing how the concept of energy arises if you will not show them much physics. However, such derivations would hardly convince most physicists.

About "what is energy", notice that there's no universal definition of it. My books usually define it as "the capacity to do work" or, as jack action said "to produce a effect". Those definitions are unacceptable, however. You may have a lot of energy, but if you are in thermodynamical equilibrium, you can't do any useful work with it. I prefer thinking of it as a number which -as far as we know- is conserved with time.

@Gerenuk: Although the internal mechanism of springs is based on electrostatics, it isn't a requirement for energy to conserve. All we need is curl F = 0, and quite a lot of forces and force fields satisfy it.

20. May 9, 2010

Gerenuk

OK, tell me one example, but it should be one where the force vanishes at infinity. Our world is made up of particles and it's natural to assume that fundamental forces should not depend on particles an infinite distance away. So which type of force between point objects is possible, which isn't inverse square?
I have the feeling curl F=0 is equivalent to inverse square in 3D.

Btw: Springs are objects that change shape. This doesn't count. We are talking about forces on test particles, which do not change any of the enviroment.