# Alternative definitions of energy?

1. Nov 17, 2011

### Millenniumf

I had an interesting challenge earlier this year in physics class, and I got a good grade on my answer, but I'd like to see what other people think about this.

Energy is defined in the dictionary as being the ability to do work, while work is defined as the application of energy (roughly speaking, of course). This is circular, so we were challenged to redefine the term energy.

I was pretty lost on this, being a first-year physics student at a community college with little more than a periphery understanding of QM, so my answer was hardly more than a best guess. I said that energy is perhaps the vibrations of cosmic strings, with vibrations in one string being transferred to another as they come into contact, which we interpret as energy transfer. Yeah I know; not that brilliant and flawed from the beginning because it relies on unproven ideas. But it was the best I had. Once we get into quantum mechanics I'll probably have a better answer.

So how would you have answered that question?

2. Nov 17, 2011

### cbetanco

Work done on a system is defined as the change in Kinetic Energy (KE) of that system. While The total energy of a system is the potential energy (PE) plus the kinetic energy, E=PE+KE.

3. Nov 17, 2011

### lustrog

I would have answered it the way Feynman explained it (see: http://student.fizika.org/~jsisko/K...sics/Vol 1 Ch 04 - Conservation of Energy.pdf). That is to say, energy is just something we have discovered in nature that is always conserved. There are many forms, so when it seems to disappear, we discover we can simply define a new form of energy to account for it. Then, the energy that appears to disappear, we see, can in fact be recovered by converting it back into another form. It doesn't "exist" in some form in the same sense that substances exist. It's just a way of accounting for the various capabilities of the various systems in nature.

A little vague. Feynman of course explains it infinitely better.

Last edited by a moderator: Apr 26, 2017
4. Nov 18, 2011

### Andrew Mason

I am not sure if you are trying to state the definition or energy or come up with a new one. How does this definition apply when, for example, a car engine does work in making the car go up a hill? Are you saying that the work done is equal to the change in Kinetic energy of the car?

Energy means the ability to do work and work means $\int \vec{F}\cdot d\vec{s}$.You cannot change this definition of energy to something that is not mathematically equivalent without changing the meaning of energy.

AM

5. Nov 18, 2011

### cbetanco

No, I am using the standard definition. The work done between the initial configuration 1, and the final configuration 2 is $W_{12} =\sum_i \int_1^2 \mathbf{F}_i \cdot d\mathbf{s}_i = \sum_i \int_1^2 m_i \dot{\mathbf{v}}_i \cdot \mathbf{v}_i dt = \sum_i \int_1^2 d \left( \frac{1}{2} m_i v_i^2\right)=T_2-T_1$ where $T=\frac{1}{2} \sum_i m_i v_i^2$ is the kinetic energy of the system.

6. Nov 18, 2011

### Andrew Mason

This is not generally true. It is only true if the applied force F is the only force acting on the body. If the force pushes the body up a hill, for example, the work done does not equal the change in kinetic energy of the body. If it pushes the body against a friction force, the work done does not equal the change in kinetic energy of the body. etc. So work and kinetic energy are not the same.

AM

7. Nov 18, 2011

### Curl

Energy is NOT the ability to do work. People who say this know nothing about physics.

The ability to do work can be destroyed. Energy cannot.

8. Nov 19, 2011

### Matterwave

If you are referring to the second law of thermodynamics, I believe you have to preface it with something like "usable work", or "useful work". Even if you dissipate all your energy into Heat - heat can still do "work", just not in any bulk fashion that would create some kind of work you could use (i.e., it can still make other molecules speed up via collisions).

9. Nov 19, 2011

### Andrew Mason

Matterwave is correct. Curl is confusing "Work" with "Mechanical Work" or "Useful Work".

It takes work to add kinetic energy to molecules. So when a Carnot heat engine adds kinetic energy to the molecules in the cold reservoir of a Carnot heat engine by delivering heat flow to the cold reservoir, work is done at the molecular level. However, that "Work" is not useful mechanical work at the macroscopic level.

AM

10. Nov 19, 2011

### juanrga

Energy is not the ability to do work {*}. Energy can be defined simply as the property

<E> = Tr{H ρ}

where H is the Hamiltonian of the system and ρ its state. Tr denotes the quantum or classical trace.

{*} And cosmic strings are a fantasy...

11. Nov 19, 2011

### Andrew Mason

I can understand why you might want to come up with another equivalent definition of energy, but it has to be mathematically equivalent to the term "energy" as it is used in physics. If it is not, it is a different quantity. In physics, energy is defined as the ability to do work and work is defined as the application of force over a distance. That is how these terms are defined. Now if you want to try to define it in another equivalent way, that is fine. It may be that the conventional definition is not the best way to think of energy in certain circumstances (eg, where the concept of force and distance are not so easy to quantify). But that does not mean that definition is wrong.

AM

12. Nov 19, 2011

### juanrga

The definition given is the usual expression to compute the energy of a given system. When the system is in a pure state it reduces to the well-known claim (mechanics) that the Hamiltonian is just the energy of the system.

Energy is not the ability to do work {*}. There is kind of energies that cannot be used to do work (ask an engineer or take a course in thermodynamics if you do not trust me).

{*} This, maybe, could be an acceptable definition in a general physics course, but not beyond.

13. Nov 19, 2011

### mrspeedybob

Energy is that which curves spacetime.

14. Nov 19, 2011

### Andrew Mason

I have studied a bit of thermodynamics. The word "useful" is used to qualify "work" in thermodynamics for a reason. You are confusing "useful work" with "work".

AM

15. Nov 19, 2011

### Matterwave

Defining the energy to be the expectation value of the Hamiltonian in your mixed state is overly narrow I think. It would also be very round-about and not practical. How do you define the Hamiltonian? If you don't want circular reasoning (e.g. H=T+V) to come into play, you have to define the Hamiltonian as the Legendre transform of your Lagrangian. How do you define the Lagrangian? Again, you can't use circular reasoning (e.g. L=T-V), so you would have to define it as the quantity which, if you integrate with respect to time between two endpoints gives you the Action. You would then have to go on to define the Action as that which, when you take the variation of to be zero, gives you the Euler Lagrange equations. And you would have to define the Euler Lagrange equations as those equations which give you the correct equations of motion for your system. You would then, finally, have to describe the correct equations of motion of your system, and thereby completely render the whole concept of "energy" to be infinitely less useful since "energy" is basically a first constant of integration of the equations of motion which we use to help us solve for the dynamics of a system. By the way, ρ there should be the density matrix, or the statistical operator, and is not the state itself.

I also ask you to give me, from that definition, a way for me to obtain a useful definition of energy in a macroscopic system, e.g. a ball rolling on the ground.

Are you saying the energy is the full stress-energy tensor, or the energies are the components of that tensor? If you take this view, then pressure and stress are also defined as "energy", even though they are conventionally not. Also, because the stress-energy tensor in Einstein's relativity is manifestly the stress-energy tensor of the matter fields (and, one should stress NOT the gravitational field itself), this definition would preclude you from being able to define any energy in the curvature itself. I.e. you would not be able to define any sort of gravitational potential energy using that definition. Granted, it is, in general, not advisable to try to attribute an energy "of the curvature", but the notion of energy can still sometimes be useful. For example, the major evidence we have for gravitational waves is due to the speeding up of the orbital period of a binary neutron star, and in that scenario we look at the "gravitational energy" carried off by the gravitational waves and confirm that this is coincident with theory.

My whole point with this post, is that physicists should not act "smug" once they've learned more sophisticated methods and definitions. The basic definition we are given in basic physics "Energy is the ability to do work" may not be the best definition of energy, and, in many cases, it is inadequate; however, to come up with a fully general definition of energy is painfully difficult. Most of the time, we just take working definitions based on which field we are studying. For basic physics, the definition "energy is the ability to do work", I think, is a nice working definition.

Last edited: Nov 19, 2011
16. Nov 20, 2011

### Curl

How about this:

Energy is something which can give rise to motion of a free particle (pick an elementary particle). The more energy, the quicker is the motion of this particle.

17. Nov 20, 2011

### Vanadium 50

Staff Emeritus
Why do we need a new definition? What's wrong with the one we have? Until you can answer that question, it will be rather difficult to improve on things.

18. Nov 21, 2011

### Andrew Mason

There is nothing wrong with the definition of energy or work. The premise of the OP is not correct:
Energy is the ability to do work. But work is not defined as the application of energy. Work is the application of force through a distance: $W = \int \vec{F}\cdot d\vec{s}$. I see nothing circular in that.

AM

19. Nov 21, 2011

### juanrga

I have said something different.

Work ('useful' or not) is only one mechanism of interchange of energy.

The concept of energy is much more fundamental. You say that you studied thermodynamics, then you would know that work is not a state function as energy. energy is a property of a system, work is not.

There are systems, with nonzero energy, whose work is zero.

By all of the above defining energy as the ability of do work must be enough for a introductory course in physics... but is not a serious def.

20. Nov 21, 2011

### juanrga

The definition is totally practical and emphasizes that energy is an observable with associated fluctuations (which is not said by the other 'definitions' given here).

The best way to introduce the Hamiltonian is as a postulate. There is not guarantee that the Hamiltonian can be derived from a given Lagrangian (already Weinberg discuss about this). Moreover Lagrangians are usually postulated. The Hamiltonian is more fundamental (specially in quantum mechanics) because is the generator of time translations.

Same about the action. The definition above is valid for dissipative and non-deterministic systems for the which its motion is not given by any action (in fact an action is not defined for this kind of systems) nor by Euler-Lagrange equations.

The more general equations of motion can be derived from the Hamiltonian, but not from the other methods that you say (valid only under certain approximations).

ρ is the state itself. The energy of a macroscopic system is easy to obtain, the definition is the same than before

<E> = Tr{H ρ}

If your macroscopic system is classical then Tr denotes the classical trace. If your macroscopic system is in a pure state and there is not instabilitites (e.g., Poincaré resonances) then

<E> = E = Tr{H ρ} = H(p,q)

And effectively as textbooks in classical mechanics emphasize the Hamiltonian is the energy of the system.

21. Nov 21, 2011

### Andrew Mason

Fine. But measure of energy is still the measure of its ability to do work. The dimensions of energy are force x distance (mass x (distance/time^2) x distance). A photon has energy $h\nu$ because it is capable of doing $h\nu$ amount of work on some element of matter. Applying a force through a distance may not be what a photon does, (and it may not be the most appropriate way to model what it does), but the measure of its energy is its ability to do work on some element of matter.
The energy of a system is still its ability to do work - that is to apply a force through a distance (at some level). This is not to be confused with its ability to do useful work, W.

Thermodynamics was developed before anyone understood that molecules existed. Heat was thought to be some kind of substance that flowed through matter. So the terminology used in thermodynamics is a bit archaic.

Heat flow, ΔQ, is actually a transfer of energy at the molecular level (molecules doing work on other molecules), W (useful work) is a transfer of energy at the macroscopic level, and ΔU is the change in the ability of the system to do both kinds of work (ie. to generate heat flow or to do macroscopic work).
W is not a property of a system so I am not sure what this means.
Can you provide a better one?

AM

22. Nov 21, 2011

### juanrga

All of this was answered before. Moreover, I only want remark that by W I am denoting work (as everyone does) not useful work, which is another concept.

Apart from the fact of that the concept of molecule had been introduced and used in science even before Carnot started the science of thermodynamics, the rest is plain nonsense, including your repetitive confusion between work and useful work, that your «Heat flow, ΔQ» is not a flow, neither is heat because heat is not a state function in thermodynamic space, and of course, that heat is not a «kind of work» as you seem to believe...

Reading what was written

It would be clear that I mean situation with a system with nonzero E that cannot do work (W=0). For instance, it would be funny to try to understand what is energy (E) from Work (W) for a system verifying ΔE = Q.

I think that I already gave a general definition beyond the limits of your pseudo-definition.

Last edited: Nov 21, 2011
23. Nov 21, 2011

### Andrew Mason

I am having difficulty making sense of your posts. Perhaps you could give us an example of the kind of energy that cannot be used to do work (either at the macroscopic level or at the microscopic level).

AM

24. Nov 21, 2011

### Matterwave

The work defined in Thermodynamics is useful work (the work to move a piston). That's why you have U=W+Q. Q is the heat, and you don't consider that work simply because you can't account for all the F*r terms in it.

1) Can you calculate for me, using this definition, the energy of a ball falling in a gravitational field? Let's say it starts at height h, in a uniform gravitational field, and free-falls.

2) If you just "postulate" the Hamiltonian, then all you've done is shift the question from "What is energy?" to "What is the Hamiltonian?". The Hamiltonian formalism is inherently difficult to work with whenever there are gauge invariances of a theory. It is even more difficult to work with in the context of general relativity where the split between time and space should not be made so artificial.

Additionally, there are explicit proofs of "when" the Hamiltonian "is" the energy of the system or not and "when" it is conserved or not. These are 2 separate questions. Goldstein goes into some detail about this. If I recall correctly, the Hamiltonian only corresponds with our usual definition of energy if the kinetic energy is dependent quadratically on the speeds, and the potential is not dependent on speeds.

3) If ρ was the state of the particle, it wouldn't make sense to take the trace of it. The trace is only good for operators (i.e. matrices), a state is a vector (more formally, a ray) in the Hilbert space. How do you take the trace of a vector? You can see this explicitly from the form of ρ:

$$\rho=\sum_i P_i |\Psi_i\rangle \langle \Psi_i |$$

That's manifestly an operator and not a state.

25. Nov 22, 2011

### juanrga

The same recommendation that for Andrew Mason: open a textbook on thermodynamics and learn the subject first. I would recommend the section «7.5 USEFUL WORK AND THE GIBBS AND HELMHOLTZ FUNCTIONS» of Klotz & Rosnberg well-known textbook (I have seventh ed.) to understand the difference between work and useful work.

E.g. the Helmholtz free energy measures the useful work obtainable from systems at a constant temperature, volume, and composition. This is not the same than work W.

From the expression obtained above $E = H(p,q)$, and as first approximation this is $(p^2/2m + V(q))$

Giving a definition of something is always shifting the question from the definiendum to the definiens. Evidently, this proccess cannot be repeated forever. Therein that formal systems contain a set of primitive elements which are not defined.

As already said, the Hamiltonian is the generator of the time-translations. All of QFT is based in obtaining a Hamiltonian, from which one obtain the S-matrix, which is tested in experiments. Weinberg has a delicious discussion about those topics.

Regarding GR, the 3+1 formalism is fundamental for an deep (and practical) understanding of dynamics. Indeed the 3+1 formalism is the foundation of most modern numerical relativity.

The problems with the usual Hamiltonian formalism of GR are more related to certain geometric deficiencies of GR than to the Hamiltonian formalism.

As said as well, the Hamiltonian formalism is fundamental when studying more general dynamics beyond QFT and GR.

I already wrote above that E=H(p,q) is valid as approximation. The rest is wrong.

If you open a textbook on QM, you will discover that the operator rho describes the general state of a quantum system (beyond the limits of |ψ>). Recall that my goal was to give a general definition of E, not one valid only in special situations.

Moreover, you are replying to a part where I said that Tr was denoting the classical trace, which means that you do not read my posts. The classical trace is an phase space integration and rho is not an operator therein but the phase space state that correspond to the classical limit.

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