Alternative definitions of energy?

[Matterwave]

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 process 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.

I'm too tired to keep arguing about this. I'll just concede that I was wrong. Obviously I made a few wrong points, but even so, I still don't see your definition as a useful definition in any sense.

Especially since your answer to my question for calculating the energy for a simple system was H=p^2/2m+V(q). V(q) is the potential energy...this has got to be the most circular definition I can think of. "Energy is equal to kinetic energy plus potential energy". That is simply a tautology.

Lastly, a 3+1 formalism of GR is impossible if the space-time is not globally hyperbolic.

 

[bbbeard]

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.

So how would you have answered that question?

I would leave out the part about string theory.

I see after 30 replies that no one has given the "conventional" response. The conventional response is that energy is the conserved charge associated with symmetry of the Lagrangian under time transations in accordance with Noether's theorem.

Other commenters have pointed out that the Hamiltonian is the generator of time translations. In my view that role is subsidiary to the relationship between energy and the Noether current. But we're risking a slide into metaphysics if we start arguing about what axioms are really fundamental.

BBB

 

[bbbeard]

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).

Heat transfer is not work. But you knew that.

If you're asking, "can internal energy u always be used to do work?" I would say the answer is contingent on the circumstances of the system and its environment. The second law places strict limits on how much internal energy can be converted to work transfer, and as you know, the conversion is always less than 100%. The other consideration is that zero-point energy of quantum systems (e.g. \frac{1}{2}h\nu for a harmonic oscillator) cannot be converted to a work transfer.

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.

I sharply disagree with that. I would say thermodynamics replaced caloric theory just as chemistry replaced alchemy and astronomy replaced astrology. The terminology in thermodynamics uses common words that have a precise and unambiguous meaning within thermodynamics. The fundamental ideas of thermodynamics are enhanced and extended by kinetic theory, but by no means require the existence of molecules. The efficiency of a Carnot cycle engine is what it is, regardless of atomic theory. And since none of the laws of thermodynamics have been overturned by modern physics, I find the terminology is anything but archaic.

Heat flow, ΔQ, is actually a transfer of energy at the molecular level (molecules doing work on other molecules)...

First, heat transfer does not require molecular motion. Heat can be transferred from one box of photons to another, for example. What is the "essence" of heat transfer? What makes an energy transfer a heat transfer and not a work transfer? One textbook I have uses an operational definition that an energy transfer is "work" if in principle all the energy can go into the lifting of a weight in a uniform gravitational field -- that is, they define a paradigmatic work transfer, and assert that other forms of work transfer can be converted to that paradigm "in principle" with 100% efficiency. This definition is not entirely satisfactory, of course. Energy transfers that cannot, even in principle, result in the lifting of a weight are deemed heat transfers.

I prefer to think of heat transfers as energy transfers that necessarily involve the transfer of entropy. That is, it is the random aspect of heat transfer that makes it heat transfer. Your formulation "molecules doing work on other molecules" doesn't really incorporate the essence of heat transfer, does it? Here are some thought experiments that show the problems with this view.

1) Consider how a vibrating wall transfers energy to a gas. Suppose the wall is the y-z plane (at t=0) and is oscillating in the x-direction. With a gas on the +x side, energy is transferred by molecular collisions between the wall molecules and the gas molecules. Does this mean that the transfer is "heat"? No, not at all. The fact that the motion of the wall is coherent means that the F dot dx applied by the wall to the gas is a work transfer.

2) Consider a wall in the same orientation as before, except that the wall is not oscillating, but is at some temperature T_w. If the gas is at some temperature T_g < T_w, then there will be a net transfer of energy to the gas. In this case the energy transfer is "heat transfer" because the "motion" of the wall is incoherent.

3) Consider the thermal radiation of a hot black body A to a cold black body B. I assume you will agree that this is a paradigmatic heat transfer.

4) Now suppose that we consider a laser beaming the same amount of power from body A to body B. Is this heat transfer? No, it is actually work transfer! Why? Because the photons are coherent.

A photon has energy hν 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.

I'd say yes and no. Certainly the paradigmatic experiment showing the quantum nature of light is the photoelectric effect, in which it was shown that the energy of emitted electrons was equal to h\nu-W, where W is the so-called "work function" of the metal (which we now understand to be an ionization energy). Clearly this fits your statement.

I would add, however, that Planck's discovery of quantization didn't really involve work transfers per se. E=h\nu was initially an ad hoc assumption that made the mathematics of the Boltzmann factor work out correctly. Prior to Planck, it was believed that all degrees of freedom were continuous, so the expectation value of the energy for each degree of freedom was exactly the same, and proportional to temperature. This meant that a collection of photons in thermal equilibrium would have an infinite energy. It turns out that by restricting the energy of photons of frequency \nu to multiples of h\nu, the expectation value of energy for low-energy photons contributed less and less (instead of a constant) to the sum in the infrared regime. To me this is a more natural, simpler interpretation of the meaning of photon quantization. But again, this is starting to drift into metaphysics, when we argue about which evidence and which paradigms are more compelling.

 

[juanrga]

The link you gave (http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper2.html) said that:



That doesn't seem right. I can have a thermal conductor at 1000C and a thermal insulator at 1000C. Yet the definition above would imply that the thermal conductor and the thermal insulator would give up energy with comparable spontaneity, which is obviously false.

Nope, textbook is rigth.

 

[juanrga]

I'm too tired to keep arguing about this. I'll just concede that I was wrong. Obviously I made a few wrong points, but even so, I still don't see your definition as a useful definition in any sense.

Especially since your answer to my question for calculating the energy for a simple system was H=p^2/2m+V(q). V(q) is the potential energy...this has got to be the most circular definition I can think of. "Energy is equal to kinetic energy plus potential energy". That is simply a tautology.

Lastly, a 3+1 formalism of GR is impossible if the space-time is not globally hyperbolic.

In your own message you quote me saying: «I already wrote above that E=H(p,q) is valid as approximation.» Therefore, it is a derived result not a tautology...

 

[juanrga]

I agree with you on this point. In Hamiltonian mechanics juanrga's expression is certainly more convenient and natural, but the whole reason that we call that expression "energy" is because, in any system where you can compute both, you find that they are equivalent.

He is plain wrong when affirms that «energy is defined as the ability to do work and work is defined as the application of force over a distance.»

The concept of work {*} can be derived from the definition given by me for energy E

<E> = Tr{H ρ}

If system is closed

d<E> = Tr{∂H ρ} + Tr{H ∂ρ} = W + Q

Work {*} is then defined as W = Tr{∂H ρ} and heat as Q = Tr{H ∂ρ}

It makes little sense try to define a basic quantity E from a derived quantity W {*}, which does not even need to exist for a specific system/process (I have given some examples where energy exists but work does not).

{*} Moreover, this modern definition of work I am introducing here is more general than the classical definition given by others in this thread as (Force x distance)...

 

[juanrga]

I would leave out the part about string theory.

I see after 30 replies that no one has given the "conventional" response. The conventional response is that energy is the conserved charge associated with symmetry of the Lagrangian under time transations in accordance with Noether's theorem.

Other commenters have pointed out that the Hamiltonian is the generator of time translations. In my view that role is subsidiary to the relationship between energy and the Noether current. But we're risking a slide into metaphysics if we start arguing about what axioms are really fundamental.

BBB

Your «conventional» response is a special case of the «standard» response given here, because when the system is stable and in a pure state {*}, L and its related symmetries can be obtained from H and its corresponding symmetries/conservations.

I emphasize again that I have tried to give the more general and rigorous possible definition of energy E, not a definition valid only under certain restrictions/approximations.

{*} Your own reference states that the theorem «does not apply to systems that cannot be modeled with a Lagrangian...» Whereas my definition is more general and valid also for that kind of systems.

 

[Andrew Mason]

To Andrew Mason: Matterwave wasn't correct, but what you said after was correct. Matterwave is incorrect in saying that that, "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)."

To matterwave: Wrong. Adiabatic expansion can convert thermal energy into non-thermal forms of energy. The result is adiabatic expansion work, and this occurs in a bulk fashion. The fact that a gas duster gets quickly cold by means other than radiating heat to the environment proves this.

I did not interpret Matterwave as you did. I just thought he was saying that heat transfer requires doing work at the molecular level (causing other molecules to speed up via random collisions requires work) although this is not (necessarily) useful work.

T\Delta S_{total} (i.e. the so called "unavailable" energy) not constant, it can even decrease in an isolated system. Although \Delta S_{total} can only increase in an isolated system, T can spontaneously decrease while maintaining this isolation. This is because T is based upon kinetic energy per degree of freedom.

I am not sure what you mean. T is based on kinetic energy in the translational mode. Temperature is not based on the energy associated with the kinetic energy of the other modes of motion.

Energy does exist in the same sense that substances exist. Energy is transported by force carriers known as gauge bosons. These are the "substances" which explain the interaction of other particles.

I have not heard anyone refer to a collection of photons as a substance. It seems to me that to call something a substance you would have to at least be able to associate a frame of reference with it. I don't think you can do that with photons or any other gauge boson.

AM

 

[Dale]

He is plain wrong when affirms that «energy is defined as the ability to do work and work is defined as the application of force over a distance.»

No, he is not. That is the usual definition taught in many introductory physics textbooks.

You may not like that definition for a variety of perfectly valid reasons, but it exists and it is simply delusional for you to claim that it doesn't. Andrew Mason is not wrong in affirming the introductory physics textbook definition of energy is the definition of energy.

The concept of work {*} can be derived from the definition given by me for energy E

<E> = Tr{H ρ}

If system is closed

d<E> = Tr{∂H ρ} + Tr{H ∂ρ} = W + Q

Work {*} is then defined as W = Tr{∂H ρ} and heat as Q = Tr{H ∂ρ}

It makes little sense try to define a basic quantity E from a derived quantity W {*}, which does not even need to exist for a specific system/process (I have given some examples where energy exists but work does not).

And here you prove his original point. The two definitions are equivalent, as you yourself have demonstrated.

{*} Moreover, this modern definition of work I am introducing here is more general than the classical definition given by others in this thread as (Force x distance)...

That is a good reason to prefer your definition. It does not make Andrew Mason's definition wrong, simply less general.

However, I am not convinced that it is, in fact, more general. I know how to apply Andrew Mason's definition to a spring pushing a mass horizontally across a surface with friction, but I don't know how to apply your definition. That could easily be a lack of knowledge on my part, but if it is correct then it seems that each definition covers situations that the other does not, making neither more general than the other.

 

[kmarinas86]

It is correct, not obviously false. If you have a hot insulator it will spontaneously give up energy to a cold conductor, and if you have a cold insulator it will spontaneously receive energy from a hot conductor, and if they are equal temperature then there will be no net spontaneous energy transfer.

If you hold the background conditions constant, a warm conductor can heat a room quicker than a hot insulator of the same dimensions. So it is unreliable to think of temperature as a measure of an object's ability to heat the surroundings spontaneously. The definition only makes sense in the limiting case of comparing two bodies in contact in a given case, but it does not make sense as a definition if we interchange one body in the first case with a second body in separate case. Such definitions constitute bad pedagogy, AFAIK.

 

[bbbeard]

I have not heard anyone refer to a collection of photons as a substance. It seems to me that to call something a substance you would have to at least be able to associate a frame of reference with it. I don't think you can do that with photons or any other gauge boson.

If you have a box of photons, then the center-of-momentum frame will usually be well-defined, will it not? Isn't that what you mean by associating a frame of reference?

On the other hand, I don't know why it's important to define "substance" in physics, kmarinas86's comment notwithstanding.

BBB

 

[juanrga]

No, he is not. That is the usual definition taught in many introductory physics textbooks.

You may not like that definition for a variety of perfectly valid reasons, but it exists and it is simply delusional for you to claim that it doesn't. Andrew Mason is not wrong in affirming the introductory physics textbook definition of energy is the definition of energy.

Wait a moment, I am not saying what you think. You are considering my words outside the context of this thread. The details that you give now already were given before by me. In #12 I wrote:

Energy is not the ability to do work {*}...

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

The following discussion with him focused about my note {*}. I am just maintaining that it is not a general definition valid everywhere.

And here you prove his original point. The two definitions are equivalent, as you yourself have demonstrated.

I have proved something totally different:

(i) Energy E has a definition which does not require the introduction of the concept of work W.

(ii) the definition of work W can be derived from the definition of E. For closed systems W is one part of the total variation of Energy: dE = d_w E + d_Q E = W + Q. This implies that whereas Energy is a state function, W is not and depends of the process.

(iii) W in my derivation is a general definition of work, whereas he used a classical definition of mechanical work as (Force x Distance). His W is not the general definition of work but follows from mine as special case: Tr{dH ρ} → ∫ F dr.

(iv) I provided examples and references of systems with a well-defined energy, whose ability to do work is null. I repeat, I makes no sense to define the energy of those systems as its ability to do work when its ability is zero.

 

[Dale]

I am just maintaining that it is not a general definition valid everywhere.

That is fine*, and if that were what you had said then I doubt anyone would have objected. What is not fine is your assertion in post 10 that "Energy is not the ability to do work", to which Andrew Mason correctly objected in post 11.

The existence of a more general definition does not negate nor invalidate a less general definition. It simply extends it to other scenarios where the less general one does not apply, and there are often historical, computational, or pedagogical reasons for using the less general definition in situations where it does apply.

*As I said in post 39, I am not certain that it is actually more general. Can you apply your definition to a spring pushing a mass horizontally across a surface with friction?

 

[juanrga]

That is fine (although, as I said in post 39, I am not certain that it is actually more general). If that is what you had said then I doubt anyone would have objected. What is not fine is your assertion in post 10 that "Energy is not the ability to do work" to which Andrew Mason correctly objected in post 11.

The existence of a more general definition does not negate nor invalidate a less general definition. It simply extends it to other scenarios where the less general one does not apply, and there are often historical, computational, or pedagogical reasons for using the less general definition in situations where it does apply.

I agree with you that my post #10 lacked explanation and even rigor; however, in #12 I added corrections/explanations, and still Andrew Mason in posterior posts (#18, #21...) continued doing the same kind of statements.

I think that I have shown (it is evident for me) that the definition given in #10 is much more general that the definition given here as «ability to do work», specially when the concept of «work» introduced by Andrew Mason (see #18) was that of mechanical classical work only.

About your question about the spring. I already answered this in #42.

 

[Andrew Mason]

Heat transfer is not work. But you knew that.

Heat transfer involves molecular forces being applied over molecular distances. What is that if it is not work?

The second law places strict limits on how much internal energy can be converted to work transfer, and as you know, the conversion is always less than 100%.

Certainly, but that applies only to useful work, which is the W in the first law: ΔQ = ΔU + W

The other consideration is that zero-point energy of quantum systems (e.g. \frac{1}{2}h\nu for a harmonic oscillator) cannot be converted to a work transfer.

Now that is an interesting point. The reason we call it energy is because we think of moving matter as being the result of a force being applied through a distance. Perhaps zero-point energy should not be called energy - because it cannot be used to actually do work. Or can it? Think of a diatomic molecule such as H₂. As a quantum harmonic oscillator it has discrete energy levels of (n+1/2)h\nu so it cannot have less energy than h\nu/2. Suppose I keep adding energy until the bond breaks. What, at least theoretically, becomes of that zero-point energy? Does it not go into the kinetic energy of the separated H atoms?

I sharply disagree with that. I would say thermodynamics replaced caloric theory just as chemistry replaced alchemy and astronomy replaced astrology. The terminology in thermodynamics uses common words that have a precise and unambiguous meaning within thermodynamics. The fundamental ideas of thermodynamics are enhanced and extended by kinetic theory, but by no means require the existence of molecules. The efficiency of a Carnot cycle engine is what it is, regardless of atomic theory. And since none of the laws of thermodynamics have been overturned by modern physics, I find the terminology is anything but archaic.

I just said it had archaic origins not that it was not still useful. The concept of "heat flow" is used all the time. "Electromotive force" is an archaic term but we still use it.

First, heat transfer does not require molecular motion. Heat can be transferred from one box of photons to another, for example.

You can have a box of photons? Are you talking about a resonant cavity?

1) Consider how a vibrating wall transfers energy to a gas. Suppose the wall is the y-z plane (at t=0) and is oscillating in the x-direction. With a gas on the +x side, energy is transferred by molecular collisions between the wall molecules and the gas molecules. Does this mean that the transfer is "heat"? No, not at all. The fact that the motion of the wall is coherent means that the F dot dx applied by the wall to the gas is a work transfer.

But you cannot have such a wall. At the molecular level, the molecules in the wall are not all vibrating the same way.

2) Consider a wall in the same orientation as before, except that the wall is not oscillating, but is at some temperature T_w. If the gas is at some temperature T_g < T_w, then there will be a net transfer of energy to the gas. In this case the energy transfer is "heat transfer" because the "motion" of the wall is incoherent.

In other words, the molecules in the wall are doing work on the molecules of gas. Not useful macroscopic work, but they are applying a (very small) force through a (very small) distance.

3) Consider the thermal radiation of a hot black body A to a cold black body B. I assume you will agree that this is a paradigmatic heat transfer.

4) Now suppose that we consider a laser beaming the same amount of power from body A to body B. Is this heat transfer? No, it is actually work transfer! Why? Because the photons are coherent.

If Body A and B consist of matter, blackbody radiation from A to B would still involve work being done by the molecules of A to produce the radiation, and work being done on the molecules of B in receiving that radiation, would it not?

AM

 

[juanrga]

Heat transfer involves molecular forces being applied over molecular distances. What is that if it is not work?

Heat transfer exists even when molecular forces are not defined. Moreover, you seem to believe that molecules are tiny iron balls moving in straight lines, but all this picture is rather inaccurate.

Certainly, but that applies only to useful work, which is the W in the first law: ΔQ = ΔU + W

I already explained you that W in the first law of thermodynamics is work, and what is the difference with the different concept of useful work (e.g. measured by A≠W). I gave a specific section of a standard textbook devoted to this topic. Moreover work W in thermodynamics is not a synonym for mechanical work W_mech.

You claimed that you had studied thermodynamics, but you show otherwise. In fact, I already explained you that heat is not a state function and thus your ΔQ makes no sense...

I think that I do not need to correct all of this again.

 

[Dale]

About your question about the spring. I already answered this in #42.

Even going back and re-reading your #42 I don't see it. What part of #42 do you believe shows how to apply the Hamiltonian definition of energy to a spring pushing a mass horizontally across a surface with friction? And can you be more explicit about the connection? It is not at all apparent to me. Does such a system even have a Hamiltonian and does taking the trace give the correct expression for the energy?

 

[cbetanco]

Does such a system even have a Hamiltonian and does taking the trace give the correct expression for the energy?

The standard model Lagrangian should (in principal) give the correct Hamiltonian. Although I admit it wouldn't be very helpful to actually solving the problem of a mass on a spring...

 

[Dale]

The standard model Lagrangian should (in principal) give the correct Hamiltonian.

Not with friction.

 

[cbetanco]

Not with friction.

Friction is an electromagnetic phenomena on the microscopic level, so yes, the SM would (in principal) give the correct Lagrangian. But, practically speaking, you are completely correct, I don't think anyone has actually taken the classical limit of the SM Lagrangian and applied it to friction (I don't think this is possible analytically, as well as numerically). But IN PRINCIPAL, the Lagrangian should give the correct physics. But no one will ever be able to solve a sliding block on an incline with the SM Lagrangian, so I guess it's a moot point.

 

[Andrew Mason]

Heat transfer exists even when molecular forces are not defined. Moreover, you seem to believe that molecules are tiny iron balls moving in straight lines, but all this picture is rather inaccurate.

So you do not agree with the model on which the kinetic theory of gases is based? It is not a perfect model and certainly does not describe what occurs at the sub atomic level in molecular collisions, but it seems to provide a pretty accurate model for macroscopic behaviour of gases.

I already explained you that W in the first law of thermodynamics is work, and what is the difference with the different concept of useful work (e.g. measured by A≠W).

I don't know if there is a language problem here but I don't understand what this means. Are you saying the W in the first law of thermodynamics means something other than useful/macroscopic work? If so, what is it? What is A?

I gave a specific section of a standard textbook devoted to this topic. Moreover work W in thermodynamics is not a synonym for mechanical work W_mech.

It need not be "mechanical" as opposed to, say, electromagnetic. But it must be macroscopic work or useful work.

You claimed that you had studied thermodynamics, but you show otherwise. In fact, I already explained you that heat is not a state function and thus your ΔQ makes no sense.

"My" ΔQ = ΔU + W as in the first law. When did I say ΔQ was a state function? ΔU is a state function but NEITHER ΔQ nor W relates to the state of a system. Those terms relate to processes in moving between states.

AM

 

[juanrga]

Even going back and re-reading your #42 I don't see it. What part of #42 do you believe shows how to apply the Hamiltonian definition of energy to a spring pushing a mass horizontally across a surface with friction? And can you be more explicit about the connection? It is not at all apparent to me.

Wait a moment. Your original question was:

I know how to apply Andrew Mason's definition to a spring pushing a mass horizontally across a surface with friction, but I don't know how to apply your definition.

In my response to that, I said you how his definition of classical mechanical work W_mech is a special case derived from the general definition of work W given by me:

Tr{dH ρ} → Tr {(∂H/∂r)dr ρ} → - Tr {Fdr ρ}

which for a pure state gives the classical mechanical (Fdr)_mech.

Therefore, if classical mechanical work is derived from the general definition of work W and you already know how to apply the classical mechanical work expression your original question would be solved.

Now you are asking me something related but different. You are asking me to obtain the specific value of a friction force from the above Trace. Sorry but I cannot give you an analytic solution here because its is not a pure state situation. However, its value is obtained solving the trace by computational methods for a given ρ. You can find specific algorithms to obtain friction, from a simple Coulomb model, in the literature {*}.

Does such a system even have a Hamiltonian and does taking the trace give the correct expression for the energy?

Precisely this is one of the advantages of my definition of energy over other definitions given here (as the definition using Lagrangians and Noether theorem).

My definition of energy is (so far as I know) valid everywhere, including dissipative systems for the which the Lagrangian does not exist (read #37 and the online reference therein).

{*} Already Feynman (Lectures on Physics, Vol 1.) remarks how friction is caused by the gradients of numerous electrostatic potentials between the atoms.

 

[juanrga]

Heat transfer exists even when molecular forces are not defined. Moreover, you seem to believe that molecules are tiny iron balls moving in straight lines, but all this picture is rather inaccurate.

So you do not agree with the model on which the kinetic theory of gases is based? It is not a perfect model and certainly does not describe what occurs at the sub atomic level in molecular collisions, but it seems to provide a pretty accurate model for macroscopic behaviour of gases.

Evidently, nowhere in the above quote I wrote that I disagree with the kinetic theory of gases. Also, evidently, molecules are not tiny iron balls moving in straight lines. Kinetic theory (specially quantum kinetic theory) does not say the contrary, neither kinetic theory says that heat is always obtained from some inexistent mechanical classical work.

I already explained you that W in the first law of thermodynamics is work, and what is the difference with the different concept of useful work (e.g. measured by A≠W).

I don't know if there is a language problem here but I don't understand what this means. Are you saying the W in the first law of thermodynamics means something other than useful/macroscopic work? If so, what is it? What is A?
It need not be "mechanical" as opposed to, say, electromagnetic. But it must be macroscopic work or useful work.

Apart from ignoring what I wrote in that quote, you also deleted an important part of my previous message. The deleted part was:

I gave a specific section of a standard textbook devoted to this topic. Moreover work W in thermodynamics is not a synonym for mechanical work W_mech.

"A" is standard thermodynamic notation and I already explained in previous posts when this thermodynamic potential gives a measure of useful work and when does not.

You claimed that you had studied thermodynamics, but you show otherwise. In fact, I already explained you that heat is not a state function and thus your ΔQ makes no sense.

"My" ΔQ = ΔU + W as in the first law. When did I say ΔQ was a state function? ΔU is a state function but NEITHER ΔQ nor W relates to the state of a system. Those terms relate to processes in moving between states.

The first law, for closed systems, is:

ΔU = Q + W

Heat is not a state function in thermodynamic space and thus your ΔQ has no meaning.

I will not correct again stuff corrected before two or three times. And I will not reply to stuff I never said.

 

[Dale]

Sorry but I cannot give you an analytic solution here because its is not a pure state situation.

I am fine with just the Hamiltonian which could then simply be solved numerically using Hamilton's equations etc.

My definition of energy is (so far as I know) valid everywhere, including dissipative systems for the which the Lagrangian does not exist (read #37 and the online reference therein).

This is essentially what I am getting at. It has been a long time since my physics courses, but I remember the professor saying that the Lagrangian formalism does not work for systems with friction (which you seem to agree with), and later saying that the Hamiltonian formalisim is just a way to turn the second-order differential equations into a system of first-order differential equations.

So, my impression has always been that the Hamiltonian formalism does not work for systems with friction either (which you seem to disagree with). We certainly never covered such systems using the Hamiltonian approach. I am open to corretion on this point.

{*} Already Feynman (Lectures on Physics, Vol 1.) remarks how friction is caused by the gradients of numerous electrostatic potentials between the atoms.

I looked in the the online reference of post 37, your statements in post 42, and in Feynman's Lectures on Physics, Vol 1 I searched for every occurence of "friction". None of those show the Hamiltonian for a simple system with friction.

You claim that your definition is more general, but it only covers situations for which the Hamiltonian formalism applies. So, based on my instruction many years ago and on the information presented here, I am skeptical that it applies to a spring pushing a mass horizontally across a surface with friction.

 

[Andrew Mason]

The first law, for closed systems, is:

ΔU = Q + W

That is one way to write it, W being the work done ON the system and Q being the NET heat flow INTO the system. This is equivalent to ΔQ = ΔU + W where W is the work done BY the system and ΔQ is the difference between the heat flow INTO the system and the heat flow OUT OF the system ie. the NET heat flow INTO the system.

Heat is not a state function in thermodynamic space and thus your ΔQ has no meaning.

It is standard terminology. It is confusing to use just Q when dealing with heat flows into and out of a system - in a cycle involving several different processes (such as a heat engine): ΔQ = Qh - Qc, for example. You would write Q = Qh-Qc. But most thermodynamic texts refer to this as ΔQ. You seem to be interpreting that as a state function for some reason.

AM

 

[juanrga]

I am fine with just the Hamiltonian which could then simply be solved numerically using Hamilton's equations etc.

As I am saying in the part that you quote, the system is not in a pure state; i.e.; the Hamilton equations do not describe its dynamics.

This is essentially what I am getting at. It has been a long time since my physics courses, but I remember the professor saying that the Lagrangian formalism does not work for systems with friction (which you seem to agree with), and later saying that the Hamiltonian formalisim is just a way to turn the second-order differential equations into a system of first-order differential equations.

So, my impression has always been that the Hamiltonian formalism does not work for systems with friction either (which you seem to disagree with). We certainly never covered such systems using the Hamiltonian approach. I am open to corretion on this point.

The dynamics of any system is given by the equation

∂ρ/∂t = L_Hρ + Dρ (1)

where L_H is the Liouvillian of the system L_H=L_H(H) and D the Dissipator, which is also a function of the Hamiltonian {*}. When there is not dissipation the equation reduces to the time symmetric equation

∂ρ/∂t = L_Hρ

and for pure states this reduces to the set of equations

dq/dt = ∂H/∂p

dp/dt = -∂H/∂t

for H=H(q,p)=E

This set of equations is what you mean by the «Hamiltonian formalism».

Of course, we can derive the Lagrangian formalism from the Hamiltonian formalism, using the Legendre transformation L(q,v) = pv - H(q,p)

Both the Hamiltonian and Lagrangian formalisms of classical mechanics are only valid for time-reversible systems in pure states. But the Hamiltonian H, the equation (1) and the expression <E> = Tr{Hρ} are much more general.

I looked in the the online reference of post 37, your statements in post 42, and in Feynman's Lectures on Physics, Vol 1 I searched for every occurence of "friction". None of those show the Hamiltonian for a simple system with friction.

As said friction is caused by the gradients of numerous electrostatic potentials between the atoms. The usual non-relativistic Hamiltonian is

H = Ʃ_i p_i/2m_i + Ʃ_iƩ_j e_ie_j/4πε₀R_ij

The last part is the Coulomb potential energy that Feynman alludes to.

This Hamiltonian can be then used (very often via further simplifications {**}) to obtain surface structures (static friction), harmonic vibrations, chemical bonds associated with kinetic friction... using different computational models.

You claim that your definition is more general, but it only covers situations for which the Hamiltonian formalism applies. So, based on my instruction many years ago and on the information presented here, I am skeptical that it applies to a spring pushing a mass horizontally across a surface with friction.

You are completely wrong. You seem to confound the equation (1) with the approximated Hamilton equations and you seem to confound the expression <E> = Tr{Hρ} with the approximated expression E = H(q,p)

{*} It is very very complex and I prefer do not to write D in explicit form.

{**} E.g. BO approximation for nuclei, approx computational model for obtaining the electronic PES, harmonic approximations in that surface, approx derivation of bond energies, distances and frequencies, canonical or some other equilibrium approx. for energy levels distribution, neglect of noise effects, neglect of fourth and higher order terms L'_H in the dissipation D, time-coarse-graining by neglect of memory effects (t_mem<<t) in the resulting kernel of D...

 

[juanrga]

That is one way to write it, W being the work done ON the system and Q being the NET heat flow INTO the system. This is equivalent to ΔQ = ΔU + W where W is the work done BY the system and ΔQ is the difference between the heat flow INTO the system and the heat flow OUT OF the system ie. the NET heat flow INTO the system.
It is standard terminology. It is confusing to use just Q when dealing with heat flows into and out of a system - in a cycle involving several different processes (such as a heat engine): ΔQ = Qh - Qc, for example. You would write Q = Qh-Qc. But most thermodynamic texts refer to this as ΔQ. You seem to be interpreting that as a state function for some reason.

AM

As said I will ignore errors corrected before as well as stuff that I have not said.

About the remaining stuff, Q in the first law ΔU = Q + W, denotes heat, not heat flux.

The recommended (IUPAP, ISO, IUPAC) notation for heat flow is \Phi. And the recommended notation for heat flux density is J_q. Of course J_q \neq \Phi \neq Q.

 

[Andrew Mason]

About the remaining stuff, Q in the first law ΔU = Q + W, denotes heat, not heat flux.

You seem to be confusing heat flux with heat flow. They are very different. Perhaps this is a language problem as English does not appear to be your first language.

What is your definition of heat? Q refers to thermal energy transfer and, for historical reasons, is usually referred to as "heat flow" not "heat". \dot Q refers to heat flow per unit time, (ie thermal energy transferred per unit time). "Heat" by itself is a term that is generally avoided in thermodynamics because it can have a variety of meanings.

AM

 

[juanrga]

You seem to be confusing heat flux with heat flow. They are very different. Perhaps this is a language problem as English does not appear to be your first language.

What is your definition of heat? Q refers to thermal energy transfer and, for historical reasons, is usually referred to as "heat flow" not "heat". \dot Q refers to heat flow per unit time, (ie thermal energy transferred per unit time). "Heat" by itself is a term that is generally avoided in thermodynamics because it can have a variety of meanings.
AM

Ignoring stuff corrected before, I will add that IUPAP, ISO, IUPAC, define the heat flux as \dot{Q}, which is not the same than heat, evidently.

 

[kmarinas86]

About the remaining stuff, Q in the first law ΔU = Q + W, denotes heat, not heat flux.

You seem to be confusing heat flux with heat flow. They are very different. Perhaps this is a language problem as English does not appear to be your first language.

What is your definition of heat? Q refers to thermal energy transfer and, for historical reasons, is usually referred to as "heat flow" not "heat". \dot Q refers to heat flow per unit time, (ie thermal energy transferred per unit time). "Heat" by itself is a term that is generally avoided in thermodynamics because it can have a variety of meanings.

AM

Juanrga is not wrong here. In thermodynamics, "heat" is not defined as a state quantity. Heat is a process quantity. Blame that on the horrendous nomenclature adopted by thermal physics. For a state quantity that is similar to what a normal person calls "heat" or "BTUs", we have the a term "thermal energy" instead. In physics, the process quantity Q is the heat, which is a transfer of thermal energy, not a transfer of heat, but in engineering, this quantity is called heat transfer. Heat therefore is a flow of energy, not the energy itself. I have some problems with this nomenclature, as I believe that heat should be synonymous thermal energy, but that is not how it is done in our sciences. In the mathematical physics sense, you cannot "store" heat (i.e. Heat (a mathematician's view)=Heat flow (a layman's view)=Heat transfer (a thermal engineer's view)). Saying that Q is the "heat transfer" or "heat flow" rather than "heat" is still misleading because: 1) the word heat (a noun or verb, but never an adjective) is modifying the following noun in both of these terms as if "heat" was what is flowing or transferring, in the layman sense of the term, and 2) neither of these two terms stand for the thing being transported but rather they refer to the measure of how much was transported, and it is not even an "accumulation of heat". But again, that is how it is done. \dot Q is the rate of heat transfer at an instant of time. Similarly, heat flux is the is rate of heat transfer at an instant of time per area.