Alternative definitions of energy?

[juanrga]

Ah, I just realized that I assumed that the term "Hamiltonian formulation" would also apply to the Liouvillian / von Neumann equation, which is probably wrong. Now your posts make much more sense to me. ;-)

Sometimes the term «Liouville formulation» is applied to the Liouville von Neumann equation. I think that I never used the term «Hamiltonian formalism» for this equation.

In my posts I only said that the Liouvillian was a function of the Hamiltonian, and showed how the «Hamiltonian formulation» was derived under certain approximations/assumptions from a more general formulation that uses the Liouvillian and the dissipator.

So the true advantage of the Hamiltonian formalism is that it generalizes easily to mixed states, while the Lagrangian formalism does not?

In #86 I gave some advantages of the classical Hamiltonian formalism over the classical Lagrangian formalism, as the computational advantages. I also explain how the quantum Hamiltonian formalism is used in QFT to obtain the S-matrix (after utilized in the lab).

Since that the Hamiltonian (not the Lagrangian) is the generator of time translations for pure states. Generalizations to mixed states, unstable states, dissipative systems, etc. utilize formalism based in the existence of a Hamiltonian.

I don't think it is the general form of dynamics for an arbitrary subsystem of a closed system. In deriving such structures from the full Liouville / von Neumann equation, one usually makes certain assumptions (Markov approximation, weak coupling) which may not be fulfilled in general.

The equation given includes non-Markovian corrections and coupling to any order. That is the reason for which the form of the dissipator D is so complex that I apologized for not writing it in explicit form. Of course applying a Markov approximation and taking coupling only up to second order (weak coupling) the dissipator simplifies a lot of.

If I know the full Hamiltonian and don't care about computational power, I could also just solve the Liouville / von Neumann equation for the whole system and trace out the environmental degrees of freedom later.

No exactly. The equation for the whole system is not the ordinary Liouville / von Neumann equation (which is unitary and time-reversible) but a generalized equation, with emergent elements beyond the Hilbert space, which is non-unitary and time-irreversible. See for instance:

1997 "The Liouville Space Extension of Quantum Mechanics" T. Petrosky and I. Prigogine Advances in Chemical Physics Volume 99, 1-120

One application of this recent formalism to an extension of scattering theory

http://prola.aps.org/abstract/PRA/v53/p4075_1

an introduction to this paper is given here

http://www.ph.utexas.edu/~gonzalo/3bgraphs.html

See also other applications of the generalized Liouville von Neumann equation:

2001 "Quantum transitions and dressed unstable states" G. Ordonez, T. Petrosky and I. Prigogine Phys. Rev. A 63, 052106

2000 "Quantum transitions and nonlocality" T. Petrosky, G. Ordonez and I. Prigogine Phys. Rev. A 62 42106

 

[kith]

The equation given includes non-Markovian corrections and coupling to any order.

You are talking about corrections and power series expansion. To me, this sounds like there are some underlying assumptions. Can you please give a reference?

No exactly. The equation for the whole system is not the ordinary Liouville / von Neumann equation (which is unitary and time-reversible) but a generalized equation, with emergent elements beyond the Hilbert space, which is non-unitary and time-irreversible. See for instance: [...]

As far as I can see, your references are talking about the dynamics of systems approaching equilibrium. From the viewpoint of fundamental dynamics, this means there is again an environment involved. So this just adds a layer of complexity. Again the equations of motion can in principle be derived from the Hamiltonian of a larger system. Namely the combined system "whole" system + relevant part of it's environment. This system evolves accordingly to the Liouville / von Neumann equation.

But I can see where this is going. We're going to end up in another discussion about the question, if the time-evolution of closed systems is unitarian. ;-)

 

[Studiot]

I haven't touched particle physics since the 1960s so I am really an (obsolete) interested bystander here.

One thing puzzles me, juanrga.

What exactly is a 'pure state' - it is not a term I am familiar with.

Further since you make the distinction what alternatives are there ie what might non pure state be and what are they called?

Thanks.

 

[kith]

I haven't touched particle physics since the 1960s so I am really an (obsolete) interested bystander here.

I don't know much about QFT myself. The last few posts were about more fundamental questions, concerning the Hamiltonian formalism which can be considered the framework for QFT and many other branches of physics.

What exactly is a 'pure state' - it is not a term I am familiar with.

A pure state is a state of maximum knowledge. In statistical mechanics, you typically only know macroscopic variables like temperature. These macro variables do not specify your micro state completely, but lead to a variety of possible states. So the fundamental objects in statistical mechanics are not pure states, but so-called mixed states.

Classically, a mixed state ρ is a probability distribution on the space of states. The Hamiltonian equations of motion generalize to the Liouville equation. Quantum mechanically, a mixed state ρ is an an operator on the space of states. The Schrödinger equation generalizes to the von Neumann equation.

If you want to include dissipation, the situation gets more complicated and that's one of the things which have been discussed throughout this thread.

 

[Andrew Mason]

The 1st statement written was

"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."

There isn't anything wrong with this, one can read that is that the NET work done on a body is equal to its change in kinetic energy.

So then, what is kinetic energy? You cannot then define kinetic energy the ability to do work (by virtue of its motion) because that is circular.

This is not so much a definition of net Work as it is a statement that net work results in a change of kinetic energy. Work is still defined as force applied through a distance and energy as the ability to do work.

Furthermore, if you read the responding post there is clearly a summation over (i)W12=∑i∫21Fi⋅dsi=∑i∫21miv˙i⋅vidt=∑i∫21d(12miv2i)=T2−T1 where T=12∑imiv2i is the kinetic energy of the system.The summation clearly takes into account all the forces and gives you the net work done on the object. So I don't know what your talking about only one force, it works for many forces.

For instance if a block slides down a plane at constant speed, two forces contribute non zero work, the work done by gravity and the work done by friction. The NET work done on the object is zero, hence its kinetic energy remains constant.

Both original posts are correct.

There is only one v. There are not a v_i for each F_i. So I am not sure what v_i means. Perhaps it should be:

W = \int_1^2 (\sum \vec{F_i})\cdot d\vec{s} = \int_1^2 \vec{F_{net}}\cdot d\vec{s} = \int _1^2 m\dot vdv = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta KE where v is the speed of the centre of mass of the body.

AM

 

[cbetanco]

So then, what is kinetic energy? You cannot then define kinetic energy the ability to do work (by virtue of its motion) because that is circular.

The Kinetic energy is KE=\frac{1}{2}mv^2. There is no mention of work in that definition. The Work done on a system between the initial and final state is defined to be the change in kinetic energy of the system in the initial state to its final state. It is NOT circular to define a new quantity to be the change of another quantity from the initial to the final state.

This is not so much a definition of net Work as it is a statement that net work results in a change of kinetic energy. Work is still defined as force applied through a distance and energy as the ability to do work.

Yes, it is still defined as the force applied through a distance, but what I had showed you is that this is equivalent to the change in Kinetic Energy of the system

There is only one v. There are not a v_i for each F_i. So I am not sure what v_i means. Perhaps it should be:

W = \int_1^2 (\sum \vec{F_i})\cdot d\vec{s} = \int_1^2 \vec{F_{net}}\cdot d\vec{s} = \int _1^2 m\dot vdv = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta KE where v is the speed of the centre of mass of the body.

AM

The F_i is the NET external force acting on the ith mass, m_i. The i is not an index over individual forces, its an index running over all the masses of the system. These are point particles with mass m_i in the strictly classical sense, but it also works with rigid bodies if you instead have a mass density where you take m_i\rightarrow\rho(\bf{x}) and integrate over the mass density \rho(\bf{x}) for the entire volume. Then, you would just have to replace m_i with \rho(\bf{x}) and \bf{v_i} with \bf{v}, where \bf{v} is the velocity of the rigid body, and then then take the sum \sum_i \rightarrow \int_V d\bf{x} in the equation in my second post and you would still get the work done on the system is equivalent to the change in KE of the system

 

[cmb]

Gosh! How complicated it all is!

And there I was thinking that energy is, simply, the essential attribute that is required to change a system from one state to another.

 

[Andrew Mason]

Yes, it is still defined as the force applied through a distance, but what I had showed you is that this is equivalent to the change in Kinetic Energy of the system

Just because the magnitude of the work is the same as the magnitude of the change in kinetic energy does not mean they are the same thing. Energy is a property that a body has - the ability to do work and work is force applied through a distance. Saying that work is defined as the change in kinetic energy confuses the fundamental difference between Work and Energy.

The difference is that Work is a transfer of energy to/from a body and Energy is a property of the state of a body.

This distinction is important. For example, in thermodynamics W=Work and Q = Heat flow are transfers of energy not properties of a state of a thermodynamic system. Energy=U, is a property of the state of a system. Saying ΔU = Q+W does not mean that ΔU is defined as Q+W. It is merely a statement that the magnitude of the change in U is equal to the sum of the heat flow to and work done on the system.

AM

 

[cmb]

Energy is a property that a body has

or a field...

...or a propagating wave...

...or...?

 

[Andrew Mason]

or a field...

...or a propagating wave...

...or...?

Fields and propagating waves are all associated with some body ie. some kind of structure that has inertia or mass. Is the energy in the field or in the body? Does a photon represent energy or the transfer of energy from one body to another?

AM

 

[kinetico]

An Alternative Definition of Energy EFor a simple particle

{\vphantom{\int}} \vec{a} = \vec{a}{\vphantom{\int}} m \, \vec{a} = m \, \vec{a}\int m \: \vec{a} \cdot d\vec{r} = \int m \: \vec{a} \cdot d\vec{r}{\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} + \; constant = \int m \: \vec{a} \cdot d\vec{r}constant = {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} - \int m \: \vec{a} \cdot d\vec{r}E = {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} - \int m \: \vec{a} \cdot d\vec{r}

For a system of n particles

E = \sum_{i=1}^n {\textstyle \frac{1}{2}} \: m_i \, \vec{v}_i^{\: 2} - \sum_{i=1}^n \int m_i \; \vec{a}_i \cdot d\vec{r}_i

 

[cbetanco]

Just because the magnitude of the work is the same as the magnitude of the change in kinetic energy does not mean they are the same thing.

Yes it does. This is the work-energy theorem. How can you say the magnitude of work and the magnitude of change in kinetic energy are the same, but then in the same sentence argue they are not the same?

Energy is a property that a body has - the ability to do work and work is force applied through a distance. Saying that work is defined as the change in kinetic energy confuses the fundamental difference between Work and Energy.

The difference is that Work is a transfer of energy to/from a body and Energy is a property of the state of a body.

I never argued otherwise. What I am saying is that the work done on the system is equal to the change in kinetic energy of that system. I never argued that the work is a property of the system. But the work done on or by the system is the same as the change in kinetic energy of that system. Please see pg 9 of the graduate level text on Classical Mechanics by Goldstein, Poole and Safko. In between Eq. 1.29 and 1.30 it is written "Hence, the work done can still be written as the difference of the final and initial kinetic energies." That is ALL I was saying. Its a pretty standard definition that is hard to argue with.

Saying ΔU = Q+W does not mean that ΔU is defined as Q+W. It is merely a statement that the magnitude of the change in U is equal to the sum of the heat flow to and work done on the system.
AM

Yes, it does mean by the definition of being equal, that the change of energy in a thermal system is defined to be the heat flowing in or out plus the work done by or on the system. And the ΔU does not have to be the magnitude in the change, it can also be negative depending if the heat is flowing in or out and the work is done by the system, or on the system.

 

[cmb]

Fields and propagating waves are all associated with some body ie. some kind of structure that has inertia or mass. Is the energy in the field or in the body? Does a photon represent energy or the transfer of energy from one body to another?

AM

I don't see how you come to make that distinction. I might equally argue that bodies are associated with some kind of field structures rather than vice versa. How many 'bodies' would exist without electrostatics, gravitational, and nuclear forces?

I do not see how the energy of a photon in free space is associated with 'a body'.

Simply, we are looking at 'configuration' here. 'Energy' is an attribute of the configuration of 'stuff'. Matter, fields, particles, relative position, relative motion.

Your argument is demonstrated incomplete because energy such as 'kinetic' is relative to other bodies, so cannot be described as 'associated with some kind of body'. A body traveling at the same speed as me has no kinetic energy in my frame, but may have KE in someone else's. Similarly, you cannot have electrostatic or gravitational energy without multiple contiguous bodies. Therefore, we must talk about the 'system' as containing the energy, not by association with a body.

 

[juanrga]

You are talking about corrections and power series expansion. To me, this sounds like there are some underlying assumptions. Can you please give a reference?

As far as I can see, your references are talking about the dynamics of systems approaching equilibrium. From the viewpoint of fundamental dynamics, this means there is again an environment involved. So this just adds a layer of complexity. Again the equations of motion can in principle be derived from the Hamiltonian of a larger system. Namely the combined system "whole" system + relevant part of it's environment. This system evolves accordingly to the Liouville / von Neumann equation.

But I can see where this is going. We're going to end up in another discussion about the question, if the time-evolution of closed systems is unitarian. ;-)

Once again, the expression for the dissipator is exact. It is only for computational reasons that it is often expanded in series expansions. Evidently, specific series expansion depends on assumptions about convergence about the expansion center, but this is a computational problem.

Contrary to what you say, the references study the general evolution of isolated systems. Some of those isolated systems approach equilibrium and others do not. Isolated systems approaching equilibrium cannot be studied with the ordinary Liouville /von Neuman equation (unitary and time reversible). That is the reason for their extension of quantum theory.

 

[juanrga]

I haven't touched particle physics since the 1960s so I am really an (obsolete) interested bystander here.

One thing puzzles me, juanrga.

What exactly is a 'pure state' - it is not a term I am familiar with.

Further since you make the distinction what alternatives are there ie what might non pure state be and what are they called?

Thanks.

This is not about particle physics but about quantum theory.

The purity of a system is given by p=Tr{ρ²}. When p=1 the system is in a pure state, otherwise it is in a mixed state.

 

[juanrga]

Gosh! How complicated it all is!

And there I was thinking that energy is, simply, the essential attribute that is required to change a system from one state to another.

Any thermodynamic book explains how an isolated system can change its state A→B (second law), whereas its energy remains constant (first law).

 

[Studiot]

How can you say the magnitude of work and the magnitude of change in kinetic energy are the same, but then in the same sentence argue they are not the same?

cbetanco I think you are missing Andrew's point.

I have emboldened the important word - magnitude.

Just because the magnitude of the work is the same as the magnitude of the change in kinetic energy does not mean they are the same thing.

Andrew's point is very succinct and stands for a much deeper question.

Work is one way for energy possessed by system A to be transferred to System B. The numerical value of the energy leaving system A equals the numerical value of that entering system B and also equals the numerical value of the work done, in consistent units.

The immediate questions are:

What is the timescale of this transfer?
Is there a time when (some of) the transferred energy has left system A and not yet entered system B?
If so where is this energy ?

 

[juanrga]

Just because the magnitude of the work is the same as the magnitude of the change in kinetic energy does not mean they are the same thing. Energy is a property that a body has - the ability to do work and work is force applied through a distance. Saying that work is defined as the change in kinetic energy confuses the fundamental difference between Work and Energy.

The difference is that Work is a transfer of energy to/from a body and Energy is a property of the state of a body.

This distinction is important. For example, in thermodynamics W=Work and Q = Heat flow are transfers of energy not properties of a state of a thermodynamic system. Energy=U, is a property of the state of a system. Saying ΔU = Q+W does not mean that ΔU is defined as Q+W. It is merely a statement that the magnitude of the change in U is equal to the sum of the heat flow to and work done on the system.

AM

Excellent! This has been my main point in this long thread. Precisely I defined energy, a state function in thermodynamics, without any appeal to work and next I defined work as one mechanism of interchange of energy (other mechanism being heat).

Only a correction. Mechanical work would not be confused with the general concept of work. Mechanical work is force applied through a distance. Other kind of work are involved in the first law

ΔU = Q+W = ΔU = Q+W_mech+Ʃ_iW_i

 

[Studiot]

The purity of a system is given by p=Tr{ρ2}. When p=1 the system is in a pure state, otherwise it is in a mixed state.

Thank you for that clarification. And also thank you Kith for your version. I will give both some consideration.

 

[juanrga]

Yes it does. This is the work-energy theorem. How can you say the magnitude of work and the magnitude of change in kinetic energy are the same, but then in the same sentence argue they are not the same?

dE_K = δW_mech

does not imply E_K = W_mech neither E = W

I never argued otherwise. What I am saying is that the work done on the system is equal to the change in kinetic energy of that system. I never argued that the work is a property of the system. But the work done on or by the system is the same as the change in kinetic energy of that system. Please see pg 9 of the graduate level text on Classical Mechanics by Goldstein, Poole and Safko. In between Eq. 1.29 and 1.30 it is written "Hence, the work done can still be written as the difference of the final and initial kinetic energies." That is ALL I was saying. Its a pretty standard definition that is hard to argue with.

Goldstein textbook is about mechanics. Therein, it is ignoring other forms of work (non-mechanical work) and also ignoring heat.

Yes, it does mean by the definition of being equal, that the change of energy in a thermal system is defined to be the heat flowing in or out plus the work done by or on the system. And the ΔU does not have to be the magnitude in the change, it can also be negative depending if the heat is flowing in or out and the work is done by the system, or on the system.

The first law ΔU=Q+W is neither the definition of internal energy U nor the definition of ΔU.

U is a state function and the definition of its change for a thermodynamic process A→B is given by ΔU \equiv U_B - U_A

 

[cmb]

Any thermodynamic book explains how an isolated system can change its state A→B (second law), whereas its energy remains constant (first law).

That's the inverse of what I said, so is not germane and is trivial to my point.

I do not suppose these books of yours show how a system does NOT change its state while energy is fed into/removed from it?

My statement is that energy is that attribute which would cause a system state to change when added/removed (not that a system state change involves a net energy flux in/out of it).

 

[Studiot]

I do not suppose these books of yours show how a system does NOT change its state while energy is fed into/removed from it?

Have you considered Fourier's Law?

Feed heat into one end of a bar and withdraw it at the other, allowing a steady state to develop.

 

[cmb]

Have you considered Fourier's Law?

Feed heat into one end of a bar and withdraw it at the other, allowing a steady state to develop.

That'd be no net energy into your bar system...(?)

 

[Andrew Mason]

Yes it does. This is the work-energy theorem. How can you say the magnitude of work and the magnitude of change in kinetic energy are the same, but then in the same sentence argue they are not the same?

The work-energy theorem simply says that the work done by a net force on a body is mathematically equal to the change in kinetic energy of that body. It does not say that work done is defined as the change in kinetic energy. I am just quibbling with the word "defined". The two concepts, work and energy, are different.

Yes, it does mean by the definition of being equal, that the change of energy in a thermal system is defined to be the heat flowing in or out plus the work done by or on the system. And the ΔU does not have to be the magnitude in the change, it can also be negative depending if the heat is flowing in or out and the work is done by the system, or on the system.

We do not seem to be agreeing on the meaning of the word "define". I am using "define" in the same sense as "explain the meaning of". I am not sure how you are using it.

AM

 

[juanrga]

Gosh! How complicated it all is!

And there I was thinking that energy is, simply, the essential attribute that is required to change a system from one state to another.

Any thermodynamic book explains how an isolated system can change its state A→B (second law), whereas its energy remains constant (first law).

I have reintroduced what you exactly wrote (bold face is from mine).

That's the inverse of what I said, so is not germane and is trivial to my point.

I do not suppose these books of yours show how a system does NOT change its state while energy is fed into/removed from it?

I replied to what you wrote then, showing how a thermodynamic system can change from a state A to other B without even mentioning energy U.

My statement is that energy is that attribute which would cause a system state to change when added/removed (not that a system state change involves a net energy flux in/out of it).

This is a new statement but again suspicious. I can imagine a system state to not change when energy is added/removed. For instance a cycle A→A where energy is lost in form of heat and added using work.

 

[cmb]

If you don't work with a description of energy based on a 'change of state' of a system, then you are going to get very confused at some point or other when you try to deal with more complex processes. Quite evidently, there MUST be energy flows within a system for it to change its state.

I agree that a 'system' may change its state without changing its net energy content - this is trivial, we only have to take the Universe as a whole to see that. So you can always find an arbitrary boundary to make that point. I think that's as much as I can say to you, if you reject that description of energy.

 

[Studiot]

That'd be no net energy into your bar system

I simply responded to your question, as written. You did not specifiy net energy flows.

Really all I was repeating was the engineering catechism:

Input = Output + Accumulation

You also did not respond to my comment, taking the idea further in respect of energy changes.

 

[cmb]

I simply responded to your question, as written. You did not specifiy net energy flows.

No problem! Then let that be a clarification! (This isn't a competition of logical argument, is it?
We're just working towards a useful definition of energy, aren't we?)

You also did not respond to my comment, taking the idea further in respect of energy changes.

Sorry, I am unclear what comment you are referring to.

Maybe it would be more precise to talk in terms of 'energy transfer'. It is the attribute that is transferred from one form and/or body to another when there is a system state change.

 

[Studiot]

Sorry, I am unclear what comment you are referring to.

I was referring to my post#107.


As regards net energy flows.
Surely any definition should include the possibility of zero for any variables used or if you like reduce to this as a special case - as for instance with forces in equilibrium or not.

So for a system that has several energy inputs and several outputs there will be, in general, a net energy flow.

Surely this net flow variable must allow for the possibility that the net flow is precisely zero?

No, I agree, this is not a competition.

 

[cmb]

Surely this net flow variable must allow for the possibility that the net flow is precisely zero?

For sure. There will always exist a boundary you can describe around any system undergoing changes of state that has no flows in nor out. You can also have the situation where a defined system is either changing towards equilibrium, or has reached equilibrium, and the net energy flow is zero.

So I think my imprecision in my original comment is the term 'system'. In hindsight, perhaps what I think I should've said as a 'working' statement is that there are energy flows within a system when it changes state.