Why is energy hard to define?

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Offline The Scientist

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Why is energy hard to define?
« on: 03/07/2010 09:47:52 »
Please answer as detailed as possible. Thank you!
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Offline Bored chemist

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Why is energy hard to define?
« Reply #1 on: 03/07/2010 16:54:17 »
Energy is the capacity to do work.

7 words is hardly difficult.
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Offline Pmb

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Why is energy hard to define?
« Reply #2 on: 05/07/2010 12:57:26 »
Quote
Energy is the capacity to do work.
That is not the definition of energy. When people use that as a definition they are trying to pll the wool over your eyes. I could very easily say that, from that definition, that the energy of a perticle must be the value E = mv. Everytime that a body is moving it has the capacity to do work. Since this quantity is conserved then people could easily confuse this with energy (where we all know its really momentum).

For a complete treatment of the definition of energy see
http://home.comcast.net/~peter.m.brown/mech/what_is_energy.htm

Note the following quote from Feynman on that page
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It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and we add it all together it gives “28” -  always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.


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Offline wolfekeeper

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« Reply #3 on: 05/07/2010 16:22:18 »
It's not that hard to define, it's basically force times distance, or anything that's convertible with that.

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Offline imatfaal

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Why is energy hard to define?
« Reply #4 on: 05/07/2010 17:24:55 »
isnt that work? ie the transfer of energy W=F.d 

Only being difficult ;-)
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Offline Pmb

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Why is energy hard to define?
« Reply #5 on: 05/07/2010 18:23:21 »
It's not that hard to define, it's basically force times distance, or anything that's convertible with that.
Force times distance is work. And work is not even a form of energy, it merely has the same units as energy. Feynman was an extremely sharp physicist and thus was not one who'd write that energy is not defined for no good reason. It's wise not to ignore anything that Feynman said. Especially when you provide no justification for ddisagreeing with him.

Feynman is not alone either. You'll see the same thing stated by other physicists too. For example, see An Introduction of Thermal Physics, by Daniel V. Schroeder. On page 17 the author writes
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To further clarify matters, I really should give you a precise definition of energy. Unfortunately, I can't do this. Energy is the most fundamental dynamical concept in all of physics, and for this reason, I can't tell you what energy it is in terms of something more fundamental. I can however, list the various forms of energy - kinetic, electrostati, gravitational, chemical, nuclear - and add the statement that, while energy can often be converted from one form to another, the total amount of energy in the universe  never changes.
« Last Edit: 05/07/2010 18:34:49 by Pmb »

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Offline Bored chemist

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« Reply #6 on: 05/07/2010 19:42:38 »
"I could very easily say that, from that definition, that the energy of a perticle must be the value E = mv. "
But, of course, you would be wrong. A little calculus would show it.

Please show me a case where my definition of energy (i.e.the capacity to do work) is incorrect.
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Offline Soul Surfer

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Why is energy hard to define?
« Reply #7 on: 05/07/2010 23:39:33 »
There is a confusion here between two things.  firstly and simply the energy that we can make use of in physical machines which is easily defined as above and secondly the fundamental source of all energy in the universe.  This second item is the really difficult bit.
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Offline Pmb

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« Reply #8 on: 06/07/2010 01:49:10 »
[quote author]
"I could very easily say that, from that definition, that the energy of a perticle must be the value E = mv. "
But, of course, you would be wrong.
[/quote]
You missed my point. The point was that saying that energy is the capacity to do work is insufficient to obtain an expression for it.
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A little calculus would show it.
Since I proposed it merely as an example of one way one could come to the wrong conclusion, your point is moot. However I'd like to see what you have in mind. Please show what you mean by "A little calculus would show it." noting that you have no definition of energy with which to invoke in your proof since this is the definition being proposed.
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Please show me a case where my definition of energy (i.e.the capacity to do work) is incorrect.
You never provided a definition because what you proposed is insufficient for an expression for energy to be constructed. In your so-called "definition" you left out one of the most important properties of energy i.e. that it is a conserved quantity. For example; As I said before I could define energy as E = mv. Regardless of what youmay think it is impossible to prove that wrong since no definition can be proven wrong (so your attempt to do so will fail). The expression E = mv means that if a body is moving it has energy. A moving body can do work. Therefore this definition works. So is any power of v and with any multiplicative coefficeint, i.e. E = mv^2/2. I could define energy as E = mv^2/2 and even though this satisfies your "definition" it is still wrong.

On the other hand, please explain how someone like Feynman could get something so basic and so important, so wrong. And why so many other physicists get it wrong? As I said, Feynman was only one, A.P. French )MIT) is yet another.

I'm afraid that you fell for one of physics most embarassing problems, i.e. that no definition exists for energy but some authors hate to leave things undefined so they give a tounge-in-cheek "definition" knowing that they'd never have to answer for their bad attempt.

The capacity to do work is one of the characteristics of energy, not its definition.

You'd be wise to find and read the article Energy is Not the Ability to do Workby Lehrman, Robert L., Physics Teacher, 11, 1, 15-18, Jan 73

Quote
Abstract - The common definition is shown to be false. A modern definition must be based on the first and second laws of thermodynamics and in terms of a set of algebraic expressions written in such a way that their sum does not change when a system is isolated. (DF)
« Last Edit: 06/07/2010 02:06:51 by Pmb »

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Offline Bored chemist

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Why is energy hard to define?
« Reply #9 on: 06/07/2010 07:06:25 »
"You missed my point. The point was that saying that energy is the capacity to do work is insufficient to obtain an expression for it."

I could look up the definition of "Mammal"; it wouldn't include an expression of how to calculate anything.
Definitions don't need to do that.


Work is done when a force F moves through a distance D. E=Fd
If you add
F=MA
then you can calculate the KE of a body by calculating the work done by bringing it to a halt.
You get 0.5MV^2
As I said, a little calculus shows that E= MV is wrong.

"You never provided a definition because what you proposed is insufficient for an expression for energy to be constructed."
As I said, that's not what a definition is.
"you left out one of the most important properties of energy i.e. that it is a conserved quantity."
Energy was energy before relativity showed that it wasn't conserved (in some sense) and it was still energy after that was sorted out.
If there were some weird circumstance where energy was not conserved, then it would still be the capacity to do work.
Conservation is an observed property of energy; it's not a definition.


I can't find the article you cited on line, could you précis it?

Please answer my question, when does my definition fail?
I will get back to you about  Feynman's point, but I have a bus to catch.
« Last Edit: 06/07/2010 07:20:23 by Bored chemist »
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Offline Pmb

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Why is energy hard to define?
« Reply #10 on: 06/07/2010 10:16:44 »
Quote
I could look up the definition of "Mammal"; it wouldn't include an expression of how to calculate anything.
Definitions don't need to do that.
On the contrary. A definition is exactly what tells you what the value of a quantity is.
Quote
Work is done when a force F moves through a distance D. E=Fd
If you add
F=MA
then you can calculate the KE of a body by calculating the work done by bringing it to a halt.
You can't do that if you have not yet defined energy. In this case the energy is energy of motion. What you've done is to take another quantity, the value of the work done, and called it "kinetic energy". If all we've done so far is to define "energy" then its premature to define "kinetic energy". One can easily define the term "kinetic energy" and calculate it as th work done in bringing a body of mass m from speed 0 to speed v. However that does not tell you what energy is. I.e. we have no idea yet of whether energy = kinetic energy.

If I were to define the term "energy" then I'd say that it's the sum of all the forms on energy such that the sum is an integral of motion (i.e. a constant of motion) for a closed system. Then we'd have to take care to find all those forms and then prove its such an integral. The problem is then reduced to finding all the forms of energy for a system. But to do that in general is not obvious and there is no general way to find all such quantities. I believe that's why Feynman said what he did.
Quote
Please answer my question, when does my definition fail?
I already did.
« Last Edit: 06/07/2010 10:35:02 by Pmb »

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Offline The Scientist

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Why is energy hard to define?
« Reply #11 on: 06/07/2010 10:27:10 »
Energy is the capacity to do work.

The concept of energy is abstract and therefore not as easy to define as the concepts of mass and volume. One definition of energy, as what Bored chemist said, is the capacity to do work.

There are two principal forms of energy: potential and kinetic.

Potential energy is stored energy.

Kinetic energy is the energy of motion.

Hope this helps.
The Scientist

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Offline Pmb

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« Reply #12 on: 06/07/2010 10:42:28 »
Quote from: The Scientist
The concept of energy is abstract and therefore not as easy to define as the concepts of mass and volume.
I agree.
Quote from: The Scientist
One definition of energy, as what Bored chemist said, is the capacity to do work.
As I said above, that is a useless definition. It doesn't allow one to determine a value for energy. In fact it tells you nothing. The purpose of calling something "energy" is so that you have a name to call a numerical quantity. "capacity to do work" is not such a definition. It's a useless definition.

Well, at this point I believe that I've said all that can be said and anything more would be me repeating myself so I'm bowing out of this thread.
« Last Edit: 06/07/2010 11:03:26 by Pmb »

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Offline Murchie85

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Why is energy hard to define?
« Reply #13 on: 06/07/2010 13:17:36 »
I agree with both Bored chemist and Pmb in certain aspects, I think defining energy as a capacity to do work is an approximate description one that would be sufficient for a member of public, although is not a true fundamental description Pmb illustrates the difficulty in actually pegging the phenomena down as the definition does not provide a value which is an essential in physic

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Offline Pmb

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Why is energy hard to define?
« Reply #14 on: 06/07/2010 14:01:28 »
I'd like to point out one area which has the potential to be problematic for those people who define energy according to the ability to do work. Some people think of the gravitational force as being non-existant and as such they see the notion of work being done by gravity as being meaningless. Therefore those people who define energy as the ability to do work will have a problem defining the energy of a particle at rest in a gravitational field for that reason.

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Offline wolfekeeper

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« Reply #15 on: 06/07/2010 14:59:36 »
Gravitational potential energy is negative. So it's the work that was done to bring an object to that position. Similarly with chemical binding energy.

Basically all of these conserved quantities seem to be symmetries or invariants of the physical laws that govern the universe. Why they happen to be conserved, that nobody knows.
« Last Edit: 06/07/2010 15:01:35 by wolfekeeper »

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Offline Pmb

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Why is energy hard to define?
« Reply #16 on: 06/07/2010 15:10:44 »
Gravitational potential energy is negative.
Only in certain cases. For example: the gravitational potential energy of a particle in a uniform gravitational field is positive.

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Offline wolfekeeper

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« Reply #17 on: 06/07/2010 15:29:57 »
Not unless you've found a source of Cavorite and are keeping it under your hat.

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Offline Pmb

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« Reply #18 on: 06/07/2010 16:11:42 »
Not unless you've found a source of Cavorite and are keeping it under your hat.
Huh? Do you understand why the potential energy for a particle in a uniform field is positive? Actually it's not as simple as that.

Choose the zero level for the gravitational potential to be zero when z = 0. Since F = -mg it follows that U = mgz > 0 for z > 0. That this is the case follows from the fact that F = -grad U. What does "Cavorite" have to do with anything (whatever that is - antigrav metal? Yeesh! Let's stick with science and not sci-fi). There's a way to construct a uniform gravitational field in a finite region of space using a finite amount of mass. Do you know how to do it? :)

The value for kinetic energy comes from the relationship for the work done, W, on a particle which is accelerated from speed v1 to speed v2. Then

W = mv2^2 - mv1^2

We can let K = mv^2/2. This gives us W = K2 - K1. However we can also define K to be K = mv^2/2 + C where C is an arbitrary constant. Then we still get W = K2 - K1.
« Last Edit: 06/07/2010 16:17:13 by Pmb »

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Offline wolfekeeper

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« Reply #19 on: 06/07/2010 16:20:23 »
There's no such thing as a uniform gravitational field. It's often a useful approximation, but the actual potentials have divergence.

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Offline The Scientist

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« Reply #20 on: 06/07/2010 16:25:18 »
Quote from: The Scientist
One definition of energy, as what Bored chemist said, is the capacity to do work.
As I said above, that is a useless definition. It doesn't allow one to determine a value for energy. In fact it tells you nothing. The purpose of calling something "energy" is so that you have a name to call a numerical quantity. "capacity to do work" is not such a definition. It's a useless definition.
[/quote]

Why is it then that energy being the ability, or capacity to do work is recognised as a worldwide definition of energy?

Since if it is a useless definition, why do most science textbooks include that as the basic definition?
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Offline Pmb

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« Reply #21 on: 06/07/2010 16:26:37 »
There's no such thing as a uniform gravitational field. It's often a useful approximation, but the actual potentials have divergence.
So your response is no. You don't know how to create a uniform gravitational field. And you're wrong. Its quite possible to create one. In fact this is a problem in a physics text I have.

I'm curious though. Why do you believe that there's "no such thing" as a uniform G-field. And exactly what do you mean by it. Do you mean that its impossible to create one or do you mean that nobody has created one for lack of needing one?
« Last Edit: 06/07/2010 16:28:11 by Pmb »

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Offline Pmb

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« Reply #22 on: 06/07/2010 16:30:10 »
Quote from: The Scientist
Why is it then that energy being the ability, or capacity to do work is recognised as a worldwide definition of energy?

Since if it is a useless definition, why do most science textbooks include that as the basic definition?
Its certainly not universally accepted, that's for sure. I've even given you two examples to show that.

Basically authors just accept what they were taught and don't think about the basics again.
« Last Edit: 06/07/2010 17:42:18 by Pmb »

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Offline wolfekeeper

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« Reply #23 on: 06/07/2010 16:34:20 »
No, in the real world you can't create perfectly uniform gravitational potentials extending over a finite space because of the way that matter consists of particles, and no number of particles arranged in any way can create a perfectly uniform potential.

If we use Newtonian gravity for the sake of simplicity (similar things happen in GR) then it obeys the Laplace equation, and it thus has divergence.

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Offline Pmb

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« Reply #24 on: 06/07/2010 17:54:12 »
Quote from: wolfekeeper
No, in the real world you can't create perfectly uniform gravitational potentials extending over a finite space because of the way that matter consists of particles, and no number of particles arranged in any way can create a perfectly uniform potential.
You've really taken this thread off track. Nobody cares that our mathematical expressions for real fields are not infinitely precise. There are no quantities in classical physics which are exact since quantum theory is the exact theory which can provide exact results. And even that might not be exact to a million decimal places. All that nonsense has nothing to do with what I referred to and whether the diverence is zero or not has nothing to do with the subject matter in this thread. Who cares is U = -GM/r is not correct for the earth or sun etc. Nobody cares about precision in this thread. It's totally irrelevant and a major distraction. I will not bother with this nonsense again in this thread. It is a waste of time. The Op has not demonstrated any interest in the precision of any expression, neither have I. And the divergence is also irrelevant. Nobdoy cares about that either.

If you didn't know how to construct such a field you should have just said so.

As far as the divergence goes, you don't seem to know what you're talking about and it's unclear why you brought it up. If g is the value of the gravitational acceleration vector at a point in space then the divergence of that vector field is zero in regions where there is no matter. So even if the field is not uniform to 600 decimal points the divergence will be exactly zero.
« Last Edit: 06/07/2010 18:00:45 by Pmb »

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Offline wolfekeeper

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« Reply #25 on: 06/07/2010 18:59:25 »
If you take the Laplacian of the field in free space it's always zero. That combined with the fact that the gravitational field is always attractive (as I say, unless you've invented Cavorite) it means that at the very least, at sufficiently close range the gravitational potential is always negative. It's extremely conventional to set the graviational potential at great distance to tend to zero. The net upshot is that the gravitational potential energy is always defined to be negative. How could you not know this?

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Offline Bored chemist

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« Reply #26 on: 06/07/2010 19:25:44 »
Quote
I could look up the definition of "Mammal"; it wouldn't include an expression of how to calculate anything.
Definitions don't need to do that.
On the contrary. A definition is exactly what tells you what the value of a quantity is.
Quote
Work is done when a force F moves through a distance D. E=Fd
If you add
F=MA
then you can calculate the KE of a body by calculating the work done by bringing it to a halt.
You can't do that if you have not yet defined energy. In this case the energy is energy of motion. What you've done is to take another quantity, the value of the work done, and called it "kinetic energy". If all we've done so far is to define "energy" then its premature to define "kinetic energy". One can easily define the term "kinetic energy" and calculate it as th work done in bringing a body of mass m from speed 0 to speed v. However that does not tell you what energy is. I.e. we have no idea yet of whether energy = kinetic energy.

If I were to define the term "energy" then I'd say that it's the sum of all the forms on energy such that the sum is an integral of motion (i.e. a constant of motion) for a closed system. Then we'd have to take care to find all those forms and then prove its such an integral. The problem is then reduced to finding all the forms of energy for a system. But to do that in general is not obvious and there is no general way to find all such quantities. I believe that's why Feynman said what he did.
Quote
Please answer my question, when does my definition fail?
I already did.

No, you didn't; the closest you got to doing so was to say "someone else says it doesn't- here's a reference which isn't readily available.

Your interpretation of the word "definition" means that you can't define anything that isn't described by an equation.
Since practically all the equations we see are approximations, that means you can't define anything.
A bit pointless really.

I have no problem with dealing with gravitational energy, nor with that associated with a point in an electrostatic field. It's the work done by a unit mass or charge "falling" from infinity to that point.
That's a force times a distance (albeit integrated because the force isn't constant).
So, yet again I'm asking where the definition fails.

Here's another way of looking at it.
There is a lot of cobblers talked about "energy" stuff about yin and yang; auoras and other dross.
A way to address that would be to replace the word energy in all science books with "Tctdw" (the capacity to do work).

In what branch of science would that not work?
I nearly forgot in all the excitement.
Feynman's point was, I believe, more or less the opposite of what you are saying.
His point was that, while we have lots of clever equations that let us calculate the paths of electrons or the speed of light, we don't really know what an electron or light actually are.
At that level it is perfectly true that we don't know what energy is; we just use t as an accounting tool- it lets us do clever things like pay the electricity bill, flatten a city or fly to the moon. But we still don't really know why.

Incidentally, since you still don't seem to get it, the capacity of a moving rock to do work is twice the capacity to do work of a rock (of the same mass) with twice as much energy. Such a pair of rocks can be shown by calculus to have velocities in the ratio 1 to √2.
A rock with twice the velocity can do 4 times as much work (if brought to a halt) and therefore has 4 times the energy.
That's why your assertion about e=mv is wrong; it gives the wrong answer.

As I recall, he likened it to a group of people looking at a bird; they can all give you the name of the bird in their various languages; but that tells you precisely nothing about  the nature of the bird.


« Last Edit: 06/07/2010 19:42:16 by Bored chemist »
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Offline Pmb

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« Reply #27 on: 07/07/2010 01:55:38 »
Since you two have been unable to follow the (quite simple and very basic) physics I've explained to you I don't believe that repeating the physics will do any good so I won't be posting anymore in this thread. I suggest that you go to a basic physics text and learn what positive and negative values of potential means and what a change in potential means and why you can add any constant value you wish to the potential function without changing the physics.

Also learn why a vauge statement like "energy is the capacity to do work" cannot serve as a definition of energy. Apparently you don't understand physics well enough to get this very simple point.

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Offline Bored chemist

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« Reply #28 on: 07/07/2010 06:57:15 »
Could someone else please tell me what Pmb claims to have already explained.
When is energy not the capacity to do work?
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Offline JP

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« Reply #29 on: 07/07/2010 08:07:54 »
Could someone else please tell me what Pmb claims to have already explained.
When is energy not the capacity to do work?

I'm not sure this is what he was getting at, but to take the zero-point energy of a quantum system as an example.  You can't extract that energy as work, but it exists and has physical meaning. 

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Offline Bored chemist

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« Reply #30 on: 07/07/2010 19:56:09 »
It's the capacity to do work that the system would have if you could extract it.

Next.
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Offline Geezer

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« Reply #31 on: 08/07/2010 04:19:47 »
Energy is only the capacity to do work if you ignore entropy, and entropy can be a very cruel master.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Bored chemist

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« Reply #32 on: 08/07/2010 20:27:42 »
Energy is the capacity of a system to do work, but in some cases this is based on the assumption that there is a "cold sink" at absolute zero. In those cases the capacity to do actual work is the Gibbs free energy  or Helmholtz free energy. The missing capacity to do work is the product of the entropy change and the temperature.
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Offline Pmb

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« Reply #33 on: 10/07/2010 04:25:36 »
Hi folks

Sorry for my bad attitude above. I have been feeling very ill and was in a terrible mood when I posted that. I really shouldn't post when I'm that sick.

Bored Chemist asked "When is energy not the capacity to do work?"

I never said that there are times when energy is not the capacity to do work. I said that that simple statement is insufficient to define energy. All it is is one property of energy and you can't define a quantity by stating merely one of its properties. For example: If I said "Sky scrapers are above ground." While that's true, and there will never be a known case where there is a skyscraper which is below ground, it doesn't tell you what a skyscraper is. I migh confuse it with the sky since the sky is also above ground.

Take another property, such as energy must be conserved. That is another property of energy and must be included in any definition of energy. Then there is the matter of the value of energy, which is really not a physically meaninful quantity. Only changes in energy have any actual physical meaning. Take a particle at rest in an inertial frame of reference. The particle is not subject to any force and there is no sources of force around. In Newtonian mechanics we could say that the kinetic energy is zero and the potential energy is zero so that the total energy is zero. However the only physically meaningful quantity is changes in those quantities so we could just as well say that the potential energy is zero and even define kinetic energy to have a constant component. This would give a non-zero value for the energy. And this would not violate the work-energy theorem.

I understand that people disagree with this and they certainly have a right to their opininon. This is my opionion (althought, like yourself, I call this "fact", but that's beside the point).

I wasn't going to post again for fear or merely repeating what I had already said and that's of no use to anybody. But if there is something I can respond to which I don't think I've already addressed then I respond (although you might think I didn't think I addressed it).

Pete
« Last Edit: 10/07/2010 05:46:41 by Pmb »

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« Reply #34 on: 10/07/2010 05:23:19 »
There is a confusion here between two things.  firstly and simply the energy that we can make use of in physical machines which is easily defined as above and secondly the fundamental source of all energy in the universe.  This second item is the really difficult bit.

Soul Surfer nailed the point being made in the rest of this thread here.

BC's definition is a good one in that if you define things properly (and aside from the fact that it may be physically impossible access some forms of quantum energy), that all energy can be described as the capacity to do work.

Of course that isn't a sufficient definition from which to derive the behavior of energy in all of physics, but it does describe energy well in terms of a simple property that all energy seems to have.  I've seen an alternative definition that it's the ability to "make things happen" or "make things change," which implies that something is occurring over a period of time (and gets to the point that conservation energy has to do with time translation invariance of the physical laws.) 

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« Reply #35 on: 10/07/2010 05:49:12 »
Could someone else please tell me what Pmb claims to have already explained.
When is energy not the capacity to do work?

I'm not sure this is what he was getting at, but to take the zero-point energy of a quantum system as an example.  You can't extract that energy as work, but it exists and has physical meaning. 
The value of energy isn't as important as changes in energy since that's really all that is observed. And I think I'd be hard to say how the energy of a virtual particle could do work. In fact I don't even think that work can be defined in quantum mechanics so in QM this "capacity to do work" can't be meaninful.

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Offline Bored chemist

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« Reply #36 on: 10/07/2010 18:48:45 »
In QM the uncertainty relationship means that energy isn't strictly conserved. Perhaps it's better not to include that in a definition. Anyway conservation is a property of energy; not a definition.
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« Reply #37 on: 10/07/2010 22:41:18 »
In QM the uncertainty relationship means that energy isn't strictly conserved. Perhaps it's better not to include that in a definition. Anyway conservation is a property of energy; not a definition.
That is incorrect. For systems which have a time indepentant Hamiltonian the energy is conserved. If you're referring to the time-energy uncertainty relation, that's the incorrect interpretation.

Last night I thought of an instance of where a particle can have energy but be unable to do work. This happens in step potentials. For example; consider the potential energy function defined by V(x) = 0 if x < 0, 1 if 0 < x < 1 and 0 for x > 1. If the particle is between 0 and one and not moving then it has a non-zero potential energy and yet it can't do work.

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« Reply #38 on: 11/07/2010 09:15:43 »
In QM the uncertainty relationship means that energy isn't strictly conserved. Perhaps it's better not to include that in a definition. Anyway conservation is a property of energy; not a definition.
That is incorrect. For systems which have a time indepentant Hamiltonian the energy is conserved. If you're referring to the time-energy uncertainty relation, that's the incorrect interpretation.

Well, you have to take the average or expectation value, since any single measurement could violate classical energy conservation...  But that's usually assumed in the definition of energy conservation, so it's fair to say that energy conservation holds in QM.

General relativity is the only place I know of where the concept of energy conservation runs into problems.  Of course, then the definition of gravitational potential energy as well as work done by that energy run into problems as well. 

But doesn't the definition of energy as the capacity to do work rely on energy conservation?  If energy isn't conserved, then some of it won't be available to do work, right?
« Last Edit: 11/07/2010 09:21:38 by JP »

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« Reply #39 on: 11/07/2010 09:44:08 »
Is this scenario "consider the potential energy function defined by V(x) = 0 if x < 0, 1 if 0 < x < 1 and 0 for x > 1. If the particle is between 0 and one and not moving then it has a non-zero potential energy and yet it can't do work." ever realised in practice.
It looks to me  like the idealised "particle in a box" system and if I'm right, then it can't exist. The particle in a box can't be stationary because it would breach the uncertainty principle. It has zero point energy and I answered that case earlier.
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« Reply #40 on: 14/07/2010 23:36:24 »
Both of you are now talking about quantum mechanics. We were discussing quantum mechanics. If you wish to discuss this from a quantum mechanical point of view then "ability to do work" is meaningless since work cannot be defined in quantum mechanics.I mentioned this above but I think it was missed/ignored(?) Some physicists define energy as the generator of time translations in Hamiltonian mechanics.

And I was not thinking of a particle in a box. I had more of a mountain in mind. E.e. picuture a mountain which has a flat top and virtical sides. Then the potential in one dimension is like the one I gave.

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« Reply #41 on: 15/07/2010 01:44:24 »
Both of you are now talking about quantum mechanics. We were discussing quantum mechanics. If you wish to discuss this from a quantum mechanical point of view then "ability to do work" is meaningless since work cannot be defined in quantum mechanics.I mentioned this above but I think it was missed/ignored(?) Some physicists define energy as the generator of time translations in Hamiltonian mechanics.
You can define work in QM if you want, in terms of expectation values, which I also mentioned above.  One of the reasons I like the definition that energy is the ability to "make things change" is the time translation operator definition in QM (and that classical work is also a change).

Quote
And I was not thinking of a particle in a box. I had more of a mountain in mind. E.e. picuture a mountain which has a flat top and virtical sides. Then the potential in one dimension is like the one I gave.
It can do work in that case, but it can't spontaneously do work because it's not rolling off the edge already.  However, it will do work if it's brought to the edge of the mountain and rolled off (it still has the same potential energy right when it starts rolling).

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« Reply #42 on: 15/07/2010 02:28:02 »
Both of you are now talking about quantum mechanics.
That will give you an expectation of work and not work.
It can do work in that case, but it can't spontaneously do work because it's not rolling off the edge already.
All that means is you have to move it which requires changing its velocity and to do that you have to do work on it. So what you're saying is that it will do work if you first do work on it.

This is getting boring and I don't think that anything new will be said after this point so I'm bowing out. Thanks for the conversation. :)

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« Reply #43 on: 15/07/2010 07:06:28 »
"That will give you an expectation of work and not work."
The capacity to do work and the expectation of a capacity to do work are pretty much the same thing.

You are still wrong about the possibility of the system you talked about; it would violate the uncertainty principle.
Do you accept that?
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« Reply #44 on: 15/07/2010 08:03:38 »
You are still wrong about the possibility of the system you talked about; it would violate the uncertainty principle.
Do you accept that?
Accept what? I don't see what you're referring to. What exactly are you claiming is violating the uncertainty principle? I don't see that I've posted anything which would violate the uncertainty principle. As I said above, most people misinterpret the expression dE dt > hbar/2. This is not a true uncertainty relation because it does not relate two uncertainties. dt is not an uncertainty, it is a time interval. To be precise it is the amount of time it takes for a system to change an observable value by one standard deviation. For a precise meaning and derivation please see the derivation I have on my website. This is the same derivation you'd find in a good text on quantum mechanics.

http://home.comcast.net/~peter.m.brown/qm/time_energy_hup.htm

To be clear - It is possible to measure the energy of a particle to arbitrary precision in an arbitrarily short amount of time, a time which is independant of the precision of the energy measurement.
« Last Edit: 15/07/2010 08:17:34 by Pmb »

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« Reply #45 on: 15/07/2010 08:29:15 »
I'm curious about something. It's clear to both of us that we have a fundamental disagreement here and we won't be seeing eye to eye on this. So what is the point of further discussion? Don't get me wrong, I like a good physics discussion. But all we've talked about so far is stuff I've given a very large amount of thought to in the past. It's not like I read a text and simply accept what the author says. Hell! That'd be a weird way to live since different texts contradict each other. Or are you hoping to come up with an argument that I've never thought of before? If so then that may end upp taking months and really? Who wants to bother with such a discussion like that? If you find such a conversation interesting and of value then I'd be glad to continue. But its rather boring to me.

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« Reply #46 on: 15/07/2010 09:26:23 »
Well, you have to take the average or expectation value, ...
since any single measurement could violate classical energy conservation... 
In any case, even if you could define work (which you can't) you'd be unable to use that to determine if the system had energy. E.g. take a free particle in an eigenstate of energy E. The energy is well defined and any measurement of energy gives the value E. But how do you propose to use the "ability to do work" to determine if it has energy? Since there is no force it follows that any attempt to evaluate any value of work (if you could) would yield zero.

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« Reply #47 on: 15/07/2010 10:45:58 »
You could have that particle hit a target, convert it's kinetic energy into some form of energy used by the target, and the move something with that energy.  Assuming no losses, you'd be using the KE of the particle to do work.  As BC pointed out when I raised some points about this definition, you have to assume that you're transferring energy to something at absolute zero (and at an absolute zero of potential energy).

The problem with your free particle and particle-on-a-mountain counterexamples is that they're examples of particles which won't spontaneously do work, but they do have the ability to do work.  In the free particle case, it could give up its KE as mentioned above.  In the particle-on-a-mountain case, you could roll it off the side of the mountain, or just remove the mountain from underneath it if you want).  In either case, the potential energy of the particle is converted into kinetic energy, which is work.

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« Reply #48 on: 15/07/2010 10:49:26 »
It can do work in that case, but it can't spontaneously do work because it's not rolling off the edge already.
All that means is you have to move it which requires changing its velocity and to do that you have to do work on it. So what you're saying is that it will do work if you first do work on it.

The requirement of having to first move the particle to the edge of the cliff has nothing to do with BC's definition.  The potential energy of the particle can be used to do work, as per his definition.  How you get it do do so is irrelevant.

Having said all this in defense of his definition, I still don't think it's ideal as a definition.  It's definitely a quality of energy, but it doesn't capture the other properties of energy, especially in models outside of classical mechanics.

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« Reply #49 on: 15/07/2010 10:56:40 »
I'm not sure how to say this so it comes across as I mean it, i.e. politely, so I'll just say it - I think at this point any posts that I might submit for reading would merely be repeating, in one form or another, what I've already said. I pretty much made my thoughts clear and, of course, if someone really wants to add something which they think is important for me to know, please feel free to PM it to me. If you think its that important then I'd love to read it. However it has to be in PM since I won't be peaking in on this thread again.

I'd like to thank everyone for sharing their thoughts with me - Thanks! :)

Best wishes

Pete