# Can Gravitational energy be derived from First Law of Thermodynamics?

1. Dec 12, 2013

### Ethan0718

I'm not sure if I'm right ..

Here's my thought:

1. The First Law of Thermodynamics is the only single law which could derive energy formula. Furthermore, we can't, solely, use Newton's law to derive Kinetic Energy. It will make us fail to explain the existence of thermal energy.(*)

2. In order to derive energy formula, we must define or determine our system. Without system, there's no meaning of energy.

3. Now, I would like to derive three kinds of common energy, especially, the gravitational potential energy:
(a)Kinetic energy of a mass point
(b)Elastic energy of an ideal spring (mass→0)
(c)Finally, Gravitational potential energy of the gravitational field

4. As you will see, we can derive any energy term of any kinds of system by the first law of thermodynamics.

Kinetic Energy of a mass point

∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)

System : a mass point

Assumption:
(a)Q=0
(b)observer is in the inertial frame so that NSL is valid.

Process: it's acted by so many forces and .. moves!

∴ΔE = ƩFi,xΔxi + Q

∴ΔE = ƩFi,xΔxi = maxΔx

∴ΔE = Δ($\frac{1}{2}mv^{2}$)

∴E$\;\equiv\;\frac{1}{2}mv^{2}+C$

For convenience, we let C equal zero so that E = $\frac{1}{2}mv^{2}$​

-----

Elastic Energy of an ideal spring

∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)

System: an ideal spring (m→0)

Assumption:
(a)Q=0
(b)the left end is connected with wall so that this end is fixed and cannot be moved.
(c)the origin of coordinate is at right end when length of the spring is original length.

Process: pull the spring, and it's stretched from (x1+L0) to (x2+ L0)

∴ΔE = Wleft end + Wright end + Q

∴ΔE = 0 + ∫(+kx)dx + 0
(integration interval: x1→x2. Drawing a FBD of the spring can help you understand why it is positive sign before kx)

∴ΔE = $\frac{1}{2}kx^{2}_{2}-\frac{1}{2}kx^{2}_{1}$

∴E$\;\equiv\;\frac{1}{2}kx^{2}+C$

For convenience, we let C equal zero so that E = $\frac{1}{2}kx^{2}$​

-----

Gravitational Potential Energy of the gravitational field

∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)

System: the gravitational field

Assumption:
(a)Q=0
(b)the positive y-axis direction is upward.

Process:
A ball is falling down from yinitial to yfinal

∴ΔE = Wball + Wearth
(the ball and earth are called surroundings)

∴ΔE = (+mg)(Δyball) + (-mg)(Δyearth)

(+mg): the force between ball and the gravitational field.
It's acted by ball, and, according to Newton's Third Law, acted on gravitational field.
It's upward because the ball is pulled downward by gravitational field.
(-mg):the force between earth and the gravitational field.
It's acted by earth, and, according to Newton's Third Law, acted on gravitational field.
It's downward because earth is pulled upward by gravitational field

P.S. You can draw the Free Body Diagram of Gravitational Field to understand what I'm talking about.. I know it's a little bit weird but I think it's plausible.

∴ΔE = (+mg)(yf - yi) + (-mg)(0)

∴ΔE = mgyf - mgyi

∴E$\;\equiv\;mgy+C$

For convenience, we let C equal zero so that E = mgy

(*) See "work and heat transfer in the presence of sliding friction" by Bruce Arne Sherwood.
-----

Am I correct?

Thank you very much.... This problem confuses me so much...

Last edited: Dec 12, 2013
2. Dec 12, 2013

### voko

You do not really derive anything from the first law of thermodynamics. You use that as a definition: energy = work.

Curiously enough, you do not define what work is explicitly anywhere, you just implicitly assume that work = force times distance. That is not contained in the first law of thermodynamics.

Finally, you use Newton's laws for the core of your derivation, despite your claim that they cannot be used as such.

All I can say to this is that you are very, very confused.

You do not need thermodynamics to deal with potential energy. The existence of potential energy follows from a property of force, namely that its work over a closed path is zero.

3. Dec 12, 2013

### Ethan0718

The definition of work is that the inner product of force and the displacement of the action point.
I use it and the first law of thermodynamics to derive every energy term.

You told me I just defined energy as work. That's not what I meant. I just assumed the heat transfer is zero. This assumption does not equivalent to the definition that "energy is work". At least, my assumption just follows that the "delta E" equals the work done on system by surroundings.

Perhaps, the definition of work does not contained in first law of thermodynamics. However, it does not contained in Newton's law, either. It seems not compelling to argue, implicitly, we can use, solely, Newton's law to derive ANY energy formula without the first law of thermodynamics although we need it to get some energy formula, like kinetic energy.

If we do not need the first law of thermodynamics to deal with ANY energy formula, then it follows a serious question : Is there any other physics law which we can use to derive the formula of energy? Newton's law only deals with center of mass term. The common work is called pseudo-work. I recommend you read this paper, "work and heat transfer in the presence of sliding friction" by Bruce Arne Sherwood.

Finally, I know what you mean by path integral. Every textbook use this to demonstrate the potential energy (especially electric potential energy). I'm trying to explain this by the First Law of Thermodynamics(F.L.T.). I need some time. However, The uncertainty about how to explain with F.L.T. does not mean my thought is wrong. Furthermore, what if we ask you the reason why "path integral" has something to do "the concept of energy"? Could you explain it without the first law of thermodynamics?

Last edited: Dec 12, 2013
4. Dec 12, 2013

### voko

Again. You cannot derive "potential energy" from the FLT. The FLT needs that "energy" and "work" be defined just so that it could be stated. "Potential energy" and "work" are trivially related by a path integral. For any force with path-independent work you can prove conservation of energy using Newton's laws. You do not need thermodynamics for any of that.

5. Dec 12, 2013

### Ethan0718

Why cannot I derive "potential energy" from the FLT?
Why cannot I define energy just solely by FLT?
How do you explain why we can prove conservation of energy solely using Newton's law?
Why don't I need thermodynamics for any of that?

Is it dogmatism?

Furthermore, how do you define "energy" and "work" when you use solely Newton's law to prove conservation of energy?

Can you explain the heat transfer in the presence of sliding friction without FLT?
If you can't do that without FLT., then you need to admit the importance of FLT in deriving energy term.

Although I don't think it's good time to discuss about what energy is and it seems like the only way you say I'm wrong is saying that I need to define what energy and work exactly mean, I think I, perhaps, need to say something about the definition of energy.

The definition of energy lies in FLT. What's energy?
Energy is the "E" in FLT! It's something that exists in every system we defined.
The only way to transfer energy between systems is work and heat transfer.

I don't know what kind of definition you want. Most of my thought comes from the feynman's block. You can see his lectures.

Last edited: Dec 12, 2013
6. Dec 12, 2013

### voko

The FLT requires that the terms "work" and "energy" be defined, because it uses them. What are your definitions of them?

7. Dec 12, 2013

### Ethan0718

The definition of energy lies in FLT.
Energy is the "E" in FLT. It's something that exists in every system we defined. (That's how we understand "E" term")
The only way to transfer energy between systems is work and heat transfer.

I don't know what kind of definition you want. Most of my thought comes from the feynman's block. You can see his lectures.

I'm not sure if I understand what you're saying.
You say the reason why we need to define energy before we use FLT is "FLT uses them".

Suppose you're right, then it follows that:

I need to define what words I'm using to talk to you because I'm using them.
Is this correct? And is this workable?

I say this apple is red. Can I define what red is?
Maybe you think this question does not make sense. Then, what about your question?
Is your question plausible? I think all of them are huge philosophical problems.

But, the more important thing is this:
As far as what you've said, the only way you say I'm wrong is give me the question which I don't think I can answer it properly. This does not really mean my thought is more possibly wrong than your thought. According to your criterion, you also need to define energy and work before you use newton's law.

If you think you don't need to define it just because you don't use "energy" in Newton's law, then........
How could you derive energy formula, or even prove the conservation of energy by Newton's law?

Last edited: Dec 12, 2013
8. Dec 12, 2013

### ShayanJ

How did you jump from $\Delta E=ma_x\Delta x$ to $\Delta E=\Delta(\frac{1}{2}mv^2)$?
Here you're assuming work=Force x distance,that's an assumption in addition to FLT!
Also you've forgot a minus sign!

9. Dec 12, 2013

### Ethan0718

Use the kinematic formula

$V^{2}_{f,x}=V^{2}_{i,x}+2a_{x}Δx$

You're right, I need to delete the minus sign in the integral. It's my typo. You can understand why I do this if you draw a free body diagram of the spring whose left end is tied with wall and right end is pulled by my hand.

And I agree with you. The definition work is not part of FLT.

Last edited: Dec 12, 2013
10. Dec 12, 2013

### ShayanJ

The spring force always opposes the change in length.So it should be negative.I meant the you should have written $\Delta E=\frac{1}{2}k(x_1^2-x_2^2)$

Anyway,in Newtonian mechanics the usual way is assuming Newton's second law and derive work-kinetic energy theorem and then it can be used to derive the conservation of mechanical energy.What you're doing,is in fact doing the reverse.You're assuming conservation of mechanical energy and using it to obtain other things.I think you'll have problem with this approach when you reach non-conservative forces.But the usual method can be used in those cases.You're just giving a reformulation which is a draw back!!!
FLT is used in thermodynamics because we have to.But in mechanics,we have Newton's law which is working well so its of no point to use FLT in mechanics!

11. Dec 12, 2013

### voko

He introduces the FLT in Vol I Chapter 44. He introduces "energy" in Vol I Chapter 4. Vol I Chapter 13 explains how "energy" and "potential energy" are related Newton's laws. Perhaps you should go back and review that.

In physics and science, yes. You and i have to know exactly what we mean. It would also help in our lives generally, but, unfortunately, that's not going to happen so we are going to stay in the permanent state of confusion.

No.

Newton's laws are not formulated in terms of energy. They are formulated in terms of force and momentum.

Because we can define "energy" and "work" in addition to Newton's laws, and then, under some additional conditions (conservative force or, more generally, homogeneity of time) prove conservation of energy.

12. Dec 12, 2013

### Ethan0718

1. The "F" in this formula, F=-kx, is acted on the "objects" tied with the ends of spring, instead of being acted on "the spring itself".
2. What I'm analyzing is spring itself, rather than the objects(my hand) tied with the right end of spring.
3. Total nergy (not mechanical energy only) is conserved in this universe according to FLT.
4. I won't have problem with this approach when I reach non-conservative forces, like frictional forces.

Suppose a block is dragged at a constant speed across a table with friction. The applied force f through a distance d does an amount of work fd. And also the friction, f, does the same work fd.

(in my opinion, it's nothing but the special case of single mass point system.)
(The result of FTL application in multiple mass points system is not just kinetic energy (of center of mass).)

$W_{total,ext}=W_{applied}+W_{friction}$

$W_{total,ext} = (+fd) + (-fd) = 0$

$W_{total,ext} = ΔE_{k}$

$0=ΔE_{k}$

That means there's no change in speed. That is consistent with Newton's second law and everything seems to be correct. However, here's the problem.

Where is the internal energy term representing the increased thermal energy of the block? The block warms up due to the friction. Is the way of explaining it just saying that "It warms up, indeed. And it's the only one explanation about the internal energy"?

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13. Dec 12, 2013

### ShayanJ

That's not a paradox.We just don't calculate the amount of heating because we're not interested in it!
Whenever that is needed to be calculated,we can do it by considering the decrease in mechanical energy!

I'm telling it again,you're just giving a reformulation of mechanics which is going to offer nothing new.Maybe it offers less too!

14. Dec 12, 2013

### Ethan0718

Yes, Feynman's block analogy is introduced in Chapter 4. Vol I. I think reading the first paragraph may help you understand what energy is.
You can also read this book: Energy, the Subtle Concept: The discovery of Feynman's blocks from Leibniz to Einstein, by jennifer coopersmith.

I don't know how to give you the definition you want. All I can say is this:
We define the change of energy of the system in the following equation:
ΔE=W+Q

Last edited: Dec 12, 2013
15. Dec 12, 2013

### Ethan0718

"Work and heat transfer in the presence of sliding friction" by Bruce Arne Sherwood
"Physics that textbook writers usually get wrong" by Robert P. Bauman

It will take me so much time to clearly explain why we need FLT to explain the increased thermal energy and NSL cannot explain it. If you insist that I should clearly explain it now, then please tell me and I'll try to show that.

I know what you mean by calculating it with the decreased mechanical energy. However, this cannot make your "Work-kinetic theorem" more plausible because the energy of that block is just "kinetic energy" by your theorem.

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16. Dec 12, 2013

### ShayanJ

I'll try to read those papers.
But there are two points:
1-In the case of the block moving on a surface with constant velocity.Let the mass of the block be m and the friction coefficient be $\mu$.The power dissipated by friction is $P=Fv=\mu m g v$ and so the energy converted to the heat in time $\Delta t$ is $E=\mu m g v \Delta t$!
2-The work-kinetic energy theorem doesn't make sense here because in Newtonian mechanics we take the objects we're studying as a single entity,sometimes as particles and that for such things,heat means only kinetic energy which is considered.Sometimes we use these theorems for extended objects too but that's just because in those cases we don't care about the heat produced and its enough for us that they're correct in calculating dynamical variables.

17. Dec 12, 2013

### Ethan0718

Thanks for trying to read those papers.

Your method has been discussed in this paper:
Thermodynamics of a Block sliding Across a Frictional Surface by Carl E. Mungan
You can also calculate the entropy in that process.

At my first glance, I wonder whether the kinetic energy of center of mass is what we usually want or not.
Afterwards, I think it is. So I agree with you.
Finally, just like what you've said, we can't use, solely, NSL to explain the energy transfer in the presence of sliding friction. The only way to explain it is using the first law of thermodynamics.

PS.
$E=\mu m g v \Delta t$ is not the heat transfer from ground to block because causing ground colder and block hotter is, obviously, impossible.

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18. Dec 13, 2013

### voko

With this sort of attitude toward people trying to help you are bound to get stuck with your misconceptions. Good luck.

19. Dec 13, 2013

### Ethan0718

All what you've said is, without telling me your reasons which are based on physics law, not your intuition, I'm wrong and very, very confused.
The final question you put toward me is what the definition of energy is.
(Did I miss some questions you've asked me?)
I've tried to tell you my answer with their sources: feynman's lectures.
If you think you're trying to help me with this sort of attitude, then you must have not help me at all.

According to your "help", you might just be a dogmatist.
(Do you really know how to judge if you're helping people or not?)

Or, all you can do is just saying you don't like me.
It's not compelling to argue that you're totally right and I'm totally wrong and very confused.

Last edited: Dec 13, 2013
20. Dec 13, 2013

### ShayanJ

Well,all you're telling is that we can't use Newton's second law in thermodynamics and using the first law of thermodynamics in thermodynamics,is inescapable which I think all of us agree!

A.P.S.
I didn't say that's the heat transferred from ground to block!!!I said that's the heat produced and that comes from the energy we're giving to block to keep it going with constant velocity.That's the energy we're spending for counteracting the friction which converts to heat!