Is conservation of energy derived from the work energy theorem?

[etotheipi]

To my mind, there are two distinct approaches to energy problems that different authors tend to use, and I wondered whether either is more fundamental than the other. The first is variations on the work energy theorem, and the second consists of defining a system boundary and setting the change in total energy of that system (kinetic, potential, thermal) to the external work done on all components of the system.

I thought about it and came up with the following. For any particle or rigid body in a system of particles or rigid bodies respectively, we can write down a work-energy equation (which follows directly from Newton II) $$\sum_{i} W_{i}= \Delta T$$ Then, we sum all of these equations for each particle in the system. For any overlapping terms due to inter-particle interactions, we can replace it with a potential energy, $W_{jk} + W_{kj} = \Delta U_{jk}$. The sum of all of the changes in kinetic energy results in the total change in kinetic energy, which can be decomposed into macroscopic and microscopic terms. This would lead to a final equation, $$W_{ext} = \Delta T + \Delta U + \Delta E_{th} = \Delta E$$which is what I often see written for conservation of energy. For instance, if we take the example of a block being pulled up a rough plank by a rope, it appears there are two ways of looking at it

1. Direct application of the WEP: $W_{tension} + W_{friction} + W_{weight} = \Delta T \implies W_{tension} + W_{friction} = \Delta U_{grav} + \Delta T$
2. Defining a system to be the block and the Earth, thus defining $E = U_{grav} + T$ and letting the external work be $W_{tension} + W_{friction}$.

The approaches are so similar, the only difference being that in (1) we consider only the block and then substitute $W_{weight} = -\Delta U_{grav}$, whilst in (2) we set out all of the energies in our defined system - which also includes the Earth.

I've read that energy conservation can be derived more fundamentally via Noether's theorem, however wondered whether the the ideas set out above are correct?

 

[Dale]

The work energy theorem is very limited and does not actually describe the energy entering or leaving an object.

 

[etotheipi]

The work energy theorem is very limited and does not actually describe the energy entering or leaving an object.

That's one problem. The approach of summing the work energy equations for all of the particles in the system seems to work fine (e.g. for lots of types of mechanics problems), generating all the correct equations for COE, until we require the transfer of heat.

Then I think, like @kuruman mentioned in his insights article, we'd need to appeal to the first law of thermodynamics as a different way in.

So is it the case that the WEP is equally as, but no more, fundamental as the law of COE?

And are there any other ways one can arrive at $W_{ext} = \Delta E$?

 

[Dale]

There isn’t a non-subjective measure of “fundamental-ness” but subjectively I would rate the fundamental-ness of the work energy theorem as less fundamental than most things. Far less than conservation of energy, maybe more fundamental than Ohm’s law or something similar, but not much more.

 

[etotheipi]

There isn’t a non-subjective measure of “fundamental-ness” but subjectively I would rate the fundamental-ness of the work energy theorem as less fundamental than most things. Far less than conservation of energy, maybe more fundamental than Ohm’s law or something similar, but not much more.

That's interesting; quite a few undergraduate textbooks (I'm looking at Kleppner and Kolenkow) use the WEP in order to define $T + U$ as a conserved quantity.

Though there do seem to be problems with this approach; like you alluded to, it doesn't include anything about heat.

And furthermore, a critical step is equating $W_{c} = -\Delta U$; however when applied to only one body, this is only valid in the limit that the source particle is stationary!

 

[Dale]

Not only does it not include heat it doesn’t include any form of internal energy. It doesn’t even work right for a spring.

 

[vanhees71]

The work-energy theorem holds for all forces and is a integral form of Newton's equation of motion:
$$m \ddot{\vec{x}}=\vec{F}(t,\vec{x}).$$
Let $\vec{x}(t)$ a solution of this equation of motion (the trajectory of the particle). Then multiply the equation of motion with $\dot{\vec{x}}$ and integrate. On the left-hand side you can use
$$m \ddot{\vec{x}} \cdot \dot{\vec{x}}=\mathrm{d}_t \left (\frac{m}{2} \dot {\vec{x}}^2 \right)=\frac{\mathrm{d} T}{\mathrm{d} t},$$
Where $T=m \dot{\vec{x}}^2/2$ is the kinetic energy. Thus you get by integrating between two times
$$T_2-T_1=\int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}}(t) \cdot \vec{F}[t,\vec{x}(t)]=\Delta W_{21}.$$
Note that you need the trajectory in this case and the integral $\Delta W_{21}$ has to be taken along this specific path.

The energy-conservation law holds if the forces are conservative, i.e., if they are not explicitly time dependent and derivable from a scalar potential,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Then the integral $\Delta W_{21}$ becomes a line integral along the trajectory of the particle, but since $\vec{F}$ is conservative it does not depend on this path but only on the initial and final point of the trajectory:
$$\Delta W_{21}=-[V(\vec{x}_2)-V(\vec{x}_1)].$$
Then you have a well-defined total energy,
$$E=T+V,$$
and
$$E_2-E_1=T_2+V_2 -(T_1+V_1)=0,$$
i.e., the energy is conserved along the trajectory of the particle.

 

[etotheipi]

Not only does it not include heat it doesn’t include any form of internal energy. It doesn’t even work right for a spring.

I'm not too clear on what you mean by not working for a spring. Suppose one end is connected to a wall, the other end being free, and we compress it through some displacement.

If we take centre-of-mass work, then the total work is zero and the centre of mass has no velocity.

If we instead sum the total work, this is evidently non-zero. However, it will just manifest itself as the internal energy of the particles in the spring. So the relation still appears to be valid. We could surely write a work energy relation for every particle in the spring and sum the results.

 

[Dale]

The work-energy theorem holds for all forces

The work energy theorem holds for all forces but it does not describe energy transfer for systems with any form of internal energy. I.e. the “work done by the net force” is the change in KE not the change in energy.

 

[etotheipi]

The work energy theorem holds for all forces but it does not describe energy transfer for systems with any form of internal energy. I.e. the “work done by the net force” is the change in KE not the change in energy.

But isn't that essentially all that it's claiming to do, the sum of the work done by all forces acting on a body equals the change in kinetic energy of that body, which we can again decompose however we like (e.g. a centre of mass kinetic energy + a microscopic kinetic energy term).

 

[Dale]

If we instead sum the total work, this is evidently non-zero. However, it will just manifest itself as the internal energy of the particles in the spring.

This part is entirely outside the scope of the work energy theorem, but this is what is relevant for the conservation of energy.

But isn't that essentially all that it's claiming to do, the sum of the work done by all forces acting on a body equals the change in kinetic energy of that body, which we can again decompose however we like (e.g. a centre of mass kinetic energy + a microscopic kinetic energy term).

Kinetic energy alone, even including microscopic, is insufficient to explain the conservation of energy in even something as simple as a spring.

 

[etotheipi]

This part is entirely outside the scope of the work energy theorem, but this is what is relevant for the conservation of energy.

Why so? For any particle in the spring I can write

$\sum W_i = \Delta T_i$

I can sum all of these equations,

$\sum W_{i,j} = \sum \Delta T_i = \Delta T$

I can now decompose $T$ into the centre of mass kinetic energy plus the random thermal kinetic energy relative to the centre of mass.

 

[vanhees71]

The work energy theorem holds for all forces but it does not describe energy transfer for systems with any form of internal energy. I.e. the “work done by the net force” is the change in KE not the change in energy.

Of course, if you apply it to the case of a single particle subject to an "external force" that's true. If you describe a closed system with conservative forces then the energy transfer between different parts of the system is included.

Of course often you also have a thermodynamical description of parts of the energy including all kinds of heat transfer and dissipation. Then again the mechanical part is described as an open system.

 

[Dale]

Why so? For any particle in the spring I can write

$\sum W_i = \Delta T_i$

I can sum all of these equations,

$\sum W_{i,j} = \sum \Delta T_i = \Delta T$

I can now decompose $T$ into the centre of mass kinetic energy plus the random thermal kinetic energy relative to the centre of mass.

You are missing the point. You can decompose $T$ thus, but $T\ne E$. And thermal energy is not the sum of internal kinetic energy, many of the internal degrees of freedom for thermal energy include potential energy as well as kinetic.

Furthermore, since you are free to define your system as you desire for the conservation of energy and you would have to apply the work energy theorem at the individual particle level to even get $T$, it is difficult to see the applicability.

 

[etotheipi]

You are missing the point. You can decompose $T$ thus, but $T\ne E$. And thermal energy is not the sum of internal kinetic energy, many of the internal degrees of freedom for thermal energy include potential energy as well as kinetic.

Furthermore, since you are free to define your system as you desire for the conservation of energy and you would have to apply the work energy theorem at the individual particle level to even get $T$, it is difficult to see the applicability.

Sure. Though I think it makes sense to say that the work-energy theorem applies always to particles, and by extension compound systems so long as we are meticulous enough in bookkeeping (which is often not physically possible). Since we can always apply it to any particle, and then add up all of these contributions.

For particles and rigid-bodies it seems to always work fine.

But perhaps as you say it is untenable to apply it to deformable bodies, precisely because of thermal energy considerations. In these cases, I assume the only way we can work is via the general statement of conservation of energy - once we have defined clearly our system boundary:

$W_{ext} + Q = \Delta E$

where $E$ is the sum of kinetic energies, internal potential energies, thermal energies etc. within the system boundaries. This doesn't appear to be derivable from the work energy theorem. In that case, is the relation above purely empirical?

 

[Dale]

If you describe a closed system with conservative forces then the energy transfer between different parts of the system is included.

I am not sure that is true. Consider a two-particle spring consisting of a left and right particle of mass $m$ with a typical quadratic potential between them with a minimum at some distance $r$. Suppose a pair of equal and opposite external forces slowly compresses the system by a distance $\Delta r$ so that there is a change in potential energy $\Delta U$.

Applying the work energy theorem at the system level: the net force is 0 so the “net work“ is 0. This correctly accounts for the macroscopic kinetic energy but fails to account for the internal potential energy.

Applying the work energy theorem at the particle level: each particle experiences an external force and an internal force. Because the compression is done slowly those forces are approximately equal. So the net force on each particle is again zero as is the “net work” on each particle. As before, this correctly accounts for the microscopic kinetic energy but fails to account for the potential energy.

 

[vanhees71]

Of course, if you have external forces the system is not closed anymore. The two particles connected by a spring of course are. It's described by the Lagrangian
$$L=\frac{m_1}{2} \dot{\vec{x}}_1^2 + \frac{m_2}{2} \dot{\vec{x}}_2^2 -\frac{k}{2} (\vec{x}_1-\vec{x}_2)^2.$$
This obeys all 10 conservation laws and is thus a closed system.

External forces could be described by adding something like $-V_1(\vec{x}_1)-V_2(\vec{x}_2)$, which violates (at least) translational invariance and thus momentum is not conserved. It's then clearly an open system.

 

[Dale]

However you want to characterize the system, my point is that the work energy theorem fails to account for the change in internal potential energy at both a macroscopic and a microscopic level.

 

[etotheipi]

However you want to characterize the system, my point is that the work energy theorem fails to account for the change in internal potential energy at both a macroscopic and a microscopic level.

However, doesn't the work energy theorem only account for kinetic energy?

We could always then set $W_c = -\Delta U$ to get out potential energy terms.

 

[vanhees71]

However you want to characterize the system, my point is that the work energy theorem fails to account for the change in internal potential energy at both a macroscopic and a microscopic level.

Can you give an example? I obviously don't understand your point.

 

[etotheipi]

@Dale Perhaps this is just a matter of abstract vs operative definitions?

I'd say that for any system, i.e. a particle, a deformable body, a collection of rigid bodies etc., the total work done by all forces (internal/external/conservative/non-conservative) equals the total change in kinetic energy. This must be true since we can simply apply the work energy theorem to each particle and sum all of them.

The only assumption is that there are no other means of transferring energy into that system, however most of the time in mechanics anyway, this is a given.

However, this is perhaps not an operative definition in lots of cases, especially when thermal energy is involved. If we have a compressed spring on a table and release it so that it jumps up, there are a few ways of looking at it:

1. System: Spring-Earth. A closed system, the total energy (kinetic/elastic/gravitational) is conserved since no external work is done.
2. System: Just Spring. The Earth exerts an upward force, however this does no work. However, the internal forces in the spring still do work on the spring and this causes a change in kinetic energy. Work-energy theorem is theoretically satisfied.
3. System: Just Spring (again). Non-zero centre-of-mass work done on the spring, so the velocity of the centre of mass is also increased (I believe Prof. Sherwood wrote an article about this)

However you want to characterize the system, my point is that the work energy theorem fails to account for the change in internal potential energy at both a macroscopic and a microscopic level.

Consequently, so long as we also include the work done by the forces which cause these changes in internal potential energy in our hypothetical work energy equation, everything will work (no pun intended...)

 

[jbriggs444]

I'd say that for any system, i.e. a particle, a deformable body, a collection of rigid bodies etc., the total work done by all forces (internal/external/conservative/non-conservative) equals the total change in kinetic energy. This must be true since we can simply apply the work energy theorem to each particle and sum all of them.

Only if you insist that deformable bodies are further decomposable into pointlike particles or rigid bodies.

 

[etotheipi]

Only if you insist that deformable bodies are further decomposable into pointlike particles or rigid bodies.

Ah; perhaps rashly I assumed that was always the case. Are there instances when deformable bodies are not modeled as systems of particles in Physics?

 

[jbriggs444]

Ah; perhaps rashly I assumed that was always the case. Are there instances when deformable bodies are not modeled as systems of particles in Physics?

Sure. Any time you have a collision with a coefficient of restitution less than 1.

And most of the time, even when the coefficient of restitution is equal to 1.

 

[etotheipi]

Sure. Any time you have a collision with a coefficient of restitution less than 1.

I thought if we had two deformable bodies, then whatever we lose as KE manifests itself as internal energy of the colliding bodies. But this internal energy is itself of the form of microscopic kinetic and potential energies, which must be due to particles that make it up.

 

[jbriggs444]

I thought if we had two deformable bodies, then whatever we lose as KE manifests itself as internal energy of the colliding bodies. But this internal energy is itself of the form of microscopic kinetic and potential energies, which must be due to particles that make it up.

You asked about models. Have you never ever in your life ever encountered a physics problem in which a ball was bounced from a wall? Seriously?

Do you seriously insist that each time such a problem is encountered that we should attack it by modelling the trajectories of 6.02 x 10²³ constituent particles?

Finite element analysis does not usually go down to that level of granularity.

 

[etotheipi]

You asked about models. Have you never ever in your life ever encountered a physics problem in which a ball was bounced from a wall? Seriously?

Do you seriously insist that each time such a problem is encountered that we should attack it by modelling the trajectories of 6.02 x 10²³ constituent particles?

Finite element analysis does not usually go down to that level of granularity.

Ah not at all, I'm speaking on a purely conceptual basis. Saying that one could theoretically measure the work done on all particles of a rigid body and hence determine the change in its KE is not to say that in practice one won't just use measures like internal energy (temperature) etc.

I'm simply making the case that the work-energy theorem applies always, to all systems. Whether or not it is useful is a different question.

 

[jbriggs444]

Saying that one could theoretically measure the work done on all particles of a rigid body and hence determine the change in its KE is not to say that in practice one won't just use measures like internal energy (temperature) etc.

If you adopt the Newtonian model, complete with instantaneous action at a distance, then I think that what you are talking about is sensible. That one could define a measure of the total "kinetic energy of all of the particles in a collection" and associate it with the total work done by all internal and external forces on particles in the collection. It's pretty much true by definition. As you say, the work energy theorem applied at the level of granularity of point particles guarantees it.

The digression we are on here started in #21 when you spoke of deformable bodies. The above model does not deal with deformable bodies. It deals with point particles.

 

[etotheipi]

The digression we are on here started in #21 when you spoke of deformable bodies. The above model does not deal with deformable bodies. It deals with point particles.

Though to my understanding, a rigid body is a system of particles whose relative distances are fixed, and a deformable body is a system of particles whose relative distances are not fixed. I might well be wrong, I haven't studied this in great detail! `But if the above is true, then I don't see why we can't treat a deformable body as a system of point particles.

I might be totally off in these definitions, however! I'm using the word deformable a little like how we would describe a gas in thermodynamics.

 

[jbriggs444]

Though to my understanding, a rigid body is a system of particles whose relative distances are fixed, and a deformable body is a system of particles whose relative distances are not fixed. I might well be wrong, I haven't studied this in great detail! `But if the above is true, then I don't see why we can't treat a deformable body as a system of point particles.

I might be totally off in these definitions, however! I'm using the word deformable a little like how we would describe a gas in thermodynamics.

A deformable body would include, for instance, a spring. Which can contain energy other than kinetic.

 

[Dale]

However, doesn't the work energy theorem only account for kinetic energy?

Yes. That is precisely the issue with treating it as something fundamental and trying to derive conservation of energy from it.

We could always then set $W_c = -\Delta U$ to get out potential energy terms.

Certainly, but then you are no longer using the work energy theorem as the fundamental rule to derive the conservation of energy. I understood that was the idea of your OP, correct?

 

[etotheipi]

A deformable body would include, for instance, a spring. Which can contain energy other than kinetic.

But in the same vein as earlier, we might conceptually consider the work done on each particle in the spring individually by internal forces between other particles, external forces etc.

This would give us our hypothetical statement of the work energy theorem for a spring. Except the conservative work terms due to the interactions between the particles which produced the "springiness" we replace with negative change in potential energy terms.

For instance, we might have two point masses attracting each other gravitationally as our system, $m_1$ and $m_2$. The system as a whole, we might think of as a deformable body. For $m_1$, $W_{12} = \Delta T_{1}$, and for $m_2$, $W_{21} = \Delta T_{2}$. Then if we sum those, since $W_{12} + W_{21} = -\Delta GPE$,

$W_{21} + W_{12} = \Delta T_1 + \Delta T_2 = -\Delta GPE \implies \Delta T_1 + \Delta T_2 + \Delta GPE = 0$

 

[etotheipi]

Certainly, but then you are no longer using the work energy theorem as the fundamental rule to derive the conservation of energy. I understood that was the idea of your OP, correct?

Ah, okay. I suppose my original question was a little misleading. Apologies!

 

[jbriggs444]

But in the same vein as earlier, we might conceptually consider the work done on each particle in the spring individually by internal forces between other particles, external forces etc.

Right. But then it is no longer being treated as a deformable body!

The point of calling it a "deformable body" and modelling it as such is that it becomes a black box that you are no longer allowed to peek inside. If you have to peek inside to see the particles and ignore the interaction forces between them, the box is no longer black. You are not treating it as a deformable body any more.

 

[etotheipi]

Right. But then it is no longer being treated as a deformable body!

The point of calling it a "deformable body" and modelling it as such is that it becomes a black box that you are no longer allowed to peek inside. If you have to peek inside to see the particles and ignore the interaction forces between them, the box is no longer black. You are not treating it as a deformable body any more.

Right, I now understand. Thank you for your patience!