How do we justify that work is a "transfer of energy" in the general case?

Question

By the work-energy theorem, we can justify that the work on a particle due to the net force equals the change in kinetic energy of the particle. In compact notation, \begin{align}\tag{1} W_{\text{net}} = \Delta KE. \end{align} This seems very useful. However, there are other contexts where work is used.

For example, we might want to find the potential energy of a certain configuration of charges in electrostatics. In that case, we imagine bringing charges from infinity together quasi-statically. By calculating the work needed to assemble the charge configuration, we say we have found the potential energy of the setup: $W = \Delta U$. But this time, we assume there is no kinetic energy is involved, because the process was quasi-static.

A more simple example is the case of trying to find gravitational potential energy, where we consider what it takes to lift something up quasi-statically (which can be done by applying a constant force $\vec{F} = mg\vec{e}_{z}$ to counteract $\vec{F}_{g}=-mg\vec{e}_{z}$), compute $W = \int_{0}^{h} mg\, dz = mgh$, and find $\Delta U = mgh$.

So now we might want to write \begin{align}\tag{2} W = \Delta KE + \Delta U. \end{align} I am aware we are no longer considering net work (which was crucial to the initial formulation of the work-energy theorem).

To make this more general, we might also consider internal energy (which will take into account thermal energy). So for example, maybe we have a system in which friction occurs internally, so there is work done by friction which "siphons" off energy into heat. In that case, if we define our system carefully, the work produced by a force external to the system will result in \begin{align}\tag{3} W_{\text{ext}} = \Delta KE + \Delta U + \Delta E_{\text{int}}. \end{align}

My question is, are there theorems or (semi-)rigorous arguments that demonstrate the validity of equations $(2)$ and $(3)$? I suppose $(2)$ might be trivial if we qualify that the forces involved are conservative (so by definition we have $\vec{F} = -\nabla U$), but it's not clear exactly what rigorous argument can be made. And what about equation $(3)$? How is that rigorously justified?

Some related questions:

• How does the work-energy theorem relate to the first law of thermodynamics?
• Does work-energy theorem involve potentials?
• Does work-energy theorem account for thermal energy?

Answer 1

Sorry, but the scheme you are looking for turns out to be a dead-end.

We can start with the celebrated work-energy theorem. You will notice that it is always missing from a discussion of relativity, because whenever a relativity textbook author wants to establish work-energy theorem in the context of relativity, he or she will quickly come to the conclusion that the whole thing is WET (pardon the joke). There is simply no acceptable equivalent of work-energy theorem in SR, because, while you definitely can get the correct relations, it is more a case of ``Einstein's energy-momentum relations are the unique solution to the work-energy integral".

Instead, your Equation (3) is really part of the statement of the $1^\text{st}$ Law of Thermodynamics (which is really better stated as ``Energy is conserved, and heat is a form of energy").

Considering modern physics, with the central importance of Noether's theorem and its wide-ranging implications, it is much better to start the opposite direction: We assume interacting Lagrangians so that Poincaré invariance is manifest, and by Noether's theorem we essentially get energy and momentum conservation for free. Looking at it another way, what we are doing, is that we are acknowledging that energy and momentum conservation is the one most foundational and well-verified and sturdy bedrock (i.e. best to be picked to comprise the postulates), from which the rest of physics is meant to follow. And then we assume half of the $2^\text{nd}$ Law of Thermodynamics, with a specific form of entropy (I like Gibbs's version), and the rest of modern classical physics follows. This is the scheme that works well with quantum theory and is the viewpoint of standard physics today.

---

Another annoying thing: Work is actually not well-defined in the thermodynamic sense, especially if you have magnetic stuff. All the more reason to try and avoid both horrible concepts of heat and work, and instead work with well-established notions, and simply assume the validity of energy and momentum conservation.

Answer 2

But this time, we assume there is no kinetic energy is involved, because the process was quasi-static.

Actually, we say there is no change in kinetic energy because the net work done is zero in bringing the charges together. To bring like charges closer together positive work must be done by an external agent while an equal amount of negative work is done by the electrostatic field which takes the energy transferred to the charges by the external agent and stores it as electrostatic potential energy of the system of charge. It doesn't matter if the process is quasi-static as long as the difference between the initial and final kinetic energy is zero.

Similarly, for gravitational potential energy (GPE), if we lift an object of mass $m$ initially at rest and bring it to rest at some height $h$ where $g$ is constant, the change in KE is zero and the work we do is stored as GPE of $mgh$ in the Earth-Object system. It doesn't matter if we carry it out quasi-statically. All that matters is the initial and final state. Of course in order to accomplish this, we must exert an upward force greater than $mg$ to initiate motion and then exert an upward force less than $mg$ to bring it to rest at $h$, so that the net applied force is zero.

My question is, are there theorems or (semi-)rigorous arguments that demonstrate the validity of equations $(2)$ and $(3)$?

Equation (2) is the general equation for the conservation of energy of a mechanical system. It only addresses KE and PE at the macroscopic level, i.e., the kinetic and potential energy of the motion and position of the system as a whole with respect to an external frame of reference.

Equation (3), by including $E_{int}$, by which I assume you mean the the change in internal energy of the system, i.e. the motions and positions of the atoms and molecules at the microscopic level, gets you (almost) into the realm of thermodynamics and the general equation of the first law of thermodynamics. In this case, energy transfer can occur by both work and heat, so heat must be included. The equation for a closed system is generally written as

$$Q-W=\Delta U+\Delta KE+\Delta PE$$

Where $\Delta U$ is the change in internal energy, $Q$ is the net heat added to the system, and $W$ is the net work done on the system. Work can be boundary work, $w_b$ or other forms of work (electrical work, etc.)

Hope this helps.

Answer 3

I will go progressively from $W_{\textrm{net}} = \Delta KE$ to $W_{\textrm{apply}} = \Delta KE + \Delta U_{\textrm{pot}} + \Delta U_{\text{int}}$.

We should be clear that the work-kinetic energy theorem applies only to net work applied to the center of mass of a particle, body, or system of particles. However, the first law of thermodynamics applies to applied work with respect to a specific contact force on a body or system of particles. The two are different.

1. For a single particle, the work-kinetic energy theorem tells us $$ \frac{1}{2}mv_{f}^{2} - \frac{1}{2}mv_{i}^{2} = \int_{i}^{f} \vec{F}_{\text{net}}\cdot d\vec{x} $$ where $\vec{v}_{i}, \vec{v}_{f}$ are the velocities of the particle, $\vec{F}_{\text{net}}$ is the net force exerted on the particle, and $d\vec{x}$ represents the differential of the position of the particle.
2. Suppose a single particle is under influence of two forces: a conservative force $\vec{F}_{0} = -\nabla U_{\text{pot}}$ (negative gradient of some potential energy function) and an external applied force $\vec{F}_{\text{apply}}$. Then we have \begin{align*} \Delta KE &= \int_{i}^{f} \vec{F}_{\text{net}}\cdot d\vec{x} = \int_{i}^{f} \vec{F}_{0}\cdot d\vec{x} + \int_{i}^{f} \vec{F}_{\text{apply}}\cdot d\vec{x} \\ &= \int_{i}^{f} -\nabla U_{\text{pot}}\cdot d\vec{x} + \int_{i}^{f} \vec{F}_{\text{apply}}\cdot d\vec{x} = -(U_{f}-U_{i}) + \int_{i}^{f} \vec{F}_{\text{apply}}\cdot d\vec{x}, \end{align*} and so $\Delta KE + \Delta U_{\text{pot}} = W_{\text{apply}}$.
3. Now let's consider a system of $N$ particles labeled $j = 1, 2, \ldots, N$. Between all the particles let us assume there is some potential energy function $U_{\text{pot}} = U_{\text{pot}}(\vec{x}_{1}, \vec{x}_{2}, \ldots, \vec{x}_{N})$. The force on particle $j$ due to this potential is denoted as $\vec{F}_{j,\text{int}} = -\nabla_{j} U_{\text{pot}}$ and an external applied force on particle $j$ is denoted as $\vec{F}_{j,\text{ext}}$. By applying the work-kinetic-energy theorem to each and every particle, we find \begin{align*} \sum_{j} \Delta KE_{j} &= \sum_{j} \int_{i}^{f} \vec{F}_{j, \text{net}}\cdot d\vec{x}_{j} \\ &= \sum_{j} \left(\int_{i}^{f} \vec{F}_{j, \text{int}} \cdot d\vec{x}_{j} + \int_{i}^{f} \vec{F}_{j, \text{ext}} \cdot d\vec{x}_{j}\right) \\ &= \sum_{j} \left(\int_{i}^{f} -\nabla_{j} PE \cdot d\vec{x}_{j} + \int_{i}^{f} \vec{F}_{j, \text{ext}} \cdot d\vec{x}_{j}\right) \\ &= -\left(U_{f} - U_{i}\right) + \sum_{j} \int_{i}^{f} \vec{F}_{j, \text{ext}} \cdot d\vec{x}_{j} \\ &= -\Delta U_{\text{pot}} + \sum_{j} W_{j, \text{ext}} \end{align*} and so $\Delta KE_{\text{total}} + \Delta U_{\text{pot}} = \sum_{j} W_{j, \text{ext}}$.
4. At this stage, we do some accounting and relabeling of energy. Internal energy is some combination of kinetic and potential energy that is left over after classifying all macroscopical forms of kinetic energy (like due to center of mass motion and rotational motion of the entire body) and potential energy (like gravitational PE, elastic PE, or electrostatic PE). This rewriting gives $\Delta KE + \Delta U_{\text{pot}} + \Delta U_{\text{int}} = W_{\text{ext}}$.

So we see that in all cases, work is a transfer of energy. All this applies to work interactions with no heat interactions. If we invoke heat, then we get to the first law of thermodynamics, which states that heat is a form of energy transfer, much like work.