I would like to pose a quick question. I attempted (which seems successful) to derive the Joule-Lenz formula. My question: Where did I make mistake? Before going into my solution, it would be appreciative to say that I searched the results on the forum and am going to read other questions, thereby gaining more acknowledgements. Also, I hereby declare that have no wish to break the rules of the forum; hence, should the question be irrelevant, then apprise me of that - I shall discard it.

To do so, I wrote the first thermodynamics law, because it involves the heat, which I wish to compute:

$$\delta Q=dU+\delta W$$

From the equation, I obtained the desired expression by setting the internal energy to be zero, because I tried to think of the internal energy as a distribution of energy inside of the system. In the case, the system is the cable within which there is the current $I = dq/dt$ or $I = q/\Delta t$. Because no electrons forsake the cable, the internal kinetic energy does not change, so its differential should be zero: $dU = 0$. It is understandable that it is in the simples case when no thermal losses are presented. Applying the idea, we arrive at the equation $$\delta Q=\delta W=-qd\varphi$$ After integration we have $Q = q \cdot \left(-\Delta\varphi\right) = qV$. Knowing that $q = I \Delta t$, we obtain the terminal expression $Q = IV \Delta t$. That is the formula, given in the Wikipedia. The derivation is based upon my understanding.

  • $\begingroup$ You are getting it backwards. The internal energy of the cable increases due to the work done by the electric field. There is no heat transfered to the cable from an external system. In real situations there is heat lost to environment. But this has nothing to do with Joule heating. $\endgroup$
    – nasu
    Nov 1 '20 at 18:02
  • $\begingroup$ And other than that, this is the standard derivation for the power dissipated, found in introductory textbook. $\endgroup$
    – nasu
    Nov 1 '20 at 18:05
  • $\begingroup$ Am I right, or wrong? Check-my-work questions are off-topic here. $\endgroup$
    – G. Smith
    Nov 1 '20 at 18:09
  • $\begingroup$ @G. Smith I shall keep that in mind in the future. By the question, I wished to verify my thoughts. In university, I had other tests - equilibrium in beams. The first comment accounts for my mistake. Would you be so kind as to provide me with a clue on further actions: should I discard question? or think a little and answer my own question based on the first comment? $\endgroup$ Nov 1 '20 at 18:17
  • $\begingroup$ You don’t need to delete your question. Answering your own question is fine here. $\endgroup$
    – G. Smith
    Nov 1 '20 at 18:24

Let us talk about the Joule heating law. The conductor is a material (for example, a conductor made up of metal) which has its own structure, meaning that there are atoms within it. These atoms are connected between themselves by bonds. The current is understood to be a fluid (liquid or gas) moving inside the conductor along an axis. In my case, the fluid is a flux of electrons (electron gas). That is what I should have discerned. Then the idea becomes clear; and clearer after reading about the law in Wikipedia. I believe I should gather more acknowledgements on the topic from the view of quantum physics.


  1. Is there any thermodynamic system?
  2. Is $dU = 0$?
  3. How to derive the formula?

In order to answer all of these questions, one ought to understand what happens within the conductor. It is natural to say that electrons move under the effect of the force $\vec{E}$, generated by the potential difference. Applying the second Newton's law provides us with some information: $$q\cdot \vec{E} = m \cdot \vec{a} \Rightarrow a=\frac{q \cdot E}{m} \ne 0$$ Hence, all electrons are accelerated by the force. That means that the velocity of the fluid should increase and therefore cause the change in kinetic energy - this would be helpful. Now we must understand from what or where the heat is generated. The clue lies behind the laws of conservations (energy and momentum). Why? Because electrons may collide with atoms of the conductor; thereafter, there arises changes in kinetic energies.

Let us consider these collissions in the simplest case. May $N$ be the number of electrons and $\tau$ be the time of the fluid's moving prior to colliding with atoms. Okay, from the equation above, we have, assuming the initial velocity equal to zero, another equation: $$u = \frac{q \cdot E}{m} \cdot \tau$$ The kinetic energy that electrons obtained from acceleration caused by the electric field is then $$ T = \frac{m \cdot u^2}{2} = \frac{m}{2} \cdot \frac{q^2 \cdot E^2}{m^2} \cdot \tau^2 = \frac{q^2 \cdot E^2 \cdot \tau^2}{2m}$$ Each electron undergoes collision at the frequency $1/\tau$; therefore, $N/\tau$ is the frequency at which all electrons collide and the total energy is then $$w=\frac{N}{\tau} \cdot T = \frac{N \cdot q^2 \cdot \tau}{2m} \cdot E^2 = \sigma \cdot E^2$$ where the conductivity is $$\sigma = \frac{N \cdot q^2 \cdot \tau}{2m}$$ The last thing to perform is to put $q = e$, because the fluid is a flux of electrons. Now what is left is to account for energy transfer. When an electron collides with an atom, it transfers some energy to the atom; this energy is $T$. So, the heat is transferred by diffussion: electrons obtain kinetic energy and give it to the conductor - that is what happened withing the conductor.

Let us answer the first question. As we observed, electrons are moving within the conductor with an obtained acceleration from the electric field $E$, from which they gained the kinetic energy as well. The enegy is given to the conductor as Joule heating. What is a thermodynamic system? It is a system (closed, isolated, open) wherein a process happens (for examply, isothermic process). The system may be made up of several objects which interacts between themselves. In result, there emits some energy (consider the engine of a car, where by means of combustion of petroleum, the heat energy from gas is converted into the moving force, so that the car would be able to move). The work of that process is given by $A = \nu \cdot R \cdot T \cdot \ln \left(\frac{V_2}{V_1}\right)$. Now return to the process of the transfer of kinetic energy to the atoms from electrons. Based on what has been said, the process happening within the wire should be thermodynamic process. Consider another idea on how to derive the formula. Prior to deriving we write out the equation of momentum conservation and the energy one, considering three type of collission. The first one is ellastic: the electron moves with $u_0$ and the atom is at rest before colliding; after that, the electron acquires velocity $u$ and the atom $v$. Then we get two equations $$ m_e \cdot u_0 = m_e \cdot u + m_a \cdot v$$ $$ \frac{m_e \cdot u_0 ^2}{2} = \frac{m_e \cdot u^2}{2} + \frac{m_a \cdot v^2}{2} $$ From these equations, we acquire $$ u = u_0 - \frac{m_a}{m_e} v$$ $$ \frac{m_e \cdot u_0 ^2}{2} = \frac{m_e}{2} \cdot \left( u_0 - \frac{m_a}{m_e} v\right)^2 + \frac{m_a \cdot v^2}{2} \Rightarrow v = \frac{2m_e \cdot u_0}{m_e + m_a}$$ From the last, we get $$ \frac{m_a \cdot v^2}{2} = \frac{2 \cdot m_a \cdot m_e}{\left(m_e + m_a\right)^2} \cdot \frac{m_e \cdot u_0 ^2}{2} \approx \frac{4 m_e}{m_a} \cdot \frac{m_e \cdot u_0 ^2}{2}$$ because $ m_a \gg m_e$. Should an electron collide with an atom along the tangent line to the surface of the atom, the transferring energy is zero, and it decreases otherwise. The average energy is the sum of minimum and maximum energies devided by two: $$E_{average} = \frac{E_{maximum} + E_{minimum}}{2} = \frac{\frac{m_a \cdot v^2}{2} + 0}{2} = \frac{2 m_e}{m_a} \cdot \frac{m_e \cdot u_0 ^2}{2}$$ Now we get to another approach. We assume that within the conductor, there are moving atoms and electrons, and we consider the colliding fluids. The idea is here that the electric field provides the electon gas with an energy. Atoms obtain the energy lesser than that obtained by electrons from the electric field. Hence, should we denote the temperature of the electron gas by $T_e$ and the atoms by $T_a$, we will have an inequality $T_e > T_a$, meaning that the energy obtained by electrons from atoms is less than that from the electic field. We have already discovered the average energy obtained by atoms from electrons, given by $E_{average}$. Now we are going to make use of the formula which connects a gas's kinetic energy and temperature: the hypothesis on the proportionality of energy to temperature: $$\frac{m_a \cdot v^2}{2} = \frac{3}{2} \cdot k \cdot T_a$$ $$\frac{m_e \cdot u_0 ^2}{2} = \frac{3}{2} \cdot k \cdot T_e$$ The resultant flux is the sum of all fluxes; it should be proportional to the difference between these fluxes, the number of collissions, and the average kinetic energy obtained by atoms from electrons: $$w = \frac{N}{\tau} \cdot \frac{2 \cdot m_e}{m_a} \cdot \frac{3}{2} \cdot k \cdot \left(T_e - T_a\right)$$ Using the first formula for $w$, we have $$\frac{N}{\tau} \cdot \frac{2 \cdot m_e}{m_a} \cdot \frac{3}{2} \cdot k \cdot \left(T_e - T_a\right) = \frac{N \cdot q^2 \cdot \tau}{2m} \cdot E^2$$ whence $$T_e - T_a = \frac{1}{6} \cdot \frac{m_a}{m_e} \cdot \frac{q^2 \cdot \tau^2}{k \cdot } \cdot E^2$$

In the first approach we derived that $w = \sigma \cdot E^2$, which can be rewritten in the vector form as $$w = \left(\sigma \cdot \vec{E}\right) \cdot \vec{E} = \vec{j} \cdot \vec{E}$$, which is true according to Ohm's law (the differential form is obtainable from the original form by using $R = \rho \cdot \frac{l}{A}$ and taking differentials). From this is not difficult to get $P = V \cdot I$, because we know that $$ V = -\Delta \varphi = \vec{E} \cdot \vec{dl}, \\ j = \frac{dI}{dS}$$ So, the quantity $w$ is power per volume! Next, $Q = I \cdot V \cdot \Delta t$, because $w$ is the power of emitting heat. Then according to the first law of thermodynamics, we get $$dU = \delta Q - \delta W = 0$$ because $\delta A = dK = -q d\varphi$, where K denotes the kinetic energy. So, $$ Q = A = I \cdot V \cdot \Delta t$$

  • $\begingroup$ Probably there is no any system but just diffusion that was described. Only in the way, not diving into quantum physics (but I wished and should have, but could not), I can understand Joule heating... But for some reason $Q = A$ and then $\Delta U = 0$. That is strange. $\endgroup$ Nov 1 '20 at 23:54

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