# Lagrangian in presence of an Electromagnetic Field

Given the following definition of the Magnetic Vector Potential $$\vec{A}$$: $$\vec{A} \ \mid \ \vec{B}=\vec{\nabla}\times\vec{A}$$ We can derive (but I don't know how) that the Lagrangian in presence of an electromagnetic field is: $$\mathcal{L}=K-V=\frac{1}{2}mv^2-q\Phi+q\vec{v}\cdot\vec{A} \tag{1}$$ where $$K$$ is the kinetic energy and $$V$$ is the potential energy. The Lagrangian is by definition the difference between $$K$$ and $$V$$, but $$q\Phi$$ is of course the potential energy of the electric field, this must mean that: $$-q\vec{v}\cdot\vec{A} \tag{2}$$ is the potential energy of the magnetic field. I now have two distinct but related problems:

• How can we prove that (2) is the potential energy of the magnetic field?1
• How can we prove that (1) is the correct formula for the Lagrangian in presence of an electromagnetic field?

Of course, these two questions are strongly linked: if you answer one of them you have also basically answered the other one.2

All this seems to me a really important and fundamental problem. But, paradoxically, I have not found any source providing a direct, simple and complete answer to it; that's what I am searching here.

[1]: Seems impossible to me to prove such a statement; because I think the statement itself is wrong. The magnetic field has non-zero curl: shouldn't this mean that its potential energy is not definable?

[2]: There is also the possibility that the second question can be answered without proving the first statement true. Maybe (2) is not the magnetic potential energy, but also (somehow) $$q\Phi-q\vec{v}\cdot\vec{A}$$ is the total electromagnetic potential energy of an electromagnetic field; in this case, I would like to understand how to prove it.

Edit: In response to mike stone's comment and Emmy's answer: In my textbook the Lagrangian, in the context of non-relativistic mechanics, is defined as: $$\mathcal{L}=K-V$$ but not only on my textbook! Take a look at this section in the related wikipedia page: citing directly:

It is nevertheless possible to construct general expressions for large classes of applications. The non-relativistic Lagrangian for a system of particles can be defined by $$L=T-V$$

Is this wrong? Seems strange.
Also: if the Lagrangian is indeed any function that gives back the correct Newtonian equations of motion once put into the Euler-Lagrange equations then why use the lagrangian formalism in the first place? The beauty of this formalism is that by knowing the energies we can derive the equation of motion; if this does not hold, at random, then I am forced to remember the arbitrary correct form of the Lagrangian in every given case or to derive it every time from Newtonian mechanics, that is not easy to do and also seems to kill half the purpose of the Lagrangian formalism.

• More on the velocity-dependent potential for the Lorentz force: physics.stackexchange.com/q/77325/2451 and links therein. Commented Mar 17, 2021 at 11:45
• The Lagrangian does not have to be $T-V$. It is whatever gives the equations of motion you desire. In particular, there is no special interpretation of the ${\bf v}\cdot {\bf A}$ term Commented Mar 17, 2021 at 11:54
• The vector potential is a kind of magnetic field momentum and the correspondint term in the Lagrangian is of kinetic rather than of potential nature. Commented Mar 17, 2021 at 12:24
• @mikestone I have made an edit to my question, explaining my perplexities. If you could give me a complete clarification it would help a lot. Commented Mar 17, 2021 at 14:51
• We use the Lagrangian formalism for non-purely-mechanical systems (for which $L\ne T-V$) mostly because Noether's theorem for constructing conserved quantities makes use of it. Commented Mar 17, 2021 at 14:57

Question 1 : you should not try to interpret this Lagrangian as kinetic energy - potential energy. In fact, the magnetic part of the Lorentz force does not come from a potential; there is no potential energy associated with the charge in the magnetic field.

Question 2 : to prove that this is the right Lagrangian, it is enough to observe that the corresponding equations of motion give you back Newton's law with Lorentz's force.

• I have made an edit to my question, explaining my perplexities. If you could give me a complete clarification it would help a lot. Commented Mar 17, 2021 at 14:51
• The best thing I can do is trying to explain mike stone's comment under the question. For a purely classical mechanical system, you have $L = T - V$ and you can write your Lagrangian very easily. But since electromagnetism is related to special relativity, this approach does not work. Yet the Lagrangian is not arbitrary. It is built from symmetry requirements. For instance, you get the one of electromagnetism just by imposing Lorentz and local gauge invariance. And if you then take its non relativistic limit, you obtain what you want Commented Mar 18, 2021 at 12:03

You are right, $$(2)$$ is not the potential energy of the magnetic field. In fact, there is no potential energy linked to the magnetic field, since the magnetic force does no work as it is perpendicular to displacement. The Lagrangian not always is the difference between kinetic and potential energy, and this is a clear example of this. The potential energy $$V$$ is just $$q\Phi$$.

Some other times, the Lagrangian can be constructed as $$L=T-U$$, where $$U$$ is a generalized potential related to its generalized force this way

$$Q_i=\frac{d}{dt}\left[\frac{\partial U}{\partial \dot{q}_i}\right]-\frac{\partial U}{\partial q_i}.$$

In cartesian coordinates, $$Q_i=F_i$$, so it can bee seen that if $$U=q(\Phi-\dot{\vec{x}}\cdot\vec{A})$$

$$F_i=\frac{d}{dt}\left[\frac{\partial U}{\partial \dot{x}_i}\right]-\frac{\partial U}{\partial x_i}$$

reproduces the Lorentz force.

The only way to see that $$(1)$$ is the correct Lagrangian for a particle in an electromagnetic field is that its Euler-Lagrange equations reproduce the known equations of motion

$$m\ddot{\vec{x}}=q(\vec{E}+\dot{\vec{x}}\times\vec{B}).$$

If you computed the Hamiltonian, it would be $$H=\frac{p^2}{2m}+q\Phi=T+V=E\neq T+U$$

I. [Cohen-Tannoudji]=Cohen-Tannoudji, C., Diu, B. & Lalo"{e}, F.: Quantum Mechanics, 3 vols, 2nd ed., Weinheim: WILEY-VCH, 2020.
II. Theorem 1. An electromagnetic field exerts the Lorentz force on a charged particle in it [Cohen-Tannoudji, vol. 2, p.1538, l.3--l.$$-$$1].
Theorem 2. Let $$\Phi=qU(\pmb{r},t)-q\dot{\pmb{r}}\cdot \pmb{A}(\pmb{r},t)$$ [Cohen-Tannoudji, vol. 2, p.1538, (58)]. Then its generalized force $$\frac{d}{dt}\frac{\partial\Phi}{\partial \dot{x}}-\frac{\partial\Phi}{\partial x}$$ is the Lorentz force [see the above 2nd solution].\ We may reorganize the proof given in [Cohen-Tannoudji, vol. 2, p.1538, l.3--l.$$-$$1] as follows not only to obtain the proof of Theorem 2, but also to make the definition of generalized force look natural.
$$\mathcal{L}(\pmb{r},\dot{\pmb{r}},t)=\frac{1}{2}m\dot{x}^2+q\dot{\pmb{r}}\cdot \pmb{A}(\pmb{r},t)-qU(\pmb{r},t)$$ [Cohen-Tannoudji, vol. 2, p.1538, (58)]
$$=T-(qU(\pmb{r},t)-q\dot{\pmb{r}}\cdot \pmb{A}(\pmb{r},t))=T-\Phi$$.
The left-hand side of the equality given in Cohen-Tannoudji, vol. 2, p.1538, (62)] results from applying the operator $$\frac{d}{dt}\frac{\partial}{\partial \dot{x}}-\frac{\partial}{\partial x}$$ to $$T$$; the right-hand side of the equality given in Cohen-Tannoudji, vol. 2, p.1538, (62)] results from applying the operator $$\frac{d}{dt}\frac{\partial}{\partial \dot{x}}-\frac{\partial}{\partial x}$$ to $$\Phi$$. Note that the latter operator is the same as the operator given in [Symon, p.385, (9.153)]. Namely, the latter operator gives a natural construction of generalized force. Thus, it can be said that the definition of generalized force is naturally based on the Lagrangian equation [Cohen-Tannoudji, vol. 2, p.1530, (15)]].