Derivation of Maxwell's equations using Lagrangian formalism [duplicate]

Question

Some time ago, I read in Landau's Theoretical Physics Course you could derive Maxwell's equations using the Lagrangian formalism, and I find this to be exciting. Unfortunately, I don't have access to the book and even if I had it, I'm not sure I could understand it, since I haven't learnt anything about tensors yet, and seemingly you need to use a stress tensor.

Could you please explain to me exactly how Maxwell's equations derive from the Lagrangian formalism? In fact, from my Classical Mechanica course, I thought Lagrangians were only used in mechanics, but I've read here they can give place to different Physics depending on your choice of the Lagrangian. How is this possible? I'm just as confused as I am thrilled about the possibilities this would open up. If you'd clarify my doubts for me, I'd appreciate it very much.

Answer 1

The principle of least action is extremely robust and has been employed in so many interesting ways.

The Lagrangian for the electromagnetic field is given by: $$\mathcal L=-{1\over 4}F_{\mu\nu}F^{\mu\nu}.$$ There are different conventions that use differing values for the constant out front, for example that which you see in J.D. Jackson's text, however, they all contain the "quadratic" form in $F_{\mu\nu}$.

Here, $F_{\mu\nu}$ and $F^{\mu\nu}$ are second order covariant and contravariant rank tensors known as the Faraday or field strength tensor, however, this is not so daunting as it may sound in this context because you can identify these quantities with matrices, albeit matrices that transform in the proper way under rotation. See Goldstein's Classical Mechanics chapter 13 for a good introduction to the Lagrangian formulation for continuous systems and fields, as that is precisely what the electromagnetic field is. At any rate, the Euler Lagrange equations for such systems has the form:$$\partial_\mu \bigg({\partial\mathcal L\over\partial (\partial_\mu \phi_\rho)}\bigg)-{\partial\mathcal L\over\partial\phi_\rho}=0.$$ $$\vdots$$ Where we have as many equations as we have fields. Notice that we have avoided differentiating with respect to the usual generalized coordinates and have instead differentiated with respect to some functions $\phi_\rho$. The functions $\phi_\rho$ are any set of functions which act as the "coordinates" of the Lagrangian, which in the continuous system is now a field or density that is defined everywhere in space, i.e. for a continuous system the lagrangian is such that: $$\mathcal L=\mathcal L(\phi_{\rho}, \partial_\mu\phi_{\rho}, x^\mu).$$ The Lagrangian may be a function of any number of fields, their derivatives and possibly the raw coordinates themselves! You are right when you say that the subject is interesting! Now back to Maxwell's theory. The field strength tensor is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu,$$ and $$F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu.$$ So for the electromagnetic system, the field involved is the 4-vector potential $A^\mu$, so inserting these into the given Lagrangian expression we get: $$\mathcal L=-{1\over 4}(\partial_\mu A_\nu-\partial_\nu A_\mu)(\partial^\mu A^\nu-\partial^\nu A^\mu).$$ So Maxwell's equations can be derived via the Euler Lagrange equations: $$\partial_\nu\bigg({\partial\mathcal L\over\partial (\partial_\nu A^\mu)}\bigg)-{\partial\mathcal L\over\partial A^\mu}=0.$$ Now, this calculation is straightforward, however it does require that you get comfortable with manipulating indices, thus I will leave off the derivation for now and offer you some good pieces of advice that I hope you will pursue as your time admits.

Firstly, read 7.4-7.6 of Goldstein's Classical Mechanics, then read chapter 13 or at least 13.1-13.2, then you will be ready to calculate the above derivatives and find Maxwell's equations from the Lagrangian formulation. This is really not a lot of material to cover, and it serve as excellent preparation for more advanced physics.

Answer 2

The use of the Lagrangian formalism is not restricted to (point) mechanics, but all fundamental interactions (electroweak, strong, gravitation) are defined by the action integral of a Lagrangian field theory.

Taking the electromagnetic field described by the (antisymmetric) field strength tensor $$F_{\mu \nu}=\partial_\mu A_\nu-\partial_\nu A_\mu= \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 &-B_z & B_x \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y &B_x & 0 \end{pmatrix} \tag{1} \label{1}$$ ($A^\mu=(\phi, \vec{A})$ is the 4-vector potential) in the presence of an external 4-current density $j^\mu=(\rho, \vec{j})$ as an example, the corresponding action integral (using the Heaviside sytem with $c=1$) is given by $$S =\int \! d^4 x \left(-\frac{1}{4} F^{\mu \nu }F_{\mu \nu}-j^\mu A_\mu \right). \tag{2} \label{2}$$ Variation of the action yields $$\begin{align}\delta S &= \int \! d^4x \left(-\frac{1}{2}F^{\mu \nu} \delta F_{\mu \nu}-j^\mu \delta A_\mu \right) \\[5pt] &= \int \! d^4x \left(-\frac{1}{2} F^{\mu \nu}(\partial_\mu \delta A_\nu -\partial_\nu \delta A_\mu)-j^\mu \delta A_\mu \right) \\[5pt] &= \int \! d^4 x \left(F^{\mu \nu}\partial_\nu \delta A_\mu-j^\mu \delta A_\mu \right) \\[5pt] &= \int \! d^4x \left( \partial_\nu F^{\nu \mu}-j^\mu\right) \delta A_\mu, \end{align} \tag{3} \label{3}$$ where partial integration and $F^{\mu \nu} =-F^{\nu \mu}$ was used in the last step. It is now obvious that $\delta S =0$ leads to the inhomogeneous Maxwell equations $$ \partial_\nu F^{\nu \mu} = j^\mu, \tag{4} \label{4}$$ being equivalent to $$ \vec{\nabla} \cdot \vec{E} = \rho, \quad \vec{\nabla} \times \vec{B} - \dot{\vec E} = \vec{j}, \tag{5} \label{5}$$ as can be seen from \eqref{1}. The homogeneous Maxwell equations $$ \vec{\nabla} \times \vec{E} +\dot{\vec B}=0, \quad \vec{\nabla} \cdot \vec{B}=0 \tag{6} \label{6}$$ are automatically fulfilled by (cf. \eqref{1}) $$F_{\mu \nu}=\partial_\mu A_\nu -\partial_\nu A_\mu, \tag{7} \label{7}$$ being equivalent to $$\vec{E}=-\vec{\nabla} \phi -\dot{\vec A}, \quad \vec B = \vec{\nabla} \times \vec A. \tag{8}$$

Answer 3

The Lagrangian density of electromagnetism is given by $$\mathcal{L}=\frac{\epsilon_0}{c}(E^2-c^2B^2)+\rho \phi + \mathbf j \cdot \mathbf A$$ where we require $\mathbf E=-\nabla \phi-\partial_t \mathbf A$ and $\mathbf B=\nabla \times \mathbf A $ (these two conditions are equivalent to Gauss' magnetic law and to Faraday's law).

Using Euler-Lagrange equations, we have $$\frac{\partial\mathcal{L}}{\partial\varphi} = \partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}\right) $$ where $\varphi\in\{\rho,\mathbf A\}$ and $\partial_\mu=(\partial_t,\partial_i)$.

Let's start with $\phi$. We have $$\frac{\partial\mathcal{L}}{\partial\phi} = \rho $$ and $$\partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}\right)=\partial_i \left(\frac{\epsilon_0}{2}\frac{\partial}{\partial \partial_i\phi}(-\partial_i\phi-\partial_t A_i)^2\right)=\epsilon_0\partial_i(-\partial_i\phi-\partial_t\mathbf A_i)=\epsilon_0 \nabla\cdot \mathbf E$$

so we get $$\nabla \cdot \mathbf E=\rho/\epsilon_0$$ which is Gauss' law.

Now let us go for $\mathbf A$. We have $$\frac{\partial\mathcal{L}}{\partial A_n} = j_n $$

and now for the hard part

$$\partial_t \left(\frac{\epsilon_0}{2}\frac{\partial}{\partial \partial_tA_n}(-\partial_i\phi-\partial_t A_i)^2\right)=-\epsilon_0\partial_t \mathbf E$$

$$\partial_m \left(\frac{\epsilon_0 c^2}{2}\frac{\partial}{\partial \partial_m A_n}(\epsilon_{ijk} \partial_jA_k)^2\right)=\frac12\epsilon_0 c^2\partial_m (\epsilon_{ijk}\epsilon_{ils} \delta_{mj}\delta_{nk}\partial_lA_s+\epsilon_{ijk}\epsilon_{ils} \partial_jA_k\delta_{ml}\delta_{ns}))=\epsilon_0 c^2 \epsilon_{imn}\epsilon_{ils} \partial_m \partial_l A_s=-\epsilon_0 c^2\nabla \times \mathbf B $$ Putting all together $$\nabla \times \mathbf B = \frac{1}{\epsilon_0 c^2}\mathbf j +\frac{1}{c^2}\partial_t\mathbf E$$

which is Ampere's law, the last of the four Maxwell's equations.