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I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the field tensor).

Substituting in for the definition of the field tensor yields $L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu)]$. I know I should be using $A^\mu$ as the dynamical variable in the Euler-Lagrange equations, which become $\frac{\partial L}{\partial A_\mu} - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)} = - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)}$, but I'm confused about how to proceed from here.

I know I should end up with $\partial_\mu F^{\mu\nu} = 0$, but I don't quite see why. Since $\mu$ and $\nu$ are dummy indices, I should be able to change them: how do the indices in the lagrangian relate to the indices in the derivatives in the Euler-Lagrange equations?

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    $\begingroup$ See in Sean carrol's book. Full of derivation there $\endgroup$ – Andika Jan 20 '16 at 1:09
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    $\begingroup$ Why is it not enough to plug the $F^{\mu\nu} = \partial{[\mu}A{\nu]}$ into Maxwell's equations and show that they hold? $\endgroup$ – Daniel Mahler Apr 29 '16 at 6:35
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Well, you are almost there. Use the fact that $$ {\partial (\partial_{\mu} A_{\nu}) \over \partial(\partial_{\rho} A_{\sigma})} = \delta_{\mu}^{\rho} \delta_{\nu}^{\sigma}$$ which is valid because $\partial_{\mu} A_{\nu}$ are $d^2$ independent components.

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  • $\begingroup$ How to actually prove it? $\endgroup$ – omehoque Jan 7 at 4:21
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We vary the action $$\delta \int {L\;\mathrm{d}t} = \delta \int {\int {\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)\mathrm{d}^3 x\;\mathrm{d}t = 0} } $$ ${\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)}$ is the density of lagrangian of the system.

So, $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }}\delta A_\nu + \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}\delta \left( {\partial _\mu A_\nu } \right)} \right)\mathrm{d}^3 x\;\mathrm{d}t = 0} } $$ By integrating by parts we obtain: $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right)\delta A_\nu \mathrm{d}^3 x\;\mathrm{d}t = 0} } \implies \frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = 0$$ We have to determine the density of the lagrangian. One terms deals with the interaction of the charges with the electromagnetic field, $J^\mu A_\mu$. The other term is the density of energy of the electromagnetic field: this term is the difference of the magnetic field and the electric field. So we have: $$\Lambda = J^\mu A_\mu + \frac{1}{{4\mu _0 }}F^{\mu \nu } F_{\mu \nu } $$ We have: $$\frac{{\partial \Lambda }}{{\partial A_\nu }} = J^\nu $$ so: \begin{align}\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} &= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}F^{\kappa \lambda } F_{\kappa \lambda } } \right) \\&= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\left( {\partial ^\kappa A^\lambda - \partial ^\lambda A^\kappa } \right)\left( {\partial _\kappa A_\lambda - \partial _\lambda A_\kappa } \right)} \right)} \right) \\&= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa - \partial ^\lambda A^\kappa \partial _\kappa A_\lambda + \partial ^\lambda A^\kappa \partial _\lambda A_\kappa } \right)} \right)\end{align} The third and the fourth are the same of first and the second terms. You can do $k \leftrightarrow \lambda $: \begin{align}\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} & = \frac{1}{{2\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right)} \right)\;.\end{align} But \begin{align}\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda } \right) &= \partial ^\kappa A^\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\kappa A_\lambda } \right) + \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda } \right) \\ &= \partial ^\kappa A^\lambda \delta _\kappa ^\mu \delta _\lambda ^\nu + g^{\kappa \alpha } g^{\lambda \beta } \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\alpha A_\beta } \right)\\& = 2\partial ^\mu A^\nu \;.\end{align}

We have:

\begin{align}\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right) &= 2\partial ^\nu A^\mu \;. \end{align}

So,

\begin{align}\partial _\mu \left( {\frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right)& = \frac{1}{{\mu _0 }}\partial _\mu \left( {\partial ^\mu A^\nu - \partial ^\nu A^\mu } \right)\\ & = \frac{1}{{\mu _0 }}\partial _\mu F^{\mu \nu } \;.\end{align} The lagrangian equations provide the non homogeneus maxwell equations:

$$\partial _\mu F^{\mu \nu } = \mu _0 J^\nu \;. $$

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  • $\begingroup$ FYI, this answer (v4) uses implicitly the sign convention $(+,-,-,-)$. $\endgroup$ – Qmechanic Dec 4 '16 at 14:54
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Dear amc, first, write your Lagrangian density as $$ L = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} (\partial_\mu A_\nu) F^{\mu\nu} $$ Is that fine so far? The $F_{\mu\nu}$ contains two terms that make it antisymmetric in the two indices. However, it's multiplied by another $F^{\mu\nu}$ that is already antisymmetric, so I don't need to antisymmetrize it again. Instead, both terms give me the same thing, so the coefficient $-1/4$ simply changes to $-1/2$.

Now, the field equations force you to compute the derivatives of the Lagrangian with respect to $A_\mu$ and its derivatives. First of all, the derivative of the Lagrangian $L$ with respect to $A_\mu$ components themselves vanishes because the Lagrangian only depends on the partiial derivatives of $A_\mu$. Is that clear so far?

So the equations of motion will be $$0 = -\partial_\mu [\partial L / \partial(\partial_\mu A_\nu)] = \dots $$ Whoops, you already got to this point. But now, look at my form of the Lagrangian above. The derivative of the Lagrangian with respect to $\partial_\mu A_\nu$ is simply $$-\frac{1}{2} F^{\mu\nu}$$ because $\partial_\mu A_\nu$ simply appears as a factor so the equations of motion will simply be $$ 0 = +\frac{1}{2} \partial_\mu F^{\mu\nu} $$ However, I have deliberately made one mistake. I have only differentiated the Lagrangian with respect to $\partial_\mu A_\nu$ included in the first factor of $F_{\mu\nu}$, with the lower indices. However, $\partial_\mu A_\nu$ components also appear in $F^{\mu\nu}$, the second factor in the Lagrangian, one with the upper indices. If you add the corresponding terms from the Leibniz rule, the result is simply that the whole contribution will double. So the right equation of motion, including the natural coefficient, will be $$ 0 = \partial_\mu F^{\mu\nu} $$ The overall normalization is important because this equation may get extra terms, like the current, whose coefficient is obvious, and you don't want to get a relative error of two between the derivative of $F$ and the current $j$.

Cheers Lubos

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  • $\begingroup$ Hey I know this is 5 years late but maybe you'll see this: Why is $\frac{\partial}{\partial (\partial_{\mu} \phi)} (\partial_{\mu}A_{\nu}F^{\mu \nu}) = F^{\mu \nu}$. Doesn't the tensor also depend on the partial derivatives? Don't we have to use the product rule then? $\endgroup$ – user17574 Jan 29 '16 at 15:36
  • $\begingroup$ Hi @user17574 - doesn't "which" tensor depend on the partial derivatives? Surely the stress-energy tensor does, and so does the Lagrangian. That's why the derivative of it with respect to the partial derivatives is nonzero. The derivative is calculated in the answer. The product rule indeed works and it's why one cancels the factor of $1/2$. Have you tried to read the answer? $\endgroup$ – Luboš Motl Jan 29 '16 at 17:31
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One method is to vary the Maxwell action (set $J^\mu = 0$ if you want, for the source-free case) $$ S = \int d^4 x {\mathcal{L}} = - \int d^4 x \left(\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + J^\mu A_\mu\right). $$ First note that $$ \begin{align} \delta \left(F^{\mu \nu} F_{\mu \nu}\right) &= 2 F^{\mu \nu} \delta F_{\mu \nu} \\ &= 2 F^{\mu \nu} \left(\partial_\mu \delta A_\nu - \partial_\nu \delta A_\mu\right) \\ &= 4 F^{\mu \nu} \partial_\mu \delta A_\nu \\ &= 4\left[\partial_\mu\left(F^{\mu \nu} \delta A_\nu\right) - \partial_\mu F^{\mu \nu} \delta A_\nu\right], \end{align} $$ where we've used the fact that $F$ is antisymmetric.

Notice also that the $\partial_\mu\left(F^{\mu \nu} \delta A_\nu\right)$ term will vanish upon converting it to a surface integral, using the standard argument that $\delta A_\mu$ vanishes at the integration boundary.

Using the above, the variation of the action is $$ \delta S = - \int d^4x \ \delta A_\nu \left(-\partial_\mu F^{\mu \nu} + J^\nu \right), $$ which, since $\delta A_\nu$ is arbitrary, yields the desired result $$ \partial_\mu F^{\mu \nu} = J^\nu. $$

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  • $\begingroup$ Instead of assuming $\delta A_\mu$ vanishing at the boundary one can assume $F^{\mu\nu}$ at the boundary. Is that wrong? See a related question here physics.stackexchange.com/questions/438277/… @EricAngle $\endgroup$ – SRS Nov 1 '18 at 17:47
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Although late in the party, I post an answer on an elemementary level. May be this proves the power of tensor calculus used in all previous nice answers.

Abstract

In this answer we'll try to derive Maxwell equations in empty space \begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \tag{001a}\\ \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} & = \mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{001b}\\ \nabla \boldsymbol{\cdot} \mathbf{E} & = \frac{\rho}{\epsilon_{0}} \tag{001c}\\ \nabla \boldsymbol{\cdot}\mathbf{B}& = 0 \tag{001d} \end{align} from the Euler-Lagrange equations \begin{equation} \boxed{\: \dfrac{\partial }{\partial t}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{\eta}_{\jmath}}\right) + \nabla \boldsymbol{\cdot}\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\eta_{\jmath}\right)}\right]- \frac{\partial \mathcal{L}}{\partial \eta_{\jmath}}=0, \quad \left(\jmath=1,2,3,4\right) \:} \tag{002} \end{equation} where \begin{equation} \mathcal{L}=\mathcal{L}\left(\eta_{\jmath}, \dot{\eta}_{\jmath}, \boldsymbol{\nabla}\eta_{\jmath}\right) \qquad \left(\jmath=1,2,3,4\right) \tag{003} \end{equation} is the Lagrangian density of the question (except a constant factor) \begin{equation} \boxed{\: \mathcal{L}=\dfrac{\Vert\mathbf{E}\Vert^{2}-c^{2}\Vert\mathbf{B}\Vert^{2}}{2}+\dfrac{1}{\epsilon_{0}}\left( -\rho \phi + \mathbf{j}\boldsymbol{\cdot}\mathbf{A}\right) \:} \tag{004} \end{equation} and $\:\eta_{\jmath}\left( x_{1},x_{2},x_{3},t\right), \:\:\jmath=1,2,3,4\:$ the components $\:A_{1},\:A_{2},\:A_{3},\phi\:$ of the EM potential 4-vector respectively. In a sense, this derivation is built on the inverse one ( : this of finding a proper Lagrangian density from Maxwell equations ) by moving backwards, see my answer here : Deriving Lagrangian density for electromagnetic field

1. Main Section

First we express $\:\mathbf{E},\mathbf{B}\:$ of (004) in terms of the potential 4-vector components $\:A_{1},\:A_{2},\:A_{3},\phi\:$ \begin{align} \mathbf{B} & = \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A} \tag{005a}\\ \mathbf{E} & = -\boldsymbol{\nabla}\phi -\dfrac{\partial \mathbf{A}}{\partial t} = -\boldsymbol{\nabla}\phi - \mathbf{\dot{A}} \tag{005b} \end{align} From (005) the Maxwell equations (001a) and (001d) are valid automatically. So the four(4) scalar Maxwell equations (001b) and (001c) must be derived from the four(4) scalar Euler-Lagrange equations (002). Moreover, it's reasonable to assume that the vector equation (001b) must be derived from (002) with respect to the components of the vector potential $\:\mathbf{A}=\left(A_{1},\:A_{2},\:A_{3}\right)\:$, while the scalar equation (001c) must be derived from (002) with respect to the scalar potential $\:\phi\:$.

From equations (005) we express the Lagrangian density (004) in terms of the potential 4-vector components $\:A_{1},\:A_{2},\:A_{3},\phi\:$ : \begin{align} \left\Vert\mathbf{E}\right\Vert^{2} & = \left\Vert - \boldsymbol{\nabla}\phi -\dfrac{\partial \mathbf{A}}{\partial t}\right\Vert^{2} = \left\Vert \mathbf{\dot{A}}\right\Vert^{2}+\Vert \boldsymbol{\nabla}\phi \Vert^{2}+2\left(\boldsymbol{\nabla}\phi \boldsymbol{\cdot} \mathbf{\dot{A}}\right) \tag{006a}\\ & \nonumber\\ \left\Vert\mathbf{B}\right\Vert^{2} & = \left\Vert\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right\Vert^{2} \equiv \sum^{k=3}_{k=1}\left[\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}-\dfrac{\partial \mathbf{A}}{\partial x_{k}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}\right] \tag{006b} \end{align} The second equation in (006b), that is the identity \begin{equation} \left\Vert\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right\Vert^{2} \equiv \sum^{k=3}_{k=1}\left[\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}-\dfrac{\partial \mathbf{A}}{\partial x_{k}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}\right] \tag{Id-01} \end{equation} is proved in 2. Identities Section. Inserting expressions (006) in (004) the Lagrangian density is \begin{equation} \mathcal{L}=\underbrace{\tfrac{1}{2}\left\Vert \mathbf{\dot{A}}\right\Vert^{2}+\tfrac{1}{2}\Vert \boldsymbol{\nabla}\phi \Vert^{2}+\boldsymbol{\nabla}\phi \boldsymbol{\cdot} \mathbf{\dot{A}}}_{\tfrac{1}{2}\left\Vert - \boldsymbol{\nabla}\phi -\frac{\partial \mathbf{A}}{\partial t}\right\Vert^{2}}-\tfrac{1}{2}c^{2}\underbrace{\sum^{k=3}_{k=1}\left[\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}-\frac{\partial \mathbf{A}}{\partial x_{k}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}\right]}_{\left\Vert \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right\Vert^{2}}+\frac{1}{\epsilon_{0}}\left( -\rho \phi + \mathbf{j}\boldsymbol{\cdot} \mathbf{A}\right) \tag{007} \end{equation}

We rearrange the items in (007) as follows :

\begin{align} \mathcal{L} & = \overbrace{\tfrac{1}{2}\Vert \boldsymbol{\nabla}\phi \Vert^{2}-\frac{\rho \phi}{\epsilon_{0}}+\boldsymbol{\nabla}\phi \boldsymbol{\cdot} \mathbf{\dot{A}}}^{\mathcal{L}_{\phi}=\text{with respect to }\phi}+\tfrac{1}{2}\left\Vert \mathbf{\dot{A}}\right\Vert^{2}+\tfrac{1}{2}c^{2}\sum^{k=3}_{k=1}\left[\frac{\partial \mathbf{A}}{\partial x_{k}} \boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}-\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}\right]+\frac{\mathbf{j} \boldsymbol{\cdot} \mathbf{A}}{\epsilon_{0}} \tag{008a}\\ \mathcal{L} & = \tfrac{1}{2}\Vert \boldsymbol{\nabla}\phi \Vert^{2}-\frac{\rho \phi}{\epsilon_{0}}+\underbrace{\boldsymbol{\nabla}\phi\boldsymbol{\cdot} \mathbf{\dot{A}}+\tfrac{1}{2}\left\Vert \mathbf{\dot{A}}\right\Vert^{2}+\tfrac{1}{2}c^{2}\sum^{k=3}_{k=1}\left[\frac{\partial \mathbf{A}}{\partial x_{k}} \boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}-\Vert\boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}\right]+\frac{\mathbf{j}\boldsymbol{\cdot} \mathbf{A}}{\epsilon_{0}}}_{\mathcal{L}_{\mathbf{A}}=\text{with respect to }\mathbf{A}} \tag{008b} \end{align}

The $\:\mathcal{L}_{\phi}\:$ part of the density contains all $\:\phi$-terms and reasonably will participate alone to the derivation of the Maxwell equation (001c) from the Euler-Lagrange equation (002) with respect to $\:\eta_{4}=\phi\:$. The $\:\mathcal{L}_{\mathbf{A}}\:$ part of the density contains all $\: \mathbf{A}$-terms and reasonably will participate alone to the derivation of the Maxwell equation (001b) from the Euler-Lagrange equations (002) with respect to $\:\eta_{1},\eta_{2},\eta_{3}=A_{1},A_{1},A_{3}\:$. Note the common term $\:\boldsymbol{\nabla}\phi \boldsymbol{\cdot} \mathbf{\dot{A}}\:$ of the parts $\:\mathcal{L}_{\phi},\mathcal{L}_{\mathbf{A}}\:$.

The Euler-Lagrange equation with respect to $\:\eta_{4}=\phi\:$ is : \begin{equation} \dfrac{\partial }{\partial t}\overbrace{\left(\dfrac{\partial \mathcal{L}}{\partial \dot{\phi}}\right)}^{0} +\nabla \boldsymbol{\cdot}\overbrace{\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\phi\right)}\right]}^{\boldsymbol{\nabla}\phi+\mathbf{\dot{A}}}-\overbrace{\frac{\partial \mathcal{L}}{\partial \phi}}^{-\frac{\rho }{\epsilon_{0}}}=0 \tag{009} \end{equation} or \begin{equation} \nabla \boldsymbol{\cdot}\underbrace{\left(-\boldsymbol{\nabla}\phi -\frac{\partial \mathbf{A}}{\partial t}\right)}_{\mathbf{E}}= \frac{\rho }{\epsilon_{0}} \tag{010} \end{equation} that is Maxwell equation (001c) \begin{equation} \nabla \boldsymbol{\cdot}\mathbf{E} = \frac{\rho}{\epsilon_{0}} \tag{001c} \end{equation}

In order to derive Maxwell equation (001b) we express it with the help of equations (005) in terms of the potential 4-vector components $\:A_{1},\:A_{2},\:A_{3},\phi\:$ : \begin{equation} \boldsymbol{\nabla} \boldsymbol{\times} \left(\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right) =\mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial }{\partial t}\left(-\boldsymbol{\nabla}\phi -\frac{\partial \mathbf{A}}{\partial t}\right) \tag{011} \end{equation} Using the identity \begin{equation} \boldsymbol{\nabla} \boldsymbol{\times} \left( \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right) =\boldsymbol{\nabla}\left(\nabla \boldsymbol{\cdot}\mathbf{A}\right)- \nabla^{2}\mathbf{A} \tag{012} \end{equation} eq.(011) yields \begin{equation} \frac{1}{c^{2}}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}-\nabla^{2}\mathbf{A}+ \boldsymbol{\nabla}\left(\nabla \boldsymbol{\cdot} \mathbf{A}+\frac{1}{c^{2}}\frac{\partial \phi}{\partial t}\right) =\mu_{0}\mathbf{j} \tag{013} \end{equation} The $\:k$-component of eq.(013) is expressed properly to look like a Euler-Lagrange equation as follows : \begin{equation} \dfrac{\partial}{\partial t}\left(\frac{\partial \mathrm{A}_{k}}{\partial t}+\frac{\partial \phi}{\partial x_{k}}\right)+\nabla \boldsymbol{\cdot} \left[c^{2}\left(\frac{\partial \mathbf{A}}{\partial x_{k}}- \boldsymbol{\nabla}\mathrm{A}_{k}\right)\right] -\frac{\mathrm{j}_{k}}{\epsilon_{0}}=0 \tag{014} \end{equation} It's sufficient to reach above eq. (014) from the Euler-Lagrange equation (002) with respect $\:\eta_{k}=A_{k},\:\: k=1,2,3\:$ :

\begin{equation} \dfrac{\partial }{\partial t}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{A}_{k}}\right) + \nabla \boldsymbol{\cdot}\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}A_{k}\right)}\right]- \frac{\partial \mathcal{L}}{\partial A_{k}}=0 \tag{015} \end{equation}

Now \begin{equation} \dfrac{\partial\mathcal{L} }{\partial \dot{A}_{k}}=\dfrac{\partial }{\partial \dot{A}_{k}}\left( \boldsymbol{\nabla}\phi \boldsymbol{\cdot} \mathbf{\dot{A}}+\tfrac{1}{2}\left\Vert \mathbf{\dot{A}}\right\Vert^{2}\right)=\frac{\partial \phi}{\partial x_{k}}+\frac{\partial \mathrm{A}_{k}}{\partial t} \tag{016a} \end{equation}

\begin{equation} \frac{\partial \mathcal{L}}{\partial A_{k}}=\frac{\partial }{\partial A_{k}}\left(\frac{\mathbf{j} \boldsymbol{\cdot} \mathbf{A}}{\epsilon_{0}}\right)=\frac{\mathrm{j}_{k}}{\epsilon_{0}} \tag{016b} \end{equation} and \begin{equation} \dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}A_{k}\right)}=\dfrac{\partial}{\partial\left(\boldsymbol{\nabla}A_{k}\right)}\left(\tfrac{1}{2}c^{2}\sum^{k=3}_{k=1}\left[\frac{\partial \mathbf{A}}{\partial x_{k}} \boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}-\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}\right]\right)=c^{2}\left(\frac{\partial \mathbf{A}}{\partial x_{k}}- \boldsymbol{\nabla}\mathrm{A}_{k}\right) \tag{016c} \end{equation} The last equation in (016c) is valid because of the identity (Id-02) proved in 2. Identities Section : \begin{equation} \dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\boldsymbol{\nabla}\mathrm{A}_{k}\right)}=\dfrac{\partial}{\partial \left(\boldsymbol{\nabla}\mathrm{A}_{k}\right)}\left(\sum^{k=3}_{k=1}\left[\frac{\partial \mathbf{A}}{\partial x_{k}} \boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}-\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}\right]\right) =2\left( \boldsymbol{\nabla}\mathrm{A}_{k}-\frac{\partial \mathbf{A}}{\partial x_{k}}\right) \tag{Id-02} \end{equation} Using the expressions of equations (016) the Euler-Lagrange equation (015) gives (014) and so Maxwell equation (001b).

2. Identities Section

If $\: \mathbf{A}= \left( \mathrm{A}_{1}, \mathrm{A}_{2}, \mathrm{A}_{3}\right) \:$ is a vector function of the cartesian coordinates $\:\left( x_{1},x_{2},x_{3}\right)\:$ then \begin{equation} \left\Vert\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right\Vert^{2} \equiv \sum^{k=3}_{k=1}\left[\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}-\dfrac{\partial \mathbf{A}}{\partial x_{k}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}\right] \tag{Id-01} \end{equation} and
\begin{equation} \dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\boldsymbol{\nabla}\mathrm{A}_{k}\right)}=\dfrac{\partial}{\partial \left(\boldsymbol{\nabla}\mathrm{A}_{k}\right)}\left(\sum^{k=3}_{k=1}\left[\frac{\partial \mathbf{A}}{\partial x_{k}} \boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}-\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}\right]\right) =2\left( \boldsymbol{\nabla}\mathrm{A}_{k}-\frac{\partial \mathbf{A}}{\partial x_{k}}\right) \tag{Id-02} \end{equation} where the functional derivative of the left hand side is defined as \begin{equation} \dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\boldsymbol{\nabla}\mathrm{A}_{k}\right)}\equiv \left[\dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\dfrac{\partial \mathrm{A}_{k}}{\partial x_{1}}\right)},\dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\dfrac{\partial \mathrm{A}_{k}}{\partial x_{2}}\right)},\dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\dfrac{\partial \mathrm{A}_{k}}{\partial x_{3}}\right)} \right] \tag{Id-03} \end{equation} Proof of equation (Id-01) : \begin{eqnarray*} && \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2} =\left(\frac{\partial A_{3}}{\partial x_{2}}-\frac{\partial A_{2}}{\partial x_{3}}\right)^{2}+\left(\frac{\partial A_{1}}{\partial x_{3}}-\frac{\partial A_{3}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{2}}{\partial x_{1}}-\frac{\partial A_{1}}{\partial x_{2}}\right)^{2}\\ %---------------------------------------- &=& \left[\left(\frac{\partial A_{1}}{\partial x_{2}}\right)^{2}+\left(\frac{\partial A_{1}}{\partial x_{3}}\right)^{2}\right]+\left[\left(\frac{\partial A_{2}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{2}}{\partial x_{3}}\right)^{2}\right]+\left[\left(\frac{\partial A_{3}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{3}}{\partial x_{2}}\right)^{2}\right] \\ %---------------------------------------- &&-2\left[\frac{\partial A_{1}}{\partial x_{2}}\frac{\partial A_{2}}{\partial x_{1}} +\frac{\partial A_{2}}{\partial x_{3}}\frac{\partial A_{3}}{\partial x_{2}}+\frac{\partial A_{3}}{\partial x_{1}}\frac{\partial A_{1}}{\partial x_{3}}\right]\\ %---------------------------------------- &=& \left[\left(\frac{\partial A_{1}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{1}}{\partial x_{2}}\right)^{2}+\left(\frac{\partial A_{1}}{\partial x_{3}}\right)^{2}\right] +\left[\left(\frac{\partial A_{2}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{2}}{\partial x_{2}}\right)^{2}+\left(\frac{\partial A_{2}}{\partial x_{3}}\right)^{2}\right] \\ %---------------------------------------- &&+\left[\left(\frac{\partial A_{3}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{3}}{\partial x_{2}}\right)^{2}+\left(\frac{\partial A_{3}}{\partial x_{3}}\right)^{2}\right] -\left[\left(\frac{\partial A_{1}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{2}}{\partial x_{2}}\right)^{2}+\left(\frac{\partial A_{3}}{\partial x_{3}}\right)^{2}\right]\\ %---------------------------------------- &&-2\left[\frac{\partial A_{1}}{\partial x_{2}}\frac{\partial A_{2}}{\partial x_{1}}+\frac{\partial A_{2}}{\partial x_{3}}\frac{\partial A_{3}}{\partial x_{2}}+\frac{\partial A_{3}}{\partial x_{1}}\frac{\partial A_{1}}{\partial x_{3}}\right]\\ %---------------------------------------- &=& \Vert \boldsymbol{\nabla}\mathrm{A}_{1}\Vert^{2}+\Vert \boldsymbol{\nabla}\mathrm{A}_{2}\Vert^{2}+\Vert \boldsymbol{\nabla}\mathrm{A}_{3}\Vert^{2}-\left(\frac{\partial A_{1}}{\partial x_{1}}\frac{\partial A_{1}}{\partial x_{1}}+\frac{\partial A_{2}}{\partial x_{1}}\frac{\partial A_{1}}{\partial x_{2}}+ \frac{\partial A_{3}}{\partial x_{1}}\frac{\partial A_{1}}{\partial x_{3}} \right)\\ %---------------------------------------- &&-\left(\frac{\partial A_{1}}{\partial x_{2}}\frac{\partial A_{2}}{\partial x_{1}}+\frac{\partial A_{2}}{\partial x_{2}}\frac{\partial A_{2}}{\partial x_{2}}+ \frac{\partial A_{3}}{\partial x_{2}}\frac{\partial A_{2}}{\partial x_{3}} \right)-\left(\frac{\partial A_{1}}{\partial x_{3}}\frac{\partial A_{3}}{\partial x_{1}}+\frac{\partial A_{2}}{\partial x_{3}}\frac{\partial A_{3}}{\partial x_{2}}+ \frac{\partial A_{3}}{\partial x_{3}}\frac{\partial A_{3}}{\partial x_{3}} \right)\\ %---------------------------------------- &=& \Vert \boldsymbol{\nabla}\mathrm{A}_{1}\Vert^{2}+\Vert \boldsymbol{\nabla}\mathrm{A}_{2}\Vert^{2}+\Vert \boldsymbol{\nabla}\mathrm{A}_{3}\Vert^{2}- \frac{\partial \mathbf{A}}{\partial x_{1}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{1}-\frac{\partial \mathbf{A}}{\partial x_{2}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{2}-\frac{\partial \mathbf{A}}{\partial x_{3}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{3}\\ %---------------------------------------- &=&\sum^{k=3}_{k=1}\left[\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}-\frac{\partial \mathbf{A}}{\partial x_{k}} \boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}\right] \end{eqnarray*} Proof of equation (Id-02) : From equation \begin{eqnarray*} && \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2} =\left(\frac{\partial A_{3}}{\partial x_{2}}-\frac{\partial A_{2}}{\partial x_{3}}\right)^{2}+\left(\frac{\partial A_{1}}{\partial x_{3}}-\frac{\partial A_{3}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{2}}{\partial x_{1}}-\frac{\partial A_{1}}{\partial x_{2}}\right)^{2}\\ %---------------------------------------- &=& \left[\left(\frac{\partial A_{1}}{\partial x_{2}}\right)^{2}+\left(\frac{\partial A_{1}}{\partial x_{3}}\right)^{2}\right]+\left[\left(\frac{\partial A_{2}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{2}}{\partial x_{3}}\right)^{2}\right]+\left[\left(\frac{\partial A_{3}}{\partial x_{1}}\right)^{2}+\left(\frac{\partial A_{3}}{\partial x_{2}}\right)^{2}\right] \\ %---------------------------------------- &&-2\left[\frac{\partial A_{1}}{\partial x_{2}}\frac{\partial A_{2}}{\partial x_{1}} +\frac{\partial A_{2}}{\partial x_{3}}\frac{\partial A_{3}}{\partial x_{2}}+\frac{\partial A_{3}}{\partial x_{1}}\frac{\partial A_{1}}{\partial x_{3}}\right] \end{eqnarray*} we have \begin{eqnarray*} \dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\dfrac{\partial \mathrm{A}_{1}}{\partial x_{1}}\right)} &=& 0 =2\left(\dfrac{\partial \mathrm{A}_{1}}{\partial x_{1}}-\dfrac{\partial \mathrm{A}_{1}}{\partial x_{1}} \right)\\ \dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\dfrac{\partial \mathrm{A}_{1}}{\partial x_{2}}\right)} &=& 2\left(\dfrac{\partial \mathrm{A}_{1}}{\partial x_{2}}-\dfrac{\partial \mathrm{A}_{2}}{\partial x_{1}} \right) \\ \dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\dfrac{\partial \mathrm{A}_{1}}{\partial x_{3}}\right)} &=& 2\left(\dfrac{\partial \mathrm{A}_{1}}{\partial x_{3}}-\dfrac{\partial \mathrm{A}_{3}}{\partial x_{1}} \right) \end{eqnarray*} So \begin{equation*} \dfrac{\partial \left( \left|\!\left| \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right|\!\right|^{2}\right) }{\partial \left(\boldsymbol{\nabla}\mathrm{A}_{1}\right)}= 2\left( \boldsymbol{\nabla}\mathrm{A}_{1}-\frac{\partial \mathbf{A}}{\partial x_{1}}\right) \end{equation*} proving equation (Id-02) for $\:k=1\:$ and similarly for the other two components $\:k=2,3$.

$\endgroup$

protected by Qmechanic Jan 20 '16 at 8:09

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