I am deriving Maxwell's equations from a Lagrange field equation and have come across something I can not figure out no matter how hard I try. The problem is in the signs. If we take the Lagrange density to be, \begin{equation} \mathcal L=\frac 12(E^2-B^2) \end{equation} Where I have cut irrelevant terms. The expression for $B$ is, \begin{equation} \boldsymbol{B}=\boldsymbol{\nabla}\times \boldsymbol{A}= \begin{vmatrix} \hat x & \hat y &\hat z\\ \partial _x & \partial _y & \partial _z\\ A_x & A_y & A_z \end{vmatrix}=\underbrace{\bigg(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\bigg)}_{B_x}\hat x +\underbrace{\bigg(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\bigg)}_{B_y}\hat y +\underbrace{\bigg(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\bigg)}_{B_z}\hat z\end{equation} When computing the relevant part of the Euler-Lagrange equation, \begin{align} &=\frac{d}{dx}\bigg(\frac{\partial(- B^2/2)}{\partial (\partial A_{x}/\partial x)}\bigg)+\frac{d}{dy}\bigg(\frac{\partial( -B^2/2)}{\partial (\partial A_{x}/\partial y)}\bigg)+\frac{d}{dz}\bigg(\frac{\partial( -B^2/2)}{\partial (\partial A_{x}/\partial z)}\bigg) \\&= \frac{d}{dx}\bigg(0\bigg)+\frac{d}{dy}\bigg(-\frac{2B_z}{2}\bigg)+\frac{d}{dz}\bigg(-\frac{2B_y}{2}\bigg) \\&= -\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z} \end{align} So I then take the curl of $B$, \begin{equation} \boldsymbol\nabla \times \boldsymbol B= \begin{vmatrix} \hat x & \hat y &\hat z\\ \partial _x & \partial _y & \partial _z\\ B_x & B_y & B_z \end{vmatrix}=\bigg(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\bigg)\hat x+\bigg(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\bigg)\hat y+\bigg(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial _y}\bigg)\hat z\end{equation} And we can see that the $\hat x$ component is almost the same, but my working has an extra minus sign up front. The proof I am working through is found in Electricity and Magnetism for Mathematicians A Guided Path from Maxwell's Equations to Yang–Mills by T.A Garrity.

I am not for a moment suggesting that the source is wrong as there are loads of other books that define the above in exactly the same way, but I cannot make the Euler-Lagrange equation equal the curl of $B$. Thank you for your time!


closed as off-topic by ACuriousMind, Kyle Kanos, Qmechanic Jun 22 '15 at 13:05

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    $\begingroup$ I strongly encourage you to try to compute again the second term of your E-L eq. Keep track of all the minuses ;) $\endgroup$ – Snaporaz Jun 22 '15 at 11:43

Your sign is wrong when computing

$$\frac{\partial{(B^2)}}{\partial{(\partial_{y} A_x)}}.$$

The only term in $B^2$ that contains $\partial_{y} A_x $ is $(B_z)^2 = (\partial_{x}A_y - \partial_{y} A_x)^2 ,$ and clearly by the chain rule,

$$ \frac{\partial{(B_z)^2}}{\partial(\partial_{y}A_x)} = -2B_z$$ which disagrees with what you have by a sign. Fixing that sign gives you the right answer.

By the way, you might wanna take a look at Einstein notation and the Levi-Civita symbol, whose use makes computing such Euler-Lagrange equations much easier.


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