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Maxwell's equations in tensor notation read:

\begin{align} \partial_\mu F^{\mu\nu} &= J^\nu \\ \partial_{[\lambda}F_{\mu\nu]} &= 0 \end{align}

Consider doing a general coordinate transformation $x^\mu \rightarrow x^{\mu'}$ on the first equation. (NB: everything that follows applies also to the second equation.) Writing the equation in primed coordinates and then expanding in terms of unprimed coordinates, we find that the equation transforms to:

\begin{equation} \frac{\partial x^\lambda}{\partial x^{\mu '}} \frac{\partial^2 x^{\mu '}}{\partial x^\lambda \partial x^\mu} \frac{\partial x^{\nu '}}{\partial x^\nu} F^{\mu\nu} + \frac{\partial x^\lambda}{\partial x^{\mu '}} \frac{\partial x^{\mu '}}{\partial x^\mu} \frac{\partial^2 x^{\nu '}}{\partial x^\lambda \partial x^\nu} F^{\mu\nu} + \frac{\partial x^\lambda}{\partial x^{\mu '}}\frac{\partial x^{\mu '}}{\partial x^{\mu }}\frac{\partial x^{\nu '}}{\partial x^{\nu }} \frac{\partial}{\partial x^\lambda} F^{\mu\nu} = \frac{\partial x^{\nu '}}{\partial x^{\nu }} J^{\nu } \end{equation}

A sufficient condition for the equation to be invariant under this transformation is that the first two terms on the left hand side vanish, and a sufficient condition for that to happen is that:

\begin{equation} \frac{\partial^2 x^{\mu '}}{\partial x^\lambda \partial x^\mu} = 0 \end{equation}

Integrating this equation, we find that this Maxwell equation will be invariant under a linear coordinate transformation:

\begin{equation} x^{\mu '} = M{^{\mu'}_{\ \ \mu}} x^\mu + a^{\mu'} \end{equation}

Here, $M{^{\mu'}_{\ \ \mu}}$ is a constant matrix and $a^{\mu'}$ is a constant vector.

Formally, this is true for all linear transformations, not just Lorentz transformations. Of course, one can appeal to the existence of a Minkowski metric field to restrict $M{^{\mu'}_{\ \ \mu}}$ to be a Lorentz matrix. However, this does not change the fact that this equation seems to be formally invariant under all linear transformations. And I didn't think it was meant to be true that Maxwell's equations were invariant under all linear transformations!

So: can someone sort me out here? Are the two equations above actually invariant under all linear transformations, or have I made an error here?

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2 Answers 2

up vote 4 down vote accepted

There is an additional condition coming from the third term on the left hand side of your transformed equation, where you have used what seemed to be the chain rule $$ \frac{\partial x^\lambda}{\partial x^{\mu'}}\frac{\partial x^{\mu'}}{\partial x^\mu}=\delta^\lambda_\mu $$ In reality however, the nontrivial positioning of the covariant and contra variant vector components makes this equation contain more than just the chain rule, and in face is what restricts $M^\mu_\nu$ to be a lorentz transformation.

Without lowering and raising, this gives us the known fact that the jacobian matrix is orthogonal. Taking the index positioning into account this gives us the condition $$ M^T \eta M = \mathbb{1} $$ Otherwise all scalar products in general relativity will be invariant under a general linear coordinate transformation rather than a lorentz transformation, when you require the transformation to be global.

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  1. Let us first replace the Minkowski metric tensor $$\eta~=~\eta_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$$ with a more general constant metric tensor $$g~=~g_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}.$$

  2. Note that the raised EM tensor $$F^{\mu\nu}~:=~g^{\mu\lambda} F_{\lambda\kappa}g^{\kappa\nu}$$ depends on the (inverse) metric. The Maxwell equations are covariant under rigid $GL(4)$ transformations if we remember to transform the (inverse) metric $g^{\mu\nu}$ accordingly.

  3. On the other hand, if the metric components $g_{\mu\nu}$ are always supposed to be equal to the Minkowski metric $\eta_{\mu\nu}={\rm diag}(\pm 1,\mp 1,\mp 1,\mp 1)$, then the rigid transformations $$\Lambda^{\mu}{}_{\nu}=\frac{\partial x^{\prime \mu}}{\partial x^{\nu}}$$ must be Lorentz matrices.

  4. Finally, let us mention that it is possible to write Maxwell equations in a general curved spacetime, so that they are general covariant.

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1  
I've never seen the notation $\odot$. Why not $\otimes$ - the metric is a tensor, after all? –  ACuriousMind Apr 14 at 19:29
2  
@ACuriousMind: The notation $\odot$ (or $\vee$) denotes the symmetrized $\otimes$ tensor product in the same way that $\wedge$ denotes the antisymmetrized $\otimes$ tensor product. –  Qmechanic Apr 14 at 19:30

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