This is simply an error in expanding tensor components on my part I'm sure but I am struggling to discover the error - where is the minus sign??

In expressing the laws of classical electromagnetism (Maxwell's Equations) in manifestly covariant form - that is to say, in a form that is fully consistent with tensor transformations - two of Maxwell's equations (Gauss's Law for Magnetism and Faraday's Law of Induction) can be expressed by the following tensor equation: $$ \partial_\mu \tilde{F}^{\mu\nu} = 0 $$

where $ \partial_\mu $ is the four-gradient $\left(\frac{\partial}{\partial(ct)},\vec{\nabla}\right)$ and $\tilde{F}^{\mu\nu}$ is the contravariant form of the dual field strength tensor (dual Electromagnetic tensor). $\mu, \nu = 0, 1, 2, 3 $. I have presented the full matrix at the end for completeness and throughout I am using a (+---) signature.

Gauss's Law for magnetism can readily be realised by evaluating $$ \partial_i \tilde{F}^{i0} = 0 $$

where $i = 1, 2, 3$.

Similarly, Faraday's Law of Induction can be realised by evaluating $$ \partial_\mu \tilde{F}^{\mu i} = 0 $$

where herein lies the problem. Upon expanding this tensor equation into its summed components, I get the following (I have gone into detail to properly explain my line of reasoning (or what is, in fact, my incorrect line of reasoning)) where each of the four equations corresponding to each value of $i$ can be summed because they all equal zero and the last three bracketed terms correspond to the three Cartesian components of the curl: $$ -\frac{1}{c}\left(\frac{\partial B_x}{\partial t} + \frac{\partial B_y}{\partial t} + \frac{\partial B_z}{\partial t} \right) + \frac{1}{c} \left(\frac{\partial E_y}{\partial z} - \frac{\partial E_z}{\partial y} \right) + \frac{1}{c} \left(\frac{\partial E_z}{\partial x} - \frac{\partial E_x}{\partial z} \right) + \frac{1}{c} \left(\frac{\partial E_x}{\partial y} - \frac{\partial E_y}{\partial x} \right) = 0 $$

where the final three bracketed terms are the $x$, $y$ and $z$ components of the curl of the $E$-field, $\vec{\nabla} \times \vec{E}$.

Now, we finally arrive at my problem. Where have I gone wrong in this equation because the minus sign on the first time-derivative term means that simplifying this equation yields: $$ \vec{\nabla} \times \vec{E} = \frac{\partial \vec{B}}{\partial t} $$ which is Faraday's law of Induction without the all-important minus sign which encapsulates Lenz's Law.

I would appreciate a keen eye who can point out my error in calculation. $\\$


As stated, below is the full matrix form of $\tilde{F}^{\mu\nu}$: $$ \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & \frac{E_z}{c} & -\frac{E_y}{c} \\ B_y & -\frac{E_z}{c} & 0 & \frac{E_x}{c} \\ B_z & \frac{E_y}{c} & -\frac{E_x}{c} & 0 \end{pmatrix} $$

  • $\begingroup$ I have not checked your signs, but you have more problems than that, You seem to have also summed over $i$ in your long equation, so that you have only one equation instead of three equations, one for each value of $i$. $\endgroup$
    – mike stone
    Commented Oct 31, 2019 at 13:53
  • $\begingroup$ @mikestone Good point, I had overlooked that initially although since they all equal zero (that is, each of the equations with a different i value) they can be summed just fine and in fact, must be in order to resolve the full vector calculus form of Faraday's law. Am I missing something? $\endgroup$ Commented Oct 31, 2019 at 15:13
  • $\begingroup$ As @mike stone said, your long equation is incorrect. You are missing the unit vectors... since it's a vector equation. Zero or otherwise, you don't add components along different directions without unit-vectors. For your sign issue: check your tensor components. As I see it right now, it's not antisymmetric. What is the form of the (non-dual) field tensor? Do a duality-transformation to recover the non-dual form. Are the signs correct there for Ampere? $\endgroup$
    – robphy
    Commented Oct 31, 2019 at 15:41
  • $\begingroup$ @robphy Ah yes, I have corrected the mistake in the matrix at the bottom. My calculations used the correct version which is anti-symmetric. And yes, thank you for pointing out the unit vector issue. I guess the unit vectors are not explicitly included in the covariant form of the equations. Therefore, each bracket will be the curl of B dotted with the corresponding unit vector then? When you sum them, then it becomes the full curl. I'm still at a loss as to the sign issue through for Lenz's Law. Thanks for the help though. $\endgroup$ Commented Oct 31, 2019 at 16:11
  • $\begingroup$ If you want your sign issues to be pinned down, you should clarify your signature conventions and sign-conventions (e.g. with the source-terms included) and show explicitly how each term in your calculation is obtained. For instance, what is $(\partial_0,\partial_1,\partial_2,\partial_3)$ and $(\partial_0F^{01},\partial_1F^{11},\partial_2F^{21},\partial_3 F^{31})$?(Unfortunately, there are no universal conventions... so you should really specify your conventions. Without this, all I can suggest is checking for consistency under transformations... like getting the field tensor from your dual). $\endgroup$
    – robphy
    Commented Oct 31, 2019 at 20:23

2 Answers 2


You didn't make an error in your calculation.

Well, okay, so you made the obvious error where you tried to flatten everything together. But if we remove that error then you derived (multiplying through by $c$) that $$ -\frac{\partial B_x}{\partial t} + \frac{\partial E_y}{\partial z} - \frac{\partial E_z}{\partial y} = 0.$$ This is a totally correct expression.

Your only error is that you got sloppy in reasoning about minus signs. In fact $$\nabla\times\mathbf E = \nabla\times\begin{bmatrix}E_x\\E_y\\E_z\end{bmatrix} = \begin{bmatrix} \partial_y E_z - \partial_z E_y\\ \partial_z E_x - \partial_x E_z\\ \partial_x E_y - \partial_y E_x\end{bmatrix}$$ so that when we see $\partial_z E_y - \partial_y E_z$ we must immediately see it as $-(\nabla \times \mathbf E)_x.$ Thus the proper promotion of the above equation is $$-\dot{\mathbf B} - \nabla \times \mathbf E = 0$$whereas you had a stray $+$ sign in the middle of those two terms incorrectly.


I use $G^{\mu\nu}$ for the dual tensor. Using the convention $(+,-,-,-)$ $$G^{\mu\nu}=\begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z/c & -E_y/c \\ B_y & -E_z/c & 0 & E_x/c \\ B_z & E_y/c & -E_x/c & 0\end{pmatrix}$$

You seem to have several errors in your derivation. First of all, in the equation $\partial_\mu G^{\mu\nu}=0$, we only sum over repeated indexes, one covarient and one contravarient. Thus, $\nu$ determines four equations, not four parts of one equation. Secondly, you should not have any unit vectors; this equation is entirely component based. Next, when you find the component $G^{\mu'\nu'}$, you first go to the $\mu'$th row (starting at zero), and then the $\nu'$th column. So, $G^{01}=B_x$, not $-B_x$. Finally, you have mixed up the curl by a negative sign: the $x$ component of the vector $\mathbf{\nabla} \times \mathbf{E}$ is $\partial_y E_z - \partial_z E_y$, not $\partial_z E_y - \partial_y E_z$, as you have. Correcting for these three things gives the correct answer. So, for example, the $\nu = 1$ equation would give $$\begin{align} 0 & = \frac{1}{c}\left( \frac{\partial B_x}{\partial t} +\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right) \\ & = \frac{\partial B_x}{\partial t} + (\nabla \times E)_x \end{align}$$ $\nu = 2, 3$ follows in the exact same manner. We can combine these three equations into one vector equation:

  • $\begingroup$ Thanks for your comment. Sadly, I don't think your $G^{\mu\nu}$ is the correct contravariant dual EM tensor. en.wikipedia.org/wiki/Electromagnetic_tensor Wikipedia and my notes have it the way I've written it? Now I get that yours brings out the minus sign but it seemingly just pushes the question back into how can lecture notes and wikipedia get it so wrong (if they have)? I'm not sure what you have written for your matrix (there is an extra $E_y$ btw). $\endgroup$ Commented Nov 1, 2019 at 7:59
  • $\begingroup$ @JamieSmith Ah, I know what went wrong. I use the $(-, +, + , +)$ convention for the Minkowski metric, while Wikipedia and your notes use $(+, -, -, -)$. $\endgroup$ Commented Nov 1, 2019 at 13:23
  • $\begingroup$ I've edited my answer to better help you. Sorry for the confusion! $\endgroup$ Commented Nov 1, 2019 at 13:47

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