Is $F^{\mu\nu}$ not the dual tensor to $F_{\mu \nu}$?

I always thought that this was the case, but now I came across the dual tensor defined as $\tilde{F}^{\mu \nu}=\epsilon^{\mu \nu \rho \sigma}F_{\rho \sigma}$.

Can somebody explain to me the difference?

  • $\begingroup$ I too have only seen $F^{\mu\nu}$ as the dual tensor of $F_{\mu\nu}$. Maybe this will help. I see people discuss what you are asking at the comments of an answer. physics.stackexchange.com/q/156857 $\endgroup$ Jan 9 '20 at 16:19
  • 1
    $\begingroup$ Since no one else has mentioned this, I believe $\tilde{F}^{\mu\nu}$ is known as the Hodge dual of $F^{\mu\nu}$. Possible duplicate of this question: physics.stackexchange.com/q/37577/157014. $\endgroup$
    – Philip
    Jan 9 '20 at 17:52

Is not $F^{\mu\nu}$ the dual tensor to $F_{\mu\nu}$?

What you are calling dual above is essentially the musical isomorphism that identifies vectors and covectors when one has a metric tensor. In more common language in Physics it is the process of "raising and lowering indices". Two tensors related in this way are sometimes called physically equivalent, see e.g. "General Relativity for Mathematicians" by Sachs & Wu for the usage of this terminology.

In your context, however, by "dual" one means more precisely the Hodge dual of a differential form. Let me be more precise. A differential form is a totally skew-symmetric covariant tensor. If it has $k$-indices we call it a $k$-form. In a local coordinate system $x^\mu$ it has therefore components $\omega_{\mu_1\dots \mu_k}$ such that $$\omega_{[\mu_1\cdots \mu_k]}=\omega_{\mu_1\cdots \mu_k}\tag{1}$$

In this context, if one is working on a $d$-dimensional manifold (for physical spacetime $d=4$) one may define the Hodge dual of the $k$-form $\omega$ to be the $(d-k)$-form $\star \omega$ which has components $$(\star\omega)_{\mu_1\cdots \mu_{d-k}}=\dfrac{1}{k!}\epsilon^{\nu_1\cdots \nu_k}_{\phantom{\nu_1\cdots \nu_k}\mu_1\cdots \mu_{d-k}}\omega_{\nu_1\cdots \nu_k}.\tag{2}$$

The $\epsilon_{\mu_1\cdots \mu_d}$ are the components of a $d$-form which is called the volume element and which, in general, reads $$\epsilon_{\mu_1\cdots \mu_d} = \sqrt{-g}\varepsilon_{\mu_1\dots \mu_d},\tag{3}$$

where $\sqrt{-g}$ is the square root of the negative of the determinant of the metric tensor and where $\varepsilon_{\mu_1\dots \mu_d}$ is the Levi-Civita symbol with $d$ indices. In Minkowski spacetime with cartesian coordinates with the flat metric $\eta_{\mu\nu}$ this factor is equal to one and therefore the components of the volume element coincide with the components of the Levi-Civita symbol.

Finally, concerning the electromagnetic field, $\tilde{F}^{\mu\nu}$ is nothing more than the Hodge dual of $F_{\mu\nu}$ with the indices raised: $$\tilde{F}^{\mu\nu}=g^{\alpha \mu}g^{\beta \nu}(\star F)_{\alpha\beta}\tag{4}.$$ Notice that since one is working in $d = 4$ dimensions and since $F_{\mu\nu}$ is a $2$-form then its dual $\star F$ is a $2$-form as well.


The covariant components (i.e. with lower indexes) of the electromagnetic field tensor expressed by the electric field $\vec{E}$ and magnetic field $\vec{B}$ (with $c$ being the speed of light) are $$F_{\mu\nu}=\begin{pmatrix} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & -B_z & B_y \\ -E_y/c & B_z & 0 & -B_x \\ -E_z/c & -B_y & B_x & 0 \end{pmatrix}. \tag{1}$$

The contravariant component (i.e. with upper indexes) of the same tensor are $$F^{\mu\nu}=\begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix}. \tag{2}$$

The transformation between the covariant and contravariant components is done by the usual index lowering/raising with the metric $\eta$. $$F_{\mu\nu}=\eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu}$$

From comparing (1) and (2) you see, the covariant and contravariant components differ only by the sign of the electric field components in the time-row and time-column.

The dual tensor is a completely different thing. Using the definition $\tilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$, the definition (1), and the definition of the Levi-Civita symbol $\epsilon$ you get $$\tilde{F}^{\mu\nu}=\begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z/c & -E_y/c \\ B_y & -E_z/c & 0 & E_x/c \\ B_z & E_y/c & -E_x/c & 0 \end{pmatrix}. \tag{3}$$ (I strongly recommend you verify this on paper.)

Comparing (3) with (2) you see, the dual tensor $\tilde{F}$ and the original tensor $F$ essentially differ by the exchange of electric field with magnetic field.


Yes $\tilde{F} ^{\mu \nu}$ is the (Hodge) dual of the field strength tensor, defined conventionally with a factor of a half in comparison to your equation, $$\tilde{F} ^{\mu \nu}:= 1/2\, \epsilon^{\mu\nu\alpha\beta} F_{\alpha \beta}. $$ So I'm wondering whether the issue is the language used. What did you understand before hand by "dual?"

If the confusion is just about raising or lowering indices note that (no tildes) $$F^{\mu \nu} = g^{\mu \alpha} g^{\nu \beta} F_{\alpha \beta} $$ Simply uses the metric to change the position of indices.


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