# Faraday tensor, antisymmetric rank two

$F^{\mu \mathcal{V}}$ is defined in

http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-7/relativistic-lorentz-force/

How to show that $F^{\mu \mathcal{V}}$$F_{\mu \mathcal{V}}, is a lorentz scalar and how to find its value in terms of vectors, Electric field (E), and magnetic field(B). This is mentioned in properties (point number 3) in wikipedia article https://en.wikipedia.org/wiki/Electromagnetic_tensor How to prove that? - Jesus, the other letter in the superscript isn't a script V but the Greek letter nu, spelled as backslash nu in TeX, just like \mu is mu. F_{\mu\nu}F^{\mu\nu} is manifestly a Lorentz scalar because the indices are nicely contracted and F_{0i} components are E_i, the electric field, while F_{ij} for spatial ij is \epsilon_{ijk}B_k, the magnetic field components. These are kind of basics of tensors and relativity and if you don't know them, it may be hard to isolate what you exactly need to be explained - there may be lots of it. – Luboš Motl Apr 25 at 9:31 add comment ## 2 Answers Geometrically, there's a reason it's also called the Faraday bivector: bivectors represent oriented planes in spacetime, and the Faraday field is just a field of these oriented planes, all with magnitudes and orientations. F^{\mu \nu} F_{\mu \nu} is just the squared magnitude of the Faraday field. This is no more exotic than talking about the magnitude of a vector. - add comment It's really easy. First use the definition of the Faraday Tensor: F_{\mu\nu}\equiv\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}, and then write down the same expression but in an another inertial system of reference, i.e. F'^{\mu\nu}=\Lambda_{\ \alpha}^{\mu}\Lambda_{\ \beta}^{\nu}F^{\alpha\beta}. And using the property of the \Lambda matrix: \eta_{\alpha\beta}=\Lambda_{\ \alpha}^{\mu}\Lambda_{\ \beta}^{\nu}\eta_{\mu\nu}, you will get that F'^{\mu\nu}F'_{\mu\nu}=F^{\mu\nu}F{}_{\mu\nu}, i.e. F^{\mu\nu}F{}_{\mu\nu} is an Lorentz invariant quantity. Then for expressing this in terms of the EM Fields, you should use the matrix expression of F^{\mu\nu}: let say F, in terms of the EM fields. And the quantity F^{\mu\nu}F{}_{\mu\nu} is, in matrix terms, the trace of the matrix FF^T. i.e.$$F^{\mu\nu}F{}_{\mu\nu}=Tr(FF^T)$$------------------------Edited-------------------------- Now, we know that the interval s^2:=x_{\mu}x^{\mu} must be Lorentz invariant, in other words if x'^{\mu}=\Lambda_{\ \nu}^{\mu}x^{\nu} is the world line of a particle measured by an inertial observer that moves with velocity \vec v=v \hat u_x, where$$\Lambda_{\ \nu}^{\mu}=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$then is must be hold that$$x_{\mu}'x'^{\mu}=x_{\mu}x^{\mu}$$where we know that x_{\mu}=\eta_{\mu\nu}x^{\nu} and the metric is$$\eta_{\mu\nu}:=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}=\mathrm{diag}(1,-1,-1,-1)=:\eta^{\mu\nu}$$then we have that$$x_{\mu}'x'^{\mu}=\eta_{\mu\nu}x'^{\nu}x'^{\mu}\eta_{\mu\nu}x'^{\nu}x'^{\mu}=\eta_{\mu\nu}\Lambda_{\ \alpha}^{\nu}x^{\alpha}\Lambda_{\ \beta}^{\mu}x^{\beta}  \eta_{\alpha\beta}x'^{\alpha}x'^{\beta}=\eta_{\mu\nu}\Lambda_{\ \alpha}^{\nu}\Lambda_{\ \beta}^{\mu}x^{\alpha}x^{\beta}$$i.e. the matrix \Lambda must satisfy$$ \eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda_{\ \alpha}^{\nu}\Lambda_{\ \beta}^{\mu} \qquad (1)$$Then using (1),$$F_{\mu\nu}=\eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta}$$and$$F'^{\mu\nu}=\Lambda_{\ \sigma}^{\mu}\Lambda_{\ \rho}^{\nu}F^{\sigma\rho}$$we are going to show that F'^{\mu\nu}F'_{\mu\nu}=F^{\mu\nu}F_{\mu\nu} (here the order does not matter due to F^{\mu\nu} is actually a scalar, the \mu,\nu component of the matrix F). OK, then we have that$$F'^{\mu\nu}F'_{\mu\nu}=F'^{\mu\nu}\eta_{\mu\alpha}\eta_{\nu\beta}F'^{\alpha\beta} =\Lambda_{\ \delta}^{\mu}\Lambda_{\ \varepsilon}^{\nu}F^{\delta\varepsilon}\eta_{\mu\alpha}\eta_{\nu\beta}\Lambda_{\ \sigma}^{\alpha}\Lambda_{\ \rho}^{\beta}F^{\sigma\rho} =\left(\Lambda_{\ \delta}^{\mu}\Lambda_{\ \sigma}^{\alpha}\eta_{\mu\alpha}\right)\left(\Lambda_{\ \varepsilon}^{\nu}\Lambda_{\ \rho}^{\beta}\eta_{\nu\beta}\right)F^{\delta\varepsilon}F^{\sigma\rho} \equiv\eta_{\delta\rho}\eta_{\varepsilon\sigma}F^{\delta\varepsilon}F^{\sigma\rho}\equiv F^{\delta\varepsilon}F_{\delta\varepsilon}$$So$$F'^{\mu\nu}F'_{\mu\nu}= F^{\delta\varepsilon}F_{\delta\varepsilon}$$Here the indices are dummy (they are being summed over), so they don't have to match. So, in conclusion the quantity$F^{\mu\nu}F{}_{\mu\nu}$(or$F{}_{\mu\nu}F^{\mu\nu}$) if is measured in any inertial frame, it would have the same value, i.e. it is an Lorentz invariant. Finally I'll write down the part of the calculation of$F^{\mu\nu}F_{\mu\nu}$in terms of the EM fields, but maybe later. You know, because of the time. But! You should read this question, I asked a few months ago the same, and here is the answer. Of course, if you have questions, ask us. - If you need more explicit expressions please tell me, I'll write them down. – Anuar Apr 26 at 0:12 Please Please write the explicit expressions . – Jishnu Ray Apr 26 at 5:11 It's actually the same. – Anuar Apr 26 at 16:36 @JishnuRay I just added some details. – Anuar Apr 26 at 18:03 As an interesting far aside, you then have the proof that any quantity$H^{\mu \nu \xi ...}H_{\mu \nu \xi ...}$is also Lorentz invariant, as long as the two quantities (called tensor)$H$have the same indices. This is called a contraction in tensor calculation. The value of$F_{\mu \nu}$in terms of the fields$E_{\mu}$or$B_{\mu}$are found using the definition$F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$with$A\equiv \left(\phi , \mathbf{A} \right)\$ the potentials. –  Oaoa Apr 26 at 19:32
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