# What does it mean that the electromagnetic tensor is anti-symmetric?

The Faraday electromagnetic tensor is anti-symmetric. What is the physical meaning of this anti-symmetry?

• If you define it via the vetor potential: $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ it is clearly antisymmetric. With this definition of the Faraday tensor one can write the maxwell equations in a very compact form: see for example here: en.wikipedia.org/wiki/Electromagnetic_tensor. With a symmetric definition the maxwell equations as well as the Lagrange density would look way more complicated. – N0va Aug 15 '16 at 7:54

Abstract

In this answer we'll show that the physical meaning of the anti-symmetry of the electromagnetic tensor is the demand for the electromagnetic (Lorentz) force to be pure, where a force is called pure if it doesn't change the rest mass $\:m_{o}\:$ of any particle moving with any (subluminal) velocity. This demand is equivalent to the Minkowski pseudo-orthogonality of the electromagnetic force 4-vector $\:\mathbf{F}\:$ to any velocity 4-vector $\:\mathbf{U}$.

The elements of the proof of this statement are interspersed in "Introduction to Special Relativity" by Wolfrang Rindler, Ed.1982.

1. The force 4-vector. Pure and Heatlike forces

In the textbook we have the relation between the force 4-vector $\:\mathbf{F}$, the momentum 4-vector $\:\mathbf{P}\:$ and the velocity 4-vector $\:\mathbf{U}$

$$\mathbf{F}=\dfrac{\mathrm{d}\mathbf{P}}{\mathrm{d}\tau}=\dfrac{\mathrm{d}}{\mathrm{d}\tau}\left(m_{o}\mathbf{U}\right) \tag{35.3}$$

Then the definition of the (relativistic) three-force by

$$\mathbf{f}=\dfrac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}=\dfrac{\mathrm{d}\left(m\mathbf{u}\right)}{\mathrm{d}t} \tag{35.4}$$ where $\:\mathbf{p},m\:$ the relativistic momentum and mass of the particle on which $\:\mathbf{f}\:$ acts. Then the expression of (35.3) in the form

$$\mathbf{F}=\gamma\left(u\right)\dfrac{\mathrm{d}}{\mathrm{d}t}\left(mc,\mathbf{p}\right)=\gamma\left(u\right)\left(c\dfrac{\mathrm{d}m}{\mathrm{d}t},\mathbf{f}\right) \tag{35.5}$$

Now the author considers a particle with 4-velocity $\:\mathbf{U}=\gamma\left(u\right)\left(c,\mathbf{u}\right)\:$ being subjected to a 4-force $\:\mathbf{F}=\gamma\left(u\right)\left(c\;\mathrm{d}m/ \mathrm{d}t,\mathbf{f}\right)\:$. Their inner product in Minkowski space is

$$\mathbf{U} \boldsymbol{\cdot}\mathbf{F}=\gamma^{2}\left(u\right) \left( c^{2}\dfrac{\mathrm{d}m}{\mathrm{d}t}-\mathbf{f}\cdot \mathbf{u} \right)=c^{2}\dfrac{\mathrm{d}m_{o}}{\mathrm{d} \tau}=\gamma\left(u\right) c^{2}\dfrac{\mathrm{d}m_{o}}{\mathrm{d} t} \tag{35.7}$$ where the second equation results from specializing the first to the rest frame of the particle, and the third is a consequence of (23.2) :

$$\dfrac{ \mathrm{d} t }{ \mathrm{d} \tau}= \left( 1 - \dfrac{u^{2}}{ c^{2}} \right)^{-1/2}=\gamma\left(u\right) \tag{23.2}$$

After these the author proceeds as follows : We shall call a force pure if it does not change a particle's rest mass, and heatlike if it does not change a particle's velocity. For observe from (35.4) that a particle which is at rest but being heated (e.g.by a candle held under it) in one frame, experiences a 'force' $\:\mathbf{f} = \left(\mathrm{d}m_{o}/\mathrm{d}t \right) \gamma\left(u\right)\mathbf{u} \:$ in every other frame.(2) The necessary and sufficient condition for a force to be pure follows at once from (35.7):

$$\mathbf{U} \boldsymbol{\cdot}\mathbf{F}=0 \quad \Longleftrightarrow \quad m_{o}=\text{constant} \tag{35.9}$$ while that for a force to be heatlike is $\:\mathbf{A}=\boldsymbol{0}\:$ (by $\:\mathbf{A}\:$ the author means the 4-acceleration vector).

2. The electromagnetic tensor and its anti-symmetry

In a next Section ( 38. The formal structure of Maxwell's theory ) the author refers that the simplest case is that of a force which everywhere depends linearly on the velocity of the particles on which it acts. In special relativity it is natural to make this requirement on the respective 4-vectors $\:\mathbf{F}\:$ and $\:\mathbf{U}\:$ :

$$F_{\mu}=\dfrac{q}{c}\;E_{\mu\nu}\;U^{\nu} \tag{38.1}$$ where the 'coefficients' $\;E_{\mu\nu}\;$ in this linear relation must be tensorial to make it Lorentz-invariant if - as is natural - we take the charge $\:q\:$ of the particle in question to be a scalar invariant.

If we next demand that the force (38.1) be pure, we need

$$F_{\mu}U^{\mu}=(q/c)\;E_{\mu\nu}\;U^{\mu}U^{\nu}=0, \nonumber$$ for all $\:U^{\mu}\:$, and hence

$$E_{\mu\nu} = - E_{\nu\mu} \tag{38.2}$$ i.e. the field tensor must be anti-symmetric.(1)

A 3-dimensional analogue of this 4-dimensional anti-symmetry is this : Let $\:\mathrm{\Omega}\:$ a $\:3\times 3\:$ real matrix $$\mathrm{\Omega}= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}, \quad a_{\imath \jmath} \in \mathbb{R} \tag{A-01}$$ with the following property $$\langle\mathrm{\Omega}\mathbf{x},\mathbf{x}\rangle=0, \quad \text{ for any } \mathbf{x}=\left( x_{1},x_{2},x_{3} \right) \in \mathbb{R}^{3} \tag{A-02}$$ What this property does mean for $\:\mathrm{\Omega}\:$?
The inner product of (A-02) is explicitly \begin{align} \langle\mathrm{\Omega}\mathbf{x},\mathbf{x}\rangle &=a_{11}x^{2}_{1}+a_{22}x^{2}_{2}+a_{33}x^{2}_{3} \tag{A-03}\\ & +\left( a_{12}+a_{21} \right)x_{1}x_{2}+\left( a_{23}+a_{32} \right)x_{2}x_{3}+\left( a_{31}+a_{13} \right)x_{3}x_{1}=0 \nonumber \end{align}
In order for this equality to be valid for any $\:\mathbf{x}=\left( x_{1},x_{2},x_{3} \right) \in \mathbb{R}^{3}\:$ we must have \begin{align} a_{11}=a_{22}=a_{33} & =0 \tag{A-04a}\\ a_{21}=-a_{12} & = \omega_{3} \tag{A-04b}\\ a_{32}=-a_{23} & = \omega_{1} \tag{A-04c}\\ a_{13}=-a_{31} & = \omega_{2} \tag{A-04d} \end{align} and so $$\mathrm{\Omega}= \begin{bmatrix} 0 & -\omega_{3} & \omega_{2} \\ \omega_{3} & 0 & -\omega_{1} \\ -\omega_{2} & \omega_{1} & 0 \end{bmatrix} =\boldsymbol{\omega}\boldsymbol{\times} \tag{A-05}$$ that is the matrix $\:\mathrm{\Omega}\:$ is anti-symmetric and represents the outer product $\:\boldsymbol{\omega}\boldsymbol{\times}\:$.
Now it's clear why (A-02) could be valid for any $\:\mathbf{x}=\left( x_{1},x_{2},x_{3} \right) \in \mathbb{R}^{3}\:$ : the 3-vector $\:\boldsymbol{\omega}\boldsymbol{\times}\mathbf{x}\:$ is always orthogonal to $\:\mathbf{x}\:$.

(2)FIGURE In the rest system $\:\mathcal{S}_{o}\:$ of a particle, see ($\alpha$), heat is transfered to the particle with rate $\:\dot{\mathrm{q}}_{o}\:$. This rate is with respect to the proper time $\:\tau\:$ and this heat power changes the rest mass $\:m_{o}\:$ of the particle: $$\dot{\mathrm{q}}_{o}=\dfrac{\mathrm{d}\left(m_{o}c^{2}\right)}{\mathrm{d}\tau}=c^{2}\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau} \tag{B-01}$$ In an other inertial system $\:\mathcal{S}\:$ moving with constant 3-velocity $\:-\mathbf{w}\:$ with respect to $\:\mathcal{S}_{o}\:$, the particle is moving with constant velocity $\:\mathbf{w}\:$, see ($\beta$), under the influence of a 'force' $$\boldsymbol{\mathcal{h}}=\dfrac{\dot{\mathrm{q}}_{o}}{c^{2}}\mathbf{w}=\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau}\mathbf{w}=\gamma(w)\dfrac{\mathrm{d}m_{o}}{\mathrm{d} t}\mathbf{w} \tag{B-02}$$ This 'force' $\:\boldsymbol{\mathcal{h}}\:$, although acting on the particle, keeps its velocity $\:\mathbf{w}\:$ constant. So, its 3-acceleration is $\:\mathbf{a}=\mathrm{d}\mathbf{w}/\mathrm{d}t =\boldsymbol{0}\:$ and consequently its 4-acceleration $\:\mathbf{A}=\boldsymbol{0}$. This 'force' is defined as heatlike.

• This is good but honestly it's a bit wordy. You could quickly summarize it as simply saying that if $\frac{d u^\mu}{d\tau}=F^\mu_{\phantom{\mu} \nu}u^\nu$, then the only way to have $\frac{d}{d\tau}\left[u^\mu u_\mu\right]=0$ is to have $F_{\mu\nu}=-F_{\nu\mu}$. – Emilio Pisanty Aug 18 '16 at 10:50
• @Emilio Pisanty : No objection. You're absolutely right. – Frobenius Aug 18 '16 at 11:23

One important conclusion that you can draw from the antisymmetry of the electromagnetic field tensor (plus the inhomogeneous Maxwell equation in covariant form) is the continuity equation:

$\partial^{\rho} J_{\rho} =\frac{1}{\mu_0} \partial^{\rho} \partial^{\nu} F_{\nu \rho} = - \frac{1}{\mu_0} \partial^{\rho} \partial^{\nu} F_{\nu \rho} \implies \partial^{\rho} J_{\rho} = 0$

Where the first equality is by the inhomogeneous Maxwell equation and the second equality comes from renaming the dummy indices and the antisymmetry of $F$.

So you could argue that antisymmetry of $F$ encodes the Continuity equation. Whether there is more significance to it I don't know - but I'm happy to learn more :)