3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – N0va Aug 15 '16 at 7:54
5
$\begingroup$

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 $\:\mathbf{U}$

\begin{equation} \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} \end{equation}

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

\begin{equation} \mathbf{f}=\dfrac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}=\dfrac{\mathrm{d}\left(m\mathbf{u}\right)}{\mathrm{d}t} \tag{35.4} \end{equation} 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

\begin{equation} \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} \end{equation}

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

\begin{equation} \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} \end{equation} 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) :

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

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):

\begin{equation} \mathbf{U} \boldsymbol{\cdot}\mathbf{F}=0 \quad \Longleftrightarrow \quad m_{o}=\text{constant} \tag{35.9} \end{equation} 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 $\:\mathbf{U}\:$ :

\begin{equation} F_{\mu}=\dfrac{q}{c}\;E_{\mu\nu}\;U^{\nu} \tag{38.1} \end{equation} 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

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

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


(1)ADDENDUM
A 3-dimensional analogue of this 4-dimensional anti-symmetry is this : Let $\:\mathrm{\Omega}\:$ a $\:3\times 3\:$ real matrix \begin{equation} \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} \end{equation} with the following property \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} 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 enter image description here 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: \begin{equation} \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} \end{equation} 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' \begin{equation} \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} \end{equation} 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.

$\endgroup$
  • $\begingroup$ 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 $F_{\mu\nu}=-F_{\nu\mu}$. $\endgroup$ – Emilio Pisanty Aug 18 '16 at 10:50
  • $\begingroup$ @Emilio Pisanty : No objection. You're absolutely right. $\endgroup$ – Frobenius Aug 18 '16 at 11:23
1
$\begingroup$

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 :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.