[NOTES/EM-12002]-Electromagnetic Field Tensor and Maxwell's equations

1.Field tensor of e.m. field 

We introduce antisymmetric second rank tensor \(F^{\mu\nu}\) defined by \begin{equation} F^{\mu\nu}= \partial^\mu A^\nu - \partial_\nu A^\mu. \end{equation} The components of \(F\) with \(\mu=\nu\) vanish due to antisymmetry in \(mu\) and \(\nu\). Other components of \(F^{\mu\nu}\) are related to the electric and magnetic fields. To see the connection we explicitly write the expressions for other components. \begin{eqnarray} F^{01} &=& \partial^0 A^1-\partial^1 A^0 = \pp[A_x]{t} + \pp[\phi]{x},\\ F^{12} &=& \partial^1 A^2 - \partial^2 A_1 =-\pp[A_y]{x} + \pp[A_x]{y} = - B_z. \end{eqnarray} Here it must be remembered that \[\partial^0=\partial_0=\pp{t},\text{ and } \big(\partial_1, \partial_2, \partial_3\big)= \left(\pp{x}, \pp{y},\pp{z}\right)= - \big(\partial^1, \partial^2, \partial^3\big).\] Similarly, we have \begin{eqnarray} F^{23} &=& \partial^2 A^3 - \partial^3 A^2 = - \pp[A_z]{y} + \pp[A_y]{z} = - B_x\\ F^{31} &=& \partial^3 A^1 - \partial^1 A^3 = - \pp[A_x]{z} + \pp[A_z]{x} = - B_y \end{eqnarray} We can think of the tensor \(F^{\mu\nu}\) as a \(4\times4\) matrix and write it as \begin{equation}\label{EQ06} F^{\mu\nu}=\left( \begin{array}{rrrr} 0\ & -E_x & -E_y & -E_z \\ E_x & 0\ & -B_z & B_y\\ E_y & B_z & 0\ & -B_x\\ E_z & -B_y & B_x & 0\ \\ \end{array}\right). \end{equation}

2.Transformation of Fields

The electromagnetic field tensor \(F^{\mu\nu}\) has been written as a \(4\times4\) matrix in \eqRef{EQ06}. We will now verify the two of the Maxwell's equations can be written as \begin{equation}\label{EQ02} \partial_\mu F^{\mu\nu}=-\frac{j^\mu}{\epsilon_0}, \quad \nu=0,..,3, \end{equation} in relativistic notation. Consider \(\nu=0 \) part of \EqRef{EQ02}: \begin{eqnarray} \partial_0 F^{00} + \partial_1 F^{10}+ \partial_2 F^{20} + \partial_3 F^{30} &=&\frac{j^0}{\epsilon_0 } \\\nonumber \partial_x E_x+ \partial_y E_y + \partial_z E_z &=&\frac{j^0}{\epsilon_0 }\\ \nabla\cdot\vec{E} &=&\frac{\rho}{\epsilon_0 } \label{EQ03}. \end{eqnarray} Next we take \(\nu=1\) in \EqRef{EQ02}. This gives us \begin{equation} \partial_\nu F^{\nu1} = \frac{j^1}{\epsilon_0} \label{EQ04} \end{equation} Writing out all terms in the summation over \(\nu\) we get \begin{eqnarray} \label{EQ05} \partial_0 F^{01} + \partial_1 F^{11}+ \partial_2 F^{21} + \partial_3 F^{31} = \frac{j^1}{\epsilon_0} \end{eqnarray} Substituting the components of \(F^{\mu\nu}\) from \eqref{\TheFile;EQ01} we get \begin{eqnarray} - \partial_t E_x + 0 + \partial_y B_z - \partial_z B_y = \frac{j^1}{\epsilon_0}\\  (\nabla\times \vec{B})_x = \frac{j_x}{\epsilon_0} + \dd[E_x]{t} \end{eqnarray} This equation can now be combined with the other two equations for \(\nu=2,3\). Restoring appropriate factors, when \(c\ne 1\), we would get \begin{equation} \nabla\times\vec{B} = \mu_0\vec{j} + {\mu_0\epsilon_0}\dd[\vec{E}]{t}. \label{EQ09} \end{equation}

Levi Civita symbol

The Levi Civita symbol \(\varepsilon^{\mu\nu\lambda\sigma}\) is a completely antisymmetric tensor under exchange of any two indices and is defined to have the value \(\varepsilon^{0123}=1\). As usual the indices can be lowered using the metric tensor and we have the tensor \(\varepsilon_{\mu\nu\lambda\sigma}\) defined by, \begin{equation} \epsilon_{\mu\nu\lambda\sigma} = g_{\mu\alpha}g_{\nu\beta}g_{\lambda\gamma} g_{\sigma \delta} g^{\alpha\beta\gamma\delta}. \end{equation} This also is completely antisymmetric under exchange of any pari of indices and has value \(\varepsilon_{0123}=1\). Using Levi Civita symbol, we define the tensor \(G_{\mu\nu}\) by \begin{equation} \label{EQ10} G_{\mu\nu} = \varepsilon_{\mu\nu\lambda\sigma} F^{\lambda\sigma}. \end{equation} By virtue of its definition the dual tensor \(G_{\mu\nu}\) satisfies the equation \begin{equation} \label{EQ11} \partial^\mu G_{\mu\nu} =0 \end{equation} We will now give a proof of the above relation. Using the definition of \(G_{\mu\nu}\) we get \begin{eqnarray} \partial^\mu G_{\mu\nu} &=& \partial^\mu \epsilon_{\mu\nu\lambda\sigma}( F^{\lambda\sigma})\\ &=& \epsilon_{\mu\nu\lambda\sigma} \partial^\mu ( \partial^\lambda A^{\sigma} - \partial^\sigma A^\lambda) \\ &=& \epsilon_{\mu\nu\lambda\sigma} \big[\partial^\mu \partial^\lambda A^{\sigma}-\partial^\mu\partial^\sigma A^\lambda\big]\\ &=& \epsilon_{\mu\nu\lambda\sigma} (\partial^\mu \partial^\lambda A^{\sigma})-\epsilon_{\mu\nu\lambda\sigma}(\partial^\mu\partial^\sigma A^\lambda) \label{EQ18} \end{eqnarray} Consider the first term. Here \(\epsilon\) is antisymmetric under exchange of the indices \(\mu\) and \(\lambda\). It is multiplied by \(\partial^\mu \partial^\lambda A^{\sigma})\) which is symmetric under exchange of \(\mu\) and \(\lambda\) \(\{\because \partial_\mu\partial_\lambda = \partial_\lambda\partial_\mu\} \). Therefore the sum over all values of \(\mu\) and \(\lambda\) vanishes. The second term in \EqRef{EQ18} also vanishes for the same reason. \MkGBx{Prove this result!}

Question For You

Write the components of \(G_{\mu\nu}\) explicitly in terms of the electric and magnetic field and show that the remaining two Maxwell's equations, \begin{equation} \nabla \times \vec{E} = -\dd[\vec{B}]{t}, \qquad \nabla\cdot \vec{B} =0, \end{equation} are contained in \eqref{\TheFile;EQ11}.

References :

1. Sec 26-3 Relativistic transformation of fieldsR. P. Feynman, Robert B. Leighton and Mathew Sands Lectures on Physics, vol-II, B.I. Publications (1964)

2. Sec 11.2 The electromagnetic field tensor

   Jack Vanderlinde, Classical Electromagnetic Theory, 2nd edition, Kulwer Academic Publishers, New York, 2004.