Find the direction of the electromagnetic fields in a frame where $A_0$ is chosen to be $0$


Let the four potential which depends on four product $\phi=kx$ be $A^\mu=A^\mu(\phi).$ In one section of the Landau and Lifshitz QED book, the authors chose a reference frame where $A_0=0,$ $\vec{A}$ is in the direction $x^1$ axis and $\vec{k}$ is along $x^3$ direction. Now they claim that the electric field is along $x^1$ axis and the magnetic field is along $x^2$ axis.

I have attempted to understand this with Maxwell's equation. Lets say we want to find the direction of the electric field and magnetic field:

$$\vec{E}=-\nabla A^0 -\dfrac{\partial \vec{A}}{\partial x^0}=-k_0 \dfrac{d\vec{A}}{d\phi}$$

$$\vec{B}=\nabla \times \vec{A}.$$

Neither of the above seems to be at a specific direction.

Best Answer

$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\mf}[1]{\mathfrak{#1}} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\clr}[1]{\left\{#1\right\}} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\tl}[1]{\tag{#1}\label{#1}}$

Reference :$^{\prime\prime}$ Quantum Electrodynamics $^{\prime\prime}$ by V.B.Beresstetskii, E.M.Lifshitz and L.P.Pitaevskii -Landau and Lifshitz Course of Theoretical Physics, Volume 4 2nd Edition 1982- $\bl\S40.$ An electron in the field of an electromagnetic plane Wave


In above Reference we read :

Dirac's equation can be solved exactly for an electron moving in the field of an electromagnetic plane wave (D. M. Volkov, 1937). The field of a plane wave with wave 4-vector $\:k\:$ ($k^2\e 0$) depends on the 4-coordinates only in the combination $\:\phi\e k\,x$, so that the 4-potential is \begin{equation} A^\mu\e A^\mu\plr{\phi}, \tl{40.1} \end{equation} and satisfies the Lorenz gauge condition \begin{equation} \partial_\mu A^\mu \e k_\mu A^{\mu\,\prime}\e 0, \nonumber \end{equation} the prime denoting differentiation with respect to $\:\phi$.

Later in the same paragraph we read :

...The significance of the components of the 4-vector $\:p$, when the field is present, is more clearly seen in a particular frame of reference chosen so that $A_0 \e 0$. Let the vector $\:\mb A\:$ in this frame be along the $\:x^1\m$axis and $\:\mb k\:$ along the $\:x^3\m$axis; the electric field of the wave is then along $\:x^1\:$ the magnetic field along $\:x^2$, and the wave itself is propagated along $\:x^3$. Then...

The OP's question is how to derive the directions of the electric field, the magnetic field and the wave propagation given above.


Given that in the textbook a 4-metric with signature $\plr{\p\m\m\m}$ is used, we have the following 4-vectors : \begin{equation} x^\mu \e \plr{x^0,\mb x}\e \plr{x^0,x^1,x^2,x^3} \qquad x_\mu \e \plr{x_0,x_1,x_2,x_3}\e\plr{x^0,\m\mb x} \tl{C-01} \end{equation} is the 4-position-vector. Note that with $\:c\e 1\:$ we have $\:x_0\e x^0 \e t$. \begin{equation} k^\mu \e \plr{k^0,\mb k}\e \plr{k^0,k^1,k^2,k^3} \qquad k_\mu \e \plr{k_0,k_1,k_2,k_3}\e\plr{k^0,\m\mb k} \tl{C-02} \end{equation} is the wave 4-vector. This 4-vector is a light-like (null) vector \begin{equation} k^2\e k_\mu k^\mu\e k^2_0\m \Vlr{\mb k}^2\e 0 \tl{C-03} \end{equation} Note that with $\:c\e 1\:$ \begin{equation} \omega \e k_0 \e \ k^0 \e \Vlr{\mb k} \tl{C-04} \end{equation} is the wave angular frequency by convention non-negative. \begin{equation} \!\!\!\!\!\!\!\!\!A^\mu \e \plr{A^0,\mb A}\e \plr{A^0,A^1,A^2,A^3} \qquad A_\mu \e \plr{A_0,A_1,A_2,A_3}\e\plr{A^0,\m\mb A} \tl{C-05} \end{equation} is the 4-potential-vector.

We have also the following scalar \begin{equation} \phi \e k_\mu x^\mu \e k_0x^0\p k_1x^1\p k_2x^2\p k_3x^3 \e k_0x_0 \m \mb k\bl\cdot\mb x \e \omega t\m\mb k\bl\cdot\mb x \tl{C-06} \end{equation} the phase of the plane wave. Note that the phase $\:\phi\:$ is a Lorentz-invariant scalar as the inner product of two 4-vectors in Minkwoski 4-dimensional space.

We consider here an electromagnetic plane wave so that the 4-potential is a function of the phase $\:\phi\:$ as by equation \eqref{40.1} of the textbook \begin{equation} A^\mu\e A^\mu\plr{\phi} \tl{C-07} \end{equation} which satisfies the Lorenz gauge condition \begin{equation} \partial_\mu A^\mu \e 0 \tl{C-08} \end{equation} From \eqref{C-08} \begin{equation} \partial_\mu A^\mu \e \dfrac{\mr d A^\mu}{\mr d\phi}\partial_\mu \phi\e \dfrac{\mr d A^\mu}{\mr d\phi}\dfrac{\partial \phi}{\partial x^\mu}\e \dfrac{\mr d A^\mu}{\mr d\phi}k_\mu \tl{C-09} \end{equation} We define the following 4-vector \begin{equation} a^\mu\plr{\phi}\stackrel{\texttt{def}}{\bl\equiv} \dfrac{\mr d A^\mu\plr{\phi}}{\mr d\phi}\e \plr{\dfrac{\mr d A^0}{\mr d\phi},\dfrac{\mr d\mb A}{\mr d\phi}}\e \plr{a^0, \mb a}\e \plr{a^0, a^1,a^2,a^3} \tl{C-10} \end{equation} the derivative of the 4-potential with respect to the phase $\:\phi$. This is a 4-vector as the ratio of the differential 4-vector $\:\mr d A^\mu\:$ and the differential Lorentz-invariant scalar $\:\mr d \phi$. It depends on the 4-coordinates as the 4-potential $\:A^\mu$(Note : in the textbook the notation $\:A^{\mu\,\prime}\:$ is used for $\:a^\mu$).

From \eqref{C-10} we have \begin{equation} a_\mu\plr{\phi}\e \dfrac{\mr d A_\mu\plr{\phi}}{\mr d\phi}\e \plr{\dfrac{\mr d A_0}{\mr d\phi},\m\dfrac{\mr d\mb A}{\mr d\phi}}\e \plr{a_0, \m\mb a}\e \plr{a_0, a_1,a_2,a_3} \tl{C-11} \end{equation}

The Lorenz gauge condition \eqref{C-08} by equation \eqref{C-09} and the definition \eqref{C-10} yields \begin{equation} \partial_\mu A^\mu \e k_\mu a^\mu \e k_0 a_0\m \mb a\bl\cdot\mb k \e 0 \tl{C-12} \end{equation}

We'll determine now the electric $\:\mb E\:$ and magnetic $\:\mb B\:$ field for the given 4-potential in equations \eqref{C-07},\eqref{C-08}.

The electromagnetic field tensor is

\begin{equation} \begin{split} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!F_{\mu\nu} \e & \quad\partial_\mu A_\nu\m\partial_\nu A_\mu \\ \e & \begin{bmatrix} \plr{\partial_0 A_0\m\partial_0 A_0}&\plr{\partial_0 A_1\m\partial_1 A_0}&\plr{\partial_0 A_2\m\partial_2 A_0}&\plr{\partial_0 A_3\m\partial_3 A_0}\vp\\ \plr{\partial_1 A_0\m\partial_0 A_1}&\plr{\partial_1 A_1\m\partial_1 A_1}&\plr{\partial_1 A_2\m\partial_2 A_1}&\plr{\partial_1 A_3\m\partial_3 A_1}\vp \\ \plr{\partial_2 A_0\m\partial_0 A_2} &\plr{\partial_2 A_1\m\partial_1 A_2}&\plr{\partial_2 A_2\m\partial_2 A_2}&\plr{\partial_2 A_3\m\partial_3 A_2}\vp \\ \plr{\partial_3 A_0\m\partial_0 A_3} &\plr{\partial_3 A_1\m\partial_1 A_3}&\plr{\partial_3 A_2\m\partial_2 A_3}&\plr{\partial_3 A_3\m\partial_3 A_3}\vp \end{bmatrix}\\ \e & \begin{bmatrix} \hp\m 0 & \hp\m E_1 & \hp\m E_2 & \hp\m E_3 \:\:\vp \\ \m E_1 & \hp\m 0 & \m B_3 & \hp\m B_2 \:\:\vp \\ \m E_2 & \hp\m B_3 & \hp\m 0 & \m B_1 \:\:\vp \\ \m E_3 & \m B_2 & \hp\m B_1 & \hp\m 0 \:\:\vp \end{bmatrix}\\ \end{split} \tl{C-13} \end{equation} So \begin{equation} E_j \e \partial_0 A_j\m\partial_j A_0\,,\quad j\e 1,2,3 \tl{C-14} \end{equation} Above equation is the tensorial version of the Maxwell's 3-vector equation \begin{equation} \mb E \e \m \dfrac{\partial \mb A}{\partial t}\m \bl\nabla A_0 \tl{C-15} \end{equation} Now \begin{equation} \partial_0 A_j \e \dfrac{\mr d A_j}{\mr d \phi}\partial_0 \phi \e \dfrac{\mr d A_j}{\mr d \phi}\dfrac{\partial \phi}{\partial x^0 }\e a_j k_0 \e \m k_0 a^j \tl{C-16} \end{equation} and \begin{equation} \partial_j A_0 \e \dfrac{\mr d A_0}{\mr d \phi}\partial_j \phi \e \dfrac{\mr d A_0}{\mr d \phi}\dfrac{\partial \phi}{\partial x^j }\e a_0 k_j \e \m a_0 k^j \tl{C-17} \end{equation} From equations \eqref{C-14},\eqref{C-16} and \eqref{C-17} we have \begin{equation} E_j \e a_0 k^j \m k_0 a^j\,,\quad j\e 1,2,3 \tl{C-18} \end{equation} so \begin{equation} \boxed{\:\: \mb E \e a_0 \mb k \m k_0 \mb a\:\:\vp} \tl{C-19} \end{equation} For the magnetic field \begin{equation} \mb B \e \begin{bmatrix} B_1\vp\\ B_2\vp\\ B_3\vp \end{bmatrix} \e \begin{bmatrix} \partial_3 A_2\m\partial_2 A_3\vp\\ \partial_1 A_3\m\partial_3 A_1\vp\\ \partial_2 A_1\m\partial_1 A_2\vp \end{bmatrix} \e \begin{bmatrix} a_2 k^3\m a_3 k^2\vp\\ a_3 k^1\m a_1 k^3\vp\\ a_1 k^2\m a_2 k^1\vp \end{bmatrix} \e \begin{bmatrix} a^3 k^2 \m a^2 k^3 \vp\\ a^1 k^3 \m a^3 k^1 \vp\\ a_2 k^1 \m a_1 k^2\vp \end{bmatrix} \tl{C-20} \end{equation} that is \begin{equation} \boxed{\:\:\mb B \e \mb k \bl\times \mb a\:\:\vp} \tl{C-21} \end{equation}

The 3-vectors $\:\mb k\:$ and $\:\mb a\:$ are not orthogonal in general. But in a frame with $\:A_0 \e 0\:$ we have also $\:a_0 \e 0\:$ and the Lorenz gauge condition \eqref{C-12} yields orthogonal 3-vectors $\:\mb k\:$ and $\:\mb a$ : \begin{equation} \boxed{\:\: A_0\e 0 \bl\implies a_0\e 0 \bl\implies \mb a\bl\cdot\mb k \e 0\:\:\vp} \tl{C-22} \end{equation} On the other hand with $\:a_0 \e 0\:$ equation \eqref{C-19} yields \begin{equation} \boxed{\:\: A_0\e 0 \bl\implies a_0\e 0 \bl\implies \mb E \e \m k_0 \mb a\:\:\vp} \tl{C-23} \end{equation} So in case of a frame with $\:A_0 \e 0\:$ the set of 3-vectors $\:\clr{\mb a,\mb k \bl\times \mb a,\mb k} \:$ or equivalently $\:\clr{\mb E,\mb B,\mb k} \:$ is a 3-dimensional complete orthogonal system. Taking $\:\mb E\:$ along the $\:x^1\m$axis and $\:\mb k\:$ along the $\:x^3\m$axis the magnetic field $\:\mb B\:$ will be along the $\:x^2\m$axis.

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