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

electromagnetismspecial-relativity

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.

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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

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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 $$$$A^\mu\e A^\mu\plr{\phi}, \tl{40.1}$$$$ and satisfies the Lorenz gauge condition $$$$\partial_\mu A^\mu \e k_\mu A^{\mu\,\prime}\e 0, \nonumber$$$$ the prime denoting differentiation with respect to $$\:\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.

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

We have also the following scalar $$$$\phi \e k_\mu x^\mu \e k_0x^0\p k_1x^1\p k_2x^2\p k_3x^3 \e k_0x_0 \m \mb k\bl\cdot\mb x \e \omega t\m\mb k\bl\cdot\mb x \tl{C-06}$$$$ the phase of the plane wave. Note that the phase $$\:\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 $$$$A^\mu\e A^\mu\plr{\phi} \tl{C-07}$$$$ which satisfies the Lorenz gauge condition $$$$\partial_\mu A^\mu \e 0 \tl{C-08}$$$$ From \eqref{C-08} $$$$\partial_\mu A^\mu \e \dfrac{\mr d A^\mu}{\mr d\phi}\partial_\mu \phi\e \dfrac{\mr d A^\mu}{\mr d\phi}\dfrac{\partial \phi}{\partial x^\mu}\e \dfrac{\mr d A^\mu}{\mr d\phi}k_\mu \tl{C-09}$$$$ We define the following 4-vector $$$$a^\mu\plr{\phi}\stackrel{\texttt{def}}{\bl\equiv} \dfrac{\mr d A^\mu\plr{\phi}}{\mr d\phi}\e \plr{\dfrac{\mr d A^0}{\mr d\phi},\dfrac{\mr d\mb A}{\mr d\phi}}\e \plr{a^0, \mb a}\e \plr{a^0, a^1,a^2,a^3} \tl{C-10}$$$$ the derivative of the 4-potential with respect to the phase $$\:\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 $$$$a_\mu\plr{\phi}\e \dfrac{\mr d A_\mu\plr{\phi}}{\mr d\phi}\e \plr{\dfrac{\mr d A_0}{\mr d\phi},\m\dfrac{\mr d\mb A}{\mr d\phi}}\e \plr{a_0, \m\mb a}\e \plr{a_0, a_1,a_2,a_3} \tl{C-11}$$$$

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

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

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

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 $$\:\mb a$$ : $$$$\boxed{\:\: A_0\e 0 \bl\implies a_0\e 0 \bl\implies \mb a\bl\cdot\mb k \e 0\:\:\vp} \tl{C-22}$$$$ On the other hand with $$\:a_0 \e 0\:$$ equation \eqref{C-19} yields $$$$\boxed{\:\: A_0\e 0 \bl\implies a_0\e 0 \bl\implies \mb E \e \m k_0 \mb a\:\:\vp} \tl{C-23}$$$$ So in case of a frame with $$\: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.