[NOTES/EM-12003]-Fields of a Moving Point Charge

The electromagnetic field tensor \(F^{\mu\nu}\) is a second rank tensor whose components transform like product of two four vectors. We shall follow Feynman and derive the Lorentz transformation property of the electric and magnetic fields. Let \(P\) be a space time point whose coordinates are \(x^\mu=(t, \vec{x})\) and \(x^{\mu\prime}=(t^\prime, \vec{x}^\prime)\) in two Lorentz frames \(K\) and \(K^\prime\). If \(K^\prime\) is moving w.r.t. \(K\) along the \(x\) axis with velocity \(v\), then we have % \begin{eqnarray} t^\prime = \gamma(t-vx), \qquad x^\prime= \gamma(x-vt),\quad y^\prime=y, \quad z^\prime=z. \label{E1} \end{eqnarray} where \(\gamma=\frac{1}{\sqrt{1-v^2}}\). The inverse transformations are % \begin{eqnarray} t = \gamma(t^\prime + vx), \qquad x= \gamma(x^\prime + vt^\prime),\quad y=y^\prime, \quad z=z^\prime. \label{E110} \end{eqnarray} % For purposes deriving transformation of electric and magnetic fields, we regard the field tensor \(F^{\mu\nu}\) as a product of two four vectors \(a^\mu, b^{\mu}\) \begin{equation} F^{\mu\nu} \sim a^\mu b^\nu - a^\nu b^\mu. \end{equation} Here we have formed antisymmetric combination \(a^\mu b^\nu - a^\nu b^\mu\) since \(F^{\mu\nu}\) is antisymmeric tensor in \(\mu\) and \(\nu\). The transformations four vectors \(a^\mu\) and \(b^\mu\) are given by \begin{eqnarray} a^{0\,\prime}=\gamma(a^0-va^1) &\quad& b^{0\,\prime}=\gamma(b^0-vb^1), \label{EQ04}\\ a^{1\,\prime}=\gamma(a^1-va^0) &\quad& b^{1\,\prime}=\gamma(b^1-vb^0),\label{EQ05}\\ a^{2\,\prime} = a^2, a^{3\,\prime} =a^3 &\quad& b^{2\,\prime} = b^2 b^{3\,\prime} =b^3,\label{EQ06} \end{eqnarray} for the components of the four vectors \(a^\mu, b^\mu\). We also need the inverse transformations \begin{eqnarray} a^0=\gamma({a^{0\,\prime} + va^{1\,\prime}}) &\quad& b^0=\gamma({b^{0\,\prime}+ vb^{1\,\prime}})\\ a^1=\gamma({a^{1\,\prime}+ va^{0\,\prime}}) &\quad& b^1=\gamma({b^{1\, \prime}+ vb^{0\,\prime}})\\ a^2 = a^{2\,\prime}, a^3= a^{3\,\prime} &\quad& b^2 = b^{2\,\prime}, b^3= b^{3\,\prime} \end{eqnarray} % Let us now start computing the transformation of components of \(F^{\mu\nu}\). \begin{eqnarray} E_x^\prime &=& F^{10^\prime} = a^{1\prime}b^{0\prime} - a^{0\prime} b^{1\prime}\\ &=&{\gamma}({a^1-va^0})\times{\gamma}(b^0-vb^1) - \gamma ({b^1-vb^0})\times{\gamma}(a^0-va^1)\\ & & \qquad \qquad \text{ using \eqRef{EQ04} and \eqRef{EQ05}} \nonumber\\ &=& a^1b^0-b^1a^0\\ &=& F^{1,0}=E_x. \end{eqnarray} Similarly, we can compute transformed magnetic field \begin{eqnarray} B_y^\prime &=& F^{13\prime} = a^{1\prime} b^{3\prime} - a^{3\prime}b^{1\prime}\\ &=& \gamma({a^1-v a^0}) b^3 - \gamma a^3 ({b^1-vb^0})\\ &=& \gamma[(a^1b^3 - a^3 b^1)- v(a^0b^3- a^3 b^0 )]\\ &=& \gamma(F^{13}-v F^{03})\\ &=& \gamma(B_x + v E_z) \end{eqnarray} Transformation of other components can be worked out in a similar fashion and we would get \begin{eqnarray} E_x^\prime= E_x, && B_x^\prime= B_x;\\ E_y^\prime=\gamma(E_y-v B_x), && B_y^\prime = \gamma(B_y+vE_z);\\ E_z^\prime= \gamma(E_z+vB_y), && B_z^\prime = \gamma(B_z-vE_y). \end{eqnarray} Putting fact factors of \(c\), we can write the transformations of the fields as \begin{equation} \begin{array}{lll} E^\prime_\parallel = E_\parallel &\qquad & B^\prime_\parallel= B_\parallel\\ E_\perp^\prime =\gamma(E+\vec{v}\times\vec{B})_\perp &\qquad & B^\prime_\perp =\gamma(B-\vec{v}\times \vec{E}/c^2) \end{array} \end{equation} The transformation property of Maxwell field tensor \(F^{\mu\nu}\) can be written in an alternate, matrix, form as follows. Let us introduce the notation \begin{eqnarray} \underline{\sf a} &=& \begin{pmatrix} a^0 & a^1 & a^2 & a^3\end{pmatrix}\\ \underline{\Lambda} &=& \begin{pmatrix} \gamma & - v\gamma & 0 & 0 \\ -v \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{pmatrix}. \end{eqnarray} For a contravariant vector \(a^\mu\) the Lorentz transformation properties are, then, given by \begin{equation} \underline{\sf a^\prime} = \underline{\Lambda}\underline{\sf a}. \end{equation} For the field tensor we have \begin{equation} F^\prime(x^\prime) = \underline{\Lambda}F(x) \underline{\Lambda}^\text{T}. \end{equation} Here \(F\) is the \(4\times4\) matrix formed from the components of the electric and magnetic fields and \(\underline{\Lambda}^\text{T}\) denotes the transpose of the matrix \(\underline{\Lambda}\).

Lorentz Force The equation of motion of a charged particle in electric and magnetic field is given by \begin{equation} \dd[\vec{p}]{t} = q (\vec{E} + \vec{v}\times \vec{B}). \end{equation} We want to know what form this equation takes in the relativistic notation. Remember that the concept of time is relative to he frame of reference. If we consider two nearby points \(A, B\) on trajectory of a particle the corresponding time intervals as seen in two frames, \(\Delta t\) and \(\Delta t^\prime\), will be different. So how are we to compare the rate of change of momentum? The answer is that the points on a trajectory can be paramterised by any parameter and that we should use an invariant parameter \(\tau\). One such parameter is proper time \(s\).  So instead of using time interval we use proper time interval \[ ds = \sqrt{c^2(dt)^2 - dx^2 -dy^2 -dt^2}.\] Then \(ds\) is same in all frames and \begin{equation} \pp[x^\mu]{s} = \frac{p^\mu}{m}, \end{equation} where \(p^\mu\) is the four momentum and the equation of motion can be written as \begin{equation} m \DD[x^\mu]{s} = v_\nu F^{\mu\nu}. \end{equation}

References

Sec 26-4 The equations of motion in relativistic notation R. P. Feynman, Robert B. Leighton and Mathew SandsLectures on Physics, vol-II, B.I. Publications (1964)