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201903030411 Exercise 4.3.1

Write out the implied sum in \mathrm{d}p^\mu /\mathrm{d}\tau =qF^{\mu\nu}\eta_{\nu\alpha}u^\alpha for \mu =x and show that it is equivalent to the x component of equation (4.22) at low velocities.

Eq. (4.22):

\mathbf{F}_{\mathrm{em}}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})

Eq. (4.15):

\displaystyle{\frac{\mathrm{d}p^\mu}{\mathrm{d}\tau}}=qF^{\mu\nu}\eta_{\nu\alpha}u^\alpha

the electromagnetic field tensor F^{\mu\nu} of which is given by Eq. (4.14):

\begin{bmatrix} F^{tt} & F^{tx} & F^{ty} & F^{tz} \\ F^{xt} & F^{xx} & F^{xy} & F^{xz} \\ F^{yt} & F^{yx} & F^{yy} & F^{yz} \\ F^{zt} & F^{zx} & F^{zy} & F^{zz} \end{bmatrix} = \begin{bmatrix} 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{bmatrix}

and the Minkowski metric tensor \eta_{\nu\alpha} of which is given by Eq. (4.6):

\begin{bmatrix} \eta_{tt} & \eta_{tx} & \eta_{ty} & \eta_{tz} \\ \eta_{xt} & \eta_{xx} & \eta_{xy} & \eta_{xz} \\ \eta_{yt} & \eta_{yx} & \eta_{yy} & \eta_{yz} \\ \eta_{zt} & \eta_{zx} & \eta_{zy} & \eta_{zz} \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

In the case \mu =x, beginning with Eq. (4.15):

\begin{aligned} \displaystyle{\frac{\mathrm{d}p^x}{\mathrm{d}\tau}} & =qF^{x\nu}\eta_{\nu\alpha}u^\alpha \\ & = q \big( F^{xt}\eta_{t\alpha}u^\alpha + F^{xx}\eta_{x\alpha}u^\alpha + F^{xy}\eta_{y\alpha}u^\alpha + F^{xz}\eta_{z\alpha}u^\alpha \big) \\ & = q \big( -E_x \eta_{tt} u^t + (0)(\eta_{xx})(u^x) + B_{z}\eta_{yy}u^y -B_y\eta_{zz}u^z \big) \\ \dots\, & \textrm{when }v\ll c\textrm{, }u^t\approx 1\textrm{, }u^x\approx v_x\textrm{, }u^y\approx v_y\textrm{, and }u^z\approx v_z\,\dots \\ & = q\big( (-E_x)(-1)(1) + (B_z)(1)(v_y) - (B_y)(1)(v_z)\big) \\ & = q(E_x + B_{z}v_{y} - B_{y}v_{z}) \\ & = q(\mathbf{E}+\mathbf{v}\times\mathbf{B})_x \\ & = \mathbf{F}_{\mathrm{em},\,x}\end{aligned}

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