= Solution
Use the inverse coordinate transformation,
$$
\mathbf r_\perp=\mathbf r'_\perp,\qquad
\mathbf r_\parallel=\gamma(\mathbf r'_\parallel+\mathbf vt'),\qquad
t=\gamma\left(t'+\frac{\mathbf v\cdot\mathbf r'}{c^2}\right).
$$
Substitution into the <wave phase> $\mathbf k\cdot\mathbf r-\omega t$ and comparison of the coefficients of $\mathbf r'$ and $t'$ gives
$$
\boxed{\omega'=\gamma(\omega-\mathbf v\cdot\mathbf k),\qquad
\mathbf k'_\parallel=\gamma\left(\mathbf k_\parallel-\frac\omega{c^2}\mathbf v\right),\qquad
\mathbf k'_\perp=\mathbf k_\perp.}
$$
The first is the <relativistic Doppler effect>; the changed direction of the <wavevector> describes <relativistic aberration>. Equivalently,
$$
\mathbf k'=\mathbf k+\left[(\gamma-1)\frac{\mathbf v\cdot\mathbf k}{v^2}-\gamma\frac\omega{c^2}\right]\mathbf v.
$$
The electric-field <wave amplitude> follows from part (b), after rotating the boost axis to $\mathbf v$:
$$
\boxed{\mathbf E'_0=\mathbf E_{0,\parallel}+\gamma(\mathbf E_{0,\perp}+\mathbf v\times\mathbf B_0).}
$$
For a vacuum <plane wave>, <Maxwell equations> give $\mathbf B_0=\mathbf k\times\mathbf E_0/\omega$ and $\mathbf k\cdot\mathbf E_0=0$. Thus an expression using only the given electric <wave amplitude> and <wavevector> is
$$
\boxed{\mathbf E'_0=\mathbf E_{0,\parallel}+\gamma\left[\mathbf E_{0,\perp}
+\frac{\mathbf v\times(\mathbf k\times\mathbf E_0)}\omega\right].}
$$
Equivalently, $\mathbf E'_0=\gamma(\mathbf E_0+\mathbf v\times\mathbf B_0)-[\gamma^2/(\gamma+1)]\mathbf v(\mathbf v\cdot\mathbf E_0)/c^2$. The <wave amplitude> is not a scalar carried unchanged by the <wave phase> transformation; its direction must remain transverse to the transformed <wavevector>.
Back to article page
