# The electric field in phasor/complex notation

The electric field in phasor notation is often written \begin{align} \mathbf{E}(x,y,z,t)&=\Re\{\mathbf{E}_0\mathrm{e}^{j\phi}\mathrm{e}^{j\omega t}\}\\ &=\Re\{\tilde{\mathbf{E}}\mathrm{e}^{j\omega t}\}. \end{align} Does it mean $\mathbf E_0$ is a constant vector, i.e. $$\mathbf E_0=E_{0x}\mathbf{\hat{x}}+E_{0y}\mathbf{\hat{y}}+E_{0z}\mathbf{\hat{z}}?$$

Or a vector field, $\mathbf E_0:\mathbb R^3\rightarrow \mathbb R^3$, i.e. $$\mathbf{E}_0(x,y,z)=E_{0x}(x,y,z)\mathbf{\hat x}+E_{0y}(x,y,z)\mathbf{\hat y}+E_{0z}(x,y,z)\mathbf{\hat z}?$$

• Regarding the two extra questions in v5 of this post: yes, and yes. – Emilio Pisanty Nov 11 '17 at 15:02
• @EmilioPisanty Great! I thought my edit was more of a math-question so I posted it at math.stackexchange instead. – JDoeDoe Nov 12 '17 at 7:33

It means a combination of the two: the complex amplitude can depend on space, i.e. it is a complex-valued vector field $\tilde{\mathbf{E}}: \mathbb R^3\to\mathbb C^3$, which gives you a space-dependent polarization, \begin{align} \mathbf{E}(x,y,z,t) &=\mathrm{Re}\mathopen{}\bigg[\tilde{\mathbf{E}}(x,y,z)\mathrm{e}^{-i\omega t}\bigg]\mathclose{}, \end{align} including spatial dependence of everything from intensity and ellipticity through the orientation of the polarization plane and of the polarization ellipse within that plane.
• I think the direct answer to the question then is that JDoeDoe's second suggestion, that $\textbf{E}_0$ is a complex vector field, is correct. Each component can be a function of x, y, and z. But not time, time dependence has been separated to the $e^{-i \omega t}$ as we are assuming the time dependence is sinusoidal. – Kthaxt Nov 8 '17 at 19:20
• @Kthaxt JDoeDoe's second suggestion hat $\mathbf E_0$ as a real vector field. – Emilio Pisanty Nov 8 '17 at 20:43