为何位于坐标原点的两个正交振荡偶极子坡印廷矢量等于两者单独存在时的和？

Awesome question—this cuts to the intersection of linearity, vector calculus, and electromagnetic energy flow, which is super important for understanding oscillating sources. Let’s break this down step by step so it makes sense:

Key Background #

• First, remember that Maxwell’s equations are linear: the total electric field from multiple sources is just the sum of the fields from each source alone. So for our two dipoles, ( \mathbf{E}_{\text{total}} = \mathbf{E}_1 + \mathbf{E}2 ), and the same goes for the magnetic field: ( \mathbf{B}{\text{total}} = \mathbf{B}_1 + \mathbf{B}_2 ).
• The Poynting vector, which describes electromagnetic energy flow, is defined as:

  \mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}
  

  For oscillating dipoles, we almost always care about the time-averaged Poynting vector ( \langle \mathbf{S} \rangle ), since the instantaneous value oscillates too rapidly to be useful for measuring net energy flow.

Why Cross Terms Cancel Out #

When we expand the total Poynting vector using the summed fields, we get:

\mathbf{S}_{\text{total}} = \frac{1}{\mu_0} \left( \mathbf{E}_1 + \mathbf{E}_2 \right) \times \left( \mathbf{B}_1 + \mathbf{B}_2 \right)
= \mathbf{S}_1 + \mathbf{S}_2 + \frac{1}{\mu_0} \left( \mathbf{E}_1 \times \mathbf{B}_2 + \mathbf{E}_2 \times \mathbf{B}_1 \right)

The critical insight here is that the cross terms (( \mathbf{E}_1 \times \mathbf{B}_2 ) and ( \mathbf{E}_2 \times \mathbf{B}_1 )) average to zero over time when the dipoles are orthogonal:

• For orthogonal dipoles (e.g., one aligned along the x-axis, the other along the y-axis), their electric field components at any point in space are mutually perpendicular.
• In the far field (where we typically measure the Poynting vector for dipoles), each dipole’s magnetic field is proportional to its electric field (( \mathbf{B} = \frac{1}{c} \hat{\mathbf{r}} \times \mathbf{E} )) and points in the same azimuthal direction.
• When calculating the time average of the cross terms, we end up with two terms that are equal in magnitude but opposite in direction (since ( \mathbf{E}_1 \times \mathbf{B}_2 = - \mathbf{E}_2 \times \mathbf{B}_1 ) when ( \mathbf{E}_1 \perp \mathbf{E}_2 )). These terms completely cancel each other out.

Final Result #

Since the cross terms vanish in the time average, the total time-averaged Poynting vector simplifies to just the sum of the time-averaged Poynting vectors from each dipole alone:

\langle \mathbf{S}_{\text{total}} \rangle = \langle \mathbf{S}_1 \rangle + \langle \mathbf{S}_2 \rangle

This only fails if the dipoles aren’t truly orthogonal, or if you’re looking at the instantaneous (non-time-averaged) Poynting vector—but for most practical applications (like measuring radiated power), the time average is what matters, and this additive property holds perfectly.

内容的提问来源于stack exchange，提问作者Annie