Is EM radiation flux an extensive or intensive property?

Question

Is electromagnetic radiation flux (measured in watts per square metre) an extensive or intensive property? Can the fluxes from two sources be somehow combined at a target and what are the rules?

Answer 1

Yes, in most cases that you would come across the radiation flux from two sources (e.g. two light bulbs, two stars, two car headlights) can just be added together to get the total radiation flux.

That is because most sources of light are sources of incoherent radiation. The light arises from multiple microscopic sources at multiple locations that have no fixed phase relationship, even on extremely short timescales.

The exception would be where you try to add together two sources of light that are coherent and have a fixed phase relationship - for example two laser beams that originate from the same parent beam. In such cases you need to add the amplitudes of the component electric and magnetic fields together before calculating the resultant Poynting vector and flux.

See https://physics.stackexchange.com/a/61323/43351

Details:

For the case of two coherent sources with a fixed phase relationship between their electromagnetic fields (e.g., a pair of lasers with identical frequencies, or one laser and a beam splitter), it is straightforward to show that the summed Poynting vector $$\vec{S} = \vec{E_1} \times \vec{H_1} + \vec{E_2} \times \vec{H_2} + \vec{E_1} \times \vec{H_2} + \vec{E_2} \times \vec{H_1}$$ $$\vec{S} = \vec{S_1} + \vec{S_2} + \vec{E_1} \times \vec{H_2} + \vec{E_2} \times \vec{H_1}\ .$$ If there is a fixed phase relationship between wave 1 and wave 2, then the time-average (which is what is measured by any "flux detector") of the "crossed terms" is not necessarily zero and could be positive or negative. This is just constructive or destructive interference.

If there is no fixed phase relationship between the two sources, which is the case for every-day light sources or light coming from different points on a macroscopic source of radiation, then the crossed terms will have the form (assuming the waves are travelling along the z-axis in vacuum) $$\vec{E_1} \times \vec{H_2} + \vec{E_2} \times \vec{H_1} = \frac{2E_1 E_2}{\mu_0 c} \sin(\omega t)\sin(\omega t + \phi)\ \hat{z}, $$ where $\phi$ is a random, time-varying phase difference. The time-average of this is zero, so $$ \langle S \rangle = \langle S_1 \rangle + \langle S_2 \rangle$$ and the flux measured is the summed flux from the two sources.

Answer 2

They CANNOT be added. As consider 2 sources , producing 2 "independant" E and B fields.

$\vec{S_{1}} = \frac{1}{\mu_0}\vec{E_{1}}×\vec{B_{1}}$

$\vec{S_{2}} = \frac{1}{\mu_0}\vec{E_{2}}×\vec{B_{2}}$,

$\vec{S_{1}} + \vec{S_{2}}$

Does not equal:

$\vec{S}_{1+2}=\frac{1}{\mu_0}(\vec{E_{1}}+\vec{E_{2}})×(\vec{B_{1}}+\vec{B_{2}})$

There are cross terms when considering the total Poynting vector as a result of 2 different E and B fields, that are not present if we just "add" the poynting vector of them up individually.

Doubling the E and B field, would NOT double the power.