Does an AC lamp radiate because of $\frac{1}{2}VI^*$ or because of $\frac{1}{2}EH^*$?

Question

let's consider an AC voltage source supplying an AC lamp.

From a circuital point of view, the power absorbed by the lamp, seen purely as a resistance, is converted into light emission and heat. But from electromagnetics, the lamp receives the power from the Poyinting vector flux across its surface, as shown in the following picture.

Now, my questions are:

1. Which are the cause and effect? The power the lamps gets through the Poynting vector is the cause, and the voltage and current on the wires adapt so that $\frac{1}{2}VI^* = \frac{1}{2}EH^*$, or viceversa? The circuital model says the lamp absorbs the power from the inside, through current flowing on it. The electromagnetic model says it absorbs the power from the external environment. They appears to be very different approaches.

2. What happens if I cover the lamp with a perfect metal surface (except for two little holes to let the wires enter the lamp? Such a situation could be approximatively realized through a tinfoil. The power transferred by the Poynting Vector will be much lower. Will its circuital absorbed power decrease as well? Will its emitted brightness decrease?

Answer 1

The power the lamps gets through the Poynting vector is the cause, and the voltage and current on the wires adapt so that 12VI∗=12EH∗, or viceversa?

They are two ways of modeling the same thing.

The circuit model is a simpler model, that can be used under more limited circumstances, but is usually much easier to calculate with when it does apply.

The electromagnetic model is more general, but is often too complex to use to make practical calculations.

Neither one causes the other, they are just two models for the same thing, or two ways of looking at the same thing.

What happens if I cover the lamp with a perfect metal surface (except for two little holes to let the wires enter the lamp?... The power transferred by the Poynting Vector will be much lower.

The power transfer won't be much lower. You can observe this is true because mains electricity is often connected to electronic devices enclosed in metal boxes and the circuits in those devices continues to work as expected.

Will its circuital absorbed power decrease as well? Will its emitted brightness decrease?

All that happens is that the fields around the wires re-arrange themselves to accommodate the new boundary conditions (the electric field goes to zero on the metal surfaces) imposed by the metal structure you added. In the regions where the wires pass through the metal sheet, the fields will be concentrated, so the electromagnetic energy density will be higher than in the regions away from the barrier.

From the circuit point of view, you can model this as two capacitors in series, connecting the two wires. They will shunt a small amount of current between the two wires, unless you deliberately design the structure to provide a substantial capacitive reactance at the mains frequency. But unless you do that, the effect on the power delivered and the brightness of the lamp is likely to be unmeasurably small.