avicenna

I think I will search for what Swansont says about "loop" detector. My problem is that my "aperture area" is very small, say 2x2cm view! I have little space to work with. I know of mini solar cells, but they dont detect IR waves.

Can you recommend a specific make of mini IR sensors. Say can detect 10,000nm IR. The sensor must not need external voltage source;, just two leads. Loop antenna is too much for me to build!!!

Say I have created a strong enough micro and ir radiation source from a certain direction. I want a mini photocell/sensor to detect it. I know there are many mini photo diodes, etc available and cheap. But I need the sensor to be directional, having only one face to detect the radiation. If I flip the face, there would not be radiations detected. Also it should be easy to use, say just two leads to detect voltage/current. Also, what electrical instruments do I need to detect a voltage/current. Can a common multimeter do the job.

Thanks. OK. The world of knowledge is vast as the universe itself.

It is commonly stated in QM that a bound electron may only absorb a photon if only there is a matching energy gap difference that matches the photon energy. This seems ridiculous. Say a photon emitted by a hydrogen atom with the typical red emission frequency of the Blamer series. If this photon meets a piece of copper, it is unlikely that a copper atom has an exact matching two energy levels matching that of the Balmer spectrum of hydrogen. The photon would just cruise through copper without being absorbed. Well, if a statement is flawed, we then decide to "patch up" our theory and say there are other means that a photon may be absorbed by matter. So the scientific method nowadays would be a series of ad hoc "patching ups".

@Carrock, I think your analysis is acceptable. Thanks.

I believe this is the real deal, but only for those with some foundations in AC circuit theory. I will see how things relate to what I am investigating. @Carrock. I am trying to reduce AC circuit to DC circuit "instantaneously". I don't want coaxial cables which complicates things. Say I have a simple ac generator that generates fairly good sinusoidal voltage source at 50 Hz (If possible at all?). I connect a long resistive wire to the terminals in a huge circular loop. When the wire is at thermal equilibrium with the environment, we know that all power will be dissipated as IR radiation loss, purely resistive loss - assume ideally. So we could always apply ohm's law of I=V/R where R is the resistance of the wire, V the instantaneous voltage. It seems that there will be the usual charge conservation along the wire as if the current is a dc current. The current should be a constant at that moment of consideration. My setup would eliminate capacitance, inductance etc. Instantaneously, we only have the magnetic fields around the wire which we assume "steady". How is such an analysis.

Thanks. This is what I want to confirm. For the AC case, I really cannot say as the real world and the ideal world may be different.

I am saying the potential difference between the ends of the wire, i.e. a long wire connected to a dc battery.

I don't understand. I = V/R, so how can there be zero current if we connect a AA 1.5 V battery to a long copper wire. My question is whether I is the same at all points of the wire.

If the voltage across a long wire is constant, is the current uniform throughout the wire length. In a 50 Hz ac voltage across a long wire, at a certain moment how does current vary along x, the wire length.

Maybe no easy answer because we know too little yet about light. The wiki says it is the bound electrons that absorbs radiation; then how the electrons energy get transferred to the nucleus kinetic energy. I think temperature not dependent on the KE of electrons, but only in the KE of the nucleus or center of mass.

I'll like to know the actual physical mechanism how matter absorbs EM radiation. Take the specific example how copper metal absorbs IR radiation and the copper having its temperature raised.

In a capacitor discharge through a plain resistor, the capacitor power supplied at any instant is VI; the power dissipated in the resistor is I²R. So VI = I²R. Consider a railgun operated with a capacitor bank. At any instant of capacitor discharge, the power supplied is VI. The total power supplied for ohmic loss is sum I²R for two rails plus the resistance of the armature. Question: Since VI = total I²R, how can the power equation include the kinetic energy supplied to the armature?

I think correct about plasma.