# DC virtual photons etc.

1. Apr 25, 2013

### Crazymechanic

Hello i have a little clarification to make for myself with the help of you:)
So we know that electricity moves so fast through the wires because the charge is being "pushed" by the EM force which is mediated by the photon.
Now AC systems no matter what the frequency have real photons mediating their respective EM fields because the voltage and current is reversing polarity constantly.
Now in a DC system we say there are virtual photons.

Now let's assume a situation you have a 12v DC source , you have a switch and a light bulb.
At the moment when you flip the switch "ON" the charge starts to flow in the wire that is after the switch , now in that moment the electrons from a random movement join in a flow of electrons in one direction.
Now it is the moment when the system get's connected that the changes happen and in that moment are there real photons involved and if so then for how long? For the moment when the voltage get's "kicked" in the circuit.Because after it is connected it just flows and being a dc circuit we would say it doesn't change polarity is not time/amplitude varying so is constant and has virtual photons right?

The same thing goes for SMPS where the high power mosfets actually rapidly switch dc ON/OFF at a very fast rate and that creates a time/amplitude varying current which has real photons and to a certain extent can be considered an AC waveform.
So when I flip a dc hand switch at the moment when the power is connected the system has virtual or real photons?

Excuse me for this messed up explanation of my question I just am in a hurry.

2. Apr 25, 2013

### Staff: Mentor

For low frequencies, photons are a bad model of the electromagnetic field. It is not impossible to consider this model, but it does not give useful results. The most fundamental theory (known) about electromagnetism treats it as a field, photons are just a way to visualize some (!) interactions of this field.

3. Apr 25, 2013

### Crazymechanic

So your saying that we are better off if we don't quantize the EM field? To which I could agree.

But still my question applies , I just really wanted to know what do we say when DC is being switched no matter what the frequency , it could be just one switch like switching something on , in the moment of the switch turning on in a simple dc circuit there must be a sudden change , the EM force which "pushes" the charge to the other side of the circuit.But we know that electrons alone would take a long long time to do that so...

My question was meant from a purely theoretical perspective about the moment of the switching on.
So if speaking about the EM from a field perspective , the field still doesn't come up all at once rather it travels with close to c in the medium (copper wire let's say)

Ok tell me is it even possible to answer my question in the way I made it?

4. Apr 25, 2013

### sophiecentaur

For a DC signal, the 'photon energy' is zero so there are an infinite (?!?!) number of them - or quanta. I guess that's a good reason for not using them in an explanation.

5. Apr 25, 2013

### f95toli

If you are turning something on (or off) it is no longer DC. At the moment you turn it on, you are generating a waveform which -in theory- is a step function. Now, if you look at the spectra of such a function you'll find that it extends up to infiniti; i.e. it is very much an AC problem.

Now, in real life there is no such thing as a step function; but if you want to turn something on quckly (fast risetime) you still need a large BW (rule-of-thumb: 3*1/risetime) or the "turn on time" will increase. This is why you need RF circuitry for fast digital electronics even though 1 and 0 are represented by "on" and "off".

My point is: even from a conventional EM point of view this is an AC problem. Hence, the usual explanation for AC transmission applies.

6. Apr 25, 2013

### Crazymechanic

The heavyweight champs of PF just kicked in.Thanks now Im actually getting closer to where I thought the answer would be.
So basically it would be fair to assume that every DC circuit in the moment when it is turned on or off is an or resembles a AC half period , because you have the initial condition at zero potential which rises suddenly depending on the switch but still slower or faster there is a rise and why that rise is happening the DC current even though not polarity switching gets amplitude varying so would it be fair then to say that at the moment of switch on or off it resembles and AC and also has the things that belong to AC ?

I guess the answer should be yes because in the most basic underlying principle that's how smps power supplies work.