There are four terms in physics/electronics that I understand in a certain context, but not at all in another. Those terms are:
So, when a conductor is placed across the two terminals, the atoms at the negative terminal want to equalize the difference in electrons. The "surplus" of electrons at the negative terminal, flow through the conductor over to populate the electrons that have a relative "deficiency". (I understand that surplus and deficiency are quantitative distinctions and not a qualitative truth. So that is how a 9V battery works. Correct?
Assuming, for this example, there are no volume or tone potentiometers, you solder one end of the pickup's coil-wire to the positive lug of the jack, and you solder the [???] end of the coil-wire to the [???] lug of the output jack. Just like a 9V battery pushes electrons across a conductive wire... through transduction, a guitar pickup can also push electrons along a conductive wire. Do they both have a positive and a negative terminal?
With a guitar pickup's transduction, I cannot envision how the "positive" end of the wire relates to the [???] end of the wire. (P.S. I keep using [???] because, I do not know if the other wire would be referred to as a negative, neutral, or ground wire.) I always imagined "transduction" kind of working like this: You have a water hose that is filled with water, if you stomp on it at some point along it's length, you create pressure that causes the water to squirt out the hose harder. If you get 20 people to all line up with hammers, and everyone starts whacking on the hose, that is kinda like the string wagging around inside the magnetic field. The magnetic force acting on the electrons inside the copper wire is analogous to the bunches of hammers waking the water-hose. That analogy breaks down in my head, however, because the copper wire has two ends, and if you "stomp" on it, it should cause the electrons to flow outwards away from the "stomp" in both directions.
In a simple circuit with a 9V battery and a light bulb, when the circuit is closed, the battery, wanting to equalize the number of electrons across it's two terminals, will push electrons from the "surplus" at the negative pole, through the light bulb, to the positive terminal, and continue doing so until the chemical reaction is spent, and the positive terminal equals the negative terminal in charge.
If you remove the 9V battery, and you replace it with a transducer (guitar pickup) that is large enough to generate 9V (much larger than a guitar pickup), and you vibrate the metal string within the mag field of the transducer, there is no "surplus" or "deficiency" at either end of the coil-wire. So... I guess... my two questions are...
- Positive
- Negative
- Neutral
- Ground
So, when a conductor is placed across the two terminals, the atoms at the negative terminal want to equalize the difference in electrons. The "surplus" of electrons at the negative terminal, flow through the conductor over to populate the electrons that have a relative "deficiency". (I understand that surplus and deficiency are quantitative distinctions and not a qualitative truth. So that is how a 9V battery works. Correct?
Assuming, for this example, there are no volume or tone potentiometers, you solder one end of the pickup's coil-wire to the positive lug of the jack, and you solder the [???] end of the coil-wire to the [???] lug of the output jack. Just like a 9V battery pushes electrons across a conductive wire... through transduction, a guitar pickup can also push electrons along a conductive wire. Do they both have a positive and a negative terminal?
With a guitar pickup's transduction, I cannot envision how the "positive" end of the wire relates to the [???] end of the wire. (P.S. I keep using [???] because, I do not know if the other wire would be referred to as a negative, neutral, or ground wire.) I always imagined "transduction" kind of working like this: You have a water hose that is filled with water, if you stomp on it at some point along it's length, you create pressure that causes the water to squirt out the hose harder. If you get 20 people to all line up with hammers, and everyone starts whacking on the hose, that is kinda like the string wagging around inside the magnetic field. The magnetic force acting on the electrons inside the copper wire is analogous to the bunches of hammers waking the water-hose. That analogy breaks down in my head, however, because the copper wire has two ends, and if you "stomp" on it, it should cause the electrons to flow outwards away from the "stomp" in both directions.
In a simple circuit with a 9V battery and a light bulb, when the circuit is closed, the battery, wanting to equalize the number of electrons across it's two terminals, will push electrons from the "surplus" at the negative pole, through the light bulb, to the positive terminal, and continue doing so until the chemical reaction is spent, and the positive terminal equals the negative terminal in charge.
If you remove the 9V battery, and you replace it with a transducer (guitar pickup) that is large enough to generate 9V (much larger than a guitar pickup), and you vibrate the metal string within the mag field of the transducer, there is no "surplus" or "deficiency" at either end of the coil-wire. So... I guess... my two questions are...
- How the hell does a guitar pickup actually work on an electron level?
- What does neutral, negative, and ground mean?