Electrons are very light so the kinetic energy required to get them moving should not be significant in any non-contrived situation I think? The energy of the magnetic field produced by the current would tend to be much more of an important effect.
My current understanding is that the electric current energy transmits through electron drift velocity (and I believe that is the standard textbook understanding?, although I admit I have some questions concerning the details). The magnetic field is just a component of the EM waves which propagate changes in electron KE between electrons (the EM waves implement the connections between masses in the equivalent mass-spring system).
I’m not sure how you got “1 kT per foot” but that seems roughly similar to the model up thread I am replying to from spxtr that got 0.05 fJ/bit/mm or 5e-23 J/bit/mm. I attempted to derive an estimate from the lower level physics thinking it might be different but it ended up in the same range—and also off by the same 2 OOM vs real data. But I mention that skin effect could plausibly increase power by 10x in my lower level model, as I didn’t model it nor use measured attenuation values at all. The other OOM probably comes from analog SNR inefficiency.
The part of this that is somewhat odd at first is the exponential attenuation. That does show up in my low lever model where any electron kinetic energy in the wire is dissipated by about 50% due to thermal collisions every τ ~ 4e-13 seconds (that is the important part from mean free path / relaxation time). But that doesn’t naturally lead to a linear bit energy distance scale unless that dissipated energy is somehow replaced/driven by the preceding section of waveform.
So if you sent E as a single large infinitesimal pulse down a wire of length D, the energy you get on the other side is E∗2−αD for some attenuation constant α that works out to about 0.1 mm or something as it’s τc, not meters. I believe if your chart showed attenuation in the 100THZ regime on the scale of τ it would be losing 50% per 0.1 mm instead of per meter.
We know that resistance is linear, not exponential—which I think arises from long steady flow where every τ seconds half the electron kinetic energy is dissipated, but this total amount is linear with wire section length. The relaxation time τ then just determines what steady mean electron drift velocity (current flow) results from the dissipated energy.
So when the wave period f−1 is much less than τ you still lose about half of the wave energy E every τ seconds but that can be spread out over a much larger wavelength section. (and indeed at gigahertz frequencies this model roughly predicts the correct 50% attenuation distance scale of ~10m or so).
There’s two types of energy associated with a current we should distinguish. Firstly there’s the power flowing through the circuit, then there’s energy associated with having current flowing in a wire at all. So if we’re looking at a piece of extension cord that’s powering a lightbulb, the power flowing through the circuit is what’s making the lightbulb shine. This is governed by the equation P=IV. But there’s also some energy associated with having current flowing in a wire at all. For example, you can work out what the magnetic field should be around a wire with a given amount of current flowing through it and calculate the energy stored in the magnetic field. (This energy is associated with the inductance of the wire.) Similarly, the kinetic energy associated with the electron drift velocity is also there just because the wire has current flowing through it. (This is typically a very small amount of energy.)
To see that these types have to be distinct, think about what happens when we double the voltage going into the extension cord and also double the resistance of the lightbulb it’s powering. Current stays the same, but with twice the voltage we now have twice the power flowing to the light bulb. Because current hasn’t changed, neither has the magnetic field around the wire, nor the drift velocity. So the energy associated with having a current flowing in this wire is unchanged, even though the power provided to the light bulb has doubled. The important thing about the drift velocity in the context of P=IV is that it moves charge. We can calculate the potential energy associated with a charge in a wire as E=qV, and then taking the time derivative gives the power equation. It’s true that drift velocity is also a velocity, and thus the charge carriers have kinetic energy too, but this is not the energy that powers the light bulb.
In terms of exponential attenuation, even DC through resistors gives exponential attenuation if you have a “transmission line” configuration of resistors that look like this:
So exponential attenuation doesn’t seem too unusual or surprising to me.
My current understanding is that the electric current energy transmits through electron drift velocity (and I believe that is the standard textbook understanding?, although I admit I have some questions concerning the details). The magnetic field is just a component of the EM waves which propagate changes in electron KE between electrons (the EM waves implement the connections between masses in the equivalent mass-spring system).
I’m not sure how you got “1 kT per foot” but that seems roughly similar to the model up thread I am replying to from spxtr that got 0.05 fJ/bit/mm or 5e-23 J/bit/mm. I attempted to derive an estimate from the lower level physics thinking it might be different but it ended up in the same range—and also off by the same 2 OOM vs real data. But I mention that skin effect could plausibly increase power by 10x in my lower level model, as I didn’t model it nor use measured attenuation values at all. The other OOM probably comes from analog SNR inefficiency.
The part of this that is somewhat odd at first is the exponential attenuation. That does show up in my low lever model where any electron kinetic energy in the wire is dissipated by about 50% due to thermal collisions every τ ~ 4e-13 seconds (that is the important part from mean free path / relaxation time). But that doesn’t naturally lead to a linear bit energy distance scale unless that dissipated energy is somehow replaced/driven by the preceding section of waveform.
So if you sent E as a single large infinitesimal pulse down a wire of length D, the energy you get on the other side is E∗2−αD for some attenuation constant α that works out to about 0.1 mm or something as it’s τc, not meters. I believe if your chart showed attenuation in the 100THZ regime on the scale of τ it would be losing 50% per 0.1 mm instead of per meter.
We know that resistance is linear, not exponential—which I think arises from long steady flow where every τ seconds half the electron kinetic energy is dissipated, but this total amount is linear with wire section length. The relaxation time τ then just determines what steady mean electron drift velocity (current flow) results from the dissipated energy.
So when the wave period f−1 is much less than τ you still lose about half of the wave energy E every τ seconds but that can be spread out over a much larger wavelength section. (and indeed at gigahertz frequencies this model roughly predicts the correct 50% attenuation distance scale of ~10m or so).
There’s two types of energy associated with a current we should distinguish. Firstly there’s the power flowing through the circuit, then there’s energy associated with having current flowing in a wire at all. So if we’re looking at a piece of extension cord that’s powering a lightbulb, the power flowing through the circuit is what’s making the lightbulb shine. This is governed by the equation P=IV. But there’s also some energy associated with having current flowing in a wire at all. For example, you can work out what the magnetic field should be around a wire with a given amount of current flowing through it and calculate the energy stored in the magnetic field. (This energy is associated with the inductance of the wire.) Similarly, the kinetic energy associated with the electron drift velocity is also there just because the wire has current flowing through it. (This is typically a very small amount of energy.)
To see that these types have to be distinct, think about what happens when we double the voltage going into the extension cord and also double the resistance of the lightbulb it’s powering. Current stays the same, but with twice the voltage we now have twice the power flowing to the light bulb. Because current hasn’t changed, neither has the magnetic field around the wire, nor the drift velocity. So the energy associated with having a current flowing in this wire is unchanged, even though the power provided to the light bulb has doubled. The important thing about the drift velocity in the context of P=IV is that it moves charge. We can calculate the potential energy associated with a charge in a wire as E=qV, and then taking the time derivative gives the power equation. It’s true that drift velocity is also a velocity, and thus the charge carriers have kinetic energy too, but this is not the energy that powers the light bulb.
In terms of exponential attenuation, even DC through resistors gives exponential attenuation if you have a “transmission line” configuration of resistors that look like this:
So exponential attenuation doesn’t seem too unusual or surprising to me.