confusion about the meaning of nc REVIEWED

Confusion about the meaning of nc PhL 9.30.14 I think this stuff is all OK, see conclusion in red at the end. Reviewed again 10.8.14, OK. Recall from Fig 1.7 that we have nc(x,ω) = (ξ1/ε1) ns(x,ω) . (1.5.17) Now, since (ξd/εd) = → (εd - jσd/ω)/εd → - j(σd/εd) (1/ω) = (Gdc/C) (1/jω) we find that nc becomes infinite as ω → 0. What on earth does that mean?? I look back at the conducting capacitor example. At DC, battery feeds constant current, why would nc go infinite? I am reading through that discussion, and it seems OK to this point: Jc2n(x,ω) = (ξ1/ε1) (jω) ns(x,ω) . ξ1 ≡ ε1 + σ1/jω = complex dielectric constant (1.5.16) because we ω→ 0, the right side here is finite, as expected. I then make this statement: "If we observe the conduction current Jc2n flowing through a unit-area loop (red in figure), we can write Jc2n = ∂tnc where nc is the total amount of conduction charge flowing through that unit-area loop per unit time. " I think something is wrong here. One could say that dnc is the amount of conduction charge flowing through the loop in time dt, but I don't think you can claim that nc is the charge flowing in 1 second. I think nc is the total integrated charge flowing through the loop from time t = 0 to time t. But maybe I can just define nc as Jc2n = ∂tnc and not try to interpret it? Consider I = dQ/dt . Here dQ flows through a wire in time dt. Q itself is the integrated charge over some time, meaning Q(t) = ∫t I(t')dt'. So maybe a better claim would be " Quantity nc is the total amount of charge that flows through the red loop from some initial time up to time t. It's units are charge/area. " Now go back to Jc2n = ∂tnc and that is then OK. Then go do Jc2n = jωnc and what does that mean? When I talk about ∂tV I don't seem to have this confusion, perhaps because V is not an "extensive quantity". I think of something like V = V0ejωt and then ∂tV → jωV. At ω = 0, I have V = V0. Maybe I have a problem with Fourier Transforms, ignoring constant things as appear in Laplace Transforms. I will just go back and review this entire Section with commentary in red: –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– Consider the situation at a general boundary between dielectric (region 1) and conductor (region 2) where there exists a surface charge density ns : Fig 1.7 In (1.1.18) it was shown that div [Jd + Jc] = 0 where Jc = σE is the conduction current and Jd the displacement current ∂tD = ε∂tE. The divergence theorem (1.1.30) then says 0 = ∫V div [Jd + Jc] dV = ∫S [Jd + Jc] dS . Comment: This "total current" is continuous at a boundary, and I really do show that in 1.1.18. There is no time element here. Applied to the blue pillbox which straddles the boundary in the figure, we find Jd1n + Jc1n = Jd2n + Jc2n where n means normal component. Writing this out, ε1∂tE1n + σ1E1n = ε2∂tE2n + σ2E2n ≈ σ2E2n = Jc2n since σ2 is huge inside the conductor. Therefore, Jc2n = σ1E1n + ε1∂tE1n . (1.5.14) The left equation is exact, but then I do make an approximation, things are in the time domain. In the ω domain I would be saying ε1jωE1n + σ1E1n = ε2jωE2n + σ2E2n ≈ σ2E2n = Jc2n (ε1jω + σ1)E1n = (ε2jω + σ2)E2n ≈ σ2E2n = Jc2n jω(ε1 + σ1/jω)E1n = jω(ε2 + σ2/jω)E2n≈ σ2E2n = Jc2n jωξ1E1n = jω ξ2E2n ≈ σ2E2n = Jc2n I don't think dropping the ε2 term is a problem here, especially near ω = 0 ! The final line just says Jc2n has the two contributions shown, and you could write that as Jc2n = σ1E1n + ε1jωE1n . (1.5.14) When ω→ 0, you are left just with the DC leakage current, as you would expect. Meanwhile, Gauss's Law (1.1.33) states that div (εE) = ρ ∫V ρ dV = ∫S εE dS . (1.1.33) Applied to the same blue pillbox we find ns = ε1En1 - ε2En2 ≈ ε1En1 since En2 ≈ 0 inside the conductor. Thus, En1 = ns/ε1 and then Jcn1 = σ1En1 = ns(σ1/ε1) . (1.5.15) Notice in the first line that it is div D = ρ. On the second line I do another approximation which is that E is very small inside conductor, I think that is OK since I always use it. Dielectric is 1. I end up with a way to relate the dielectric conduction current to the surface charge ns. Note that (σ1/ε1) = G/C from 4.4.10, so the last equation says Jcn1 = ns(G/C). I think this is correct and OK. Then (1.5.14) can be written as Jc2n = σ1E1n + ε1∂tE1n = (σ1 + ε1∂t)E1n = (1/ε1)(σ1 + ε1∂t)ns or, writing out the arguments, Jc2n(x,t) = (1/ε1)(σ1 + ε1∂t)ns(x,t) . This is the conductor's current expressed in terms of surface charge ns and ε1 and σ1. All is OK here. The current has two components as expected. The leakage is (σ1/ε1)ns as shown in (1.5.15). The second term is the capacitative current. In the frequency domain with rules (1.5.2) this becomes Jc2n(x,ω) = (1/ε1)(σ1 + ε1jω)ns(x,ω) = (1/ε1) (jω)(ε1 + σ1/jω) ns(x,ω) = (ξ1/ε1) (jω) ns(x,ω) . ξ1 ≡ ε1 + σ1/jω = complex dielectric constant (1.5.16) Nothing new added here, it is OK. If we observe the conduction current Jc2n flowing through a unit-area loop (red in figure), we can write Jc2n = ∂tnc where nc is the total amount of conduction charge flowing through that unit-area loop per unit time. Thus we have ∂tnc(x,t) = (1/ε1)(σ1 + ε1∂t)ns(x,t) or jω nc(x,ω) = (ξ1/ε1) (jω) ns(x,ω) or nc(x,ω) = (ξ1/ε1) ns(x,ω) . (1.5.17) where is our result claimed at the start that ns = (εd/ξd) nc. Note that: OK, here is where I may have a problem of sorts. I think I am defining nc(t) in this manner really, nc(x,t) ≡ !Syntax Error, IJc2n(x,t') dt' where here I arbitrarily integrate from some time t = 0. There is nothing wrong with such a definition, so it is certainly then OK to say Jc2n(x,t) = ∂tnc(x,t) . We then end up with (1.5.17) which is the key result of this section, and all is well. The only confusion is in the interpretation of nc . In the time domain, nc is the total charge passing through the unit red rectangle from time t = 0 to time t = t. There is no confusion about what that means. The confusion is how you translate that concept into the frequency domain. I now end this section of commentary and return to the main doc. _____________________________________________________________________________ At this point I wrote some notes called " Those constant terms in Laplace Transforms.doc". I consider Fourier Transforms as well and I show that: Fact: Let F(x) have a Fourier transform f(ω). Let G(x) = ∂xF(x). Then: g(ω) = jω f(ω) provided that F(±∞) = 0. So there is a condition on the little ∂t → jω rule! I don't think I ever mention this in lines doc. Suppose we have nc(t) = !Syntax Error, I dt' Jc2n(t') Jc2n(t) = ∂tnc(t) Let's now assume that Jc2n(t) = Jc2n(t=0) ejω1t = Jc2n0 ejω1t as a monochrome probe (I guess it runs all the time). Then we have Jc2n(ω) = Jc2n0 2πδ(ω-ω1) for the official Fourier Transform. The spectrum is all at the one frequency ω1, and ω1 = 0 is as good as any other value of ω1. Now consider nc(t) = !Syntax Error, I dt' Jc2n(t') = !Syntax Error, I dt' Jc2n0 ejω1t' = Jc2n0 [ejω1t - 1]/ (jω1) nc(t=ε) = Jc2n0 [1 + jω1ε - 1]/ (jω1) = Jc2n0 ε nc(t=0) = 0 This all seems reasonable since integration of nc starts at t = 0. The second term "-1" is in there exactly to cause the integrated charge to be 0 at t = 0, as shown. We shall now compute nc(ω) as a standard Fourier Transform, using nc(t) shown above, nc(ω) = !Syntax Error, Idt nc(t) e-jωt = !Syntax Error, Idt {Jc2n0 [ejω1t - 1]/ (jω1)}e-jωt = Jc2n0 / (jω1) * !Syntax Error, Idt [ejω1t - 1] e-jωt = Jc2n0 / (jω1) * [ 2πδ(ω-ω1) - 2π δ(ω) ] Notice that the "second term" has created a second term here as well which is the δ(ω) term. Comment: The second term "- 1" is really illegal for a FT for functions, but is legal for distribution theory. It creates δ(ω). A spectrum δ(ω) always corresponds to a constant in the t domain. Let's go the other direction to see what this dual term spectrum means: nc(t) = (1/2π) !Syntax Error, Idω e+jωt nc(ω) = (1/2π) !Syntax Error, Idω e+jωt{ Jc2n0 / (jω1) * [ 2πδ(ω-ω1) - 2π δ(ω) ]} = Jc2n0 / (jω1) * !Syntax Error, Idω e+jωt{ [ δ(ω-ω1) -δ(ω) ]} = Jc2n0 / (jω1) * [ e+jω1t - 1] and as expect, the second term in ω space creates that second term in t-space. We can also compute ( we really already did this above) Jc2n(ω) = !Syntax Error, Idt Jc2n(t) e-jωt = !Syntax Error, Idt { Jc2n0 ejω1t }e-jωt = Jc2n0 2πδ(ω-ω1) So we now have this interesting pair of equations nc(ω) = Jc2n0 / (jω1) * [ 2πδ(ω-ω1) - 2π δ(ω) ] Jc2n(ω) = Jc2n0 2πδ(ω-ω1) As long as we avoid ω = 0, we can say that Jc2n(ω) = jω nc(ω) where we cancel out the two delta functions. I think this is pretty much OK. Lets back up and replace the t = 0 endpoint with t = -∞ as a start time. We then get these equations, nc(t) = !Syntax Error, I dt' Jc2n(t') Jc2n(t) = ∂tnc(t) Jc2n(t) = Jc2n0 ejω1t going on forever Jc2n(ω) = Jc2n0 2πδ(ω-ω1) nc(t) = !Syntax Error, I dt' Jc2n(t') =!Syntax Error, I dt' Jc2n0 ejω1t' = Jc2n0 [ejω1t - 0]/ (jω1) = Jc2n0 [ejω1t]/ (jω1) In this last line, we do our usual trick to say that ejω1(-∞) = 0 by adding a tiny imag part to ω1. nc(ω) = !Syntax Error, Idt nc(t) e-jωt = !Syntax Error, Idt {Jc2n0 [ejω1t ]/ (jω1)}e-jωt = Jc2n0 / (jω1) * !Syntax Error, Idt [ejω1t] e-jωt = Jc2n0 / (jω1) * [ 2πδ(ω-ω1) ] and now that extra δ(ω) term is gone. Let's go the other direction to see what this dual term spectrum means: nc(t) = (1/2π) !Syntax Error, Idω e+jωt nc(ω) = (1/2π) !Syntax Error, Idω e+jωt{ Jc2n0 / (jω1) * [ 2πδ(ω-ω1) ]} = Jc2n0 / (jω1) * !Syntax Error, Idω e+jωt{ [ δ(ω-ω1)]} = Jc2n0 / (jω1) * [ e+jω1t] and we recover where we started. Finally nc(ω) = Jc2n0 / (jω1) * [ 2πδ(ω-ω1 ] Jc2n(ω) = Jc2n0 2πδ(ω-ω1) => Jc2n(ω) = jω nc(ω) So either way, I think I am "comfortable" with this last equation. Interpret in terms of ω? Well, suppose we define script variables as unscripted over 2πδ. Then scripted are the "line strengths" (but only one line, so line strength): Jc2n(ω1) = jω1 nc(ω1) = jω1* Jc2n0 / (jω1) = Jc2n0 Or writing ω as the name of the single line (as I always do in lines doc) Jc2n(ω) = jω nc(ω) = jω* Jc2n0 / (jω) = Jc2n0 So, as ω → 0, Jc2n(ω) = Jc2n0. Comment and Question: Consider our starting point in the time domain Jc2n(t) = (1/ε1)(σ1 + ε1∂t) ns(t) . In the ω domain this reads Jc2n(ω) = (1/ε1)(σ1 + ε1jω) ns(ω) = (ξ1/ε1) (jω) ns(ω) ≡ jω nc(ω) where nc(ω) is defined as shown. This is then precisely our equation above that Jc2n(ω) = jω nc(ω). The comment is this: if you just look at Jc2n(ω) = jω nc(ω), you might be tempted to say that in the time domain this reads Jc2n(t) = ∂tnc(t) . But the real equation is Jc2n(t) = (1/ε1)(σ1 + ε1∂t) ns(t) as shown above. The question then is this: what is going on here? Are these consistent? ∂tnc(t) = (1/ε1)(σ1 + ε1∂t) ns(t) ∂t[nc(t) - ns(t)] = (σ1/ε1) . nc(t) - ns(t) = (σ1/ε1)t + K nc(t) = ns(t) + (σ1/ε1)t + K So nc has the capacitative ns(t) component, and then it has a linear-in-time component due to the ongoing constant leakage. It we were to magically "start the leakage" at time t = 0, then we would have nc(t) = ns(t) + (σ1/ε1) t I think this is in fact OK, and therefore these two equations are consistent with each other, Jc2n(t) = ∂tnc(t) Jc2n(ω) = jω nc(ω) Jc2n(t) = (1/ε1)(σ1 + ε1∂t) ns(t) Jc2n(ω) = (1/ε1)(σ1 + ε1jω) ns(ω) = jω(ξ1/ε1) ns(ω) nc(ω) = (ξ1/ε1) ns(ω) ξ1 = ε1 + σ1/jω Now what does all this say as ω → 0 in the ω domain? Look at each of the three equations, starting with the middle one: Jc2n(ω) = (1/ε1)(σ1 + ε1jω) ns(ω) → (σ1/ε1) ns(ω) as ω→ 0 This says in the time domain that Jc2n(t) = (σ1/ε1) ns(t) = (σ1/ε1) ns = constant leakage. Next, nc(ω) = (ξ1/ε1) ns(ω) = [ 1 + (σ1/ε1)/jω ] ns(ω) Now I think of ns(t) = ns0 ejωt with ns0 being some finite quantity (capacitor problem). This means that the line strength ns(ω) is also some finite number, in fact just this ns0. So nc(ω) → ∞ as ω→0. How should I interpret that fact? We go back to Jc2n(ω) = jω nc(ω) → (σ1/ε1) ns(ω) = (σ1/ε1)ns0 As ω → 0, the only way for Jc2n(ω) → finite is if nc(ω) → ∞, otherwise we could get Jc2n(ω) → 0. So this says that it is a requirement that nc(ω) → ∞ as ω→ 0. You can see what the plot looks like: nc(ω) = [ 1 + (σ1/ε1)/jω ] ns0 I just altered lines doc to add the notion of a line strength. Then consider ns(t) = ns0 ejωt Then the Fourier transform will be ns(ω') = ns0 2πδ(ω'-ω) and then the "line strength" is just ns0 . Then (*) has line strength nc(ω) on the left. nc(ω) = [ 1 + (σ1/ε1)/jω ] ns0 = [ 1 - j (σ1/ε1)/ω ] ns0 = [ 1 - jA/ω ] B Notice that Im(nc(ω)) only blows up if σ1 ≠ 0. I guess I no longer have a problem with nc blowing up in this way. As ω→ 0, we know (line strengths) Jc2n(ω) = (1/ε1)(σ1 + ε1jω) ns0 → (σ1/ε1) ns0 as ω→ 0 approaches its leakage value which is a small constant, as ω→ 0. Since Jc2n(ω) = jω nc(ω), we must have nc(ω) blow up as 1/ω. It's "line strength" gets stronger and stronger as ω → 0 in order to maintain Jc2n(ω). Conclusion of this doc: Everything about nc in lines doc is OK except the paragraph about interpretation which I will right now fix. OLD If we observe the conduction current Jc2n flowing through a unit-area loop (red in figure), we can write Jc2n = ∂tnc where nc is the total amount of conduction charge flowing through that unit-area loop per unit time. Thus we have NEW If we observe the conduction current Jc2n in the conductor just below the surface and flowing through a unit-area loop (red in Fig 1.7), we can write Jc2n = ∂tnc where nc(t) is the total amount of conduction charge flowing through that unit-area loop from some initial time to the current time t: Jc2n(x,t) = ∂tnc(x,t) nc(x,t) = !Syntax Error, Idt' Jc2n(x,t') Thus we have from the first line of (1.5.16),