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> Your initial response to me here didn't address the exterior field at all My initial response to you was to your statements about a subthread where I address
by T-A 7y ago
> Your initial response to me here didn't address the exterior field at all
My initial response to you was to your statements about a subthread where I addressed the exterior field.
> your linked calculations (which you called "the full derivation") were explicitly just about fields between the plates
That was in response to your invocation of angular momentum.
> and in fact assumed the plates were infinite.
No. This is what I said in that response: "Once it's through the screen, the force changes sign, but quickly drops to zero (for infinitely large screens, it is exactly zero everywhere outside the space bounded by the two screens)."
The statement about infinite screens is something which any physicist would understand immediately: a reminder that there is a simple case (the limit of infinite size) which tells you how things must go as you scale up the device. If your calculations don't reproduce the known result in that limit (and ballenarosa's don't; they actually blow up in that limit) you know you're doing something wrong.
There is no assumption that the plates are infinite. In fact, thanks to plasma screening, the electric field outside the plates falls off much quicker than you'd expect even from the correctly calculated vacuum expression (which I linked to in that thread). The Debye length in low Earth orbit is measured in millimeters [1][2], so for any device significantly larger than that, the far field which you keep invoking is completely negligible.
> You also claimed there that I had "failed to apply conservation of momentum correctly in the general case of two different masses", even though my objection had nothing to do with the mass difference.
A look at the derivation should tell you that the mass difference is crucial for the device to work. If the ions being ejected in opposite directions were of equal mass, there would indeed be no net momentum transfer to the device.
In other words, I was actually trying to make the most charitable interpretation possible of your argument. (The more severe one being that somebody who professes to teach E&M does not even understand basic properties of a multipole expansion.)
> the specific functional form of that field can't matter, since we know from fundamental principles that its integral (the electric potential) must go to zero at infinity
No. What matters is how fast it goes to infinity. That determines what you get when you integrate over it, from the plate out to infinity. If the potential's gradient falls to zero right at the plate surface, and then stays there, all you get is zero. If it falls off slower, you get more. How much? Enough to give back all the momentum picked up inside the device? To answer that, you certainly need to know the functional form.
> You claimed there that if one neglects plasma screening, "you get a finite contribution which can be made arbitrarily small relative to the work done inside the capacitor, where the force is constant, simply by increasing the size of the capacitor." This claim is incorrect.
This "claim" is correct, and easily verified by simply looking at the referenced closed form expression for the field outside a plate capacitor.
> Again, the easy way to prove that is just to note that the electric potential goes to zero at infinity.
Again, this is nonsense. It is the functional form of the potential out to infinity which determines how much momentum the ions transfer back to the device on their way out. The potential gradient being zero, at infinity or at any other point, simply tells you that no momentum can be transferred at that point. It tells you exactly nothing about how much momentum was transferred on the way to that point, unless you know the full functional form of the potential along that way.
> if you want to actually consider the electric field due to a finite parallel plate capacitor, you can compute that easily enough along the axis of a pair of circular plates
Yes, that's done in the reference I provided in the ballenarosa subthread.
> As it must be, the integral from the midpoint to the positive plate is exactly equal and opposite to the integral from the positive plate to infinity (independent of the plate radius or the separation).
Are you integrating the potential?!? That only tells you that energy is conserved (surprise!). It doesn't tell you how it's distributed. Look again at the derivation: kinetic energy of the ion (exchanged with the potential) is proportional to v^2, momentum is proportional to v (no square). That's what makes the device work: energy and momentum of the system (device + ions) are both conserved (of course) but not transferred between its components at the same rate.
You want to integrate the equations of motion of the ions as they move in the potential, not the potential.
As for this,
> In particular, if we imagine a positively charged plate in a plasma, I can easily imagine a negative screening charge accumulating right next to the plate, which would in a very short distance bring the exterior field to zero. But at the same time, that nearby accumulation of negative charge would strengthen the field right next to the plate, almost like a second parallel plate capacitor. The effect would be that positive ions outside of the positive plate would be strongly repelled
it also makes no sense at all. First you say (correctly) that plasma screening "would in a very short distance bring the exterior field to zero", but after just one more sentence you go "positive ions outside of the positive plate would be strongly repelled", a complete contradiction.
I am sorry, but at this point I must retract my previous claim that I don't find it hard to believe your claimed profession.
[1] https://en.wikipedia.org/wiki/Debye_length#Typical_values https://en.wikipedia.org/wiki/Debye_length#Typical_values
[2] https://arxiv.org/abs/0906.3884 https://arxiv.org/abs/0906.3884
- Steuard 7y agoI suspect that we're only talking to ourselves at this point, so perhaps it's not worth a lot more time on this conversation. But a few notes: Regarding the infinite plate limit: Locally, it is absolutely true that as you increase the size of the plates (relative to the distance between them) the fields must approach the result for an infinite plate. But as soon as you want to consider global questions like the behavior of the potential at infinity, there is a true qualitative difference between a large-but-finite capacitor and an infinite one. Any finite charge distribution will result in a potential of zero at infinity (or rather, will allow you to consistently make that choice), but an infinite charge distribution will not in general allow that. For the case of an infinite parallel plate capacitor, if you set the potential to zero out to infinity on one side, it will be non-zero (and constant) out to infinity on the other. One upshot of this is that it's misleading at best to use the infinite plate approximation for any sort of scattering analysis (like what we're discussing here). Regarding multipole expansions: I love multipole analysis, it's a beautiful way to study charge distributions. That said, I think it's massive overkill to actually use it in this case, since at least for circular plates along the axis that process is far more complicated than just using the exact fields themselves. Regarding momentum transfer: You've claimed that "It is the functional form of the potential out to infinity which determines how much momentum the ions transfer back to the device". I agree that if you want to know the rate of momentum transfer at any given point, you need to know the exact functional form of the potential (and/or the field). But if we're only interested in the total momentum transferred in the interaction, we don't need to know the details of where and when the momentum transfer happened: all we need to know is the net result. So in particular, if your ion starts far in front of the device with (relative) velocity v0 in the -x direction, and we can show from energy conservation that it ends far behind the device with the same velocity v0 in the -x direction, that's sufficient to prove that the ion has experienced no net change in velocity and thus that no net momentum transfer has occurred (even though there may have been substantial transfers along the way). With that in mind, if we specialize to the simple case of an ion moving along the axis of symmetry of two circular plates, we can find exactly this result explicitly from an energy calculation. So to heck with it, let's just do this. (And apologies if this is repetition, but I don't think I was able to get your link in that other thread to work, or perhaps I tried to follow the wrong one.) Along the axis of a uniformly charged circular disk of radius R, E_x(disk) = K * (1-|x|/sqrt(x^2+R^2)) * sign(x), where K is a constant (K=sigma/(2 epsilon_0) for surface charge density sigma). Assembling two of these with opposite charges a distance D apart (with the + plate at x=0 and the - plate at x=-D) gives for x>0: E_x(x>0) = K * [(x+D)/sqrt((x+D)^2 + R^2) - x/sqrt(x^2 + R^2)] This has the properties you'd expect: for example, it approaches D/R->0 as R approaches infinity, matching the infinite plate limit. (The expression's clearly different between the plates.) You can integrate this (or rather, its negative) to get the electric potential: phi(x>=0) = K * [sqrt(x^2 + R^2) - sqrt((x+D)^2 + R^2) + D] The constant +D was chosen to maintain the convention phi=0 at infinity. But then we find that the potential at x=0 is phi(x=0) = K * [R - sqrt(D^2 + R^2) + D], or approximately K * D in the limit of large R (and this actually decreases as R gets smaller). [This is consistent: by symmetry, phi=0 at the midpoint between the plates (x=-D/2), and in the large R limit the electric field inside is a constant sigma/epsilon_0 = 2K, so 2K * D/2 = K * D.] No matter how large you make the plates, the total change in the incoming ion's energy and thus the total effect of the exterior field on the ion's speed will not approach zero, even though the magnitude of the field at any given point does approach zero as R grows (as it must, to match the infinite plate limit). And no matter how large you make the plates, the total effect of the field on the speed of an ion that moves from infinity to the midpoint of the plates will be zero (and similarly on the negative/outgoing side). You're welcome to solve equations of motion to find the exact functional form of x(t) for this particle, but that's a lot of work, and energy conservation guarantees that there will be no net change in speed. Finally, regarding plasma screening: My comment about positive ions being strongly repelled outside of the positive plate was perhaps unclear, and certainly I'd welcome corrections if I'm misunderstanding screening somehow. I was speaking specifically of ions that came closer to the positive plate than the length scale over which the screening brings the outside field to zero. The screening charges can't change the electric potential at the position of the plate itself (unless you allow them to discharge the whole thing), so rather than falling off to zero very slowly over a long distance as in the unscreened case (giving rise to a small but wide electric field as the gradient), the potential must fall to zero very rapidly over the short screening distance (giving rise to a large but short-ranged electric field). Either way, the height of the potential energy barrier at the plate (relative to infinity) can't change. So again, in the screened case I would think that incoming positive ions would feel essentially zero field for most of their approach but would then feel a very strong repulsion in the narrow region immediately before the plate.