I had the same experience. I was essentially going to say "meta is only useful insofar as it helps you and others do object-level things, so focus on building object-level things..." oh.
This whole post seems to mostly be answering "who has the best ethnic restaurants in Europe/America?" along with "which country has the best variety of good restaurants?" and not "who has the best food?" I think that's an important distinction. Clearly, Indian, Chinese, and middle eastern foods are the best.
I haven't heard of ECL before, so I'm sorry if this comes off as naive, but I'm getting stuck on the intro.
For one, I assume that you care about what happens outside our light cone. But more strongly, I’m looking at values with the following property: If you could have a sufficiently large impact outside our lightcone, then the value of taking different actions would be dominated by the impact that those actions had outside our lightcone.
The laws of physics as we know them state that we cannot have any impact outside our light cone. Does ECL (or this post) require this to be wrong?
From the summary you linked,
Many ethical theories (in particular most versions of consequentialism) do not consider geographical distance of relevance to moral value. After all, suffering and the frustration of one’s preferences is bad for someone regardless of where (or when) it happens. This principle should apply even when we consider worlds so far away from us that we can never receive any information from there.
...
Multiverse-wide cooperation via superrationality (abbreviation: MSR) is the idea that, if I think about different value systems and their respective priorities in the world, I should not work on the highest priority according to my own values, but on whatever my comparative advantage is amongst all the interventions favored by the value systems of agents interested in multiverse-wide cooperation.
Is the claim (loosely) that we should take actions we think are morally inferior according to us because ... there might be other intelligent beings outside of our light cone with different preferences? I would want them to act a little bit more like me, so in turn I will act a little more like them, in a strange game of blind prisoner's dilemma.
This is obviously hogwash to me, so I want to make sure I understand it before proceeding.
No you're right, use 2 or 3 instead of 4 as an average dielectric constant. The document you linked cites https://ieeexplore.ieee.org/abstract/document/7325600 which gives measured resistances and capacitances for the various layers. For Intel's 14 nm process making use of low-k, ultra-low-k dielectrics, and air gaps, they show numbers down to 0.15 fF/micron, about 15 times higher than .
I remember learning that aspect ratio and dielectric constant alone don't suffice to explain the high capacitances of interconnects. Instead, you have to include fringe fields -- turns out they're not actually infinite parallel plates (gasp!).
Again, it's not a big deal and doesn't detract much from your analysis. I somewhat regret even bringing it up because of how not important it is :)
This is an excellent writeup.
Minor nit, your assertion of is too simple imo, even for a Fermi estimate. At the very least, include a factor of 4 for the dielectric constant of SiO2, and iirc in real interconnects there is a relatively high "minimum" from fringing fields. I can try to find a source for that later tonight, but I would expect it ends up significantly more than . This will actually make your estimate agree even better with Jacob's.
Active copper cable at 0.5W for 40G over 15 meters is ~J/nm, assuming it actually hits 40G at the max length of 15m.
I can't access the linked article, but an active cable is not simple to model because its listed power includes the active components. We are interested in the loss within the wire between the active components.
This source has specs for a passive copper wire capable of up to 40G @5m using <1W, which works out to ~J/nm, or a bit less.
They write <1 W for every length of wire, so all you can say is <5 fJ/mm. You don't know how much less. They are likely writing <1 W for comparison to active wires that consume more than a W. Also, these cables seem to have a powered transceiver built-in on each end that multiplex out the signal to four twisted pair 10G lines.
Compare to 10G from here which. may use up to 5W to hit up to 10G at 100M, for ~J/nm.
Again, these have a powered transceiver on each end.
So for all of these, all we know is that the sum of the losses of the powered components and the wire itself are of order 1 fJ/mm. Edit: I would guess that probably the powered components have very low power draw (I would guess 10s of mW) and the majority of the loss is attenuation in the wire.
The numbers I gave essentially are the theoretical minimum energy loss per bit per mm of that particular cable at that particular signal power. It's not surprising that multiple twisted pair cables do worse. They'll have higher attenuation, lower bandwidth, the standard transceivers on either side require larger signals because they have cheaper DAC/ADCs, etc. Also, their error correction is not perfect, and they don't make full use of their channel capacity. In return, the cables are cheap, flexible, standard, etc.
There's nothing special about kT/1 nm.
Indeed, the theoretical lower bound is very, very low.
Do you think this is actually achievable with a good enough sensor if we used this exact cable for information transmission, but simply used very low input energies?
The minimum is set by the sensor resolution and noise. A nice oscilloscope, for instance, will have, say, 12 bits of voltage resolution and something like 10 V full scale, so ~2 mV minimum voltage. If you measure across a 50 Ohm load then the minimum received power you can see is This is an underestimate, but that's the idea.
Maybe I'm interpreting these energies in a wrong way and we could violate Jacob's postulated bounds by taking an Ethernet cable and transmitting 40 Gbps of information at a long distance, but I doubt that would actually work.
Ethernet cables are twisted pair and will probably never be able to go that fast. You can get above 10 GHz with rigid coax cables, although you still have significant attenuation.
Let's compute heat loss in a 100 m LDF5-50A, which evidently has 10.9 dB/100 m attenuation at 5 GHz. This is very low in my experience, but it's what they claim.
Say we put 1 W of signal power at 5 GHz in one side. Because of the 10.9 dB attenuation, we receive 94 mW out the other side, with 906 mW lost to heat.
The Shannon-Hartley theorem says that we can compute the capacity of the wire as where is the bandwidth, is received signal power, and is noise power.
Let's assume Johnson noise. These cables are rated up to 100 C, so I'll use that temperature, although it doesn't make a big difference.
If I plug in 5 GHz for , 94 mW for and for then I get a channel capacity of 160 GHz.
The heat lost is then Quite low compared to Jacob's ~10 fJ/mm "theoretical lower bound."
One free parameter is the signal power. The heat loss over the cable is linear in the signal power, while the channel capacity is sublinear, so lowering the signal power reduces the energy cost per bit. It is 10 fJ/bit/mm at about 300 W of input power, quite a lot!
Another is noise power. I assumed Johnson noise, which may be a reasonable assumption for an isolated coax cable, but not for an interconnect on a CPU. Adding an order of magnitude or two to the noise power does not substantially change the final energy cost per bit (0.05 goes to 0.07), however I doubt even that covers the amount of noise in a CPU interconnect.
Similarly, raising the cable attenuation to 50 dB/100 m does not even double the heat loss per bit. Shannon's theorem still allows a significant capacity. It's just a question of whether or not the receiver can read such small signals.
The reason that typical interconnects in CPUs and the like tend to be in the realm of 10-100 fJ/bit/mm is because of a wide range of engineering constraints, not because there is a theoretical minimum. Feel free to check my numbers of course. I did this pretty quickly.
Ah, I was definitely unclear in the previous comment. I'll try to rephrase.
When you complete a circuit, say containing a battery, a wire, and a light bulb, a complicated dance has to happen for the light bulb to turn on. At near the speed of light, electric and magnetic fields around the wire carry energy to the light bulb. At the same time, the voltage throughout the wire establishes itself at the the values you would expect from Ohm's law and Kirchhoff's rules and such. At the same time, electrons throughout the wire begin to feel a small force from an electric field pointing along the direction of the wire, even if the wire has bends and such. These fields and voltages, outside and inside the wire, are the result of a complicated, self-consistent arrangement of surface charges on the wire.
See this youtube video for a nice demonstration of a nonintuitive result of this process. The video cites this paper among others, which has a nice introduction and overview.
The key point is that establishing these surface charges and propagating the signal along the wire amounts to moving an extremely small amount of electric charge. In that youtube video he asserts without citation that the electrons move "the radius of a proton" (something like a femtometer) to set up these surface charges. I don't think it's always so little, but again I don't remember where I got my number from. I can try to either look up numbers or calculate it myself if you'd like.
Signals (low vs high voltages, say) do not propagate through circuits by hopping from electron to electron within a wire. In a very real sense they do not even propagate through the wire, but through electric and magnetic fields around and within the wire. This broad statement is also true at high frequencies, although there the details become even more complicated.
To maybe belabor the point: to send a bit across a wire, we set the voltage at one side high or low. That voltage propagates across the wire via the song and dance I just described. It is the heat lost in propagating this voltage that we are interested in for computing the energy of sending the bit over, and this heat loss is typically extremely small, because the electrons barely have to move and so they lose very little energy to collisions.
I downvoted, because
... are not low status fun, but long-term life decisions that should not be taken lightly.
... are just "be rude to friends," which I consider immoral.
... are probably illegal.