Maximum information transmission rate using an electrical wire Suppose that we're trying to transmit information at a distance $ L $ using an electrical wire made out of a material with known properties. The circuit setup is as follows: there's a wire that's hooked up to a battery with a controllable voltage on one end and to an ideal amperemeter on the other end. The entire setup is assumed to be at room temperature (so 300 kelvin) throughout.
I'm interested in the question of the theoretical maximum bit transmission rate per watt of heat dissipated that we can achieve using this setup. I have calculated a rough bound myself, but I may have made mistakes in this calculation. I want to know if my logic here is sound, and if not, how else we might go about calculating a bound of this nature. Note that I'm only interested in the order of magnitude so I may drop small constant factors throughout the calculation.
My calculation goes as follows: at room temperature, free electrons have a thermal velocity $ v $ of $ \approx 10^6 $ meters per second. If the material has an electron density per unit volume of $ n $ and the cross section of the wire is $ A $, we'll get that the drift velocity of the electrons in the wire is equal to
$$ v_d = \frac{I}{enA} $$
where $ I $ is the current through the wire, $ A $ is the cross sectional area and $ e $ is the elementary charge. On the other hand, because $ v \gg v_d $, the amperemeter sees electrons at a rate of $ vnA $.
Since $ v $ is roughly the standard deviation of the velocities of the electrons the amperemeter interacts with, and the Kullback-Leibler divergence of a distribution with itself shifted by some small multiple of its standard deviation is second order in the multiplier, the maximum information per electron should scale as $ \sim (v_d/v)^2 $ when $ v \gg v_d $ up to some negligible constant factor. Multiplying this by the rate of electrons arriving gives us an information transmission rate of
$$ = vnA \cdot (v_d/v)^2 = \frac{v_d^2 n A}{v} = \frac{I^2}{e^2 v n A} $$
To link this to the heat dissipated, we can rewrite it by introducing the resistivity $ \rho $ of the material as follows:
$$ = \frac{I^2 R}{R e^2 v n A} = \frac{P}{\rho e^2 v n L} $$
In other words, we get a bound $ r = (\rho e^2 v n)^{-1} $ on the information transmission rate at room temperature, which has units of bits times distance per energy.
We can compute this explicitly for some materials. For instance, for copper we get roughly
$$ r = ((1.72 \cdot 10^{-8}) \cdot (1.6 \cdot 10^{-19})^2 \cdot 10^6 \cdot (8.49 \cdot 10^{28}))^{-1} = 2.67 \cdot 10^{10} \ \textrm{bits} \cdot \frac{\textrm{meters}}{\textrm{joules}} $$
As a concrete example, modern transatlantic internet cables often have a transmission capacity on the order of $ 10^{14} $ bits per second. Nowadays they are all fiberoptic and not made of copper, which makes them significantly more heat efficient; but if they had been made of copper, my bound says that given the $ \approx 5000 \ \textrm{km} $ distance, we would need
$$ r^{-1} \cdot (5 \cdot 10^6) \cdot 10^{14} = 1.87 \cdot 10^{10} \ \textrm{W} $$
of power to operate such a cable, which would amount to around 0.1% of global primary energy consumption. Does that sound like the right order of magnitude for this quantity?
Edit: I'm adding this paper to lend some credibility to the above calculation, since if nothing else the answer I get looks like it has the right dimensions and order of magnitude. I get that
$$ r^{-1} = 3.74 \cdot 10^{-11} \ \frac{\textrm{J}}{\textrm{bits} \cdot \textrm{m}} = 37.4 \ \frac{\textrm{fJ}}{\textrm{bits} \cdot \textrm{mm}} $$
which is a very good match to the order of magnitude of the bit transmission energies quoted in Table 1 of the linked paper. This could just be the magic of dimensional analysis but I think it suggests this calculation has something going for it.
 A: According to this reference, the information bit-rate capacity $B$ maximum ideal theoretical limit of a dielectric insulated round wire cable depends largely by the aspect ratio (i.e. assuming that the transmission line is not using repeater amplifiers):
$$
\frac {1}{B} \propto \frac {\ell} {\sqrt{A}}  
$$
Where $\ell$ is the length of the cable and $A$ the total cross-section area of all the wires inside the cable (i.e. including dielectric insulation thickness) thus the total cross-section area of the cable.
For given dielectric and conductive materials used by the cable,  this interconnection aspect ratio is largely independent of the cable type technology (e.g. strip or coaxial).
So, for a copper coaxial cable with polyethylene dielectric
and rated impedance at $Z=50 Ω$, we get an ideal limit of $B$ (bits/s):
$$
B<9.1 \times 10^{15} \times A / \ell^2
$$
In particular for example, a $10 cm$ radius cross-section and $5000$ Km length interatlantic submerged cable, will result to an ideal  limit value of only $B \approx 11 b/s$ ($11$ bps)! Of course this limit will increase substantially when repeater amplifiers are used along the transmission line however the above calculation demonstrates how theoretical and practical impossible the transmission of high speed information on very long distances is using electrical wires, limited by the electrical properties of the wire, without any further power signal degradation and signal to noise considerations necessary.
This is one of the main reasons why very long distances high speed information wired transfer is only possible today with optical wires.
Also, for a transmission electrical line without the use of repeaters the signal amplitude voltage must be adjusted within the acceptable signal degradation value due to the ohmic resistance thermal loss of the line for the successful transmission of information. For example, an 'ideal' calculation without taking into consideration any skin effect, first estimate for our 5000Km 20cm thick copper cable case results to a total 2.7 ohms ohmic resistance. A $12$V $240$mA (i.e. A 50 ohm termination ohmic load is used in the line for impedance matching) rms amplitude FSK or PSK modulated signal thus $2.9$W power signal would generate a total of approximate $65$ mW rms thermal noise on the cable. Nevertheless, thermal noise will be much larger when taking into consideration also skin effect of the wires. This can be compensated partially by using  multi threaded wires cable instead of single core per signal wire cable since this type has a much larger total surface area and therefore also reduced skin effect that would allow a higher bit rate capacity of the cable.
