[dcrazy]

Energy cannot be created or destroyed, but it can enter and leave an open system. When I lift a 10kg box 1 meter in the air, I don’t raise its temperature at all, and I only raise mine a tiny bit, yet I have still done work on the box and therefore have imparted it energy. The energy came from food I ate earlier, and was ultimately stored in the box as gravipotential energy.

Is this not analogous to storing energy in the EM fields within the CPU?

[lmm]

CPUs don't store nontrivial amounts of energy, and even if storing a 1 was a significantly higher energy level than a 0 (or vice versa) there's no plausible workload that would be causing the CPU to switch significantly more 0s to 1s than 1s to 0s (or vice versa).

[db48x]

Yes, but only briefly. When you study the thermodynamics of information you’ll discover that it’s actually erasing information that has a cost. Every time the CPU stores a value in a register it erases the previous value, using up energy. In fact, every individual transistor has to erase the previous state on basically every clock cycle.

Curiously there is a minimum cost to erase a single bit that no system can go below. It’s extremely small, billions of times smaller than the amount of energy our CPUs use every time they erase a bit, but it exists. Look up Landauer’s Limit. There is a similar limit on the maximum amount of information stored in a system which is proportional to the surface area of the sphere that the information fits inside. Exceed that limit and you’ll form a black hole. We’re no where near that limit yet either.

[imtringued]

>In fact, every individual transistor has to erase the previous state on basically every clock cycle.

This is incorrect in both directions.

Only transistors whose inputs are changing have to discharge their capacitance.

This means that if the inputs don't change nothing happens, but if the inputs change then the changes propagate through the circuit to the next flip flop, possibly creating a cascade of changes.

Consider this pathological scenario: The first input changes, then a delay happens, then the second input changes so that the output remains the same. This is known as a "glitch". Even though the output hasn't changed, the downstream transistors see their input switch twice. Glitches propagate through transistors and not only that, if another unfortunate timing event happens, you can end up with accumulating multiple glitches. A single transistor may switch multiple times in a clock cycle.

Switching transistors costs energy, which means you end up with "parasitic" power consumption that doesn't contribute to the calculated output.

[db48x]

My apologies if I wasn’t clear enough. I was only intending to make a statistical statement that the number of erasures is of similar order to the number of transistors, not that every single transistor changes its state exactly once per cycle. Some don't change their state this cycle, others end up changing multiple times before settling. In fact, some are completely powered off! (Because you’re not using the built–in GPU right now, or you’re not doing AVX512 right now, etc, etc.)

Note also that discharging the internal capacitance of a transistor, and the heat generated by current through the transistor’s internal resistance, are both costs over and above the fundamental cost of erasing a bit. Transistors can be made more efficient by reducing those additional costs, but Landauer discovered that nothing can reduce the fundamental cost of erasing a bit.