Based on the fundamental constants, how big does a CPU register ever need to be?
Let's start with distance. I'll be playing fast and loose with the numbers here because I'm not a physicist, and it's also just a thought experiment. If any experts wish to chime in and correct me, I'd be happy to keep revising this.
There's plenty of dispute about this, but for this thought experiment let's take the Planck length as the smallest possible distance that makes sense in our current understanding of physics. This is 1.6x10-35 meters, or about 10-20 the diameter of a proton.
At the other end of the distance scale, the observable universe is estimated at about 1.7x1026 ± 20%. Let's assume ~25% more to be safe and say 2.1x1026. The size of the entire universe (observable and unobservable) is larger than this, but for lack of knowledge we'll stick with what is observable.
If we divide up the observable universe into a Planck length ruled grid, how many grid lines is this? The observable universe is 2.1x1026 x 1.6x1035, or 3.4x1061 Planck lengths in diameter. Or about 2205 Planck lengths. So you'd need three 205 bit numbers to "address" a single point in the observable universe. Let's round that up to the nearest power of 2: 256 bits. And now we have about 20% bit width to spare in case we need fixed-point arithmetic.
However, if you wanted to measure the entire volume, you'd need 2205 cubed, or 615 bits. Let's round that one up to the nearest power of two: 1024 bits.
So with 1024 bits we can represent the full range of measurable distances and volumes in the observable universe. As an added bonus this is also large enough to hold a 3-dimensional 256 bit distance vector. (of course, this doesn't take into account the need for the representation of real numbers or the many places where extra precision in helpful in calculation)
Got corrections for my very loose interpretation of physics or my math? Drop me a line!