Hacker News new | ask | show | jobs
by Strilanc 4314 days ago
Correct. There's spatial dimensions, where you can put the points, and degrees of freedom, the number of numbers you specify when setting up a system. The word dimension can refer to either definition (and others), depending on the context.

What I was trying to communicate to the gp is that they were mixing the two concepts. Constraints reduce the number of degrees of freedom, not the number of spatial dimensions, so the intuition that you get the holographic principle by adding a constraint to 3d space simply doesn't type-check.

1 comments

No, they really do reduce the number of spatial dimensions. Take EM fields. You have a 6-tuple relation, x×y×z×t×e×m. Maxwell's equations aside, this relation is four-dimensional, because two of its dimensions (e and m) are dependent on the other four (x, y, z, and t), making the relation a function: x×y×z×t→e×m. Meaning, for any values of x, y, z, and t, they are related to exactly one value of both e and m in the relation. Without any constraints, e and m may be freely chosen for each value of x, y, z, and t in the relation.

Now, if I assume Maxwell's equations, I am free to rewrite this relation as dependent on only three variables; let's say x, y, and z, but it really doesn't matter: x×y×z→t×e×m. To further refine the example, if I took t=0 for all x, y, and z, I'd have initial conditions at t=0 (very standard).

Given Maxwell's equations however, I can derive exactly one relation of the form x×y×z×t→e×m from my x×y×z→t×e×m initial conditions. Hence P(x×y×z×t→e×m) and P(x×y×z→t×e×m) are in bijection, |P(x×y×z×t→e×m)| and |P(x×y×z→t×e×m)| have exactly the same cardinality, and therefore the same spatial dimension, three.