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0 pointscolanderman11y ago0 comments

EDIT: jamii seems to understand that we are working with very different definitions of dimensionality.

My (and I suspect most readers') understanding of dimensionality is as follows. If I have a graph with x and y axes with six RGB-colored points on it, that does not make it a 30-dimensional graph. It's two dimensional because there are two independent variables in the 5-tuple relation that graph is representing.

Number of particles and number of dependent attributes do not affect dimensionality.

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Strilanc11y 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.

colandermanOP11y ago

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.

jamii11y ago

I believe Strilanc is talking about dimensions in phase space rather than in physical space. If you plot all the possible configurations of those points on a graph, that graph would have 30 dimensions. Whether or not that is a valid point is beyond me.

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