[kgwgk]

Physical entropy is defined from the probability distribution over states. Velocities or squared-velocities are not states, they are derived quantities. Points in a phase space would describe states. Physical states are discrete anyway when you consider quantum physics :-)

As for the entropy of probability distributions in general, I think relative entropy is invariant under reparametrizations because both the probability of interest and the reference probability transform in the same way [1]. But I don't remember what does it mean exactly. [And I am not sure if that makes ogogmad wrong, I may not have understood well his comment.]

([Edit: forget this aside. You probably were talking about speeds as positive magnitudes.] By the way using an example analogue to yours discrete entropy wouldn't be invariant either: if you have a distribution {-1,1} and square it it collapses to a zero-entropy singleton {1}.)

[1] https://en.wikipedia.org/wiki/Kullback–Leibler_divergence#Pr...

[omegalulw]

+1. The commenter above also wanted cared about bijective mappings, and squaring a random variable in [-1, 1] is not bijective. Squaring a random variable defined over positive real numbers would lead to a bijective mapping and the distribution would still remain uniform.

Actually, I find it hard to come up with a bijective mapping that leads to a non uniform distribution that's useful for anything practical.

[im3w1l]

Ok so first to have a uniform distribution we have to have a bounded set. Maybe you can do something clever with limits but lets not overcomplicate things. Lets say we have 0 <= v < 10. Define E = v^2. Then 0 <= e < 100

Uniformity of v would mean that p(0 <= v < 1) = 1 / 10

Uniformity of E would mean that p(0 <= E < 1) = 1 / 100

But by construction p(0 <= v < 1) = p(0 <= E < 1). So it's not possible for both to be uniform.

[omegalulw]

It's not necessary to have p(0 <= v < 1) = p(0 <= E < 1). Only that P(f(X)) is uniform.

But this does bring up a good point. H(X||Y) = H(f(X)||f(Y)) for any bijective f if the distributions are discrete. When they are continuous this is not true, even with a bijective f. For example f = x^2 doesn't work even though it yields a binary distribution. Interestingly however, affine transformations work.

[im3w1l]

(For non-negative v) v^2 is less than one if and only if v is less than one. That's why the probabilities have to agree in our specific case.