Surely that might be naive but the entire issue is that they want to stick to the original contract, which is of course the purpose of a contract in the first place.
5,032 karma · joined February 8, 2012
I was previously a staff scientist at Forschungszentrum Jülich [1] and a postdoc at Lawrence Livermore National Lab [2] doing computational nuclear [3] and atomic physics on absolutely gargantuan computers.
[0] https://en.wikipedia.org/wiki/University_of_the_Virgin_Islands
[1] https://en.wikipedia.org/wiki/Forschungszentrum_J%C3%BClich
[2] https://en.wikipedia.org/wiki/Lawrence_Livermore_National_Laboratory
[3] https://en.wikipedia.org/wiki/Lattice_QCD
Surely that might be naive but the entire issue is that they want to stick to the original contract, which is of course the purpose of a contract in the first place.
Feynman's writing of course is stellar. The order is a bit unusual and not really designed for a "standard" university-level course. I can pick and choose, but I wish I could easily reorder the material.
> Summers fade and roses die
> The answer came, the wind and rain
> [...]
> Circle songs and sands of time
> And seasons will end in tumbled rhyme
> And little change, the wind and rain
Fare thee well, Bob.
3) there's a misunderstanding about ordinary least-squares.
1/3 = 1/4 + 1/16 + 1/64 + ...
https://evanberkowitz.com/images/2014-03-15-quarters/SquareA...https://evanberkowitz.com/images/2014-03-15-quarters/Triangl...
1/7 = 1/8 + 1/64 + 1/512 + ...
https://evanberkowitz.com/images/2014-03-16-eighths/EighthsA... 1/8 = 1/9 + 1/81 + 1/729 + ...
https://evanberkowitz.com/images/2014-03-17-ninths/NinthsAni...On the other hand, it is common to need a metric, which is actually the set of weights in the dot product. If `g` is the metric,
dot(a, g, b) = np.einsum('x,xy,y->', np.conj(a), g, b)
g doesn't have to be diagonal, but if you want the dot product to be symmetric in a and b it ought to be self-adjoint. Then you can find a basis where g is diagonal with real diagonal elements, which you can interpret as the weights.I think this is slightly inaccurate. The butterfly effect is about the evolution of two nearby states in phase space into well-separated states. But the parameter a is not a state. To see the butterfly effect by changing a we would need to let the system settle down, give the parameter a small change, and then change it back. The evolution during the changed time acts as a perturbation on states.
Instead, showing that the attractor changes qualitatively as a function of the parameter is more akin to a phase transition.
While the video doesn't touch on this explicitly, the discussion of the different path lengths around 25:00 in is about the trigonometric effect of the different distances of the beam from the camera. Needing to worry about that is the same grappling with the limitation on the one-way speed.
But it's really so---according to GR, black holes don't have global charges. So even if you see a star made out of baryons collapse into a black hole, once the BH settles down into a steady state you can't say it's "really" got baryons inside: the baryon number gets destroyed.
(Of course, a different model of gravity that preserves unitarity might upset this understanding.)
Or, more to the point, suppose someone came with a different system of units and said: look, one of our standard lengths is 10^-6 of one of yours, but one of our standard areas is defined just like yours: 1 ha = (100m)^2 and (1 of our areas) = (100 of our lengths)^2.
In other words, their standard length = 10^-6m; their area = 10^-12 ha = 10^-8 m^2.
Is their standard area bigger or smaller than their standard length?
Notice that in proportion the lengths and areas are the same.
Another fair comparison is between dimension-dependent lengths is the ratio of the (hyper)volume to the surface (hyper)area V(n)/A(n). This monotonically decreases from n=1.
In the article 2π(d) = the ratio of the circumference to the radius. This is dimensionless, in the sense that the circumference and the radius are both lengths (measured in meters, or whatever), so 2π(d) is really just a number.
But the (hyper)volumes you're talking about depend on dimension, which is exactly why you say "hyper". In 2 dimensions the volume is the area, πr^2, which has dimensions L^2 [measured in m^2 or whatever]. But in 3 dimensions the volume is 4/3 πr^3, which has dimensions L^3. The 5 dimensional (hyper)volume has dimensions L^5, and so on.
So, "comparing" these to find out which is bigger and which smaller is not really meaningful---just like you shouldn't ask which is the bigger mass: a meter or a second? Neither is, they aren't masses.