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For a square dance, it would be an impressive choreography. I guess, E8 is still important in theoretical high energy physics. In math it's the largest exceptional compact Lie group.

This is a valuable contribution to the story. I hadn't thought about this.

@@Meewee466 I guess, it's not part of the standard program, but at a lot of physics departments you can take courses on astrophysics or cosmology.

What’s your email?

physik.martinATgmail.com

Actually the spin quantum number does not arise from the Laplace operator. It has its origin from the transformation properties of the particle's wave function with respect to rotations or more generically Lorentz transformations. Therefore it is not connected to the spherical harmonics.

There is a claim, that the hat tile also can be obtained from projections of higher dimensional lattices. "Direct Construction of Aperiodic Tilings with the Hat Monotile", Ulrich Reitebuch, 2023. But I haven't had time to study it yet. It's on my agenda. arxiv.org/abs/2306.06512

It's a "realtime" speed of sound. It's very sensitive to the proton density. It's that high because there are a billion more photons than baryons. Photons dominate the plasma.

Thank you. I really appreciate it.

I agree and I think you should support this creator but you also might like PBS Space Time

The first Super-thanks I've ever received. I'm flattered:-)

The only unpleasant part of this story is that 95 percent of the content of the universe is dark, or in other words not really understood😉

Yes that's right. I got this wrong unfortunately. Thank you for clarifying this.

​@@Number_Cruncher Whats the difference ultimately? Stellar Novae make new stars. It's like saying this isn't a human nursery, it's just the place where human mothers give birth.

Also pretty sure humans didn’t first settle in Africa and Asia 6,500 years ago.

Well, I was thinking about Mesopotamia, which is dated between 6000 and 1500 BC, as well as the Ancient Egypt, whose early period is dated around 3000 BC. I think, I have worse glitches than that in my introduction.

I'm not sure. From 1 second to 378000 years all perturbations will be tiny and completely controlled by linear evolution. When the background and the high energy physics is under control, it should be possible to understand. But I agree in principle, there can be a few surprises, phase transitions, etc

What about gravitational wave background ? I know it will require enormous size of interferometer, but in space it can be made

@Cyber_Nomad you probably need at least two big ones, if you want to get some angular resolution (get an idea where the signal comes from). I have no idea about the scale of sophistication. But it's already cool to know that the information is out there and maybe we will be able to know at some point in the future 😉

To be honest, I don't know enough about the data analysis to give a qualified answer. I can only refer to the paper I cited in the comments, where they perform the best fit evaluation. I also think that it is a highly elaborate task.

I took the music from the TH-cam library. The piece is called Requiem No. 8. th-cam.com/video/DLQZelThXgQ/w-d-xo.html

@ Thanks!

Wut?

@drdca8263 ohhh.... the Newtonians and einsteinians are fools dude. Sure in the west they larp like they have it all figured out... but it's just science fiction written with numerology lol Gravity isn't even massbased and spacetime is the aether rebranded with rubber sheet geometry lol

Yeah, elementary math is the hardest part. I don't know, where I got the distance of 800 ly from. That's what I was calculating with. But I see there is some debate about the distance and wiki says 433 ly. And afterall, what's a few centuries in comparison to the age of the universe:-)

Thank you, for some reason, your comment was blocked. That's why I only found it now.

@Number_Cruncher extra comment for the TH-cam algorithm so that it can see that this is a nice video and recommend it to other

@@elia0162 ASTRONOMERS ASSEMBLE!!!! COMING IN HOT WITH COMMENTS AND REPLIES!

That's true. I was worried that I would loose them during video compression.

Clearly a visual representation. Don’t be daft

@@ongrys2000 this is the most remarkable result. I tried to highlight it in the one-dimensional setup. And by the way, the projection operation is non-linear and non-invertible.

@@pedroth3 it's the projection of the vertices and edges of an eight dimensional polytope. The plane of projection is rotating in the indicated principle directions.

I don't think that the shown projection is a stereographic one, though. In the worst case, one vertex would end up infinitely far away. But maybe I don't understand the point you are making.

​@@Number_Cruncher You would only get a vertex at infinity in the projection (infinitely far away from the origin) if you choose the center of the projection to be located at a vertex in the polyhedron that is being projected (then that vertex would end up at infinity), but from what you show in your video that doesn't seem to be the case, at least not for the projections you show at 14:12. I'm not completely sure you use a stereographic projection. However, in the polyhedron that you show at 14:12, all faces are regular polygons, so all vertices are going to be located on a sphere (i.e. there exists a sphere that contains all vertices). If the point in which all the green lines intersect-the center of projection-also is located on that sphere, then the projection of the vertices is a stereographic projection. Then the 4D case is very similar to the 3D case; you would have a hypersphere instead of a sphere, but a stereographic projectionb in 4D space would project down to R^3 instead of down to the plane (R^2) (If you project down from R^4 to R^2, you're not using a stereographic projection, though). If the center of projection is not located on the hypersphere, but all vertices are still located on a hypersphere, you should be able to calculate the inverse projection by considering the projection line emanating from the center of projection that intersect the point in the projection and finding where that line intersects the hypersphere, although depending on how the projection was performed initially, this might give you two points on the sphere that would be projected to the same point on the projection plane. I don't know if that made any more sense?

@kristoferkrus yes, I understand it. But I don't think that one can call the projection at 14:12 a proper stereographic projection. If it was a stereographic projection, the sphere would have to be the circumsphere of all the vertices. The focal point also would have to lie on this sphere, which I didn't care about. Also, usually the plane of projection goes through the middle of the sphere or the sphere sits on it. But I agree in so far as that the projection might be invertible, if you know all parameters.

@@Number_Cruncher If the center of projection is not on the (hyper)sphere, then, indeed, you don't have a stereographic projection. On the other hand, it doesn't matter if the projection plane passes through the middle of the sphere, if it tangents the sphere (which is the way the projection is typically described), or if it passes outside of the sphere, never intersecting it (as long as it doesn't pass through the center of projection itself). The only thing that changes is the scale of the projection, which you can easily change afterwards by simply scaling it up or down to your liking. However, the projection plane (or the subspace that you're projecting onto, if you're performing the projection from R^4 or higher) needs to be orthogonal to the line that passes through the projection center and the center of the sphere. As long as it is that, though, there shouldn't be a problem.

@@Number_Cruncher If you try to solve for a point on a sphere, though (no matter what the original projection is), I recommend placing the origin the center of the sphere to make life easier for you.

What about the common faces? I could try a color gradient through the dimensions. For three dimensions there would obviously be the RGB space. Any idea for the remaining two dimensions? Or, I just distribute the hue values across all the cubes of the sample lattice. I think about it.

@@Number_Cruncher I actually hadn’t considered that these are lower-dimensional faces being projected rather than 5D volumes. So maybe this isn’t such a good idea, I don’t know

@@jakobr_ It probably wouldn't look exciting. If the color was somehow "smoothly" distributed across the lattice, the coloring of the tiling would be a color gradient depending on the orientation of the plane.

It's nice to see that you really try to imagine these rotations. Unfortunately, I'm not really sure what you mean by "flipping over". Let me add instead some of my observations hopefully addressing your concern. The rotations are smooth and represented by the squares on top. Nevertheless, the change in the tiling is not smooth. As long as the tiles are just changing size, everything is fine. But at some point, tiles get replaced by others from different dimensions. There is no way that this can be done smoothly. Maybe this is what you mean by flipping. This can also be seen in the three dimensional analogy. It's also nice to shift the lattice. Then you "flip" from one Penrose tiling to another. When you shift the unit cube you have to make sure that all five components of the shift vector add to zero to flip from one Penrose rolling to another one. This all can be explored in the blender files. Let me know when we talk about different things.

@@Number_Cruncher The thing I’m talking about is a little hard to describe. Imagine you’ve got a 3D cube, colored so that opposite faces match, and you’re looking at one edge, so you see two faces. Rotate the cube until one of those faces disappears, and keep going until the same color reappears on the other side.

It happens on rotations 2, 3, and 7

This is very cool. I didn't realize these flips before. But now I can see them. And I agree, it's most likely that tiles get replaced by tiles with the same color from the opposite side of the cube. And the rotations 2, 3 and 7 go beyond 90 degrees, which seems necessary for the flips to occur. I'm not sure, whether it is possible to go from the parallely aligned plane to the slope for the Penrose tiling without flips. It was not completely trivial to find these primitive rotations in the first place. I wonder whether some people have special mental skills which allows them to understand these rotations and the related 5d geometry better than others.

@@Number_Cruncher If those people do exist, I’m definitely not one of them, haha. I guess I’m just more used to recognizing that sort of pattern, having dipped my toes into some 4D games. I don’t have the full picture of the geometry, but I’m occasionally able to see how some weird visual movements come about by extrapolating from my 3D experience.

There is no fundamental understanding for the area. There are only approximate results.

Yeah, in eight dimensions these transformations would just be boring rotations. Due to the projection, wild trajectories occur.