Only for those who are not conversant with knot theory.Originally Posted by Profpat
If you can prove that then the Clay Mathematical Institute will award you a $1M dollars.Originally Posted by Max™
Only for those who are not conversant with knot theory.Originally Posted by Profpat
If you can prove that then the Clay Mathematical Institute will award you a $1M dollars.Originally Posted by Max™
Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c˛
Oops!
I forgot one of the implications of my own rules!
If you have to start the loops with any positive (upwards) charges present, then the one I had identified as an electron was actually a meson!
Silly me, confusing electrons and mesons, the super nerds will find that funny.
Also: it is mathematically in question whether or not a knot has to intersect with itself or not?
Not Knot Nottingham, lol, so many nots and knots I do not know which knot to not at!
Emily: Has anyone really been far even as decided to use even go want to do look more like?
Stage Manager: No. *pauses* The physicists and mathematicians, maybe they do some.
The LHC will not find a Higgs particle, rather it will discover a sign of a dark matter particle, which I am tentatively calling the Photino. I believe this particle will be a tetraquark, đuūd, with a mass of roughly 1.25 GeV, 0 Charge (1/3, 2/3, -2/3, -1/3), 0 Spin (-˝, ˝, ˝, -˝), fulfilling the role of dark matter very well.
There ya go, a testable prediction of Simply Relativity.
Wish me luck.![]()
Emily: Has anyone really been far even as decided to use even go want to do look more like?
Stage Manager: No. *pauses* The physicists and mathematicians, maybe they do some.
Supersymmetric theorists see themselves in multiple mirrors reflections and discovered that Moebius bands are 2 1-dimensional loops linked together but they refused to believe what they see.
Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c˛
Well, not sure if it's a supersymmetric particle, though there is a certain elegance to supersymmetry theories, I just liked the name photino, seen it bandied about in a few sci-fi books, and it would be the first "even" particle after the photon.
Emily: Has anyone really been far even as decided to use even go want to do look more like?
Stage Manager: No. *pauses* The physicists and mathematicians, maybe they do some.
Evenness has to do with the even and odd functions having space-time reflection symmetry meaning their antiparticles are not distinguishable being themselves their own antiparticle.Originally Posted by Max™
Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c˛
Well, the photino I described would have two quark/anti-quark pairs, does that count?
Emily: Has anyone really been far even as decided to use even go want to do look more like?
Stage Manager: No. *pauses* The physicists and mathematicians, maybe they do some.
No one have ever directly seen any of the six flavors of quarks. Although in high energy experiments there are circumstantial evidence that they exist by using a technique called asymptotic freedom and infrared slavery.Originally Posted by Max™
Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c˛
Again, I know this.
Quarks are just a label I apply to "ends" of the loops in that drawing above.
The gluons are the ties from the ends to the other ends, and accordingly, where most of the mass comes from.
Emily: Has anyone really been far even as decided to use even go want to do look more like?
Stage Manager: No. *pauses* The physicists and mathematicians, maybe they do some.
Actually they are gauge symmetry Lie groups from a theory of gauge invariance and not the same as coordinate labels. I think gluons are simply field directions of degrees of freedom not particles.Originally Posted by Max™
Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c˛
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