LuboŇ° Motl

Czech Republic

motls.blogspot.com

Age: 40

Hi, I am a string theorist and a publicist.

17h
awarded Good Answer
17h
comment Existence of Tripoles?
Right, firtree. I am personally always using $z=r\cos\theta$. In that convention, $\sin 3\theta$ isn't smooth near the poles and the function is a combination of infinitely many spherical harmonics (all odd $l$ $Y_{lm}$). If one uses the "latitude" $\theta$ going from $-\pi/2$ to $+\pi / 2$, then $\sin 3\theta$ i.e. roughly $\cos 3\theta$ in my normal language is smooth near the poles and the function is a simple combination of $Y_{10}$ and $Y_{30}$.
17h
reviewed Approve suggested edit on How do you add temperatures?
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answered Commutator algebra in exponents
20h
comment Existence of Tripoles?
It is definitely not orthogonal to the other spherical harmonics, pretty much to none of them. The inner product with $Y_{l0}$ is nonzero for all odd values of $l$ - it's the integral over $\theta$ from zero to pi of the product of the functions times an extra $\sin\theta$. The spherical harmonics form a basis so they're complete - orthogonal to each other and you can't find any other that would be orthogonal to them.
20h
comment Existence of Tripoles?
Sorry, I was inaccurate. You chose the function so nicely that the laplacian vanishes almost everywhere - but it doesn't on the $z$-axis i.e. for $\theta=0$ or $\theta=\pi$. Or maybe it's an eigenstate of the Laplacian with a nonzero eigenvalue? You may calculate those things.
20h
comment Existence of Tripoles?
The particular potential you wrote down doesn't obey $\nabla^2 \phi = 0$ almost anywhere. It's easy to check. You can't fix this problem even by choosing a different dependence on $r$ exactly because the angular part isn't an eigenstate of the angular part of the Laplacian, i.e. the $L^2$ operator.
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revised Existence of Tripoles?
added 391 characters in body
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answered Existence of Tripoles?
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awarded Enlightened
1d
awarded Nice Answer
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answered How do you add temperatures?
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answered Do massive particles exchange Higgs bosons?
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comment Is the Standard Model consistent (UV complete)?
Dear CuriousOne, the neutrino masses were once considered not to be a part of the "Standard Model" but the theory with these extra terms added - and we need to add them, as experiments show - is consistent up to the very high (GUT etc.) scales, too. The experimentally demonstrated ninimal ones are the Majorana masses for the left-handed neutrinos. They may arise as effective terms from the see-saw mechanism from a completely consistent, gauge-invariant grand unified or similar theory at the GUT scale. There are arguments that this GUT-scale physics has to exist for the neutrino masses
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comment Is the Standard Model consistent (UV complete)?
No, Prof Wen, the Standard Model has no inconsistencies at scales 200 GeV or at least a few orders of magnitude above it. All the would-be inconsistencies are cured by having added the right massive gauge bosons with the right coupling to the Higgs boson, and so on. This theory with the electroweak scale as its characteristic scale also has nonperturbative effects like instantons but they are effects, not inconstencies.
Aug
25
comment Spontaneous symmetry breaking and time-reversal symmetry
Dear CMFT, it's unphysical to be looking for eigenstates of the symmetry generator that is spontaneously broken because they're superpositions of states from different superselection sectors. An example is a superposition of a bar magnet magnetized in one direction with the state of the bar magnet magnetized in a different direction. A typical Schrodinger's cat state. But yes, if you don't mind that this is a superposition across superselection sectors, you may find eigenstates of the generators. What's your problem with that?
Aug
25
comment Is the Standard Model consistent (UV complete)?
And yes, at low enough energies below the instability scale, and perhaps even above it, the Standard Model is consistent even nonperturbatively. I am sure that I have already answered that question. You may put the Standard Model on a lattice, for example (ignoring technical issues with fermion doublings etc. which are basically solvable). And yes, the perturbative consistency with a perturbative analysis of the RG flows etc. is enough to prove the nonperturbative consistency, too. The perturbative expansions know about "almost everything".
Aug
25
comment Is the Standard Model consistent (UV complete)?
Dear akrasia, the quartic Higgs coupling doesn't diverge assuming the (now) known value of the Higgs mass, below 200 GeV. So there are two inconsistencies, but they're the Landau pole and the Higgs instability. The latter is something else than a divergent quartic coupling - to some extent, it's the opposite problem because it occurs because the measured mass of the Higgs boson is too low for the Standard Model. ... I personally consider metastability of this kind to be an inconsistency.
Aug
25
answered Is the Standard Model consistent (UV complete)?
Aug
25
comment an example where changing the frame of reference of an observer changes the outcome of events!
This extra magnetic field $B$ will indeed contribute a positive attraction you may calculate - it will reduce the original repulsive force even further. If you combine all these things, you will see that the charges are accelerating as expected from a transformation of the trajectories.
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