Tuesday, February 08, 2011 ... Deutsch/Español/Related posts from blogosphere

Why is the Wigner-Eckart theorem useful?

...and other questions...

The Wigner-Eckart Theorem

Wikipedia on the theorem
is a formula that tells us about all "simple constraints" that group theory - the mathematical incarnation of the wisdom about symmetries, especially in the SO(3) ≈ SU(2) case (rotations in a three-dimensional space) - implies about matrix elements of tensor operators - those that transform as some representation of the same symmetry.




The dependence on the indices labeling basis vectors of the representations - m_1, m_2, m_3 in the SU(2) case - is totally determined. Instead of thinking that some matrix elements depend on many variables, physicists may realize that the symmetry guarantees that the matrix elements only depend on a few labels labeling the whole "multiplets" of the states and operators rather than on all the labels identifying the individual components. It's always critically important to know how much freedom or how much uncertainty there is about some observables - for example, experimenters don't want to repeat their experiment (2J+1)^3 times without a good reason - and we would get a totally wrong idea without this theorem.

To see that the theorem is used all the time, check e.g. these 5700 papers
Google Scholar on the theorem
many of which are highly cited ones. The topics of the papers include optics, nanotubes, X-rays, spectroscopy, condensed matter physics, mathematical physics involving integrable systems, quantum chemistry, nuclear physics, and virtually all other branches of physics that depend on quantum mechanics. In many other cases, the theorem is being used without mentioning its name, and its generalizations are being used all the time in advanced theoretical physics, e.g. in the contexts with groups that are much more complicated than SO(3) ≈ SU(2).

It's interesting to note that in some sense, quantum mechanics allows the symmetry to impose as many constraints as classical physics. In classical physics, we could start with an initial state labeled by numbers I_i, apply some operations depending on parameters O_j, and we would obtain a final state described by parameters F_k. Classical physics would tell us Yes/No - whether the process may occur. That's the classical counterpart of the probability.

The rotational SO(3) symmetry - which has 3 parameters (3 independent rotations - or the latitude and longitude of the axis, and the angle of the rotation) would only tell us that if we rotate all objects I_i,O_j,F_k by the same rotation, we obtain a valid proposition again. So the dependence on 3 parameters - corresponding to the 3 rotations - is eliminated. In quantum mechanics, we also eliminate the dependence on 3 parameters - in this case m_1,m_2,m_3, the projections j_z for the two state vectors and for the operator sandwiched in between them.

Why it takes a longer time to fly from Melbourne to Perth than back?

The speed of an aircraft during the flight is ideally kept constant relatively to the air surrounding the aircraft. However, in the moderate zone - between 30 and 60 degrees of latitude on both hemispheres - the dominant winds are called westerlies
Wikipedia on westerlies
for a good reason. They blow from the West to the East. So it's faster to get from Perth to Melbourne because it's in the direction of the wind; but it takes a longer time to fly from Melbourne to Perth because it's against the wind and, again, the natural speed of the aircraft is measured relatively to the air (i.e. wind).

Why do the westerlies have this direction?

It's because of the Coriolis force. The Coriolis force - a "fictitious" force in a rotating frame that adds to the centrifugal one - steers winds (and rivers) in the clockwise direction on the Northern Hemisphere and counter-clockwise direction on the Southern hefmisphere.
Wikipedia on Coriolis force
Why does the Coriolis force create westerlies - which are particularly strong on the Souther Hemisphere e.g. in Australia? It's because the normal air that heats up near the equator expands and it would travel to one of the poles. However, this air going towards the poles gets steered by the Coriolis force, so before it reaches e.g. the Southern pole, it's already moving in the East direction.

The westerlies are weaker on the Northern Hemisphere as well as on the summer-having hemisphere because when it's so, the wind from the equator is strong enough (amplified by the land on the Northern Hemisphere, or by the summer) so that it usually manages to reach the pole (without drifting to the East), anyway.

It's very, very far but my uncle lives in Melbourne while a generous IT expert who lives in Perth sent me an iPod Touch for a relatively modest and doable help with a project half a year ago. ;-)

How was CMB created?

The radiation was "produced" about 380,000 years after the Big Bang, and it was produced at every point of the Universe. From the very beginning, it's been (almost) uniform (the same at all places) and isotropic (the same in all directions). Since that time, the radiation was moving in all directions, essentially without any interactions.

The cosmic microwave radiation "decoupled" - separated - from the rest of the matter in the Universe in this era we call "decoupling". Before the decoupling, the temperature of the Universe was so high that electrons and protons were largely separated in a plasma filling the Universe. Plasma carries a lot of random electric charge that severely interacts with photons all the time - so the plasma was opaque for the radiation.

However, after the "decoupling", the atoms were formed for the first time - mostly Hydrogen atoms. Hydrogen atoms are neutral and their interactions with the photons are much weaker so the Universe became essentially transparent. The photons - and everything else in the Universe - at the moment of decoupling had a certain high temperature (thermal equilibrium, about 3000 Kelvin) which means that their spectrum was Planck's black-body thermal spectrum corresponding to the temperature.

From that moment, photons were moving without any interactions and their wavelength was increasing proportionally to the size of the Universe. That also means that the energy of each photon was decreasing by the same factor; the temperature of the black body radiation did the same thing. That's why the current CMB temperature is just 2.7 Kelvin. You may see that the Universe's linear dimensions expanded about 1,000 times from the decoupling.

(Note that 13.73 billion years over 380,000 years is substantially more than 1,000. That's because in the early stages, the expansion of the Universe was "decelerating" as a function of time. Only in recent few billions of years, the expansion got actually accelerating because of dark energy that gradually became important.)

When the WMAP probe detects a photon of the cosmic microwave background, this collision with the telescope is the first interaction of this photon since the moment when the Universe was 380,000 years old. This fact allows you to to deduce how far is the point when the photon was born - or when it last interacted with another object. The birth place is clearly a point in the direction where the photon is coming from. The distance is always the same so all the photons we see here today had to be produced at a particular spherical shell in spacetime. The center of the shell is "our place in the past" and the radius is such that the photons from the shell, when travelling inwards, exactly needed those 13.7 billion years of the cosmic time to get here.

Why two objects get charged by rubbing?

It is always told as a fact without explaining the reason. WHY two objects get charged by rubbing? Why one object get negative charge and other get positive charge?

The effect in which two objects get charged by rubbing and remain charged is called the triboelectric effect,
Wikipedia on the effect
where the root "tribo" means friction or rubbing in Greek (and tribere in Latin). Friction is actually unnecessary: contact is enough in principle.

This effect shouldn't be confused with the (Volta or Galvani) "contact potential" between metals which only exists as long as the two metals remain in contact, and especially not with "contact electrification" which was a name of a scientifically incorrect theory of electricity at the end of the 18th century that attempted to overgeneralize the interpretation of the triboelectric effect. "Electrophorus" was a gadget, first produced by Volta, that used the triboelectric effect.

The cause of the triboelectric effect is adhesion - the atoms on the surface literally form chemical bonds. Materials such as fur are ready to lose electrons and become positively charged while the materials such as ebonite or glass gain electrons and become negatively neutral. To get some idea about which atoms are likely to lose or gain electrons, it's useful to know their electronegativity:
Wikipedia on electronegativity of elements
The redder atom, the higher electronegativity, and the more likely it is for the atom to gain electrons and become negatively charged. That's especially true for light halogens (fluorine, chlorine) and oxygen. That's partly why glass - with lots of SiO2 - likes to get negatively charged in the triboelectric effect. Even sulfur (40% of ebonite) has a higher electronegativity than e.g. carbon and hydrogen that are abundant in the fur which is why fur loses electrons and becomes positively charged.

Of course, the actual arrangement of the atoms in the molecules matters, too. So this overview of the periodic table was just an analogy, not a reliable way to find out the results of the triboelectric effect.

Does the density of a Bose condensate have limits?

It's a good question. The answer is that the bound on the density is given by the requirement that the interactions between the bosons have to remain weak for the Bose-Einstein condensate to exist. In practice, the helium-4 atoms have to be further away from each other than their radius.

Why it is so? Well, if you're talking about the bosons occupying the "same state", it really means that you are constructing a multi-particle state in the multi-particle theory. If you want the energy of this state to be simply given by the sum of the energies of the individual bosons - i.e. N times the energy of the one-particle state - you must guarantee that you have the right Hamiltonian which is essentially the Hamiltonian for the bosons in an external potential, without any significant interaction term in between the bosons.

A sufficiently strongly interacting Hamiltonian for the bosons couldn't be solved that easily.

If you try to push the composite bosons really close to each other, i.e. by lowering the temperature extremely close to the absolute zero, the interactions between them will start to matter which will prevent you from approximating the Hamiltonian by a sum of many one-particle terms. Consequently, the right description is in terms of the component particles - which are often fermions.

It's believed by many condensed matter physicists that the ultimate state of any bound matter very near the absolute zero is a superconductor (not a Fermi liquid, I used to think for a while) - and I don't know. Consider helium-3 as an example. I am actually not sure what one gets at superextremely low temperatures.

Why spacetime points become lines in the twistor space?

The ordinary twistor space is parameterized by (λ^α,μ_α'). Here, the α is a 2-valued SL(2,C) spinor index of one chirality and the dotted or primed index is its complex conjugate, the index of the opposite chirality.

At the level of spinors, vectors are equivalent to "spintensors" with one undotted and one dotted index.
Vμαα'μVαα'.
This is a basic fact about the Lie algebras. SO(3,1) is locally isomorphic to SL(2,C) - they're the same 6-dimensional Lie groups - and the 4-vector is the tensor product of 2 and 2bar.

If you're unfamiliar with this equivalence of vectors of "spintensors" with two indices, notice that the components of a 4-vector may be organized as the matrix
vαα' =
((v0+v3, v1+iv2),
(v1−iv2, v0−v3))
Note that the determinant of this matrix - a natural function of the matrix elements - is simply v_μ.v^μ. The whole matrix may be understood as the appropriate combination of the three Pauli matrices - plus the "time-like Pauli" (identity) matrix multiplied by the time component of the vector. Sorry if my signs deviate from the prevailing convention.

So far, it has only been a story about the vectors or spinors. What about the twistors? Well, it's a simple one-line formula. If I have a point x^μ which is equivalent to the x^αα' matrix - as any 4-vector - I may simply write an equation
λα=xαα'μα'
Note that it is a set of two complex linear equations - for α=0,1 - so it defines a linear object. For a different value of x^μ, I get different equations. Moreover, the equations link the λ and μ objects which are coordinates on the twistor space.

Now, I have to explain why the equations above define a line.

First, the λ and μ objects are pairs of complex numbers, so in total, we have four complex coordinates. However, the twistor space is a projective space. Note that if the equations above are satisfied for some λ and μ - four complex numbers - they will also be satisfied if you multiply both λ and μ by an arbitrary complex number (the same one for both). This projectivity holds universally: μ is naturally scaled in the same way as λ because it may be understood as a dimensionally inverse object to λ^α' (an object that appeared for the first time in this answer) that scales in the inverse way relatively to λ^α in order to keep e.g. the vector λ^α λ^α' (which is e.g. the momentum of external gluons that scatter) constant.

So the twistor space is really a complex projective space, CP^3, and the equations above are two complex conditions, so we're left with a one(-complex)-dimensional object, a complex line. Spacetime points are in one-to-one correspondence to complex lines in the twistor space.

Many more things may be translated. For example, if two spacetime points are separated by a null interval, the corresponding two lines in the twistor space intersect. The intersection - a point in the twistor space - may be identified with a null line in the Minkowski space, and I could derive many other things of this kind.

See more answers at Physics Stack Exchange

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