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18

Yes, the dimensionless spin such as $0.2$ in this case is simply the ratio $$ a= 0.2 = \frac{|\vec J_{\rm measured}|}{|\vec J_{\rm max}(M)|}$$ where the denominator is the maximum angular momentum allowed for the same value of the mass (as the measured mass). For the $d=4$ Kerr black hole, the maximum (the angular momentum of the extremal Kerr black hole) is ...


8

Is this simply the ratio of the angular momentum that the blackhole is observed to have as a ratio of the maximal angular momentum as limited by some Physics (Kerr Metric?)? Yes, exactly. For a spinning black hole there are two event horizons, an inner and an outer horizon. The positions of the horizons are given by: $$ r = \tfrac{1}{2}\left(r_s \pm \sqrt{...


7

This question leads to some subtleties. There are at least two distinct notions of "revolution" that could be meaningful in physics. Namely, "to revolve" can mean: To have angular momentum; To transform by a particular kind of Euclidean isometry (a rotation) (or, to be broader and more technical, a representation of that Euclidean isometry). As far as we ...


4

Angular momentum, or a measure of rotation, in very large astrophysical or cosmological bodies and energy can become relativistic, and must be treated using General Relativity. In General Relativity (GR) angular momentum is not too different from momentum and energy. Different quantities, the last two are contained in the stress energy momentum tensor and ...


3

All of your claims are essentially true. The angular momentum of light, in both its orbital and spin varieties, is indeed angular momentum that can be transferred to matter to make it spin and give it the garden variety of mechanical angular momentum. This is well explained in the relevant Wikipedia section, with good references for experiments that show it. ...


3

Spin was assigned to elementary particles so that conservation of angular momentum would hold in the quantum mechanical framework of elementary particles and nuclei. The Stern–Gerlach experiment involves sending a beam of particles through an inhomogeneous magnetic field and observing their deflection. The results show that particles possess an ...


3

Electrons And Spin From Scientific American Unfortunately, the analogy breaks down, and we have come to realize that it is misleading to conjure up an image of the electron as a small spinning object. Instead we have learned simply to accept the observed fact that the electron is deflected by magnetic fields. If one insists on the image of a spinning ...


3

In quantum mechanics, the relevant representations of symmetry groups on the space of states are not our usual linear representation, but projective representations on the Hilbert space. The projective representations of a semi-simple Lie group - such as the rotation group $\mathrm{SO}(n)$ - are in bijection to linear representations of its universal cover. ...


2

The spin group is related to spin-half objects, called spinors. If you rotate a spinor by 360 degrees, you get back the negative of the spinor you started with. Now it would be nice if you could represent the action of this rotation by saying that an element of $SO(n)$ is acting on the spinor. However, this cannot be done because a rotation by 360 degrees is ...


2

There are two quantum numbers to consider, the total spin = 1/2 and the azimuthal or projection quantum number on an axis, say z with angular momentum +-hbar/2. This quantum number is also 1/2 for electrons. (The x and y components are undefined by the uncertainty principle as spin quantum number and spin z component are defined) Thus for 2 electrons there ...


2

The issue is that the "spin representation of $SO(3)$" is not a representation of $SO(3)$ at all, but a representation of its double cover $SU(2)$ (see https://en.wikipedia.org/wiki/Spin_group). Since we sometimes write down representations in terms of infinitesimal generators (in other words, as a representation of the Lie algebra of the Lie group in ...


2

$$ [P_i,L_j]=\varepsilon_{jkl} ([P_i, X_k]P_l+X_k[P_i,P_l])=-\mathrm{i}\hbar \varepsilon_{jkl}\delta_{ki}P_l = -\mathrm{i} \varepsilon_{jil} P_l = \mathrm{i}\varepsilon_{ijl} P_l$$ Which shows that as expected $P_k$ is a vector. The lesson here is that you should use not use the same letter twice as an external index and a dummy index. Here $i, j$ are ...


2

Okay, so: Unfortunately, the word "mass" has been used in two different ways in physics. One was the way Einstein used it in E=mc^2, where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is sometimes called the inertial mass. It's also the ...


2

We don't need to separate electrons out in order to observe them. The structure of an atom, as revealed in electron transitions (atomic spectroscopy) is clearly based on orbitals at specific energy levels, with a two-electrons-per-orbital limit. And, the collective behavior of unpaired electrons that gives rise to ferromagnetism, and subtle spectroscopic ...


1

In quantum mechanics you cannot "compute the total spin" of a certain quantum state: all you can compute are the expected values of the spin components. (I'm going to use Dirac notation because I find it much clearer) To compute expected values in your case it is convenient to use the $\mid s,m_z\rangle$ representation. If $\vec S = \vec S_1 + \vec S_2$ is ...


1

In general, $\frac{\partial L}{\partial \dot{q}}$ is the canonical (or generalized or conjugate*) momentum, and $m\dot x$, for $x$ the actual position, is kinetic momentum. Likewise, the cross product of the former with the generalized coordinate vector $q$ might be called "canonical angular momentum", and the cross product of the latter "kinetic angular ...


1

The Wikipedia page has been modified since the question was posed and the cited text disappeared, which is a good thing as it does not make much sense. Now it can be interesting to note that the preceding sentence was Action is a general physical concept related to dynamics and is most easily recognized in the form of angular momentum. So I take it ...



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