New answers tagged

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 ...


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 ...


0

galaxy come in many different sizes: some of the small-er ones do rotate ["orbit"] around the edge of a large galaxy ... one can also visualize galaxy-clusters, in which the entire cluster rotates .....


0

First of all, I shall assume that the system is kept on a horizontal plane so as to simplify the calculations. Secondly, I shall assume that the line joining the mass $m$ and $M$ is perpendicular to the direction of motion of mass $3m$. Thirdly, I shall assume that the spheres to be point objects (in other words, you can consider it to be a perfect head-on ...


0

When we talk about a (non-spinning) black hole we normally mean the spacetime geometry called the Schwarzschild metric. This is one of the simpler solutions to the Einstein equations, but it's important to understand that the solution is based on a number of assumptions and as a result it is only an approximation to a real black hole. The assumption that is ...


0

The total angular momentum of the disk doesn NOT change. There is an outflow of angular momentum due to the shearing of matter rotating at different velocities. Faster inner annuli get a spin down torque from slower outer ones, so they lose angular momentum in favour of the angular of momentum of the outer annuli. I can express this better and answering ...


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 ...


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

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 ...


0

The shape of the Sun tells you something about it's mean rotation rate. The faster it spins, the more oblate it gets. See Why is the Sun almost perfectly spherical? Of course you are not sensitive to the nuances of latitudinal or radial differential rotation. For the former you really do need to "see" the surface, for the latter you need helioseismology ...


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{...


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 ...


0

You want to read the classic paper by Richard Beth, Mechanical detection and measurement of the angular momentum of Light, Physical Review 50 115 (1936). Beth used bright circularly polarized light to drive a torsion pendulum in a vacuum chamber, and was able to observe torques due to circular polarization of order $10^{-16}\rm\,N\,m$. This was with a ...


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 ...


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 ...


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. ...


0

I know the final kinetic energy is 0.1566J How did you find the final kinetic energy? To find the final state of the system you have to assume that there are no external torques acting on the system and that angular momentum is conserved.


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 ...


0

To begin with, it is important to clarify the scenario. When two masses are attracted to each other, regardless of their magnitudes, the gravitational force (and hence acceleration) vector acting on each is in the direction of the other. If two masses were sitting motionless in a vacuum, they would accelerate toward each other in a straight line until such ...


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 ...


0

The $\vec f_i$ are the external forces acting on the cylinder, i.e. gravity, reaction force and friction force. The reaction and the friction act at a single point, the contact with the hill. The torque about the center of the disc due to the reaction force is zero since the force is anti-parallel to the lever arm. The torque due to the friction $F$ is ...


0

Why is it the case that $C\ddot{θ}=aF$ as shown on page 87? How does one leverage the formula above to arrive at this equation? Conceptually speaking, one sums the moment of the forces about the center of mass for each point over the entire body, so what exactly are the forces (i.e. what are $f_i$?) at various points on the cylinder (gravity, reaction force, ...


0

Yes. An easy way to think about this, is that the mass moves as a whole. So when balancing, it doesn't "know" whether the triangular space above the left weight is filled or not. The attachment point doesn't matter. It could even be attached to the right of the support. What matters in mechanical terms is the total moment around the support (it'll rotate ...


0

Geometry says no. Since the orbital (DISREGARD - direction reverses) (INSERT - period changes) in a horseshoe orbit, the rotation of the body around its axis would have to change as well, and this is not going to happen over the course of a single orbit. Tidal locking takes a long time.


0

To perhaps give an answer in a slightly different way, it's complicated. One must keep in mind that the decomposition of the angular momentum into spin and orbital components is model dependent. In the language of the renormalization group, it is scheme and scale dependent. If you probe a hadronic system at a given length scale (or, equivalently, energy) ...


0

Quarks as elementary particles are quantum mechanical entities. The same is true of electrons and protons. The hydrogen atom has the simplest quantum mechanical solution of how quantum mechanical particles are bound by attractive forces. This solution is a wave function, and it tells us that the electron is in a quantum mechanical probability locus around ...


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 ...


5

Phenomena in quantum mechanics may be expressed using any basis (that's the English word for the set of vectors, not a "base"). It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the ...



Top 50 recent answers are included