Tag Info

New answers tagged

2

In the absence of angular momentum, then material would be able to follow radial paths to be accreted (because of their mutual gravitational attraction) and thus have a spherically symmetric distribution. For many reasons (see linked questions to the right), the "circum-object" material does have angular momentum, which must be conserved. In the co-rotating ...


1

The comments seem to provide the answer. $$ L = p \times r $$ $$ L = m ( v \times r ) $$ Thus if r (the distance to the focus point of the orbit) changes, the velocity can change without the angular momentum changing. This is not possible for circular orbits where v is always perpendicular to r and the magnitude of r is constant. However, in elliptical ...


0

Yes. If any of the two nuclei has spin $j=0$, then there will be no J-coupling between them. This is made clear by the J-coupling hamiltonian, $$H\propto\sum_{ik}I^{(1)}_i J_{ik} I_k^{(2)}.$$ If either spin is zero then all the spin components $I_i$ will be zero as operators, and $H$ will be zero. You should note that atoms with even mass numbers need not ...


6

The Earth's rotation rate and the location of the rotation axis change over time. These collectively are called the Earth orientation parameters. On very short time scales, a day or less, the changes in the Earth orientation parameters result predominantly because of the ocean tides. On the scale of decades to a century or so, the dominant driver is exchange ...


0

As David Hammen points out, individual interactions and collisions are extremely important in determining the individual spins of planets and the explanation involving a very large moon-forming collision is almost certainly a major factor in determining the Earth's original spin. However, there are more fundamental reasons why all the planets should spin and ...


1

Yes, but only very slightly as the Earth is much heavier than the atmosphere. If you were accelerating at the equator you would exchange angular momentum with Earth, slowing or accelerating it's rotation, depending which way you accelerate. Same goes for air masses, as they accelerate when they warm up or cool down and move more one way as the other way is ...


2

The dominant hypothesis regarding the formation of the Moon is that a Mars-sized object collided with the proto-Earth 4.5 billion years ago. The Earth is rotating now because of that collision 4.5 billion years ago. As the linked question shows, angular momentum is a conserved quantity. Just as something has to happen to make a moving object change its ...


0

Electrons don't spin. It is just what they decided to call a certain intrinsic property. They could of called it the X factor or magnetic factor but they called it spin for some reason


1

Lie algebras are not a group w.r.t. to the commutator (the Lie bracket). The first reason is that the commutator is not associative. Another is that they almost always lack an identity element, since the identity matrix is, for example, not in $\mathfrak{su}(2)$, and Schur's lemma would, in the fundamental representation, guarantee that only multiples of ...


0

Your question asks why there is a rotatory force that turns the coin to the right. The gravitational force is perpendicular to surface and so the vector cross-product of the gravitation force and the angular momentum (which is not exactly parallel to the surface) is perpendicular to both those vectors. That doesn't actually help (for me anyway). However, if ...


1

We can start with the definition of angular momentum $\vec{L} = \vec{r} \times \vec{p}$. Differentiate both sides with respect to $t$ to get \begin{equation} \frac{d\vec{L}}{dt} = \vec{v} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} \end{equation} The first term on the right hand side is zero because $\vec{v}$ and $\vec{p}$ are parallel to each other. ...


10

Spin is best understood as an intrinsic angular momentum. It is probably easier to understand the concept for a charged particle. A classical charged particle moving along a circle has an angular momentum and the "circuit" has a magnetic moment. Further, the two are proportional to each other. It is experimentally found that a charged particle like an ...


3

Spin arises from the need to represent the rotation group $\mathrm{SO}(3)$ upon our Hilbert space of states. We need such a representation because the rotations (together with space translations) correspond to the non-relativistic changes of reference frames. Since states are only determined up to rays in the Hilbert space, the true space of states on which ...


2

Apparently it's a historical quirk. Characterizing spectral lines as principle, sharp, or diffuse dates back to the 1870s with the works of George Liveing and Sir James Dewar. Living and Dewar also noted that these lines appear in series. Arno Bergmann discovered a fourth series in 1907, which he labeled as the fundamental series. If Arnold Somerfeld had ...


1

The 'triangle rule' seems to be (as used, e.g. in Molecular Spectroscopy by D.J. Millen) the application of the triangle inequality to the addition of angular momentum. More specifically, if you are adding momenta with magnitudes $j_1$ and $j_2$ and magnetic quantum numbers $m_1$ and $m_2$, then the result will only have nonzero amplitude along the state ...


3

This is due to spinning forces, such as centrifuged force. Remember that galaxies form (at least the regular ones) from the spinning of matter around a galaxy nucleus. So, you're not wrong. Gravity is isotropic, but in this cases the spinning forces are the definition to the galaxy form.


0

The "math works it out" indeed. I try to write it as accessible as I can. A 0-domensional Euclidean space is just a point. The 1-dimensional is a line. The 2-dimensional is a plane. The 3-dimensional is the space as we know it. This can be continued to 4-5-6 whatever dimensions. On a plane you can draw lines and point, but not planes. In space you can ...


0

$\sigma_y$ isn't an observable, but $S_y$ is, so let's focus on that. If you know that $S_y | \psi \rangle$ can be written $a |\psi\rangle$, where $a$ is a number, then it only tells you that $| \psi \rangle$ a measurement of the $y$-component of spin will definitely yield $a$. That's all. I don't think you can consider $S_y|\psi\rangle$ a measurement of ...


1

My quantum mechanics teacher, always say to me, that those models for "see" the spin and the angular momentum, are not correct. the reason is that the picture of an electron precessing associated with the spin in't correct because the spin is'nt associated with any spacial coordinate. in quantum mechanics is better abandon those pictures, the important ...


0

Any operator, $A$, acting one of its eigenvector, $| \psi_i \rangle$, will give, $$ A | \psi_i \rangle = \lambda_i | \psi_i \rangle, $$ where $\lambda_i$ is the corresponding eigenvalue. The eigenvalues of the Pauli matrices are $\pm 1$ corresponding to either spin up or down in the corresponding direction. The eigenvalues are possible results of measuring ...


0

This is a great question I too have been pondering about. I think theoretically it has to do with good quantum number. The optical pumping relies on defining circular polarization transition selection rules as delta m = +1 for a RCP polarization. Why is that? It comes from the fact that B defines the quantization axes. Imagine the experiment in which B and ...


1

The paper addresses the case of a circular, finite quantum well with different electron effective masses inside and outside the well. The equation you point to is the result of a derivation from the time-independent Schrodinger equation, effective mass approximation and BenDaniel-Duke boundary conditions. It is also, however, dependent on the statement ...


2

You know Newton's second law: $$\vec F = m \vec a~.$$ Now you multiply both sides by "$\vec r \times~$", then you get $$\vec r \times \vec F = \vec r \times m \vec a = \frac{d}{dt} (\vec r \times \vec p)~.\tag{1}$$ (If you carry out the time derivative, the first term from the product rule is $\vec v\times\vec v$, which of course is $0$.) This is very ...


0

Your intuition about inertia is essentially correct. The spin state of an electron does not change instantly. If the electron is in the spin-up state then the z-component of its angular momentum is $\frac{1}{2} \hbar$. If it is in the spin-down state then its angular momentum is $-\frac{1}{2} \hbar$. Classically, angular momentum cannot change instantly, ...


1

I don't want to give away the answer immediately, since it's homework. Try to visualize one huge big wheel and one very small wheel. Try to imagine what would happen if at each instant the tangential velocity is not the same.


2

Spin of an electron is measured as a magnetic property. You should not visualize it as an electron "spinning" around its axis, which is what you seem to indicate if I'm not mistaken. Electrons are considered to be point particles. Also, the spin of an electron never changes instantaneously. For example, changes in the electron's spin in the Stern-Gerlach ...


1

I'm trying to give a less technical answer. It's not rigorous but should give you the idea how spin and the regular rotation related. Maxwell's equations say in order to have magnetic field, you need a ring current. This can be achieved by giving angular momentum to charged particles. This can be orbital or simply because the particle is spinning. This was ...



Top 50 recent answers are included