# Tag Info

## Hot answers tagged angular-momentum

6

OP is essentially pondering if commutativity is a transitive relation, ie. if three normal operators$^{1}$ $A$, $B$, and $C$ satisfies $$[A,B]~=~0\quad \wedge\quad [B,C]~=~0 \quad\stackrel{?}{\Rightarrow} \quad [A,C]~=~0 .\tag{T}$$ The answer is No, but OP argues via the existence of a common basis of eigenvectors for two commuting normal operators that ...

5

The Earth's tilt doesn't change so much as it's position around the sun changes. Notice the North Pole sees more sun in summer than in winter cause it's tilted towards the sun in summer but not in winter. The north pole always points in the same direction into space pretty much. It wobbles slightly, and quite slowly, completing a full wobble every ...

5

For a single particle, yes they're parallel. For a system of particles, $$\sum_i \frac{{\bf r}_i\times \dot{\bf r}_i}{\|{\bf r }_i\|^2}\neq \alpha\sum_i{m_i{\bf r}_i\times \dot{\bf r}_i}$$ (you can come up with a specific counterexample but it should be obvious the two sides don't have to be proportional/collinear -- each vector in the sum is weighted ...

4

The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic ...

4

Two operators may be simultaneously diagonalized if and only if they commute. As you can see, $L_z$ commutes neither with $L_x$ nor with $L_y$ – and not with any other linear combinations different from a multiple of $L_z$ – so there's no way to diagonalize two different components of $L_i$ at all. However, $L_z$ (and similarly other components) commutes ...

4

The answer is basically, angular momentum. The collapsing proto-solar nebula has some angular momentum. Whilst dissipative processes can allow the nebula to collapse along the axis of rotation, there is still the problem of how to shed angular momentum in order to allow gas/dust to orbit closer to the rotation axis. This is just a basic application of ...

3

I don't think this is as dumb a question as everyone downvoting you seems to think. The definitions of angular velocity ($\omega$), angular momentum ($L$), and moment of inertia ($I$) ARE defined in order to perfectly mirror Newton's laws. Angular momentum is the analogue of momentum, angular velocity is the analogue of velocity, and moment of inertia is ...

3

The eigenspaces of the quadratic Casimir $L^2 = L_x^2+L_y^2+L_z^2$ of the Lie algebra of infinitesimal rotations $\mathfrak{so}(3)$ are precisely the irreducible representations of $\mathfrak{so}(3)$ - we usually label a representation by its highest weight $l$, which is in this case just a number telling you what the largest possible value for any of the ...

2

The $L_i$ has many eigenspaces corresponding to many eigenvalues. Each of those eigenspaces is also an eigenspace of the Casimir operator. So they share common eigenspaces in the sense that there are eigenspaces that are eigen to both. But they don't share them in the sense that they are the same. Look at the hydrogen atom. There are energy eigenspaces and ...

2

The penny is usually not balanced when you spin it - it is precessing like a gyroscope. That means that as long as it has significant angular momentum, the torque due to gravity will not be able to topple it. As friction slows the penny down, the gyroscope effect becomes weaker until it drops. If the penny were spinning on a low friction surface about a ...

2

I was not able to convince him that this propulsion drive cannot work due to conservation of momentum. Am I wrong about that? No, you are not wrong. It's clear that the engine cannot work because of momentum conservation. It's basically just a fixed double pendulum. Why should there be any positive momentum in any direction after one full circle? ...

2

My professor defined the angular momentum operators in quantum mechanics as the infinitesimal rotations in the Hilbert space of quantum states. Your professor likely did not mean what you think they meant. Rotations in the Hilbert space are performed by unitary transformations. If you choose the position representation then your Hilbert space is a set ...

2

Jac's answer excellently lays down how the mechanics of rotating bodies work. Since you still seem a bit confused, I thought I'd try and expand a bit about what this means for gyroscopes in particular. Let's say you have a gyroscope with its angular velocity $\vec \omega$ pointing horizontally, and ignore gravity for now. The set-up will look something ...

1

A simple answer is to look in the Heisenberg picture with the Heisenberg equation of motion. In this picture, operators evolve instead of states as in the Schrodinger picture. Here the evolution of the operator is chosen so that its expectation value has the same time evolution as in the Schrodinger picture. To do so its time evolution is governed by: $... 1 Faced by the same question and a background that includes courses in vector calculus, I have sought a simpler answer. My answer is much that same as to why one can easily balance on a typical bicycle. Bicycles are constructed so that the point where the extension of the front fork pivot would hit the ground is in front of the point where the front tire ... 1 For a body rotating around an axis, we can reformulate the second law of Newton as follows: $$\frac{d\vec L}{dt}=\vec \tau$$ with$\vec L$the angular momentum and$\vec \tau$the torque applied to the body, both calculated around the axis of rotation. Furthermore, as $$\vec L=I \vec \omega$$ with$I$the moment of inertia and$\omega $the angular ... 1 Given the persistence of this kind of thing, maybe determining the misconceptions underlying it are not so easy to determine and resolve. [Given that we're talking about physics students, I guess it's fair to them to sort this out rather than ignore as one might otherwise.] For my money, I'd ask whether within each part of the mechanism Newton's Third Law ... 1 As one of the comments mentions, it is simpler to consider a linear case. Dropping a body of mass$m$on one moving with mass$M$and velocity$v$is essentially considered the instantaneous transformation$M \to M + m$. Momentum must be conserved in the collision, but the mass of the system effectively increases, producing a smaller kinetic energy: $$... 1 Energy is conserved, but if you ignore some kinds of energy then it will look like it isn't conserved. You can imagine a really big disk with some radial pointing two by fours attached at the one o'clock, two o'clock etcetera positions then attach springs to each two by four with the spring pointing in the clockwise/counter-clockwise directions. Add a nice ... 1 If the spin is an actual magnetic moment, then its behavior under time reversal is simply similar to that of classical magnetization, which changes sign. Think of magnetic fields and dipoles as generated by electric currents. Under time reversal the currents reverse direction and so do the corresponding magnetic fields or dipoles. At quantum level, spin ... 1 This is pretty niche notation, and it is indeed not defined in the paper, but the name "vector-coupled product" does seem to be used by a few people beyond Varga and Suzuki. In essence,$$ [\mathcal Y_{l_1}(\mathbf x_1)\mathcal Y_{l_2}(\mathbf x_2)]_{LM} $$is a coupled wavefunction with total angular momentum L that's made up of the single-particle ... 1 There are solutions to Einstein's Field equation (General Relativity) where the ratio of angular momentum to mass is so large that the singularity is visible to the outside instead of being shielded by an event horizon. But no known astrophysical black hole has a ratio that high. And it looks like when you try to give more angular momentum to an existing ... 1 L_z is related to an infinitesimal rotation, and you seem to think you have rotated you coordinates by 90^\circ. A rotation around the z axis by a small angle \phi is given by$$ R(\phi) = 1 + \imath \phi L_z + O(\phi^2)$$To see how$L_z$acts on a wavefunction we can apply$Rto it, expand and compare terms \begin{align} ... 1 If you want to define yourJ_\pm$in terms of$J_x$and$J_y$then you'll need to know those matrices. If you consider$J_x=\frac{\hbar}{\sqrt 2}\left(\begin{matrix} 0&1&0\\ 1&0&1\\ 0&1&0\end{matrix}\right),$and$J_y=\frac{\hbar}{\sqrt 2}\left(\begin{matrix} 0&-i&0\\ i&0&-i\\ 0&i&0\end{matrix}\right)\$ then ...

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