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## Hot answers tagged angular-momentum

7

There is a identity for the derivative of the cross-product of two vector functions $\mathbf A(t)$ and $\mathbf B(t)$; \begin{align} \frac{d}{dt} (\mathbf A \times \mathbf B) = \frac{d\mathbf A}{dt}\times \mathbf B + \mathbf A\times \frac{d\mathbf B}{dt} \end{align} Using this rule with the computation you're considering, we obtain \begin{align} ...

4

The vector product of a vector $\vec{a}$ with itself is alwals zero: $\vec{a} \times \vec{a} = 0$ For two smooth vector-valued functions $\vec{a},\vec{b} \colon \mathbb{R} \to \mathbb{R}^3$ the product rule holds: $$\frac{d}{dt} (\vec{a} \times \vec{b}) = \frac{d}{dt} \vec{a} \times \vec{b} + \vec{a} \times \frac{d}{dt} \vec{b}$$ You can see this for ...

4

An angular momentum eigenstate can be rotated using, $$\left| J , m \right\rangle \rightarrow e ^{ i {\vec S} \cdot {\vec \theta} } \left| J , m \right\rangle$$ where ${\vec S}$ is the $2J+1$ dimensional Pauli matrices. For spin $1/2$ for example, ${\vec S}$ are just the ordinary Pauli matrices, $\frac{1}{2} ... 3 If you read the wikipedia article on orbital angular momentum of light you will see that in the first place it is a classical electromagnetic concept, where the light has a vorticity, i.e. a helical motion around the axis of the vortex. When one goes to the quantum detail of photons one can define an OAM against this classical axis for each photon in this ... 3 The two particles$m_s$and$m_I$live in different vector spaces, so you are actually not picking the same basis vectors (because the basis vectors of the different particles belong to two separate vector spaces). Secondly, the tensor product between the basis vectors of the two different vector spaces will form the basis vectors of a new$3 \times 3 = 9$... 2 Elementary particles differ in flavour from their antiparticles, so conservation laws do, indeed, restrict whether particles or antiparticles can be produced in certain processes. (Compare, e.g., the photon, which has zero for all its flavour quantum numbers, and is identical to its antiparticle.) For example, when a neutron decays, the result is a proton, ... 2 Good work and a good idea. d = L/2 would correspond to the moment of inertia for a point mass M at distance L/2. Momentum p = M$\omega$/2 L/2. What would you get if the mass of the rod was concentrated at the two end points, each 1/2 M ? One point zero, the other M/2$\omega$L. In other words d = 1. So the mass distribution along the rod plays a role. ... 2 First of all I try to restate your question into a more clear form. Consider$\mathbb R$equipped with the equivalence relation:$x \sim y$if and only if$x-y= 2k\pi$with$k \in \mathbb Z$. The space${\mathbb R}/ \sim$of equivalence classes$[x]$is$\mathbb S^1$also as a topological space using the quotient topology. Next consider the standard ... 1 There are two parts to angular momentum that both contribute at the same time. In vector form (where × is the cross product) $$\vec{H}_A = I_{cm} \vec{\omega} + \vec{r}_A \times m \vec{v}_{cm}$$ For a horizontal rod rotating about end point A you have$\$ \begin{aligned} \vec{\omega} & = (0,0,\Omega) & \vec{v}_{cm} &= \vec{\omega} \times ...

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