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An ideal black hole with non-zero angular momentum is described by the Kerr metric. The singularity of such a black hole is not a point.

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They rotate because they are produced by matter that has net angular momentum, and angular momentum is conserved in axially symmetric space-time. So, there's nothing unusual making them rotate that's different from any other physics. However, you are absolutely right to object that rotation of an infinitesimally small point wouldn't make much sense. In ...

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I'm willing to gamble that I'm wrong on this, but it seems to me that you should be able to use the same method of solving the Cartesian $\langle x|p\rangle$ for this polar coordinate one. In order to get $\langle x|p\rangle$, we used $$\langle x|\hat p|p\rangle=p\langle x|p\rangle=-i\hbar\frac{\partial}{\partial x}\langle x|p\rangle$$ which has a ...

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The kinetic energy possessed by the weight depends on its mass (given) and its velocity. The rotational kinetic energy of the cylinder depends on its moment of inertia (found from given information) and its angular velocity. But, because of the cable, the angular velocity of the cylinder is related to the linear velocity of the block, through the radius ...

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Yes the skater does increase the angular momentum by doing work; pulling her arms in. You do work on a swing (sitting up and down) to increase your angular momentum likewise.

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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, $\frac{1}{2} ... 1 As I am not allowed to comment for lack of reputation and cannot find a way to message I would like to point you to a reasonably recent source on the Kerr metric. I am by no means an expert but from what I have read and from what I understand the "Lines" of space time do twist and become unstable at the Cauchy Horizon. From what I gather the rotation isn't ... 3 The two particles$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, ... 0 The problem as is stated is somehow ambiguous, but using some simplifications we can manage to get something: if we assume that forces don't change depending of the angle (i.e. there is no "correcting trayectory rocket" that acts depending of its orientation), and that the center of mass is fixed, then you can express net force F' and net torque T' with ... 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 ... 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} ... 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 ... 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. ...

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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 ...

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