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No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the ...

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Your group theory text probably betrayed you if it did not spend much time contrasting the two cases. A possibly related question is 254461. People use loose language and symbols that aggravate the confusion. Talking abstractly without explicit hands-on formulas clinches it (the confusion)! Let me stick to your 4-dimensional matrices and vectors, all tensor ...

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Spin is connected to the intrinsic magnetic dipole moment of the particle, this is what makes the particle capable of interacting with an external magnetic field. Namely, the intrinsic dipole magnetic moment $\vec\mu$ of a particle with spin $\vec{S}$ can be found through: $$\vec\mu = g\left(\frac{q}{2m}\right)\vec{S}$$ where $g$ is the g-factor, and $q$ ...

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Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ...

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It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results. Let $\rho_m(\boldsymbol r)$ and $\rho_e(\boldsymbol r)$ be the mass and charge densities of the electron. The $g$ factor is given by $$g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ ... 2 You are right that the magnetic field, or rather the apparatus used to generate it, absorbs some angular momentum from the quantum spin. This apparatus includes, for example, some coils of wire containing more than 10^{23} electrons, as well as many other macroscopic components. The upshot is that increasing the angular momentum of this apparatus by an ... 1 Let´s go: [1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition. We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations: From a cross-section of longitudinal (scalar) ... 1 The surrounding system is in a very large very mixed state, and that washes out the distinction in angular momentum. Trivial Example Consider the trace distances between the following nearly-maximally-mixed density matrices:$$\begin{align} D \left( \frac{1}{2} \begin{bmatrix} 1& & \\ &1& \\ & &0 \end{bmatrix}, \frac{1}{2} ...

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First, let's clarify the expression of $|\phi\rangle$. The kets $|+\rangle_z$ and $|-\rangle_z$ are eigenvectors of $\hat{S}_z$ such that $$\hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z$$ This means that in the $\{|+\rangle_z = ... 1 The system can be separated, but not necessarily in nice form. For instance, the time derivative of the first eq. reads $$i\hbar {\ddot c}_1 = - B {\dot c}_1 - {\dot V}c_2 - V {\dot c}_2$$ Now remove$c_2$using again the first eq., $$c_2 = -\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1$$ and${\dot c}_2$using the second eq.,${\dot c_2} = ...

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The Spin of a particle is better understood from a group-theoretic point of view. It is just telling you how a particle, i.e. an asymptotically free state of your theory, transforms under the symmetry of your theory, Lorentz symmetry. Well, actually under its double cover as Weinberg explains in his first book, that is why we are allowed to have spinors. ...

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Neutral charged objects are made of charges. Although net charge is zero, imperfections in spatial arrangement of charges, causes them to interact with electromagnetic fields. Edit: 1) Neutrino's are formed during nuclear reactions. Once created, they sustain their direction of spin which does not change during course of time. 2) Photons don't change their ...

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The multiplicity $2S+1$ actually tells you how many degenerate spin states there are, each labelled with the total spin projection quantum number $M_S$ (this is from the total spin projection operator $\hat{S_z}$(conventionally taken to be in the z-direction) whose eigenvalues are $\hbar M_S$). The possible values of $M_s$ are $-S\le M_S\le S$ in integer ...

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