# Tag Info

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There is no alignment between the Sun or the Solar System's net angular momentum and the "spin axis" of the Galaxy. Think for a moment about whether the line of the ecliptic (which marks the "equatorial line" of the Solar System) and the Milky Way (which roughly marks the plane of the Galaxy) are lined up? If this were so, then you would always see the ...

13

Various data from Kepler and stellar modeling allows the inclination of the orbit to be determined. Several SETI Seminar videos goes into this in some detail. Whether the eclipse cuts a little bit or straight through the middle of the star and the statistics that indicate complete misses are consistent with random orientations without enough data to see if ...

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This is a tricky bit of intuition to get right. In essence, having a lower angular momentum expands the radial range that the electron is allowed to span - the inner turning point moves inward and the outward turning point moves outward - but the electron is moving much slower at the outward turning point, which means that it spends more time there and ...

6

The expression for the spin density is almost correct, but it only involves the rotational part $A_\text{rot}$ of the vector potential $A$, which can be Helmholtz decomposed into $$\vec A = \vec A_\text{grad} + \vec A_\text{rot}$$ where $\vec A_\text{grad}$ is a gradient (hence curl-free) while $\vec A_\text{rot}$ is a curl. Since a gauge transformation is ...

5

For the angular momentum there is no lower bound for the product $(\Delta L_a)_\psi (\Delta L_b)_\psi$ differently from $x$ and $p$. Indeed there are states $\psi$ such that $L_a\psi =0$ for $a=x,y,z$ simultaneously. I am referring to the states with $L^2\psi =0$ which imply $$(\Delta L_a)_\psi (\Delta L_b)_\psi=0$$ So you cannot write something like $$... 3 Let's look closer to home. Axial tilt gives the axial tilt of the more familiar bodies in the Solar System. An axial tilt of greater than 90 degrees implies the body is rotating backwards. So we see Venus, with little axial tilt, rotating very slowly backwards (due to a tidal resonance with Earth) and Uranus and Pluto with pronounced tilts exceeding 90 ... 3 Let \chi be the spinor defined as follows:-$$\chi=\begin{pmatrix} a\\b\end{pmatrix}$$then for measuring S_x we need to find the eigenspinors of S_x which are$$\chi_{+}^x= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \end{pmatrix} ,\hspace{1cm} \chi_{-}^x= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ -1 \end{pmatrix}$$Now the spinor \chi can be ... 2 The norm of your spin wave function is 5 so they divided by 25 to normalize it, then multiplied it by hbar over 2 to get 1/50 2 You have to be a little careful what you mean by "spin" in a system like the one you're talking about. Ideally, we'd measure everything in an inertial reference frame (one that is fixed with respect to the "fixed" stars at infinity), but often that is not actually how we talk about things in an everyday sense. When you talk about the spin of the earth, for ... 2 A skydiver is not an isolated system, and therefore a skydiver's angular momentum is not conserved. Indeed, the drag forces on a skydiver are very great: terminal velocity for someone falling in a standard atmosphere is between 50{\rm m\,s^{-1}} to 60{\rm m\,s^{-1}}. By shaping one's body, hands, or feet, one can control the moment of the system of these ... 2 Emilio Pisanty has already given a good answer. Here we offer a qualitative (as opposed to quantitative) proof of the angular momentum dependence. Recall first of all that the energy-levels$$\tag{2} E_n ~=~-\frac{R_{\mu}}{n^2}$$in the non-relativistic hydrogen atom without spin-orbit interactions are linked to the principal quantum number ... 1 In Einstein summation notation, we'd write$$\vec U = \vec r \times(\nabla\times\vec F) - \nabla\times(\vec r\times\vec F),$$using a Levi-Civita symbol \epsilon_{ijk} as:$$ U_a = \epsilon_{abc} ~r_b ~\epsilon_{cde} ~\partial_d ~F_e - \epsilon_{abc} ~\partial_b ~\epsilon_{cde} ~r_d ~F_e.$$Since \epsilon is not varying with space we can commute it with ... 1 The spin indicates the length (=2s+1) of the vector that a real world particle rotates like. They do not all rotate like pencils (3-vectors). Your questions are not silly! Part of Quantum mechanics involves 1) making a correspondence between a symbol (a |ket>) that you write on piece of paper and an object in the real world, and 2) making a ... 1 My question is whether individual photons also carry orbital angular momentum? Yes. See https://en.m.wikipedia.org/wiki/Orbital_angular_momentum_of_light If yes, what are the values of orbital angular momentum in one-particle states? To quote the wikipedia page In particular, in a quantum theory, individual photons may have the following ... 1 Yes, single photons can have orbital angular momentum. However, unlike spin, they are not required to have any. Just like in the classical case, the orbital momentum of single photons is determined by the shape of their EM mode- roughly speaking, the wavefront must have a helical aspect to it. In particular, this means that the eigenmodes of light in a 3D ... 1 I) The main point is that when we apply Noether's theorem for a field theory, the total angular momentum Noether current$$J^{\mu,\nu\lambda}~=~L^{\mu,\nu\lambda}+S^{\mu,\nu\lambda}$$splits in an orbital angular momentum current$$L^{\mu,\nu\lambda}=x^{\nu}T^{\mu,\lambda}-(\nu\leftrightarrow \lambda)$$and an internal spin angular momentum Noether ... 1 So the state is clearly$$\left(\begin{matrix} 1 \\ 0\end{matrix}\right)_1\otimes\left(\begin{matrix} 1 \\ 0\end{matrix}\right)_2. Or $\left|s_1 s_{z1} s_2 s_{z2}\right\rangle=\left|\frac{1}{2} \frac{+1}{2} \frac{1}{2} \frac{+1}{2}\right\rangle.$ So really you are just trying to write it in the total angular momentum basis. This is what Clebsch-Gordan ...

1

To get $\langle \chi|\chi \rangle =1$, you find that $A=\frac{1}{5}$. Then, when you calculate $\langle S_x \rangle$, you have to use that value for A and then you find the right result of $\hbar/50$. You if didn't get it, I'll do explicitly for you :)

1

The adjoint eigenspinor you multiplied by was the unit length eigenvector of $\sigma_z$ with positive eigenvalue. If you you want a spin up result for the direction $(n_x,n_y,n_z)$ find a unit length eigenvector of $n_x\hat\sigma_x+n_y\hat\sigma_y+n_z\hat\sigma_z$ with positive eigenvalue. And use that instead. If you wanted to do an interaction in the x ...

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