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An electron revolving in the atomic orbit gives rise to an electric current which is equivalent to a magnetic shell of magnetic moment $\mu_l$. We can calculate this magnetic moment in this way:- Time period of electron is $T=\frac{2\pi r}{v}$; So, equivalent current $I=\frac{q}{T}=\frac{qv}{2\pi r}$ Hence magnetic moment, $\mu_l=I.A=\frac{qv}{2\pi r}.\pi ... 0 You can take a look on explanations like this one. Yes, it is caused by the interaction with the magnetic field. "Tt suggests the possible existence of an angular momentum like quantity that can take on half-integer values." As a consequence, there may be a singlet-type fine structure spectrum as well. 0 It is definitely possible. Take a two-planet system where the planets orbit in opposite directions. Have the planets' masses be equal, but small enough so that their interaction is swamped by their interaction with the central sun, thus giving them stable orbits. The outer planet's angular momentum will be greater than the inner planet's, and the sun will ... 4 Given the orbital angular momentum operator$L$on the "spatial space"$\mathcal{H}_1$and the spin angular momentum operator$S$on the "spin space"$\mathcal{H}_2$, we have the total angular momentum operator on the combined space$\mathcal{H}_1\otimes\mathcal{H}_2$given by $$J = L \otimes \mathbf{1} + \mathbf{1} \otimes S$$ where$\mathbf{1}$is the ... 2 Now, acceleration, a = radius ( r ) x angular acceleration (α) Now as there is, a = gsinθ so there must be nonzero value of α. That is only true for rotational motion. As it stands there is nothing in your problem to suggest that any sort of rotational motion will happen. Since the plane is "FRICTIONLESS" (your words), it does not do the normal "rolling ... 2 One has to start from what "laws of nature are". They are a distillation of observations. The basic observation is that objects do not change if thrown or spinned . A golf ball may be spinning , but its shape is invariant , it does not change shape because it is spinning. An airplane travels with supersonic velocity, but in the plane the objects on the ... 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 ... 0 For some experimental evidence of photons with orbital angular momentum, consider the gammas emitted in$2^+\rightarrow 0^+$transitions in even-even nuclei. I'm not sure of the superposition question, but experimentally the coulomb excitation of a$4^+$state from a$0^+$might be explained as via two virtual$E2$photons. (Comments from other nuclear ... 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 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 ... 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 =0which imply $$(\Delta L_a)_\psi (\Delta L_b)_\psi=0$$ So you cannot write something like $$... 6 The expression for the spin density is almost correct, but it only involves the rotational part A_\text{rot} of the vector potential A, which can be Helmholtz decomposed into$$ \vec A = \vec A_\text{grad} + \vec A_\text{rot}$$where \vec A_\text{grad} is a gradient (hence curl-free) while \vec A_\text{rot} is a curl. Since a gauge transformation is ... -2 The tilt of the Earth relative to its orbit plane is definitely caused by a force: the evidence is that the tilt oscillates. Space is frictionless and so a force by a collision or a huge volcanic blast pushes the tilt in one direction and then the oscillation starts with the tile going to maximum in one direction and then back in the other direction, like a ... 0 This answer possibly isn't at the level that you would like, but I'm inclined to write it anyway because it's a good introductory answer. In the event that that this question does merge as duplicate, I'll probably just move my answer over there (as this is slightly different than the posted answers). First, watch this Minute Physics video as it provides a ... 1 The spin indicates the length (=2s+1) of the vector that a real world particle rotates like. They do not all rotate like pencils (3-vectors). Your questions are not silly! Part of Quantum mechanics involves 1) making a correspondence between a symbol (a |ket>) that you write on piece of paper and an object in the real world, and 2) making a ... 0 The spin of a particle is a number that describes its angular momentum. The earth orbits the sun, making years- that is angular orbital momentum. The earth spins on its own axis, making days- that is angular rotational momentum The spin of a particle is analogous to the latter of those two. Not exactly alike due to the quantum nature of spin, but ´same ... 1 In Einstein summation notation, we'd write$$\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.$$Since \epsilon is not varying with space we can commute it with ... 2 Emilio Pisanty has already given a good answer. Here we offer a qualitative (as opposed to quantitative) proof of the angular momentum dependence. Recall first of all that the energy-levels$$\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 ... 9 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 ... 0 I am not sure if this is the answer you are looking for, but you could say it is because we defined p (the lowercase one) already with a factor of \gamma: p = \gamma m v. If this \gamma weren't there, then it would need to be present in your formula: P = m v \cdot \gamma(v). The reason that we put the gamma in for p is because that is the only ... 0 It is true that angular momentum is conserved in all frames, but the actual value of the angular momentum will, in general be different. If you look at the extra terms you will find that they correspond to the change in the angular momentum of the centre of mass about your chosen origin. \begin{align} \mathbf{r}_{CM}\times M\mathbf{V} & = ... 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 ...