# Tag Info

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There exists a very simple and concise semi-classical explanation of the electron's spin angular momentum, without the notion of rotation of any material object: Qualitatively speaking, the electron’s spin angular momentum is the electromagnetic field's angular momentum resulting from the combined electromagnetic field surrounding an electron just in such a ...

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The issue here is whether or not you have a sum over $l$. If you want $r_l^2$ to mean any of $r_1^2, r_2^2, r_3^2$, then when you write $r_l^2 = r_l r_l$ you should not be summing over $l$. So $[r_l^2,L_i] = 2i\hbar \epsilon_{ijl}r_jr_l$ is correct as long as you sum over $j$ but not over $l$. On the other hand, if in $r_lr_l$ you sum over $l$ you get the ...

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For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ...

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First of all, you mention that when the instrument disconnects, it is no longer rotating. Why would that be? However, the bigger issue is that if you want to conserve angular momentum, you have to measure it in the same way each time. For angular momentum, that means you have to retain the same axis each time. The instrument may no longer be rotating ...

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In a sense this is an assumption, however it's an assumption that has a basis in the history of QM at the time. I attach images of four pages from http://www.amazon.com/Einstein-Quantum-Quest-Valiant-Swabian/dp/0691168563, which I'm currently reading and recommend if you're interested in the history, that give something of the flavor. Bohr can be said to ...

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Its not that tough. You can work it out by using just two equations. But the one thing you should keep in mind is that when the comet is at the minimum distance from the sun, its velocity must be perpendicular to the radial vector (sun to comet). So the minimum distance is itself the minimum perpendicular distance used in the angular momentum formula at ...

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The angular momentum $L$ of an object is always defined with regards to the axis about which rotation takes place: $$L=I\omega,$$ where $I$ is the inertial moment and $\omega$ the angular speed. The inertial moment is defined and can be calculated as shown here. Where it's necessary to calculate $I$ with respect to another axis parallel to the first one, ...

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Angular momentum is a vector defined by $$\vec{L}_{cm} = I_{cm} \vec{\omega}$$ where $I_{cm}$ is a 3×3 mass moment of inertia about the center of mass. When the rotational velocity vector $\omega$ is about one of the principal inertial axes then the angular momentum vector is parallel to the motion axis and $\vec{L}_{cm} =m \rho^2 \vec{\omega}$ where $m$ is ...

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Comments to the question (v1): In Newtonian mechanics with Newtonian gravity, a body can have orbital angular momentum wrt. a reference frame. A non-point-mass can also have spin angular momentum. Bodies can exchange angular momentum via tidal forces. In GR, it possible to assign angular momentum to certain space-time regions (such as e.g. the Kerr ...

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They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally ...

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There are plenty of satellite galaxies orbiting larger galaxies. The question is how long are you willing to wait for an orbit? The Milky Way has a mass $M$ of something like $6\times10^{11}$ solar masses, or $10^{42}\ \mathrm{kg}$. The small Magellanic Cloud is at a distance $R$ of $2\times10^5$ light years, or $2\times10^{21}\ \mathrm{m}$. A test mass ...

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I think the first two things that is transformation of position and momentum operators are defined from definition...because by parity transformation sign of the position coordinates changes and time coordinate remain unchanged...so accordingly we get change in sign for position and momentum...once you do that then angular momentum should remain unchanged ...

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Interesting question which even made me laugh. As already pointed out the Power of an hurricane is too high to be connected to any grid. My laugh came about this; "Giant heaters out to sea which dump heat into sea water?" -Why? Because it's basically the heat of the sea which feeds the energy to the Hurricane, and thus this kind of system would not do ...

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(sorry, I couldn't write this in the comment section) Have you met the postulates of quantum mechanics? Here is a summary of them http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html Postulate 3 says if an observable has associate a (hermitian) operator, the only values we would observe for one photon the spin-angular momentum are the eigenvalues ...

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Regarding your question in your comment: consider the hamiltonion of a particle in a central force field: $$H=-\frac{\hbar}{2m}\nabla^2+V(r)= \frac{1}{2m}P_r^2+\frac{L^2}{2mr^2}+V(r),$$ where $P_r=-i\hbar~ r^{-1}\partial_r r$ is the radial momentum operator. Let's operate with this Hamiltonion on some wavefuncton $\psi$: $$\left[ ... 0 An astronaut dressed in a magnetic suit inside a room or environment of copper or similar element in characteristics of conductivity or alloy (copper chamber) is a human magnet (magnet man) than will face a force that is opposed to his displacement inside the "chamber". If one of six sides or of all the sides than compose the chamber, one of the sides is the ... 1 It depends on what you want how you choose the polarization of the light. The polarization of your light determines the recoil of your electron and your ion. In photoelectron vmi you would like to see the angular distribution of how the electron detaches from the molecule, so you should select the polarization of your light such that the velocity vector of ... 1 Spin and orbital angular momentum are two different things, as already pointed out in Aniket's answer, but there is a good reason why we still call spin a "spin". This is because the Einstein-de Haas-Richardson experiment shows that electron spin is indeed of the nature of an angular momentum, although not exactly due to a "spinning electron". In fact, ... 1 In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to ... -2 There is a very interesting Am J Phys paper by Ohanian, titled "What is spin?". You can find free PDF copies on google, in case you don't have academic access. He points out that ALL forms of angular momentum, even spin, arise from linear momentum via the relationship \vec r \times \vec p. In other words, even spin is orbital angular momentum. This is ... 1 You are quite correct that if you have items floating freely inside your space station they won't experience any artifical gravity as the station starts spinning. The artificial gravitational acceleration of an object is a consequence of its tangential velocity v and is given by:$$ g = \frac{v^2}{r} $$where r is the distance to the axis. The freely ... 1 A quick Google search turns up a couple of additional resources besides the free paper abstract. Stanford has a summary of the paper, and the University of Delaware has slides associated with a talk on the paper. Offshore wind turbines currently exist and do act the opposite of an aircraft propellor: wind energy is converted into electricity, slowing the ... 0 Wait...wait... what? No, no, no, no... No. If a body is spinning, supposing non frictional surfaces and all of that, the energy would not decrease (if there is no external force). The energy is always constant. In that kind of problems there is only rotational energy (there is no other energy):  E_{rot}={1\over 2}I\,\omega^2, ... 1 Yes, the rotational kinetic energy decreases. The extra energy is converted to thermal energy in the wheel and environment. If you imagine letting the weight go, it will slide across the surface of the wheel as it moves towards the edge. This sliding is motion against friction, so energy is lost there. Then the weight might bang into whatever holds it at ... 0 The energy is dissipated when the mass stops at the perimeter. Work has to be done to stop the radial motion of the mass. 0 You have mixed indices in the end of your line. Correctly:$$-\epsilon_{iab}\epsilon_{jcd}(x_ap_d\delta _{bc} - x_cp_b\delta _{ad}).$$So further, -\epsilon_{iab}\epsilon_{jcd}(x_ap_d\delta _{bc} - x_cp_b\delta _{ad})=\epsilon_{iab}\epsilon_{jcd}x_cp_b\delta _{ad}-\epsilon_{iab}\epsilon_{jcd}x_ap_d\delta _{bc}=\\=\epsilon_{idb}\epsilon_{jcd}x_cp_b - ... 1 You need to use vectors. Since L \neq r \times p, you need to use \vec L = \vec r \times \vec p instead, where the \times is the vector cross product of vectors, not the scalar multiplication of scalars. So you have$$\vec L= \left[(v_o t \cos \theta) \hat x+ (v_o t \sin \theta - \frac{1}{2}gt^2)\hat y\right] \times m\left[(v_o \cos \theta)\hat x+ ...

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$m \frac{v^2}{r} = Force$ $m \omega^2 r = Force$ $Force = ma$ $\omega = \sqrt{\frac{k}{m}}$ $2 \pi f = \omega$ $f= \frac{1}{2 \pi} \omega$ From here you can make substitutions and arrive at the result. Let me know how it goes.

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Arturo's answer above is great, I just want to add mine because after working on understanding it, I think I have a simple explanation that might help someone. In the first equation listed in the original post, force and radius are inversely related if velocity and mass are held constant - this means that the frequency must change since velocity depends on ...

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