# Tag Info

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L(angular momentum) = (moment of Inertia) x (angular velocity)

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The angular momentum $L_{A/B}$ of a rigid body $A/B$ about its center of mass is $$L_{A/B} = I_{A/B} \omega_{A/B},$$ where $I_{A/B}$ is the inertia matrix of $A/B$ about its center of mass in the world frame and $\omega_{A/B}$ is the angular velocity of $A/B$. The angular momentum $L_{A/B}^0$ of a rigid body $A/B$ about the origin of the world frame is ...

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Some bits and pieces on angular momentum: Angular momentum is that which is conserved in rotationally invariant systems, just like energy is that which is conserved in time translation invariant systems and momentum is that which is conserved in space translation invariant system. This is the essence of Noether's theorem. The analogue in QFTs are ...

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If you are looking for quantitative values, look to further than the equations of motion \begin{align} \sum_{i} \left( \vec{F_i} \right) & = m \vec{a}_{cm} \\ \sum_i \left( \vec{\tau}_i + (\vec{r}_i-\vec{r}_{cm}) \times \vec{F}_i \right) & = I_{cm} \vec{\alpha} + \vec{\omega} \times I_{cm} \vec{\omega}\end{align} where the left hand side is ...

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The angular equivalent of the impulse-momentum theorem states The change in angular momentum of a system is equal to the product of the (average) external torque time the time it is applies In math that is (finite version): $$\langle \vec{\tau}_{ext} \rangle \,\Delta t = \Delta \vec{L} \,,$$ or (infinitesimal version): $$\vec{\tau} \, \mathrm{d}t = ... 1 Some conservation laws are related to conservation of angular momentum. There is a famous example (from Feynman if I recall correctly), where you assume an infinite flat space and conservation of angular momentum about any point, and then you get conservation of linear momentum for free. Intuitively, to get say the x component of linear momentum is ... 0 Answer to first question: At the instant being considered, the space and body axes are identical, so at that moment the matrix a that relates the two sets of axes is simply the identity matrix. dG' is a vector, so a_{ji}dG_j' = dG_i' is simply equivalent to the statement that with I the identity matrix and V an arbitrary vector, IV=V. Answer ... 0 Electromagnetic field of light has two kind of angular momentum first spin angular momentum (SAM) and secondly orbital angular momentum (OAM). former one represent the dynamical rotation of electric (or magnetic) field of around propagation direction and indicate the polarization of beam. Later one represent the rotation of light around beam axes. The ... 0 By no means not a complete answer, more a criticism of @luksen’s one. It is posted here because the text is too long to fit in the comment field. First of all, the spin is not a well-defined concept for composite particles. More precisely, whether the spin of a particle is defined depends on how the “particle” is defined. Look at an atom: it has the nucleus ... 1 There is no law of physics that requires an external force to be present when some double time derivative of a parameter in a system changes. What you were taught in high school, that is$$F = m \cdot \dot v$$and$$M = I \cdot \dot \omega$$are actually the simplified version of$$F = \frac{d}{dt}m \cdot v = m \cdot \dot v + \dot m \cdot v$$and$$M ...

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A simple picture here would be that dissipation leads loss of mechanical energy. But angular momentum has to be conserved. The lowest possible mechanical energy with a given value of the angular momentum of fluid inside a container is that of rigid body rotating around the axis of largest moment of inertia.

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You apparently mean "simulate" when you used the word "model". You'll need two things to accomplish this: A better rotational integrator than the one presented in this answer to your other question, and A physical model of a system that loses energy while conserving angular momentum. Regarding the first item, that integrator is not bad. It has the ...

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For massive particles the intuition of thinking of spin as a rotation is correct. In the rest frame, a massive ($M^2>0$) particle has momentum $$p_0^\mu=(M,0,0,0).$$ Remember that the quantity $p^2=p_\mu p^\mu$, for arbitrary $p_\mu$ is invariant under Lorentz transformations. In the case above the subgroup of the Lorentz group leaving $p_0^\mu$ ...

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The best way to understand spin is actually to consider the Dirac Equation $$i\hbar \frac{\partial }{\partial t}\Psi=\left[c\sum_i{\alpha_i p_i}+mc^2\beta\right]\Psi$$ or more compactly: $$(i\gamma^\mu\partial_{\mu}-m)\psi=0$$ The solutions to the Dirac equation are collections of complex valued fields called spinors. The spinor solution actually ...

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For what it's worth, I've always had the same feeling that the spin should have some sort of reason behind it. It seems so unsatisfying to be told more or less that "it just came that way." Is there any deeper sort of explanation at all? I recall a paper in AJP from years ago called "What is spin?" by Ohanian, but I didn't put in the effort to follow it. I ...

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Spin is a wave property. It exists in classical relativistic wave theories as well. A circularly polarized wave carries an angular momentum that's related to the spin of the field. A gravitational wave (spin-2) can carry twice the angular momentum of a classical electromagnetic wave (spin-1). Being "pointlike" is a particle property. You can think of the ...

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You are correct. The total angular momentum remains constant. As the objects are not approaching head-on, there must be angular momentum. That momentum remains in the joined object as it spins around its COM. The speed of the spin for the joined object will be proportional to the perpendicular distance and speed of approach, and inversely proportional to ...

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Spin is not about stuff spinning. (Confusing, I know, but physicists have never been great at naming things. Exhibit A: Quarks.) Spin is a purely quantum mechanical phenomenon, it cannot be understood with classical physics alone, and every analogy will break down. It has also, intrinsically, nothing to do with any kind of internal structure. ...

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"The electron has no known internal structure", but since it does have a spin, does that mean that we know the electron has an internal structure but we just don't know what it is? An electron has no known internal structure simply means that nobody knows if the electron has an internal structure. So far they know none and therefore they suppose it ...

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Answer expected by following author's hints. \begin{align} \mathbf{L}_{eg} &= \dfrac{1}{4\pi}\int \mathbf{r'} \times \left [\mathbf{E} \times \mathbf{B} \right] d^3r'\\ & = \dfrac{1}{4\pi} \int \left [ \left (\mathbf{B.r'} \right)\mathbf{E} - \left (\mathbf{E . r'} \right) \mathbf{B} \right] d^3r'\\ & = \dfrac{1}{4\pi} \int \left [ \left ...

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One does not need to switch "gravity off" to make a system like that. Replace gravity with a simple string, and cut the string. Physically that's completely equivalent to your problem, as far as I can see. And while this may seem counterintuitive, the angular momentum in a system of a mass rotating at the end of a string is, indeed, conserved, when you cut ...

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I see you have ticked Muphrid's answer but I still think a less abstract reply could be useful. 1. Yes, angular momentum is conserved. 2. Yes, They will move off at constant velocity at a tangent to their circular motion. 3. Yes, their angular momentum will be of equal magnitude. 4. "but opposite sign." - No - they are travelling in opposite directions, ...

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The question is "Why are the possible outcomes the same for all directions?" It happens also for observables of classical physics! QM does not matter here, the truly relevant idea is the fact that in a inertial system physics appears to be isotropic. In practice, it is not possible to physically distinguish different directions with physical experiments. ...

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Although I see already two answers, I’ll add mine. The situation, as original poster described, is such for a massive spin-½ particle and is not symmetric in this sense for a massless spin-½ particle (some people say this case is properly named “helicity”, but it is a question of terminology). Why does it matter? Because a massive particle has its ...

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Short intro to ladders As you say, they're ladder operators. Let's get rid of the annoying $\hbar$ by setting it to one, and call them more systematically $L_{-1},L_0,L_1$ instead of $L_-,L_z,L_+$. Then, the commutation relations take the uniform form $$[L_n,L_m] = (m-n)L_{m+n}$$ If we had countably many of these, we'd have a Witt algebra, if there was ...

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Just remember that a Black Hole doesn't have infinite gravity - it just has however much mass created it in the first place. Yes, anything that gets within the event horizon is trapped forever, but that event horizon will actually be smaller than the size of the equivalent amount of mass composed of ordinary matter. This is also why "microscopic black holes" ...

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Consider the following diagram: This shows a mass $m$ moving past a point $P$ in a straight line. Note that the mass isn't connected to $P$ in any way - it's just moving past in a straight line. The angular momentum of $m$ about $P$ is given by: $$\vec{L} = \vec{r} \times m\vec{v}$$ So the direction of $\vec{L}$ is normal to the screen and the ...

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Ball A has linear momentum pointing in the direction of its flight path. But it is not conserved, because you have a force acting on it (centripetal force), that as no counter part (unless you specify where the axle is mounted and allow for the mount to move as well). Observe, that you can ascribe an angular momentum to B as well even when it does not move ...

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Just like $j=1/2$ (spin-half) particles have two-complex-component wave functions, spinors, particles with spin $j$ have $(2j+1)$-dimensional wave functions describing the spin degrees of freedom. The dimension is what it is because $j_z$ always goes from $-j$ to $+j$ with the spacing equal to one. All the transformation rules under rotations may be ...

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As you know, any operation, which can be continuous or discrete, in quantum mechanics is represented by an operator say $\hat{O}$. Under the operation, a physical quantity $\hat{X}$ transforms as $\hat{X}' = \hat{O}\hat{X}\hat{O}^{-1}$. In case $\hat{O}$ is continuous (forming a Lie group), one can often write it as $\hat{O} = e^{ir\hat{T}}$ (there is more ...

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You already received several answers. However the fundamental physical reason is elementary: In classical, quantum and relativistic physics the physical laws describing an isolated physical system in an inertial reference frame are the same (are invariant) if you rotate (with an element of $SO(3)$) the system (there are many other symmetries depending on the ...

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One can make sense of the introduction of $\mathrm{SO}(3)$ into quantum mechanics as follows: Consider a physical system in three spatial dimensions which we'll think of as residing in a box sitting on a table, like a table-top experiment. Suppose that we prepare the system in a particular way, so that the system is measured to be in a (pure) state ...

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The very important group in physics is the Lorentz group $SO(3,1)$. It is built from the $SO(3)$ group and Lorentz boosts. The algebra of the Lorentz group generators (boosts $L_{i}$ and 3-rotations $R_{i}$) doesn't have the separation on $SO(3)$ and boosts' parts. But by introducing the "new" generators $J^{i}_{\pm} = \frac{1}{2}(R_{i} \pm iL_{i})$ we may ...

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Spin in (non-relativistic) QM is fairly ad hoc, it has no deep reason. The underlying reason is that, in a relativistic setting for QM/QFT our states/fields must transform in some representation of the Lorentz group $\mathrm{SO}(1,3)$ if we want the amplitudes/Lagrangians to be Lorentz invariant (since something cannot be invariant if there's no rule how its ...

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Physically, $R$ is just a rotation. However, when you write $R \psi(x)$, the general result is $R \psi(x) = (R\psi) (R^{-1}x)$. The $R^{-1}x$ takes in account that the rotation has an effect on the space-time coordinates, and it corresponds to orbital angular momentum , but you have to take in account that $\psi$ itself may be not a scalar (an invariant) ...

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