# Tag Info

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First you need to define the operator whose eigenvalues are j. As your avatar invites me to, I skip the fine formal fussing and just calculate instead. Such operators are routine, e.g., Curtright and Zachos (1990) PhysLettB243. For the Casimir $\vec J \cdot \vec J$, with eigenvalues j(j+1), define $$L_0\equiv \frac{\sqrt{1+4\vec J \cdot \vec J}-1}{2},$$ ...

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The force vector acting in the ball by the cue is a sum of a normal and a friction one. If there were no friction between cue and ball, that would follow a straight line to the right, without spinning. The friction force is to the left. Its is certainly small for that material, so the composite force is to the right. If the kick is just above the ball ...

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The spin of the red ball has two components: the clockwise you mention, and the forward roll. Imagine the red ball is not moving forward but only spins clockwise, as you draw. You then roll it forward by a quarter revolution. Now it will role to the left. This is what gives the left curving.

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No the angular momentum of the particle would remain zero (it is still a constant of motion). The angular velocity of the particle, on the other hand, would steadily increase as the particle approaches the Kerr black hole (i.e. the Newtonian relationship between angular momentum and angular velocity does not hold in GR!) Note that the frame dragging does ...

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Angular Acceleration is not the same as Translation Acceleration. Translational Acceleration in an object cannot come without any external force, but the angular acceleration in an object may come without any external torque. External torque is NOT the only source of angular acceleration. Angular acceleration in a rotating body may come even when the moment ...

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A more mathematical way to think of it is that there are two relevant vectors: the initial velocity $\vec v$ and the initial acceleration $\vec a$. These span a two dimensional subspace (regardless of the number of dimensions in the full space). The solution will stay within this subspace. There is the degenerate case when $\vec v$ and $\vec a$ are not ...

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As answered elsewhere, yes, the motion is confined to a plane. For me, analogy to cross products is less intuitive in higher dimensions. What makes it planar? Force dictates the second derivative. You always have a second order differential equation. All solutions to "nice" second order differential equations are defined by two functions. Below is a kinda-...

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Tesseract's answer is correct. More conceptually, in general dimensions the angular momentum is what is known as a "two-form", which you can think of a plane with a magnitude and orientation of in-plane rotation. This works in any dimension, and in any dimension the conservation of this plane means that motion under a central force is confined to that plane. ...

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In general, the angular momentum is defined as $\mathbf{L}=\mathbf{r}\wedge\mathbf{p}$. In our problem, switching to index notation, we have, in the CM frame, $$L_{\mu\nu}=r_\mu p_\nu-r_\nu p_\mu=\mu(r_\mu\dot{r}_\nu-r_\nu\dot{r}_\mu)$$ Now, since there is no external torque, angular momentum is constant. If we define a Cartesian coordinate system $(w,x,y,z)... -1 If you want$S^{i0}$to be zero, then the spin$S^{\mu\nu}$is the angular momentum of the body about its energy centroid in the lab frame. Let's see how this comes about. Suppose that the we have a conserved and symmetric energy-momentum tensor: $$\partial_\mu T^{\mu\nu}=0, \quad T^{\mu\nu}=T^{\nu\mu}$$ which is non-zero only within the body of ... 0 The spin of a particle is defined by its intrinsic angular momentum and as such is defined in the particle's rest frame. So your spin four vector in the rest frame must take the form $$S^\mu = (0, \mathbf S)$$ Where the 3-vector$\mathbf S$is your familiar non-relativistic spin. The four vector will transform precisely thus, in the vector representation ... 1 Physically yes but technically the order matters in the sense that the phase of the state of "good"$Jactually depend on the ordering. Writing \begin{align} \vert J M_J\rangle =\sum_{m_1m_2} C_{j_1m_1;j_2m_2}^{JM_J}\vert j_1m_1\rangle \vert j_2m_2\rangle \end{align} whereC_{j_1m_1;j_2m_2}^{JM_J}$is a Clebsch-Gordan coefficient, and using the symmetry ... 2 In essence yes, the only difference is in the labels you give to each subspace - 1 or 2. This could mean simply giving different labels to particle 1 and particle 2, each of which has certain angular momentum. Or if the system is one particle with orbital and spin contributions to its angular momentum its just a different assignation of labels one and two ... 0 Newton's second Law does in fact apply to rigid bodies. There is no restriction on applying it to only point masses. The torque of the net force (which is acting on the skater's body minus the arms, and the arms themselves) is not zero. A net non-zero torque will be produced by the tangential forces$F_t$which will speed up both the skater's armless body ... 0 There is a thing called moment of inertia of extended bodies. It is the fundamental leap that allows us to still do mechanics, and we treat it sort of like a mass. Look up the definition, it is a summation or integral of the mass points times their position squared about the axis of rotation. When the skater has her arms extended then$I$is much bigger. ... 0 It may not be obvious, but it is definitional. For both J1 and J2, we define the z axis to point in the same direction (is it's an external axis) and from your definition of J it follows that$J_z=J_{z1}+J_{z2}. Does that help? 2 So how do we compute the angular momentum precisely with these constraints? We do not compute the angular momentum precisely with these constraints. Conservation of momentum is like an accounting principle. The money in your bank account consists of money that is yours plus money that you owe. Maybe you don't know exactly how much money is in the account, ... 0 Conservation laws , either of quantum numbers or of energy, momentum,angular momentum need a "before and after",usually an interaction is allowed or not according to the conservation laws. The example: an electron in the 1s orbital of the hydrogen atom. There is no before and after in this case. The atom is eternal, unless it interacts. The orbital ... 2 This last line makes use of the fact that $$\sum_{n=0}^N n=\frac{N(N+1)}2.$$ This expression is called a triangular number and there are many great proofs for it. It also splits the sum in two parts \begin{align}\sum_{n=0}^N (n+1/2)&=\sum_{n=0}^N(n)+\sum_{n=0}^N(1/2)\\&=\frac 1 2N(N+1)+\frac 1 2(N+1)\\&=\frac 1 2(N+1)^2\end{align} Obviously you ... 9 Angular momentum itself does not contribute to the invariant mass, however an object with angular momentum will have KE in its center of momentum frame, and this KE is part of the invariant mass. It is the KE that provides this “additional” mass, not the angular momentum. So, for example, two objects with the same angular momentum but different amounts of ... 1 Although the static friction does no work it does apply a force to the bottom of the rod (which accelerates the center of mass horizontally), and a torque. The normal force also produces a torque relative to te center of mass. Both of these torques change as the rod falls, and cause changes in the angular momentum. (The angular acceleration is a variable, ... 1 The stool can rotate freely only about the vertical axis. It can and does provide counteracting torque to any component of L perpendicular to that. That allows it to provide the horizontal components of torque necessary to cancel the L2 in the diagram. Physically, the stool is what’s preventing the person from rotating in pitch (head forward, feet back)... 3 Your orthogonal matrix $$R(\phi,\theta)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ \sin(\theta)\cos(\phi) & -\cos(\theta)\cos(\phi) & \sin(\phi)\\ \sin(\theta)\sin(\phi) & -\cos(\theta)\sin(\phi) &-\cos(\phi) \end{bmatrix}$$ must have antisymmetric generators. To find ... 1 The parametrization you've given just isn't of the exponential form you're after. The matrix you've written down is parametrized by the location of thex$axis after the rotation, which has polar angle$\theta$and azimuthal angle$\phi$in polar spherical coordinates drawn around the old$x$axis. It is not a rotation by angle$\phi$about an axis at ... 3 The spin of a particle is its intrinsic angular momentum, a vector quantity unrelated to any actual rotation of the particle. As you learned in quantum mechanics its magnitude is quantized as$\sqrt{s(s+1)}\hbar$and any of its three components as$m_s\hbar$. Here$s=0,1/2,1,3/2,2,...$is the principal spin quantum number and$m_s$, which ranges from$+s$to ... 0 Correction: Clearly M=0. There are three such states: $$S=2: 2|0,0\rangle+|-1,1\rangle+|1,-1\rangle; \\ S=1: |1,-1\rangle -|-1,1\rangle; \\ S=0: |-1,1\rangle+|1,-1\rangle- |0,0\rangle;$$. The probability that S=1 is zero. 2 When you pull your arms in you aren't pulling them directly towards the centre, because you're rotating as you're pulling them in. This is where the force comes from that actually makes you spin faster. You should definitely watch this video where he explains exactly this. Skip to 10m in if you're in a hurry but the whole video is well worth watching. I ... 1 Can we explain the above case with the help of just forces without using any result like "Angular Momentum Conservation? I think that will give us more insights into what is exactly happening. Well, yes and no. Because with angular conservation laws, you can just say that$L = L$, which means that$I_1 \omega_1 = I_2 \omega_2$, and then you can do some ... 0 You have to compute $$|1,0 \rangle \otimes |1,0 \rangle = \sum_{J,m} \langle J,m|1,0;1,0 \rangle |J, m\rangle$$ where$\langle J,m|1,0;1,0 \rangle$are Clebsch-Gordan coefficients. Indeed there are selection rules, and your comment shows that$J = 0,1,2$. Likewise you can easily show that only$m=0$can appear. So you need to look up 3 possible CG ... 2 so we can conclude that the man has angular acceleration without any external torque, which is an apparent contradiction of the terms, so how do we reconcile the case with the concept? We reconcile it with the law of conservation of angular momentum. The angular velocity of the skater increases when drawing in the arms in order to conserve angular ... 13 The definition of torque is not$\tau=Id\omega/dt$. We can't even define things like$I$and$\omega$for rotation that isn't rigid. The definition of torque is$\tau=dL/dt$. So yes, it is possible to have an angular acceleration without an external torque. Your example shows correctly that this can happen. 1 It is not conservation of angular momentum, it is Newton's third law of motion that should concern you. For every action there is an equal and opposite reaction. The motor shaft spins the prop against air resistance, the motor housing spins the craft with an equal force in the opposite direction. This is why helicopters have the tail rotor, to push against ... 2 Although increasing an object's spin will increase its total energy, it will not increase its tendency to collapse to a black hole. The physical reason for this, is that conservation of angular momentum will work against any possible collapse. From a slightly different perspective, it can also be seen from the fact that the event horizon radius of a spinning ... 1 This question is not quite phrased in the way that a specialist would phrase it, but it still sort of makes sense, and basically the answer is that for a typical astrophysical black hole, the spin's contribution to the mass is pretty big. Although a black hole has spin, the standard models of black holes that we study are vacuum solutions. There is no ... 4 "The same rule" is associativity, but everything else is quite different. That is, the group commutator is, in some convention, $$[R_{x}(\alpha),R_{z}(\beta)]= R_{x}(\alpha)R_{z}(\beta) R_{x}(-\alpha)R_{z}(-\beta),$$ where$R_{x}(\alpha)= e^{-i\alpha J_x}$, etc... For notational simplicity, let's call$-i\alpha J_x =A$and$-i\beta J_z =B$, antihermitean, ... 1 It is well-known that the finite dimensional irreps$V_{\ell}$of the Lie algebra$so(3)$are classified by spin$\ell\in\frac{1}{2}\mathbb{N}_0. To rule out half-integer representations for the orbital angular momentum (OAM) Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3),$$ one should use the fact the OAM operators $$L_j~=~\sum_{k,\ell=1}^... 2 ShoutOutAndCalculate, Please take a look on this paper: Proposed physical explanation for the electron spin and related antisymmetry Cetto, A.M., de la Peña, L. & Valdés-Hernández, A. Quantum Stud.: Math. Found. (2019) 6: 45. https://doi.org/10.1007/s40509-017-0152-8 You can read it free here: https://arxiv.org/pdf/1707.08674.pdf The paper presents ... 0 You can add up the angular momentum L of the motor-propeller assemblies and the main body to get a total L. It doesn’t matter that they’re in different places; angular momentum is independent of position. But that’s not entirely why a drone rotates. The much larger effect is “prop torque” due to the propellers pushing on the air. This effect is why ... 2 Aside of that, there is another assignment for this problem. I would like to reword the solution of this assignment here in my own words because maybe it can help someone else to understand. Also, if I ever forget how they got those results, I can simply look back at this answer. I'm sure it can be of use to the site and that other users can profit of it. ... 1 Q1. In the standard theory the total angular momentum of the field is {\bf r} \times {\bf P} , where \bf P is the Poynting vector. Note that this depends on whatever origin of the coordinate system you choose. Q2. No, as the Poynting vector vanishes in this case. 0 So, basically, my question is: Does anyone have a decisive argument on why do we exclude the half-integer values from the orbital operator spectrum? Lets go back to basics. To start with the eigenvalues of angular momentum are not integer . Not even an integer multiple , as a square root gets in the way. This was discovered while solving the ... 1 In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT)$$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$cf. NessunDorma's answer. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation & creation operators$$ \begin{align} a^j~=~& \sqrt{\frac{m\omega}{2\hbar}}... 2 He’s going to take a superposition of such solutions at a later step and sum over all values ofm$. That will pick up the parts of the general solution that you’ve identified. 1 It's just Euler's$\cos \Phi+i\sin \Phi= e^{i\Phi}$compbined with the usual forms of Pauli's$\sigma_x$and$\sigma_y$. 1 Given $$[ L_z, L_x ] = i \hbar L_y$$ just multiply both sides by$L_x$: $$L_x \bigg( L_z L_x - L_x L_z \bigg) = i \hbar L_x L_y$$ which gives $$L_x L_x L_z = L_x L_z L_x - i \hbar L_x L_y$$ 1 What you are looking for is a photon OAM switch. The OAM of light is the component of angular momentum of a light beam that is dependent on the field spatial distribution and not the polarization. It can be further split into an internal and an external OAM. The internal OAM is an origin-independent angular momentum of a light beam that can be associated ... 2 No. Expectation values of observable operators are always real. 9 From$\mathbf{L}=\mathbf{Q}\times \mathbf{P}$we have$L_z=Q_xP_y-Q_yP_x$. Then, introduce the following new operators (assuming units of$\hbar=1): \begin{align} q_1=\frac{Q_x+P_y}{\sqrt{2}},\\ q_2=\frac{Q_x-P_y}{\sqrt{2}},\\ p_1=\frac{P_x-Q_y}{\sqrt{2}},\\ p_2=\frac{P_x+Q_y}{\sqrt{2}}. \end{align} It is immediate to check that\$ [q_1,q_2]=[p_1,p_2]=0,\...

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I was wondering how the stationary flywheel scenario worked. As Adrian said if the flywheel were not to spin, it would fall over (i.e. you would not observe what is called precessional motion). Why? Because there's no angular momentum due to the spin of the flywheel and thus no torque. Why is it that the gyroscopic falls along a "circular path?" This is ...

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The diagram is a little confusing, but when the flywheel is not spinning, it has no gyroscopic effect, it will fall just as anything else. The dashed arrow that says "path of free end" is meant to show that the end is pulled down, but in a curve, because the other end of the rod is held up by the pivot point. Say you put one end of a horizontal pencil on the ...

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