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Hot answers tagged angular-momentum

6

Imagine that you have just two particles with the same mass and same speed, but going in opposite directions. They have opposite momenta, so the total momentum is zero. But they each have energy, and the total energy is not zero. The reason is because kinetic energy is just $\frac{1}{2} m v^2$. That square means that the kinetic energy can never be ...

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The different definitions you mentioned are NOT definitions. In fact, what you are describing are different representations of the Lorentz Algebra. Representation theory plays a very important role in physics. As far as the Lie algebra are concerned, the generators $L_{\mu\nu}$ are simply some operators with some defined commutation properties. The choices ...

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I believe you can, if you try to follow the path of finding representations of the $SO(n)$ group over a given Hilbert space. I really haven't done the calculation, but if it is the same, you would have something like this: $H=L_2(\mathbb R^n,\mathbb C)$ would be the Hilbert Space that would correspond to spin 0 particles, and the representation of the ...

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Am I right to say that some of the kinetic energy can be converted to angular momentum[?] No, angular momentum is a conserved quantity. In any isolated interaction you get out exactly as much as you put in. But you may have intended to ask Can a ball that is not spinning when I toss it at the ground come off with spin? to which the answer is ...

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The velocity of the orbiting space junk is a vector, with both a radial and a tangential component. $$\vec{v}_f = \dot{r}_f\hat{r} + r_f\dot{\theta}_f\hat{\theta}$$ (my $r_f$ is your $r$) The equation for conservation of angular momentum involves only the tangential component of velocity, because it comes from the cross product of the radius vector and the ...

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Here we will only discuss the case of finite-dimensional irreducible representations (irreps) of a complex semisimple Lie algebra $L$. Recall that the set $Z$ of Casimir invariants is the center $Z(U(L))$ of the universal enveloping algebra $U(L)$, cf. e.g. this Phys.SE post. OP's question is answered without proof on p. 253 in Ref. 1: Theorem 2. For ...

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Ladder operators are usually constructed to form a Lie algebra (we want them to have specific conmutation relations). The mathematical basis is weight theory. The important thing of Lie algebras is that they are a vector space and their elements, which are called generators obbey this conmutation rule: $$[X_i,X_j]=f_{ijk}X_k$$ Where we have used the ...

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An isolated body that doesn't exchange any angular momentum with the outside universe will never stop spinning (by conservation of angular momentum). There is no way to absorb angular momentum within the body in internal degrees of freedom; the angular momentum must be transported away if you want to stop. For example: if something is not rigid, you could ...

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This looks like an example of the Tennis Racket Theorem. Some axes of rotation for a rigid body are more stable than others. If the initial rotation axis does not correspond to one of the principal axes, a wobble can grow and cause the rotation axis to move to a principal axis. This is a result of Euler's Equations of Motion and the moments of inertia. The ...

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Without friction, the forces during the collision (glancing or head-on) are applied exclusively through their centres of mass. (Illustration available on Wikipedia.) The torque is given by $\tau=\mathbf r \times \mathbf F$ - but if the forces are applied through the centre of mass, then $\mathbf r$ and $\mathbf F$ are parallel, and hence $\tau=0$. Without ...

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First of all, I want to comment that, at least in 3D case, the statement that we impose on some momentum-like commuting operators $S_i$ the relation $S_x^2+S_y^2+S_z^2=S(S+1)I$ is more or less tautologic, since in general it follows from the commutation relations that the RHS can be cast (by a change of basis) to block-diagonal form, where each block has ...

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Angular momentum is a bivector, $J = x\wedge p$, and since the exterior/wedge product is antisymmetric, you do indeed get $n(n-1)/2$ independent components for any bivector. In general the Hodge dual provides an isomorphism between $k$-vectors and $(n-k)$-vectors. In three dimensions, $\star(x\wedge p)$ is an axial vector which and we call this ...

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How can we apply angular momentum conservation when friction is present? Why not? If we have a closed system, momentum and angular momentum are conserved. In this case, the full system is disk A and disk B, and there are no external forces, so the system is closed. There are internal forces, namely in this case, friction, but that doesn't matter. You ...

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Why do they have that form and not some other? I suppose one answer is "the form of the Hamiltonian". Because of the form of the Hamiltonian for the QHO, there is a "number" basis for the states. Suppose you don't use the ladder operator algebra to solve for the energy eigenstates of the Hamiltonian. You still find that the energy eigenvalues are of the ...

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You may recall from high school algebra that $x^2 + y^2 = (x + iy)(x - iy)$. Because the way the adjoint operator works, you could define an operator $\hat a = x + iy$, and its adjoint becomes $\hat a^\dagger = x - iy$. The hamiltonian for the quantum oscillator is just this relation with some constants. You have to be careful because the ladder operators ...

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Conservation of angular momentum means that the sphere will continue to spin forever. In order to change the angular momentum you need to apply an external torque. Note that this treats the sphere as a rigid body. If you consider just a small part of the sphere there are forces acting on it in such a way that the sphere remains undeformed. On a microscopic ...

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You don't need to apply Steiner's theorem onto the point mass. The point mass finds itself at a distance (apparently) $R$ of the x-axis. Since the moment of inertia is an extensive value, you can simply add all moments of inertia. There's the moment of inertia of the solid disk with respect to it's diameter. You have to 'Steiner' that away from a distance ...

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It depends on the friction of the contact. With a frictionless plane the top would precess around its center of gravity and the contact point will prescribe a circle. Add friction, and the friction force translates the center of gravity the same way tire traction translates a car. Here you have the cases of a) pure rolling, or b) rolling with slipping. ...

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