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0

I want to give an answer for the case, that the mentioned particles are electrons. Let us consider that the magnetic dipole moments of this two electrons are aligned in a straight line through the points (0,a,0) and (a,a,0). Since both electrons are moving their magnetic dipole moments begin to turn when the electrons leave the mentioned points. Perhaps it ...


3

This decay (occurring via the strong interaction) violates the charge conjugation since $J^{PC}(\pi^0) = 0^{-+}, J^{PC}(\rho^0) = 1^{--}, J^{PC}(\eta'^0) = 0^{-+}$. The charge conjugation transforms a particle in its anti-particle. In the case of the 3 particles involved in this decay, they are all their own anti-particle, and the effect of the charge ...


0

Not all galaxies are spiral, there are elliptic and irregular shaped ones also. Search for them on Google and images abound of non spiral shaped ones. A more interesting question is: why are galaxies, in general, laid out in a flat plane, like a dinner plate? According to studies dating back over 40 years ago, spiral galaxies are gravitationally unstable ...


3

The theorem does not apply, as we do not have spherical symmetry. All we have is rotational symmetry about a preferred axis. In fact, the gravitational field outside a rotating object will be Kerr, which only reduces to Schwarzschild in the case of no rotation. Otherwise, there will be time-space terms in the metric, making it not static. Still, Kerr is ...


0

Key to understanding this question is the fact that a "football", in this case, is a hand-egg: When it is given spin about its long axis, it will act a bit like a gyroscope: it tends to maintain its attitude. So the question becomes - when is that desire of the ball to keep pointing in the same direction greater than the torque due to airflow that ...


1

Since this is a homework problem, I won't provide a full solution, but here's a nudge in the right direction. Take a look at these two plots of the effective potential: k = -1, $\alpha$ = 1, L = 0.25 k = -1, $\alpha$ = 1, L = 1 What's different about these two effective potentials? We only changed $L$ between the two graphs; what does that imply about ...


2

Firstly, $m$ does not have to be an integer, it is entirely possible for $m$ to be 1/2 for instance. Your points ,1-3 are fine. There are is a maximal and a minimal value of $m$. Call the maximal value $M$ (we have to call it something). Now we can apply the lower operator any number of times, each time it lowers the value of $m$ by a full integer ...


0

The cited diagram depicts but a small part of the motions that occur during a dive. That appears to be an overly simplified diagram of a rather simple springboard dive, a forward 1 1/2 somersault tuck. It's overly simplified because it misses the complex motions a diver undergoes at the start of the dive. There are two mechanisms by which a diver turns ...


0

Let's consider two ships passing each other. When they pass, a rope is thrown from a ship to the other ship. Then then rope is pulled sharply. That causes the ships to collide, the rears of the ships hit each other and the ships start to spin. In the previous scenario part of the energy used to pull the rope became rotational energy of the ships, that ...


3

How does the kinetic energy of a ballerina increase? Conservation of angular momentum: $$L_1=L_2 \implies I_1\omega_1=I_2\omega_2\quad\quad (1)$$ Pulling in your arms reduces moment of inertia $I$, since the same mass is now distributed over a volume closer to the spin centre, $I=\sum mr^2$. As you say, reducing $I$, so $I_2<I_1$, implies ...


2

Everything depends on how your fields (vectors and spinors are fields in the classical theory, and when you quantize in QFT, they become operator-valued fields) transform when you make a Lorentz transform: An scalar is a field that doesn't change at all: $\phi'(x') = \phi(x)$. Examples are the Higgs and pions. A vector field is a field that transform like ...


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Magnetic moment, in classical physics, is related to current in a loop, which in turn can be connected to angular momentum of a charged particle. Thus, in classical physics, magnetic moment and angular momentum are connected. In fact, they are proportional with the constant of proportionality being the gyromagnetic ratio. Moving to quantum mechanics, some ...


6

That spin follows the angular momentum algebra is no accident - like angular momentum, it is part of the conserved quantity - the Noether charge - associated to rotations. The reason why the $\mathfrak{so}(3)$ transformations of spin should be indeed those associated to the $\mathfrak{so}(3)$ of spatial rotations is not answerable in QM alone - you have to ...


1

The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle. For a flat planar loop of current, it's true that $\mu = IA$, with the direction of the dipole normal to the loop. However, the more general case is that of a a volume current $\vec{J}$ in some finite region of space. In this case, the the ...


0

The direction of the magnetic moment is perpendicular to the plane of the loop. Seeing that the angular momentum is also perpendicular to that plane, and having shown that their magnitudes are proportional, is all it takes to show that two vectors are proportional. If you insist, we can still formally go through all that. One way of enforcing an area vector ...


1

Angular momentum depends on the axis about which you calculate it. That means you'll get a different number depending on the axis, but no matter the choice, it's always conserved (assuming no external forces act on the system). For example, imagine a ball of mass $m$ and velocity $v$ hits and sticks to the rod a distance $d$ from its center: If you choose ...


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Angular momentum should be conserved in any inertial frame of reference - if you move with the center of mass, the motion you see will be rotation about the center of mass; if you move with a different frame of reference, you will see rotation about a different axis. So the short answer is "it doesn't matter". Whether they "give the same answer" depends on ...


4

You appear confused by how spin is introduced in ordinary QM. It is rather ad hoc: Given a Hilbert space without spin degrees of freedom of a particle $\mathcal{H}_0$, and the spin $s$ of the particle, we take the total space of states of the particle to be $\mathcal{H}_0\otimes \mathcal{S}_s$, where $\mathcal{S}_s$ is a $2s+1$-dimensional complex Hilbert ...


7

Ernie is close to the correct answer, but the fundamental thing that needs to be considered is how the internal energy of the body flows. I researched this in a very interesting book I''m still reading, Principles of Animal Locomotion . Chapter 7 addresses running and section 7.5 discusses Internal Kinetic Energy . Limb accelerations can store kinetic energy ...


6

Speculating here... I suspect that for light weights the answer is yes - with the right technique. Your center of mass moves up and down which requires energy being absorbed and expended by your legs. Moving your arms with small weights should allow you to even out the motion, lowering the peak stress on your legs so they tire more slowly. In a sense you ...


1

Consider your arms as pendulums. The period of a pendulum is determined by its length and by gravity. Though the period is affected neither by the weights nor by the amplitude of your arms, these two quantities affect your balance and efficiency. When you run you get into a rhythm of arms and legs. Your legs are do the work; your arms are along for the ...


5

The selection rules for atomic transitions are entirely governed by how much angular momentum the photon can carry away. A dipole transition (in the nomenclature, E1 or M1, depending on its parity) carries away angular momentum $\hbar$, so the atom's initial and final state must have angular momentum quantum numbers $J$ differing by zero or one. A ...


1

Angular momentum is a conserved quantity (in a closed system) and this is true also for the angular momentum that is carried by the electromagnetic (EM) field. This conservation is a manifestation of rotational symmetry and the azimuthal part of the EM field emitted must be single valued. In other words, when rotating the EM field in the azimuthal ($\phi$) ...


2

Even if the original dust cloud only had a relatively small angular velocity (which it might have had for all sorts of reasons), the process of collapsing would have amplified it. That is, the collapse process preserves the angular momentum, but it translates to a much larger rotational speed in the newly-collapsed system. Think of what happens to a spinning ...


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Your final result looks right to me. Everything should be half-integers. A basic rule of combining two quantized angular momenta is that the quantum number of the resultant can be anywhere between the sum of the original quantum numbers and the absolute value of the difference of them, in integer steps. Consider $\ell_1$ = 1 combining with $\ell_2$=3. The ...


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$\newcommand{\ket}[1]{\left| #1 \right>}$ Note that $ l=0$ has only one state $m=0$. Therefore the tensor product of $l=1$ and $l=0$ can be written as: $$ (l=1)\otimes (l=0) = \left\{ \begin{array} &\ket{l= 1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \end{array} \right\}=(l=1) $$ As ...


1

All the stars would be attracting each other and hence the would stick to each to attain equilibrium. Why doesn't this happen? You are forgetting angular momentum. Consider a binary star pair. Ignoring the expansion of spacetime, and in the absence of some mechanism that removes angular momentum from the system, those stars will orbit one another ...


1

Clebsch-Gordan coefficients let you treat n spins (or generaly - any n particles with arbitrary angular momentum) as a single composite system. The coefficients are simply the matrix element of basis transformation from seperated to composite system. For 2 particles with total angular momentum eigenvalues $ l_1,l_2 $, such that for example $ ...


2

From this paper$^1$ we have the equations for a particle of zero total energy on an infalling trajectory in the equatorial plane: $$\begin{align} \Sigma\frac{d\theta}{d\tau} &= 0 \\ \Sigma\frac{dr}{d\tau} &= -\sqrt{2Mr(r^2 + a^2)} \\ \Sigma\frac{dt}{d\tau} &= -a^2\sin^2\theta + \frac{(r^2 + a^2)^2}{r^2-2Mr+a^2} \\ \Sigma\frac{d\phi}{d\tau} ...


1

A spin-spin interaction is really a magnetic moment - magnetic moment interaction, where the magnetic moment of each particle is proportional to spin. [Of course, it might be a chromomagnetic moment - chromomagnetic moment interaction if two quarks are interacting, as they are here.] In any case, the interaction term goes like $\vec S_1 \cdot \vec S_2$. ...


2

If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ...


1

All the stars would be attracting each other and hence they would stick to each to attain equilibrium. Why doesn't this happen? This is an old question. Even Newton himself had thought about this question. His idea was that in long distances or separations (say, inter-galactic distances) the force of gravity might appear to be repulsive. That's why not ...


-1

If there would be initial rotation(as we see most of all objects today are rotating around another) gravitational force is accounting for the centripetal force. I'm new here so, I don't know how to type the equations, but I hope you get my point. Further many objects are there which have many other forces, like Coulombic force(when charged bodies are ...


1

I) Perhaps it is helpful to point out that even if the physical system $S$ has no rotational symmetry (e.g. if the system $S$ is a 3D an-isotropic harmonic oscillator), then the Lie group $G=SO(3)$ of rotations still has a group action $G \times S \to S$ on the system. See also e.g. this Phys.SE post. In particular the Hilbert space ${\cal H}$ of the system ...


0

My understanding of this limited, but this might help (too long for a comment): The state space is spanned by the set of simultaneous eigenstates of the Hamiltonian, $ \hat L^2$, and $ L_z $. In fact, they form an orthonormal basis of a Hilbert space $ H $ which is the state space. Out of convenience, we denote the eigenstates by the quantum numbers, ...


2

For orbital angular momentum, indeed, $L = x\times p$ even as a quantum operator, see this question. When writing a ket $\lvert l,m \rangle$, this is meant to live in the $2l+1$-dimensional space $\mathcal{H}_l = \mathbb{C}^{2l+1}$ on which the representation of the angular momentum algebra labelled by $l$ exists ($m$ is the eigenvalue of the ket for ...


3

It would appear that the object is in circular motion at any given instant, which implies that the tension force provided by the string is always perpendicular to the velocity of the mass. Mostly true, but not quite. If the string is held steady, then the object is in circular motion. But by pulling the string harder than that, it moves off of ...


1

The two questions are slightly different. Each individual measurement of $L^2$ or $L_z$ will return an eigenvalue. In this case, you have only one possible measurement for $L^2$ (corresponding to $l=1$), but you have two possible measurements for $L_z$; 2/3 of the time you'll get $m=1$, and 1/3 of the time you'll get $m=0$. The expectation value, on the ...


2

You're probably used to the convention where a hat is used to denote that something is an operator. But that convention is not universal. In many cases, when it's clear from the context whether something is an operator or not, we just write it without a hat either way. For this case in particular, $\boldsymbol{J}$ is defined to be an operator. The fact that ...


0

The 'quantum' angular momentum, in the classical limit, does reduce to the 'classical' angular momentum. In this sense, they are the same thing. Where the classical angular momentum occurs in the classical Hamiltonian, we replace that with the quantum angular momentum in the quantum Hamiltonian. Of course, 'QM being QM', brings with it some new aspects. In ...


3

It is a misconception to say that In QM [...] we know that the the electron does not radiate EM-Waves because it is not actually circling around the nucleus. It is sometimes here and there. In QM the notion of "circling round the nucleus" does indeed fail to make sense, but this is not why electrons don't radiate EM waves. Instead, an atom in its ...


-1

$$\begin{align} \left[\hat{A}_{i}, \hat{B}_{j} \right] & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i} & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3}\epsilon_{ijn}\left( \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i}\right) & = ...


6

I have read somewhere that commutation relations of the form \begin{equation} [a_i,b_j]=\epsilon_{ijk} c_k \end{equation} admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is: $$ ...


4

First, let us look at the current rate at which the moon slows down. I have a few different sources, and they don't all give me the same answer. First, there is this claim that Earth slows down at a rate of about 0.005 seconds per year per year. A year has approximately $365.25 \cdot 24 \cdot 3600 = 3.15\cdot 10^7 \mathrm{sec}$, so 0.005 seconds change ...


2

The Moon rotates around the Earth slower than the rotation of the Earth itself. That's why, from a fix point on the Earth, the Moon appears to be moving. The Moon creates the tide on Earth. So the tide "follows" the Moon. However as the Earth rotates faster than the Moon it will tend to carry the tide with itself "forward". The Moon pulls the tide toward ...


1

You are really asking about the reaction forces felt on a rod when you push on its end. For a rod, you can work with a quantity called the reduced mass (see for example this excellent answer for the derivation). As long as the rod is balanced on its end, the reduced mass tells you exactly how much greater the inertia of the cart appears to be: $$m_r = ...


6

It's not possible to derive the orbital angular momentum $L = r \times p$ from the $\mathfrak{so}(3)$ commutation relations alone, since the spin operator $S$ also fulfills the same commutation relations, but certainly is different from $r \times p$.


2

You say you get the physical justification for having some ladder operator(s) give $\hat L_+ \left|\text{state}\right> = 0$. For angular momentum both the raising and lowering operators eventually terminate; for the harmonic oscillator only the lowering operator terminates, at the ground state. Here's a mathematical argument that the termination must end ...


0

The raising operator has to raise something. It can't operate on an eigenstate of $J_z$,i.e. $|\alpha,\beta_{max}\rangle$, and give the same thing back. Think of it as raising the eigenstate to a new eigenstate, but with a 0 multiplying the resulting eigenstate to ensure an unphysical value of angular momentum doesn't exist. And this is not a scalar 0. It is ...


3

Angular momentum is that which is conserved under rotations. Equivalently, the angular momentum operators are the generators of rotations. This holds both classically and quantumly by (versions of) Noether's theorem. Defining "angular momentum" as $\vec x \times \vec p$ classically and then showing that it is conserved is doing it the wrong way around from ...



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