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-1

Any angular momentum in QM is quantized. Spin is not an exclusion. A hydrogen atom may also have spin even though "constructed" with spinless particles. And a free atom is not localized in space, just like a free electron. So, the angular momentum is a property of elementary and non elementary "particles" in QM. Spin is the angular momentum in the system ...


0

To say, in non-relativistic QM, that a state has spin $\frac{1}{2}$ means that it transforms in the representation of $\mathrm{SU}(2)$ with highest weight $\frac{1}{2}$, which is a two-dimensional space. In general, to say that a state has spin $s$ means to say that it transforms in the representation with highest weight $s$.


78

This whole question is a mistaken premise. There are spherical (or at least nearly spherical) galaxies! They fall into two basic categories - those elliptical galaxies that are pseudo-spherical in shape and the much smaller, so-called "dwarf spheroidal galaxies" that are found associated with our own Galaxy and other large galaxies in the "Local Group". Of ...


5

You mentioned elliptical galaxies, which the other answers haven't touched upon. Contrary to your statement about the galaxies being 2D, elliptical galaxies are "3 dimensional" in the sense that the stars are not confined to one plane; You could think of them as being "egg shaped". So why are elliptical galaxies not confined to a plane? Mostly because they ...


1

Let's take the spin, it's the simplest case, $Q = \mathbf{s}$. The operator $\mathbf{s}$ is a vector, \begin{equation}s = \mathbf{i}s_x + \mathbf{j}s_y + \mathbf{k}s_z\end{equation} while the operator s^2 is a scalar \begin{equation}s^2 = s_x^2 + s_y^2 + s_z^2\end{equation}. The operators $s_x$, $s_y$, and $s_z$ don't commute two by two, but $s_x^2$, ...


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Actually, there are parts of a galaxy that extend beyond the galactic plane: Galactic halo: This is actually the primary part of a galaxy that is not in the main galactic disk. It's made up of multiple sections, and is composed or an array of objects. Dark matter halo: This is a section of the galaxy's dark matter that exists in a semi-spherical shape. ...


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Every galaxy has to rotate so that a centrifugal force acts. Without the centrifugal force, all matter contained in the galaxy will collapse into the center of the galaxy due to gravitation. For there to be rotation however, there needs to be an axis, a line about which all matter revolves in the galaxy. Now, the manner in which all the matter revolves ...


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It is due to the combined effect of rotation and "dissipation". A rotating cloud of gas consists of particles which interact strongly with each other (colliding physically) on relatively short timescales can radiate away some of their energy and momentum by emitting photons. For both of these reasons, a dense cloud of rotating gas will collapse to form a ...


0

To find the spin eigenstates corresponding to a multi-particle state, all one needs to do is build the appropriate multi-particle spin operator using the direct product, e.g. $$J_z^{(2)}=J_z \otimes 1 + 1\otimes J_z \\ J_z^{(3)}=J_z \otimes 1 \otimes 1 + 1\otimes J_z \otimes 1 + 1\otimes 1\otimes J_z $$ and then diagonalize the resulting operator to find ...


3

Since the early 2000s, Matthew Bate and collaborators have been producing smoothed particle hydrodynamic (SPH) simulations of collapsing clouds. The clouds have an initial uniform density, no net angular momentum, but a turbulent velocity field. These clouds begin from rest and collapse under their own gravity to form hundreds of stars including many with ...


2

In this scenario, the dust in over-dense regions will fall on radial trajectories to the gravitational centre of their over-density. Assuming density inhomogeneities are continuous (meaning no abrupt change in density), we can always model this in a symmetric way local to each over-density. This means that in the frame of reference of the centre of gravity ...


2

Why does the angular momentum depends on the position? Angular momentum is always defined relative to a reference point, say $\mathbf r_0$, (which is often, but not necessarily the origin). If the system is invariant under rotation around this reference point the quantity that we call "angular momentum with respect to $\mathbf r_0$" is conserved. (Note, ...


8

I physically understand it as the momentum of an object rotating around something given a certain position. However, I can't give a physical explanation to the formula. Why do we multiply the linear momentum by the position? Why does the angular momentum is a function of the position? Angular momentum $L = mv * r$ (This is a late answer, but I ...


2

The spacetime geometry around a rotating uncharged black hole is described by the Kerr metric. I'll give this below, and it will look terrifying, but bear with me because there's only one small bit of the equation we need to see why the horizon disappears. Anyhow, the Kerr metric is: $$\begin{align} ds^2 &= -(1 - \frac{r_s r}{\rho^2})dt^2 \\ ...


2

This is an abstract answer, but I find it extremely helpful to the kind of "basic nature" question you seem to be groping for. Think of two things: Noether's theorem and a thought experiment "what if we had evolved as unsighted but clever beings?". As in David Hammen's Answer, it is Noether's theorem that would tell us that if our physical laws are ...


3

There are several ways to describe a particle's motion. For example, in 2 dimensions, you could use cartesian $x,y$ coordinates or polar $r,\varphi$ coordinates. To each coordinate, we can associate a 'quantity of motion' or 'generalized momentum'. If a given coordinate corresponds to a symmetry of the system, the corresponding quantity is conserved by ...


1

The angular momentum is a concept analogous with the linear momentum p = mv, in which m is the mass of the body and v its velocity. Now, see where the angular momentum comes from. Consider for simplicity a body moving on a circle around some axis, and let ω be the angular velocity, i.e. the angle by which the object rotates, in a unit time. The ...


8

Ultimately, what's special about angular momentum is this: Look up in the sky. A certain set of physical laws pertain in that direction. Look to the north. A certain set of physical laws pertain in that direction. Look to the west. A certain set of physical laws pertain in that direction. Those physical laws: They're the same in all directions. There's ...


4

Consider something like a door. A piece of wood with a hinge on one edge. Maybe it is one meter tall and three meters long. Now say that you're trying to hold the door in place, at the position half a meter from the hinge, while someone else throws a baseball at the other side of the door. If the baseball hits the hinge, you don't have to push at all. If ...


0

Maybe you can see it this way: The modulus of a vector multiplication is like this: $$|\mathbf{L}|=|\mathbf{r}\times\mathbf{p}|=rp\sin{\hat{rp}}$$ where you can see the main feature of angular momentum: position and linear momentum of the matter considered need to be both proportional to $L$ and inversely related to each other. That is how you guarantee ...


1

Have a look at the video here I hope that when you see the man spinning around and moving the weights (changing $r$) you can see that $r$ is important. Remember $r$ is the distance from each part of a rotating object to the axis of rotation, (which is not exactly the same as the position)


4

Here's a paper with a proof that the ground state must be l=0 for spherically symmetric potentials for a single particle, assuming there's a bound state. Abstract: The variational principle is used to show that the ground-state wave function of a one-body Schrödinger equation with a real potential is real, does not change sign, and is nondegenerate. As a ...


2

You could start from the premise that there was not net angular momentum in the universe at all; but it would still be the case that everything of interest was spinning. On the scales of stars and planets there are (at least) two important mechanisms that result in individual systems having angular momentum. The first is turbulence. If you take a parcel of ...


2

There is an angular momentum problem with regard to star formation, but you have the sense of the problem completely backwards. The problem is not where the angular momentum arises. The problem is where does it go. Gas clouds a tenth of a parsec across have been routinely been observed to rotate at about one revolution every five or ten million years or so ...


2

I asked the question because I did not believe in the accepted answer that has been sitting for more than 3 years. I have my own understanding, but since it is not good practice to put it with the question, I am posting it as one possible answer. My problem is that I do not believe the first statement quoted in the question which is contradicted by the ...


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Objects in orbit tend to lose their spin on their own axis. However they do not completely lose their rotation and end up rotating with a period that is the same as the orbital period, so that they face always the same side towards the other body. The best know example is the Moon that shows always the same side to Earh. The phenomenon is called tidal ...


0

Consider (at least perturbative) contribution of the additional effective "potential" in the radial equation when $l\ne 0$, analyze its sign and judge correspondingly. The physical explanation is simple - it is an additional kinetic energy of a rotational motion.


0

Using $\hbar=1$, your matrix $$j_x=\frac{1}{\sqrt 2}\pmatrix{0 & 1 & 0\\ 1 & 0 & 1\\ 0 &1 &0}$$ has the igenvalues $1,0$ and $-1$ as you stated. But when you wnat to get the eigenvectors, you must use the factor $\dfrac{1}{\sqrt 2}$. That is, the system you have to solve is $$\pmatrix{-1 &1/\sqrt 2 & 0\\1/\sqrt 2 &-1 ...


1

I hope that you are familiar with the calculus with vectors. Then you can find the answer in Wikipedia, "Angular momentum in classical mechanics", the entry "Angular momentum simplified using the center of mass". I had a look at it right now, it is very simple and clear. You will find that, no matter if the object is symmetrical or not, you can do the ...


0

The dot product is between the same unit vectors. You get |n|^2 * cos theta, where theta is 0. That leaves you with that magnitude of n squared. Magnitude is 1. The cross product becomes zero because the cross product of the same vector is zero. This is because n x n is |n|^2 sin theta. Here sin theta evaluates to zero.


3

To show that $$ \left(\sigma\cdot\mathbf{n}\right)^2=\mathbf n\cdot\mathbf n+i\sigma\cdot\left(\mathbf n\times\mathbf n\right)\tag{1} $$ consider writing the above as \begin{align} \left(\sigma\cdot\mathbf a\right)\left(\sigma\cdot\mathbf b\right)&=\sum_j\sigma_ja_j\sum_k\sigma_kb_k\\ ...


3

First of all $\vec{J}$ is not an Hermitian operator from a Hilbert space to the same Hilbert space, its three components separately are. Therefore nothing requires that there must exist an orthonormal basis of eigenvectors of $\vec{J}$, that is a orthonormal basis of simultaneous eigenvectors of $J_x$, $J_y$, $J_z$. Otherwise these operators would commute ...


0

As the components of angular operator $\hat J$ does not commute, you couldn't find a vector which is an eigenstate of all three components $\hat J_x, \hat J_y, \hat J_z$, talking about the eigenstates of angular operator usually need to define a reference direction such as "eigenstates of $\hat J \cdot \vec x,\,\, \hat J \cdot \vec z$ or $\hat J \cdot \vec ...


0

I think the answer is this: when we say the angular momentum doesn't have an eigenstate, we actually mean it doesn't in "Cartesian coordinate system". The reason is that you know, i.e., the components can't commute. However, $\vec{J}$ is Hermitian and since any Hermitian matrix can be diagonalised, so it must have an eigenstate, but we don't know how we can ...


-1

Since $v=r\omega$ Where $v$ is velocity and $r$ is radius and $\omega$ is angular velocity So $\omega=v/r$ This equation shows that if $r$ decreases $\omega$ increases


2

a)Find the total angular momentum of the skater and the masses both before and after the arm movement. Explain any difference b) Find the total kinetic energy of the skater and the masses both before and after the arm movement. Explain any difference. The moment of inertia is 50 + 10 = $I_i = 60 kg m^2$ (50*24/21) Since $L = I \omega = 60 *10, ...


2

Start with the force felt while holding weights and spinning with arms at full extension. Ask if it is easier or harder than when not spinning. Here you are forcing the weights to move from a straight line and to go in a circle, the force has to be felt all the time to keep pulling the weights into a circle. To make the circle smaller requires even more ...


-1

Because the conserved angular momentum is the product of radius and angular velocity, thus in order to remain conserved (a compatibility condition) the velocity has to increase when radius decreases. it is similar in a way to the lever principle (based on conservation of energy) Did not notice the intention of the question (updating with other answer) ...


0

The angular momentum of the two masses is computed independent of the skater - you were given the total angular momentum of the skater (including arms and hands which are normally considered part of the person) and ONLY have to compute the moment of inertia / angular momentum of the masses. A point mass at the end of a string has $$I=mr^2$$ as you know. The ...


0

In the first case it is taken that the masses of the hands are small compare to the masses that holds the hands. Hence we can neglect the masses of two hands and the case becomes two masses of weight 5 kg moving about the rotational axes. That's why we calculate the moment of inertia by In second case we have to add the moment of inertia skater and the ...


2

Many planets have been found where their orbital axes do not align with the rotation axis of their star. This is achieved using measurements of the Rossiter-McLaughlin effect in transiting systems or by observing planets transit over spotted features on a star's surface. As the stellar rotation axis is highly likely to coincide with its protoplanetary disk ...


0

We don't know, but there what I gather is the typical pattern of hypothesis, accretion disc theory, which attempts to find a model dynamic that could explain what we see, particularly in our own solar system, which is the only one we have very complete information about. In addition to seeing that our own solar system's planets all orbit in the same plane, ...


0

The spin of a particle is not created by a rotation of the particle around itself. It is a relativistic effect. Please ask yourself why the electron in a hydrogen atom can be found at certain distances from the nucleus, and NOT at ANY distance. It is because the wave-function interferes with itself and there are forbidden distances because there, the ...


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The Poincare group has two Casimir Invariants - namely $p^2$ and $W^2$ where $$ W_\mu = \frac{1}{2} \epsilon_{\mu\nu\rho\sigma} J^{\nu\rho} p^\sigma $$ is the Pauli-Lubanski pseudo-vector. Thus, representations of the Lorentz group are labelled by the eigenvalues of both $p^2$ and $W^2$. When $p^2 = -m^2$, we have the property $W^2 = -m^2 {\bf J}^2$. ...


1

It is first important to note that it doesn't matter if you write a state in the $| l \; m_l\; s \; m_s \rangle$ basis or the $| l \;s\; j \; m \rangle$, you'll always have that $m = m_l + m_s$. This means that $$| l = 2, m_l = 2,s=1/2, m_s=1/2 \rangle = | l=2, s=1/2, j=5/2, m=5/2 \rangle$$ because these are the only kets in each of the bases that have $m ...


1

There are two other interpretation of the Pauli matrices that you might find helpful, although only after you understand JoshPhysics's excellent physical description. The following can be taken more as "funky trivia" (at least I find them interesting) about the Pauli matrices rather than a physical interpretation. 1. As a Basis for $\mathfrak{su}(2)$ The ...


1

I can't quite follow your diagram. The angle can be measured from any line in the plane of motion to the line from the origin to either mass. The angle $\phi = 0$ represents the line you choose as your datum. The positive direction can be chosen arbitrarily as long as the choice doesn't change. The differential equation of motion doesn't care where the ...


10

Let me first remind you of (or perhaps introduce you to) a couple of aspects of quantum mechanics in general as a model for physical systems. It seems to me that many of your questions can be answered with a better understanding of these general aspects followed by an appeal to how spin systems emerge as a special case. General remarks about quantum states ...



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