# Tag Info

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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 ...

16

In your problem, "Earth" is not an isolated system. The combined "Sun-Earth" system, however, is, so we can know that the angular momentum of the Sun-Earth system is conserved. As the earth's mass is accelerating the sun, you have to take its angular momentum into account as well. While the mass and size of the sun mean that we can ignore its motion with ...

15

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. ...

13

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 ...

11

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 ...

11

You are a point in the circle. The torque is: $$\mathbf\tau = \mathbf r\times\mathbf F$$ Where $\mathbf r$ is the position of Earth and $\mathbf F$ is the force (radially towards the sun). Notice when your reference point is somewhere in Earth's orbit, as you said, and your object is earth, the force will not be parallel to the position. Therefore, 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 ...

9

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? 1) - Angular momentum $L = mv * r$ (p * r = arm of the ...

9

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 ...

7

It is great that you "think differently" about problems - that is at the heart of all innovation. When it comes to the rotation of planets, you have to go back to the origins of the solar system: Planets are formed by accretion: a large cloud of debris starts to experience some gravitational pull, and as one "lump" becomes bigger than the others, it starts ...

6

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 $W^2 = -m^2 {\bf J}^2$. ...

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 ...

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 ...

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 ...

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 ...

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 ...

3

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, ...

3

Groups are abstract mathematical structures, defined by their topology (in case of continual (Lie) groups) and the multiplication operation. But it is almost impossible to talk about abstract groups. That is why usually elements of groups are mapped onto linear operators acting on some vector space $V$: $$g \in G \rightarrow \rho(g) \in \text{End}(V),$$ ...

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 ...

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\\ ...

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 ...

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

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

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 ...

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 ...

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 $L = I \omega = 60 *10, ... 2 In this scenario, the dust in over-dense regions will fall on radial trajectories to the gravitational centre of their over-density. Assuming density inhomogeneities are continuous (meaning no abrupt change in density), we can always model this in a symmetric way local to each over-density. This means that in the frame of reference of the centre of gravity ... 2 This is an abstract answer, but I find it extremely helpful to the kind of "basic nature" question you seem to be groping for. Think of two things: Noether's theorem and a thought experiment "what if we had evolved as unsighted but clever beings?". As in David Hammen's Answer, it is Noether's theorem that would tell us that if our physical laws are ... 2 The spacetime geometry around a rotating uncharged black hole is described by the Kerr metric. I'll give this below, and it will look terrifying, but bear with me because there's only one small bit of the equation we need to see why the horizon disappears. Anyhow, the Kerr metric is:$\$\begin{align} ds^2 &= -(1 - \frac{r_s r}{\rho^2})dt^2 \\ ...

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 ...

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