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83

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 ...


19

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 ...


16

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. ...


14

All matter in the galaxy has to rotate (not necessarily in the same direction) 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. The rotation happens about an axis, a line about which all matter revolves in the galaxy. Now, the manner in which ...


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 ...


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

To quote Stephen Gasciorowicz, Before evaluating these quantities to obtain an idea of their magnitude, we will introduce some notations that will be very useful. First, it is $h/2\pi$ rather than $h$ that appears in most formulas in quantum mechanics. We therefore define $$\hbar=\frac{h}{2\pi}=1.0546\times10^{-34}\,{\rm J\cdot s}$$ So basically it's ...


6

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

As explained in this article by Neill DeGrasse Tyson, the tidal forces between the Earth and the moon do indeed slow down the rotation of the Earth each year, the same process that caused the moon's rotation to become tidally locked with its orbit of the Earth. This effect would eventually cause the Earth's rotation to be tidally locked with the moon as ...


4

I have read on science forum that electron in orbital s has no angular momentum and would fall into nucleus, so hydrogen atom would not be possible. In this sentence you are encapsulating the reason quantum mechanics was "invented". The planetary like theory of the Bohr model imposed the "not falling", quantized stable orbits with a minimum binding ...


4

Rather than beating your student over the head with facts, try to approach the problem the way scientists did in the first place, by following the scientific method. Your student should come up with a hypothesis, and use known theory to make a prediction (calculate the momentum transfer in some idealized model), and then build a model to test the prediction. ...


4

The rotation of the Earth causes a force called the Coriolis force. This does have an effect on ocean currents, but the effect is only significant on length scales of hundreds of miles. Over the diameter of a shell, even a big one, the Coriolis force is completely swamped by other effects like tides, local currents, random thermal fluctuations or whether a ...


3

The uncertainty principle may be stated more generally for two observables $A$ and $B$ as: $\begin{equation}\Delta A \Delta B \geq \dfrac{1}{2}\left|\langle\left[\hat{A},\hat{B}\right]\rangle\right|,\end{equation}$ where $\langle \hat{C}\rangle$ is the expected value of the observable $C$ and $[\cdot\,,\cdot]$ is the commutator (see here for details). From ...


3

Perhaps some additional information is in order to shed additional light... The whole discussion begs the question: If $\hbar$ is so convenient, why do we have $h$ around? As usual, "historical reasons". Planck originally invented $h$ as a proportionality constant. The problem he was solving was blackbody radiation, for which the experimental data came ...


2

The discussion is mostly semantic. They are both calculated relative to a point, in the case of the torque the point has the additional meaning that if you put an axle trough the point, the object will start to rotatte around it if the net torque is not zero. It happens also that the torque will be the same if you chose any other point along the axis ...


2

Spin of an elementary particles is not necessarily the result of a movement of the particle around itself i.e. around some rotation axis that passes through the particle.If there were such an axis, the projection of the spin in the plane perpendicular to that axis were zero. But, this is not the case. So, along whatever axis we would measure the spin, we ...


2

Let $\mathcal{H}$ be the Hilbert space for one particle. Then, $S_{x}\in\mathcal{B}(\mathcal{H})$ is a bounded, self-adjoint operator. Now, if you want to have the Hilbert space for two particles, remember that this is the tensor product, i.e. $\mathcal{H}=\mathcal{H}_1\otimes \mathcal{H}_2$ (where $\mathcal{H}_1$ is the Hilbert space of the first and ...


2

In the cases of Bottle A and Bottle C, they are full and empty. So the amount of water in them (or not) can be considered to be a complete system along with the bottle, since there is no possible way in which the fluid in the full bottle would reduce its volume or overall distribution, certain properties like the system's center of mass,center of gravity and ...


2

The first equation is not normalized, which is the result of the lowering operator. While the second equation is the normalized state, you can check it easily by using $\langle 10|10 \rangle$. Or you can use the normalization condition $\langle \psi|\psi \rangle = 1$ where $|\psi \rangle = Z \hbar | \uparrow \downarrow + \downarrow \uparrow \rangle$.


2

In general the planes of solar systems are not aligned with the plane of the Galaxy, but are oriented in all different directions. The size of a solar system is so much smaller than the size of the Galaxy, that the Galaxy's structure has no impact on the orientation of a solar system. What determines their orientations is the direction of the angular ...


2

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 $s_x^2$, ...


1

before I answer your question, let me put forward another scenario. Imagine a mass less rod with two massive spherical objects (of mass $m$) is placed at either of its ends. Further imagine that the rod rotates about an axis. Let's define the axis to be the center of the rod. Let's choose the reference point at a distance $x$ from the axis and on the rod. ...


1

Electron has mass and momentum, but we don't know anything about how the electron "moves" between the various points within that orbital. Stated another way, we can observe that the probability of finding the particle around the nucleus has a particular form (the s-orbital), and the "motion" if it could be defined for a wavelike dispersed electron in an s ...


1

Just two cents: I assume you already introduced Newton's laws, you can say that is something like "When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force" yu can explain that from the other two Newtons laws it can be shown that the first law naturally ...


1

I suggest a way: bring a toy of a gyroscope form, put it on a table, and give it a brief torque. Although you don't act anymore on the toy, it continues to rotate. Ask your students WHY does it happen. I assume that they learned about the conservation of LINEAR momentum. So, we have an analogy: a body in linear movement keeps moving as long as no force ...


1

Yes, for the hydrogen atom Hamiltonian, $L^2$ commutes with the Hamiltonian because it commutes with $r^2$.


1

Generally speaking, the choice of what the $z$-axis (equivalently $x,y$) is is arbitrary. You can choose any direction to be your $z$-axis, as long as you do the calculations consistently with this choice. If the system has a priviliged direction (like that imposed by the magnetic field in the Stern-Gerlach case) that is usually choosen to be the $z$-axis. ...


1

A central force does not perform work only if the motion is tangential. When the particle moves radially, it has a component of the speed that is not tangent. The cetripetal force and the velocity are no longer paralell so there the centripetal force actually does work on the system. The work is actually $W=\int_{r_i}^{r_f}F_{centripeta}(r) dr$ NOTE: for ...


1

It's a well known identity that $\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ Therefore, if $$J_{i} = -\frac{1}{2}\epsilon_{ijk}M_{jk}$$ then we have: $$\begin{align} \epsilon_{lmi}J_{i} &= -\frac{1}{2}\epsilon_{lmi}\epsilon_{ijk}M_{jk}\\ ...


1

I'm essentially interpretting your question as "are there any cannonical commutation relationships (CCRs) where the Lie bracket is a different scaling constant of the identity matrix: $$[\hat{x},\,\hat{p}] = i\,\alpha\,\mathrm{id}\tag{1}$$ or, equivalently, are there any pairs of canonically commuting observables where the scaling constant $\hbar$?" ...



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