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2

In short, it continues to rotate simply because it has no reason to stop rotating. Imagine I have a puck and I whack it across an ice-rink. The puck will continue to travel at the same speed until it hits a wall; i.e it will stay the same unless there is a reason not to. This is conservation of linear momentum as shown here. The same thing happens with ...


2

The $p$ orbitals that you refer to are actually certain linear combinations of the ones we usually use with the magnetic quantum number, i.e., the wave=functions are of the form $$\begin{eqnarray}p_z &=& p_0\\ p_x &=& \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right)\\ p_y &=& \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} ...


2

The magnetic quantum numbers $m$ give you the projection of the angular momentum $L$ in units of $\hbar$ on one specified axis, normally the $z$- axis. So it's not one $m$ for every axis, but multiple $m$'s for one axis. For the $P$-state for example - that is $L=1$ - the projection on the $z$-axis can take on the values $$\hbar,0,-\hbar~,$$ so ...


0

For spin-1/2 particles, the entire wavefunction has to be antisymmetric under particle exchange. Also, the spatial component has parity $(-1)^\ell$, so if $\ell$ is even, the spin component must be odd under exchange.


1

I can't quite fathom the source of your confusion (I think it might have something to do with a focus on the notion of rotation here---angular momentum does not require rotational motion), so I'm having trouble writing a really clear response. For the moment I would rather offer a program for practicing the right skills rather than reinforcing the mistaken ...


0

...causes the rock to rotate in a clockwise direction I think that means the velocity vector of the rock is in a direction such as it were thrown out of a clockwise motion of the disc. In that case, the disc itself, along with the girl will rotate in an anti-clockwise direction. The total angular momentum is $0$. The rocks angular momentum as it is ...


0

I don't work in astrophysics, but I make a simple guess based on general mechanics. Assume that in the beginning (billions of years ago), the galaxy was spherical, and that was rotating around some axis, maybe the same as today. Then, where is the biggest centrifugal force? The centrifugal force on a piece of material of mass $m$ in the galaxy is $F_c = m ...


0

A related question would be: why most of the Solar System planet lie in the same plane? That is, why don't the planets close enough to the Sun rotate at random angle with each other (such as Pluto which rotates in a plane at some angle to the planets' plane)? Similarly, why do the stars in a galaxy tend to rotate around the center of the galaxy roughly in ...


0

Research in 2013 produced a paper (here:http://www.sciencedirect.com/science/article/pii/S0043164813000732) In the abstract, they agree that frictional forces are not the cause: the curling motion has been attributed to an asymmetrical distribution of the friction force acting on the sliding stone, such that the friction on the rear of the stone (as ...


1

I have considered the wet friction idea too. Ice when compressed will form a liquid layer where the pressure is applied. Yet one thing many seem to overlook is that the surface of the ice that the curling stones move over is not smooth. to prepare the ice for curling we first use a scraper to smooth it and level it. the scraper is a very sharp knife with a 5 ...


1

You have conservation of angular momentum and parity in electromagnetic interactions. The vector boson must be massive to be able to decay into two photons due to energy-momentum conservation. So we can consider its decay in its rest frame. So if it decays in two photons their momenta must be back-to-back. We assume those vectors to be $\pm k \vec{e}_z$. The ...


1

Isotropy isn't a property of a particle; it's a property of spacetime. A coupling between the direction of the vector particle's spin and the direction of the linear momentum of a decay photon would be a parity-odd observable; all such observables are forbidden in purely electromagnetic decays. (Parity-odd observables are allowed in weak decays, but you ...


0

You seem to be neglecting the change in the moment of inertia of the planet in the definition of angular momentum: $$ \mathbf L=I\boldsymbol\omega $$ Since $\boldsymbol\omega$ increases, then $I$ must decrease to satisfy angular momentum conservation. There isn't any need for angular momentum to somehow "reach out" and take it from a nearby planet.


1

You say: But for the observer on the planet, since the total angular momentum of the star about its axis is zero it should remain zero. But the observer on the planet does not occupy an inertial frame. An observer in a rotating frame measures fictitious forces. So there is no reason why angular momentum should be conserved.


2

The difference between the whole system properties and its constituents can be explained on two-particle system. Consider positronium (electron+positron) in the state $l=1$. The quantum number $l$ describes the relative motion of constituents. This state has a magnetic moment, which belongs to the whole system. But when you consider the angular momentum ...


5

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 ...


1

The only outright requirement is that you compute all the angular momenta in your problem around the same center (modulo applying the parallel axis theorem to break the angular momentum of extended bodies into of-and-around the CoM parts). So you can freely chose any single point to use Now, as with most such "free" choices in physics there are generally ...


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 ...


0

Consider the simplest case, the movement in a spherically symmetrical potential. The Hamiltonian becomes, $$-\frac {\hbar^2}{2m}[\frac {1}{r^2} \frac {∂}{∂r} (\frac {r^2∂}{∂r}) + \frac {1}{r^2sin\theta} \frac {∂}{∂\theta} (sin\theta \frac {∂}{∂\theta}) + \frac {1} {r^2sin^2\theta}\frac {∂^2}{∂\phi^2}] + V(r)$$ There is nothing in this Hamiltonian to hint ...


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 $\hbar$?" ...


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 ...


0

I now know (as pointed out by 'Javier Badia') that as this inequality is like any other normal inequality, the units on both sides of the inequality must be equal.


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 ...


0

Contract both sides of the first equation with $e^{ijk}$. In three dimensions, we have the identity for the Levi-Civita, $$\epsilon_{ijk} \epsilon^{imn} = \delta_j^m \delta_k^n - \delta_j^n \delta_k^m.$$ Using the antisymmetry of the $M_{jk}$, you obtain the second equation.


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}\\ ...


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 ...


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 ...


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. ...


0

Any, the Hamiltonian whos potential nergy is a function of $r$ only has rotational invariance. Since the components of $\mathbf{L}$ are generators of rotation, it can be shown that $[\mathbf{L},V]=[L^2,V]=0$


0

Angular momentum is defined as, $$ L=r\times p\equiv r\times\frac{\hbar}{i}\nabla $$ From this it follows that $$ \left[L_x,\,x\right]=\left[L_x,\,p_x\right]=0 $$ where $$ L_x\sim yp_z-zp_y $$ Noting that $$ \left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right] $$ the expected results can be determined.


1

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


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 ...


0

So we have than with a mixture of water and air we have more energy dissipation? One possible reason is this: the B bottle is assymetric, with the heavy watered side down. While it turns, water shpuld turn sides. Because of the viscosity of the water nd n relation with the glass, water has to exert a work, thus losing energy


0

What you did seems fine to me, and indeed you can show that $\frac{\partial}{\partial{p}^k}(a_\vec{p}^\dagger{a}_\vec{p})=0$, just take the definitions $$a_\vec{p}=\int{d^3x}\,e^{-i\vec{p}\cdot\vec{x}}\left[\frac{E_\vec{p}}{2}\phi(\vec{x})+\frac{i}{E_\vec{p}}\dot\phi(\vec{x})\right]\\ ...


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 ...


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 ...


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 ...


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

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 have yet to find a physics book that doesn't make this really confusing. If one has a vector fixed in inertial space, its components as viewed in a moving frame are obtained by the dot product of the vector with the moving unit triad fixed to the body but moving relative to inertial space. While the inertial frame would measure its components as constants ...


-1

As someone who has spent some time playing with gyros, I am very familiar (though on a non-mathematical level) with these objects. I have six battery powered high precision Gyros found here: http://www.gyroscope.com/d.asp?product=SUPER2 I have been intensely fascinated with the way they behave when spinning at maximum speed. I feel hesitant to comment ...


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. ...


0

The answer to this is really very close to Ted Bunn's answer to this question, but I think there may be one additional point .... a disk the most stable configuration Maybe it could be thought of as the 'most stable configuration', but I think it should rather be thought of as the natural state that accreting matter will form for the following ...


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 ...


0

Electron is not like a ball, as it has no volume at all. So it can not spin like a ball. Magnetic moment comes "as is" from quantum mechanics, which do not explain its nature.


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 $|\psi \rangle = Z \hbar | \uparrow \downarrow + \downarrow \uparrow \rangle$.


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 ...


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

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

About your reference point, torque acting on earth is not zero. But total torque of sun and earth is zero. That's why total angular momentum of earth and sun remains conserved.



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