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13

Nothing is happening. At least, nothing except that a new generalized quantity suggestively called "angular momentum" was defined and subsequently measured. But nothing we know about the usual angular momentum of photons is changed by this in any way. Standard total angular momentum is $J = L + S$, where $L$ is the orbital and $S$ the spin angular momentum. ...


6

It probably does not mean anything. That paper concerns the quantization of electromagnetic waves in less than three spatial dimensions. In fact, there are a number of decades-old results showing that it is often possible to evade the spin-statistics relationship in lower-dimensional systems. While these kinds of results (including this new one) may be ...


5

Phenomena in quantum mechanics may be expressed using any basis. It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the observable that was just measured. That's why the basis of the ...


3

Just as a supplement to ACuriousMind's answer, it is worth noting that buried in the bottom of their paper they actually show what the "spin 1/2" eigenstates are in terms of the regular basis: $|j=1/2\rangle=\frac{1}{\sqrt{2}}(|1, -1 \rangle + |0,1\rangle$) $|j=-1/2\rangle=\frac{1}{\sqrt{2}}(|-1, 1 \rangle + |0,-1\rangle$) where $|l, \sigma\rangle$ is the ...


3

The air flow around the car creates a vortex. There is an image of the air flow around a car: which was used by this earlier answer - that rotating air behind the car is what spins the wheels of the bike.


3

Classically, angular momentum is only conserved in a central potential by considering the torque (correct me if I am wrong). In quantum mechanics, it is also true, isn't it? In QM, an operator is conserved iff it commutes with $H$, because $$ i\dot {\mathcal O}=[H,\mathcal O] $$ Therefore, the angular momentum is conserved iff it commutes with $H$. As ...


3

I think the two big factors would be that the Earth would 'want to' become tidally-locked to the Moon and the Sun. The Moon would win here, which is easy to see because tides are caused more strongly by the Moon than by the Sun. So in due course the Earth would end up tidally-locked to the Moon with a rotation period which would be the same as a lunar ...


3

If $|\pm\rangle$ are the eigenvectors of ${\hat \sigma}_x$, ${\hat \sigma}_x |\pm\rangle = \pm |\pm\rangle$, then a rotating $x$-basis is defined as $$ |+\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|+\rangle = e^{-i\omega t/2 } |+\rangle $$ $$ |-\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|-\rangle = e^{i\omega t/2 ...


2

To answer the generic dimension part, in case it is not self-evident from Valter's answer: In D dimensions, the rotation matrix is the exponential of an angle θ times a matrix K, a normalized generator of the corresponding rotation group SO(D) around some unit axis D-vector k, in the vector representation, so the matrix is D×D. The eigenvectors of these ...


2

Spin is connected to the intrinsic magnetic dipole moment of the particle, this is what makes the particle capable of interacting with an external magnetic field. Namely, the intrinsic dipole magnetic moment $\vec\mu$ of a particle with spin $\vec{S}$ can be found through: $$\vec\mu = g\left(\frac{q}{2m}\right)\vec{S}$$ where $g$ is the g-factor, and $q$ ...


2

It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results. Let $\rho_m(\boldsymbol r)$ and $\rho_e(\boldsymbol r)$ be the mass and charge densities of the electron. The $g$ factor is given by $$ g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ ...


2

If you place the astronauts in a centrifuge, then at some point you're actually increasing the G's that they experience. Think about it. If the centrifuge spins about a horizontal axis, then when the astronaut reaches the bottom of that spin, they'll feel extra force downward, even though they might be weightless at the top of the spin. If the centrifuge ...


2

The angular momentum is conserved in central force motion (like what we have in the case of Earth-Moon system). In such a case, the force $\vec{F}$ and the radius vector $\vec{r}$ are parallel so that the resultant torque is zero $$\vec{\tau}=\vec{r}\times\vec{F}=o$$ This means the angular momentum ($\vec{L}$) of the of the Moon about the center is a ...


2

You know that the normalisation of the inner product is 1, that is, $$ \langle n\, l\, m\ |\ n\, l\, m\rangle = 1 $$ you can use this information to find the value of $\langle n_x=1\, n_y=0\, n_z=0\ |\ n=1\, l=1\, m=1\rangle$ as, $$ \langle 1\,1\,1|1\,1\,1\rangle= 1 $$ leaving some of the algebra for you as part of the exercise*, you will obtain, $$ 1 = ...


1

In general, $\frac{\partial L}{\partial \dot{q}}$ is the canonical (or generalized or conjugate*) momentum, and $m\dot x$, for $x$ the actual position, is kinetic momentum. Likewise, the cross product of the former with the generalized coordinate vector $q$ might be called "canonical angular momentum", and the cross product of the latter "kinetic angular ...


1

The expression you derived seems quite correct to me. I'd say the reason why you don't have an explicite dependence on the angle $\theta$ is that if there is no observable nutation (that is, if the top’s angular momentum due to precessional motion is small compared to its spin angular momentum), then the torque due to the earth’s gravitational field is ...


1

This is one of those three part dynamics questions. For the first part you need to use energy conservation to work out the horizontal speed of the person just before hitting the pole. The second part is the application of the conservation of angular momentum about the pole's pivot point when the person grabs hole of the pole. Note that the collision between ...


1

The reverse motion is easier to understand, assume he first has his arms stretched, and then pulls the weight to the center. What happened? He did work. The man feels centrifugal force, and against this force (which is not constant, so you cannot just calculate $W=F\cdot s$) he does work. Now the other direction of the motion is less intuitive. Because, ...


1

It's worth thinking about why the tidal bulges1 are not lined up with the line between the bodies (which is where you would naively expect them to) and then thinking about how that affects the gravitational interaction between them. Because the moon takes about one month to orbit and the Earth takes one day to turn, the naive location of the tidal bulges ...


1

The left hand: rotate the state $|JM\rangle$ by applying a rotation $R$ on it. Right hand side: insert completeness condition $\sum_{M'} |JM'\rangle\langle JM'|$ $D$ is the matrix representation of rotation matrix $R$ in basis ${|JM\rangle}$. The rotated state is expanded in terms of basis ${|JM\rangle}$ with coefficient $D$ in terms of rotation matrix.


1

As MSha has explained, centrifuges won't work. The only way to protect astronauts from g-forces is to transfer the normal force that act on their bodies in a different way. The problem is that normal forces only act on the boundary surface. Obviously the entire volume of the astronauts will be accelerated, this means that that the force acting on each volume ...


1

The error lies in the theoretical confusion between forces/torques and energy. The kinetic energy is linked to the motion generated by forces and torques which are the causes of the motion itself. Understanding the energy value in the two situations results impossible without knowing the time course of the applied force, being the energy, and so the work, ...


1

The Spin of a particle is better understood from a group-theoretic point of view. It is just telling you how a particle, i.e. an asymptotically free state of your theory, transforms under the symmetry of your theory, Lorentz symmetry. Well, actually under its double cover as Weinberg explains in his first book, that is why we are allowed to have spinors. ...


1

Neutral charged objects are made of charges. Although net charge is zero, imperfections in spatial arrangement of charges, causes them to interact with electromagnetic fields. Edit: 1) Neutrino's are formed during nuclear reactions. Once created, they sustain their direction of spin which does not change during course of time. 2) Photons don't change their ...


1

One promising way of implementing arbitrary physics simulations, is by programming in terms of 'constraints'. I highly recommend reading this article that covers constraints very well: http://gamedevelopment.tutsplus.com/tutorials/simulate-tearable-cloth-and-ragdolls-with-simple-verlet-integration--gamedev-519 I was very surprised when I first saw this ...



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