Tag Info

0

You are right that the argumentation for the existence of ortho and para hydrogen is identical to that for the existence of ortho and para helium, that is, applying the permutation operator on a wave function for two identical particles should result in an eigenvalue of $+1$ or $-1$ if one considers bosons or fermions, respectively. Whereas in helium you ...

0

In the second example, only the 3d shell is not completely filled. Using the aufbau principle to fill this shell with the 7 valence electrons leaves three unpaired electrons resulting in a total electron spin of $S=3/2$, so you arrive at a quarted state. The electron orbital angular momenta of the five possible states in the d orbital are -2,-1,0,1, and 2. ...

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 ...

0

Of course it can. Angular momentum (the rotational analog to linear momentum) can be expressed as an axial vector (sometimes called pseudovector), which means a quantity that is similar to a regular vector, except for the fact that it changes sign under improper rotations (such as reflections). I suppose you're refering by "wobble" to nutation, i. e., a ...

0

The linear momentum of the system is always conserved when there is no external force acting on it, so you should always consider it as a whole. And as said in the comments, it doesn't matter which point you chose for calculating angular momentum.

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 ...

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 ...

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 ...

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 ...

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. ...

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 ...

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, $$1 = ... 1 The Spin of a particle is better understood from a group-theoretic point of view. It is just telling you how a particle, i.e. an asymptotically free state of your theory, transforms under the symmetry of your theory, Lorentz symmetry. Well, actually under its double cover as Weinberg explains in his first book, that is why we are allowed to have spinors. ... 1 The reverse motion is easier to understand, assume he first has his arms stretched, and then pulls the weight to the center. What happened? He did work. The man feels centrifugal force, and against this force (which is not constant, so you cannot just calculate W=F\cdot s) he does work. Now the other direction of the motion is less intuitive. Because, ... 0 The torque derivation is as follows:$$ \sum \vec \tau = \frac{\mathbb{d}L}{\mathbb{d}t} $$The magnitude of the torque (measuring \theta with respect to the horizontal) is:$$ |\vec{\tau}| = mgl\cos\theta $$The angular momentum L is given by:$$ L = mr^2\omega = ml^2\frac{\mathbb{d}\theta}{\mathbb{d}t} $$So the rate of change of angular momentum ... 2 Spin is connected to the intrinsic magnetic dipole moment of the particle, this is what makes the particle capable of interacting with an external magnetic field. Namely, the intrinsic dipole magnetic moment \vec\mu of a particle with spin \vec{S} can be found through:$$\vec\mu = g\left(\frac{q}{2m}\right)\vec{S}$$where g is the g-factor, and q ... 0 What you are missing is that when charged particles are accelerated away from the magnetic poles of the neutron star, they are tied to helical paths along the field lines. Since the particles are highly relativistic, the consequent curvature and synchrotron radiation is doppler boosted and doppler beamed in the forward direction. The beaming opening angle ... 0 From FURTHER EVIDENCE FOR COLLIMATED PARTICLE BEAMS FROM PULSARS AND PRECESSION (2007): "An overwhelming majority of the observers and theorists interpreting these observations seem to suggest and endorse the following basic picture. The jetlike features nearly along the symmetry axis, bisecting the arcs and the diffuse glow spread about them, are ... 0 Magnetic fields. The beams are generated from charge particles moving in magnetic fields. The pulsar is the result of a large star collapsing and the magnetic field of the original star is compressed, making a very strong field. 1 Neutral charged objects are made of charges. Although net charge is zero, imperfections in spatial arrangement of charges, causes them to interact with electromagnetic fields. Edit: 1) Neutrino's are formed during nuclear reactions. Once created, they sustain their direction of spin which does not change during course of time. 2) Photons don't change their ... 3 If |\pm\rangle are the eigenvectors of {\hat \sigma}_x, {\hat \sigma}_x |\pm\rangle = \pm |\pm\rangle, then a rotating x-basis is defined as$$ |+\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 ...

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 ...

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 ...

0

In an isotropic space (which means it's equal in all directions), Noether's theorem tells us that angular momentum is conserved. For the Earth this means that it keeps the angular momentum that it has now. This can be calculated from $$L = I \omega$$ where $I$ is the moment of inertia and $\omega$ the angular speed. You'll see that if $L$ remains constant ...

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 ...

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.

0

You can think, on the left is a short hand notation for a (2J+1) x (2J+1) matrix R applied to a (2J+1) component vector |J,M> with the components labelled by M. On the right, the matrix elements are explicitly shown, and the sum over M' is the matrix multiplication. Actually on the left is an abstract rotation operator R that will rotate any J. When ...

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 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 ... 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 ... 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 ... 0 A pure moment cannot be created by one force. In real life, producing pure torques is almost impossible. There will always be a non-zero net force applied also. In your examples the torque about the center of mass is \tau_C = 2 r F in both cases, but on the second case you also have a net force applied 2 F that changes linear momentum also. So your ... 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 www.physicspages.com/2013/04/11/magnetic-dipole-moment-of-spinning-spherical-shell/ My search gives \mu = \frac{e\omega R^2}{3} This gives g = 5/3 = 1.667 Did not you provided link given below? https://en.wikipedia.org/wiki/Electron_magnetic_moment#The_classical_theory_of_the_g-factor Which explains that non-uniform charge distribution can explain ... 0 Kinetic energy is$$K = \frac{1}{2} \vec{\omega} \cdot [I] \vec{\omega} $$when the 3×3 mass moment of inertia matrix [I] is expressed in world coordinates. Remember$$[I] = [R] [I_{body}] [R]^\top$$is how body inertias is transformed into world inertias. You seem to apply a scalar mass moment of inertia to a vector rotation. If you are careful with the ... 1 The multiplicity 2S+1 actually tells you how many degenerate spin states there are, each labelled with the total spin projection quantum number M_S (this is from the total spin projection operator \hat{S_z}(conventionally taken to be in the z-direction) whose eigenvalues are \hbar M_S). The possible values of M_s are -S\le M_S\le S in integer ... 2 In general, the change in angular momentum resulting from a change in moment of inertia depends on how the change is implemented, and to some extent your perspective. In physics, you can think of global conservation laws as constraints that feed into your interpretation of a system. Consider the simple problem of determining the change in linear momentum ... 4 The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as:$$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain ... 2 Remember that the variation of the angular momentum equals the external torque. If there are no external torque (as in your case), the angular momentum is conserved. 2 The notation (\vec S_1\otimes 1) \cdot (1\otimes \vec S_2) is interpreted as a contraction over spatial indices. Hence, \vec S_1\cdot \vec S_2=\sum_{ij} \delta_{ij}S_{1i} \otimes S_{2j}. By contrast, (S_{1x}+S_{1y}+S_{1z})\otimes(S_{2x}+S_{2y}+S_{2z})=(\vec S_1\cdot [1,1,1])\otimes (\vec S_2\cdot [1,1,1]). 0 The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction \mu gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ... 0 You have my sympathy; precession is little understood. Try it this way! i) remove the front wheel from a bicycle and mark a spot on the tyre. ii) hold it by the ends of the axle between your hands with the mark at "top dead centre". iii) set the wheel spinning as though the machne was progressing forward (very slowly) Note that the spot travels ... 2 Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ... 1 MRI signal is always complex and it is related with signal demodulation. The detected signal is multiplied by a sinusoid or cosinusoid with frequency equals to \omega_0 +\delta \omega, respectively leading to the real and imaginary channels. You can find the complete algebra at Haacke,\ Magnetic\ resonance\ imaging chapter 7.3.3 Phase is really useful ... 0 First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ... 0 The space ship should undergo both linear motion and angular (rotational motion). Linear motion will take place in the direction in which the force is applied, and rotation about its centre of gravity. So the centre of gravity will act as an imaginary pivot. 0 a Stern Garlach apparatus does not rotate the state of the particle, what it does is to split a beam or, if you have a single particle, it chooses" a state in the desired direction. What you might be looking for is how to write the eigenstate in terms of Z basis. For a 1/2 spin it is going to be:$$|\pm \rangle_y = \frac{1}{\sqrt{2}}\left( |+\rangle_z ...

4

The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows: $$\left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle$$ where the second factor is a Clebsch-Gordan coefficient and $q=-1,0,1$ indicates the type of transition. For ...

Top 50 recent answers are included