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1

You will need to make use of the following: $$\cos^2\beta + \sin^2\beta\cos^2\gamma = \cos^2\beta + (1-\cos^2\beta)(1-\sin^2\gamma)$$ $$ = \cos^2\beta+1-\cos^2\beta-\sin^2\gamma + \cos^2\beta\sin^2\gamma$$ $$ = \cos^2\gamma + \cos^2\beta\sin^2\gamma$$ We therefore have $$\sqrt{v_\theta^2+v_\phi^2} = r\dot\gamma$$


2

A collapsing gas cloud is an open system. It loses mass, energy and angular momentum as it collapses. Even if the net angular momentum of the cloud is zero, after the collapse the final planetary disk can have a significant net angular momentum, and the ejected material will have the opposite angular momentum. What can not happen, and that's where your ...


1

Does $$L_+ L_- Y_{lm} $$ ,where $Y_{lm}$ is a spherical harmonic function, equals to zero. If so, why? It may or may not equal zero depending on the value of $m$. If $m$ is equal to $-l$ then yes, otherwise no. If $m=-l$ then applying the lowering operator annihilates the state (i.e., gives zero) since there is no state with an $m$ lower than $-l$. ...


3

You are right. If $q$ is a generalized coordinate then $\dot{q}$ is the generalized velocity and hence the generalized momentum is \begin{equation} p = \frac{\partial L}{\partial\dot{q}} \end{equation} Therefore, your sequence looks correct. Further, equations (20) and (21) of the article you have referenced also tell that the $p_{\theta_i}$ are indeed ...


0

The plane of polarization is perpendicular to the direction of propagation. If the plane of polarization is perpendicular to the detector, the light is parallel to the detector; therefore, none will hit the detector.


0

The difficulty of the prior experiments consisted in isolating the relativately weak gyro magnetic effect against the background of the purely magnetic forces acting on the studied rod... In order to avoid this difficulties, in the variant of the experiment proposed by Einstein, the magnetic field of the coil acts on the iron rod ... for a very short ...


1

In that case, $\hat{\sigma}$ here refers to a vector formed by $\hat{\sigma}_x$, $\hat{\sigma}_y$ and $\hat{\sigma}_z$ as its Cartesian components. The individual components of the expectation value of the magnetic moment vector would then be obtained using the corresponding components of the Pauli spin operators.


0

Let $$|s\rangle = \alpha|\uparrow\rangle + \beta|\downarrow\rangle$$ We assume that $s$ is normalized i.e. $\langle s | s\rangle = 1 \implies |\alpha|^2+|\beta|^2 = 1$. Then the expectation value of $\hat{\mu}_e$ is: $$\langle\hat{\mu}_e\rangle = \langle s|\hat{\mu}_e|s\rangle$$ $$\implies \langle\hat{\mu}_e\rangle = |\alpha|^2\langle\uparrow| \hat{\mu}_e ...


1

You chose the $\lvert \pm \rangle$ to be an eigenvector of $S_z$ with eigenvalue $\pm\frac{1}{2}$ - that's what the $m_s$ is: The eigenvalue of the state w.r.t. the $z$-spin. Since $S_x$ and $S_y$ do not commute with $S_z$, $\lvert \pm \rangle$ is not an eigenvector of them, hence the state cannot stay the same after they are applied to it. That the spin ...


0

Spin is an intrinsic quantity of a particle similar to a particle's mass or charge, and has units of angular momentum. It is an observable, and the corresponding operator has a matrix representation with respect to a particular basis.


2

If we treat the Earth as an isolated system then both its linear and angular momenta will remain constant. To answer your question you need only consider the angular momentum. The angular momentum is given by: $$ L = I\omega $$ Since $L$ is a constant, if the moment of inertia changes from $I_1$ to $I_2$ then we have: $$ I_1\omega_1 = I_2\omega_2 $$ and ...


0

Neither the linear momentum or the angular momentum will change (they are conserved quantities), unless you apply a force or a torque respectively. In this case, the change in radius will likely change the moment of inertia. And as the (constant) angular momentum depends on the product of moment of inertia and angular velocity, then the angular velocity ...


1

you are using the wrong command. Use MatrixPower[J/h,3] to calculate cube of J/h. In[7]:= MatrixPower[J/h, 3] Out[7]= {{0, -(i/Sqrt[2]), 0}, {i/Sqrt[2], 0, -(i/Sqrt[2])}, {0, i/ Sqrt[2], 0}} Clearly, this is equal to J/h.


2

First, two remarks: The melting of the ice in the Arctic ocean would have zero effect. What matters is the ice over Greenland and Antarctica. There is a very long-term secular slowing of the Earth's rotation rate due to the recession of the Moon from the Earth. This answer ignores that effect. (Alternatively, this answer has this secular effect subtracted ...


5

To a first-order effect, there would be no change. But one consequence of melting is that the water moves to other places. Water that moves from the poles to other areas on the surface of the earth would serve to (slightly) increase the moment of inertia of the planet. This is because the mass of the water would be farther from the rotational axis. The ...


2

When you calculate $\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$ of a particle of mass, m, having a linear momentum of $\vec{\mathrm{p}}$ in an inertial frame via a rotating frame or rotating body where the acceleration is directed towards the origin, you get ...


1

Hmmm, an old question without a satisfactory answer. I'll have a go. The spins of the two $B$ may combine as \begin{align} \text{singlet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> - \left|\uparrow\downarrow\right\rangle}{\sqrt2}, & \text{or triplet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> + ...


2

For most of the trials the racket head reached peak speed just at the time of impact. The racket head showed great acceleration just before impact where the racket head speed went from around 10 m/s to its peak value [35 m/s] in less than 0.1 seconds Probably tou have misinterpreted the text, it nowhere says that the acceleration gives more power, ...


1

Several things here. First - how efficiently you transfer momentum depends in part on whether you hit the "sweet spot". The location of the sweet spot is a function of the instantaneous center of rotation - an almighty swing from the shoulder puts the center of rotation further back so the same velocity of the racket might be less efficient. That depends on ...


1

They are the same thing. Parts of the racket though produce more force than others, suppose I completely missed the strings and hit the metal part (that's why without technique is so important). Usually, when you play you hit in what is called the sweet spot which is a vibration node and although it may not give you the biggest pop or bounce it gives you ...


0

What he means is: Sum of forces is $\sum \limits_i \vec{F}_i = \sum \limits_i [F_x,F_y]$ Sum of torques is $\sum \limits_i \vec{r}_i\times\vec{F}_i = \sum \limits_i (F_y r_x-F_x r_y)$ So it is not a circular argument because you can have net forces zero but net torque not zero (two equal and oppsite forces a distance apart, or a force couple). In fact, ...


1

The definition of being "in balance" for a rigid body is the absent of acceleration: both laterally and angularly. For a rigid body, the condition of to be "in balance" is the sum of all torque has to be zero with respect to any axis, not just to any particular one. If this condition is not held i.e. there exists an axis that the net torque is not zero, ...


1

Not necessarily. Some operators representing other physical quantities can be transformed so that they have the same algebraic structure of the angular momentum operators. For example, the inverse of "Jordan-Wigner transformation". Of course, you can think of them as effective angular momentum.


4

No. The commutation relation merely means that the $T_i$ form the Lie algebra $\mathfrak{su}(2)$. There are $\mathrm{SU}(2)$s (and consequently $\mathfrak{su}(2)$s) which have nothing to do with angular momentum, e.g. the $\mathrm{SU}(2)$ in the electroweak symmetry group $\mathrm{U}(1)\times\mathrm{SU}(2)$.


2

Yes, we expect all astrophysical black holes to have nonzero rotation. If nothing else, a rotating black hole that absorbs even a single particle with net angular momentum will then have nonzero angular momentum.


1

Gyroscopes depend on the conservation of angular momentum. Orientation and navigation gyroscopes are finely balanced/symmetrized so that gravitational fields will not exert external torque and modify the angular momentum. As the container which holds the gyroscope moves, a gimbal mount allows the gyroscope to maintain a constant rotational axis ...


2

Why is angular momentum conserved when a planet revolves about sun in an elliptical orbit? Why is linear momentum not conserved in this case? $$\rm \text{no external }\color{red}{torque}\to\color{red}{angular}\text{ momentum conserved}\\ \text{no external }\color{red}{force}\to\color{red}{linear} \text{ momentum conserved}\\$$ There is no external ...


1

Both are conserved if you consider the whole system: If the earth looses linear momentum, the sun will gain it and vice versa. Subsystems may violate conservation laws (e.g by transfering energy/momentum). This is called local violation. But globally conservation laws will always hold. The question why they hold globally in the first place, can be answered ...


2

I'll make an example, to make things clear. Take a two body system, in which the particles are seperated by a constant distance $d$ and have mass $m_1 = m_2 = m$. This is a holonomic constraint, since $$ | \vec{r}_1 - \vec{r}_2 | = d $$ with the particle-positions $\vec{r}_1$ and $\vec{r}_2$. This system is therefore reduced to 5 degrees of freedom (6 minus ...


2

The reason the $x$ and $y$ components lie on circles is because the expectation values for the components are zero (here you should check the math in the book you're learning from): $$\langle L_x\rangle=\langle L_y\rangle=0$$ When one talks about expectation values then one should always consider an experiment. In your case an experiment would be that we ...


1

I think I'm essentially supposed to show... Why do you think this? Is this a homework question? If so then it should be tagged as such. If this is not a homework question and you are interested to read an explanation involving no explicit equations then consider the following: The operator $\vec L$ is the generator of rotations. Therefore, any ...


0

There are states that correspond to eigenvalues of $\hat{L}_x$ or $\hat{L}_y$. But they are (almost always) of no interest: By the convention of what we mean by x,y, and z, at best only $\hat{L}$ and $\hat{L}_z$ are simultaneous eigenstates of the sort of systems usually considered (for example eigenstates of an atomic Hamiltonian). Hence for such ...


6

I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become $$F\delta t$$ (from ...


3

The value $m_{l}$ is the eigenvalue of the operator $L_{z}$ determined by seeing the action of this operator on the eigenstate $|~l,m_{l}>$ or in other words $$ L_{z} |~l,m_{l}> = m_{l} |~l,m_{l}> $$ While $l$ is related to the total angular momentum operator $L$ and it acts on the same eigenstate giving you $$ L^2 |~l,m_{l}> = l(l+1) ...


1

Acceleration of the center of mass is always $F/m$, so if force and mass are the same, the center of mass will accelerate the same way, regardless of the point where the force acts. After the same time of experiencing the same force, the body in the rotating case has greater kinetic energy than in the non-rotating case. This is due to greater work done by ...


1

When you have orbital angular momentum and spin angular momentum for an electron then the resulting magnetic moment $\mu_j$ is $$\vec \mu_j=\vec \mu_l+\vec \mu_s= -\frac{e}{2m_e}(\vec l+g\vec s)=-\frac{\mu_b}{\hbar}(\vec l+g\vec s),$$ where $\mu_b=\frac{e\hbar}{2m_e}$ is Bohr's magneton, $l$ denotes the orbital angular momentum, $s$ the spin angular ...


0

I'm writing this assuming you're talking about something like a disk that cracks and the pieces move without interacting with each other. You need to conserve energy: $m_0v_0^2 = m_1v_1^2 + m_2v_2^2 $ $I_0w_0^2 = I_1w_1^2 + I_2w_2^2 $ where $v_0$, $v_1$, and $v_2$ are the magnitudes of the velocities $v = \sqrt{v_x^2 + v_y^2}$ You also need to ...


2

Your expression "the angular momentum is $m_j \hbar$" (where $\hbar = h/2 \pi$) is incorrect. This quantity is the projection of the angular momentum on the $z$-axis; it represents the direction of the spin. This is why it corresponds to the $m$ quantum number, not the $\ell$ quantum number in the $| n \ell m \rangle$ basis. Which angular momentum to use ...


2

This comes from the representation theory of the rotation group $\mathrm{SO}(3)$. Quantum mechanics takes place in a vector space, and observables are operators on this space. The total amount of angular momentum is obtained from the angular momenta $L_x,L_y,L_z$ about the three axes of space as $$ L = \sqrt{L_x^2+L_y^2+L_z^2} $$ since that is the length of ...


0

Let's solve this by considering the time-reversed process: two spinning objects collide inelastically. We know that linear and angular momentum are conserved: $$ m_1\vec{v}_1 + m_2\vec{v}_2 = m_0\vec{v}_0 $$ $$ m_1\vec{r}_1\times\vec{v}_1 + m_2\vec{r}_2\times\vec{v}_2 = m_0\vec{r}_0\times\vec{v}_0 $$ The latter can also be written (assuming you have the ...


2

Yes, that's what a commutator is. Remember that the momentum operator is $\vec p = -i\hbar\vec\nabla$, so $\nabla^2 = -p^2/\hbar^2$; also that $\nabla^2$ is a scalar, not a vector. Usually when you find a commutator you need a "test function" for the operators to operate on; that's how you find for instance that $[\hat x, \hat p_x]=+i\hbar$.


0

Assuming it is an elastic collision, you will have, right after it, a ball with the same rolling motion (anticlockwise) but translating in the opposite direction at $v_0$. Now there is friction because the ball is "sliding", and the new equilibrium movement will be when both are moving without sliding again at $v_k$. Let me know if you do not know how to ...


-2

No, there are just spin-0, spin-1 and spin-1/2 particles, even the Higgs-Boson is 0.


1

Meaning that L is in the "position representation". But isn't J also in the position representation since it is still angular momentum? Or is it that since J = L + S (where S is the spin) then it is no longer in "position representation"? Then what "representation" is it in? If $L$ is in the "position representation" (call its eigenstates ...


1

With current technologies the Kerr parameter could not be precisely estimated yet. The best result today is a Kerr factor $a=0.52$. Genzel, R., Schoedel, R., Ott, T., Eckart, A., Alexander, T., Lacombe, F., Rouan & Aschenbach, B., Near-infrared flares from accreting gas around the supermassive black hole at the Galactic Centre (2003), ...



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