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1

I think (but I'm not sure) that if you do it like if the photon has three physical polarization states then you get, as you said $S=0,1,2$. You can compute easily all of these states. But remember, you can express the states of the coupled basis in the uncoupled basis (via a Clebsch-Gordan decomposition). Then, you can delete the terms containing a non ...


1

So the key point to understanding this problem is to understand that it is the modes that contain information about the physical parameters of your photons (such as the momentum or angular momentum), and quantization is just a description of excitation of those modes. For instance the canonical quantization of the plane-wave expansion which you've ...


2

See also Simple Harmonic Motion - What are the units for $\omega_0$ ? and https://en.wikipedia.org/wiki/Joule#Confusion_with_newton-metre Here's a somewhat shorter explanation reflecting my own (possibly incorrect) intuition: Radians aren't "real" units; they're just a trick to keep track of which quantities involve angles and which don't, since it's ...


3

In answer to the edit, any transitions due to single-graviton exchange will involve energies that are just impossibly small. To convince yourself of this, remember that the energy levels of the hydrogen atom are given by: $$E = \frac{\mu k^{2}e^{4}}{2\hbar^{2}n^{2}} = \frac{13.6\,\,{\rm eV}}{n^{2}}$$ If you do the same for two solar mass neutron stars ...


0

Check ✓ except the the speed depends on the center of rotation, not the choice of the origin. The general rule is that if the origin as velocity $\vec{v}_O$ then a point A located at $\vec{r}_A$ has speed $\vec{v}_A = \vec{v}_O + \vec\omega \times \vec{r}_A$. It happenstance that your origin does not move. Check ✓. In general, torques and angular momental ...


21

Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves. Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its ...


3

Your mistake is your definition of the angular velocity $$ \omega=rv \tag{not correct} $$ is incorrect. We know that $\omega$ has units of $1/s$, but your assertion gives it units of $m^2/s$. The correct definition is $$ \omega=\frac vr \tag{correct} $$ which gives the correct units. Using this: $$ \left[L\right] = [m]\left[r^2\right]\cdot\left[v\cdot ...


3

The easiest thing for this exercise is to use Levi-Civita symbol for the vector product: $$\vec{a} \times \vec{b} = a_i b_j e_k \varepsilon_{ijk},$$ where I denote by $e_i$ the canonical basis of $\mathbb{R}^3$. Using this notation, we have: $$[L_j,p_i]=[r_k p_l \varepsilon_{klj},p_i]= i \hbar p_l \varepsilon_{ilj}.$$ and $$[L^2,\vec{p}]=e_i[L_j ...


0

It is best to read a book on dynamics to get all the details you need, but here is an overview. I assume all the rotation axes intersect at the origin, and the payload center of mass is located at a distance $\vec{c}$ from the last gimbal (in local coordinates). The mass moment of inertia of the payload is ...


0

As pointed out in the comments, the 'machine' in the movie Contact is just the gimbal part without the gyroscope. The point of this gimbal is to allow the thing being held at the center to be completely free to rotate in all directions. So, if they are working well they have no effect on the system except in the translational degrees of freedom, and they ...


1

It seems like you know what the answer is, but you just don't know how to prove it. You are right though. To make things simpler, just view things in the center of mass frame. Then the total momentum is zero, and, like you said, the total angular momentum is just the sum of the orbital momentum of each planet plus the sum of the spin angular momentum of ...


-5

Black holes pull matter in thus creating a pulling of matter or rotation (gravity) force that matter rotates or spirals around thus creating our shape form and rotational direction of our galaxy. If our sun doesn't die first then yes one day we will be pulled to the center of our galaxy. Every galactic year our solar system speeds up and grows closer to the ...


0

Case a) Body with uniform motion (no rotation), with C the center of the disk and A a point on the edge (for example below, at a distance $R$). $$ \begin{aligned} \vec{v}_A & = (v,0,0) \\ \vec{\omega} & = (0,0,0) \\ \vec{v}_C & = \vec{v}_A + (0,-R,0) \times \vec{\omega} = (v,0,0) \\ \vec{L} & = m \vec{v}_C = (m v,0,0) \\ \vec{H}_A & = ...


0

From your description, I assume the disk only only translating, not rotating. Is this correct? If so, read on. If not, I'll delete. I'm uncomfortable with the first method that uses $L=I\omega$. In this equation, it is assumed that every point on the rigid body can be characterized by the same angular velocity $\omega$. From your description of the motion ...


0

The first method is correct. The second method is wrong because the equation you use only applies to point particles, not continuous masses with volume (such as a disk). You're incorrectly treating the disk as a point particle located at the disk center. If you want to use the second method, you'll need to use this equation for angular momentum of ...


1

The question is essentially just asking you to perform Clebsch-Gordan (CG) decomposition, namely a change of basis on the Hilbert space $\mathcal H_{j_1}\otimes\mathcal H_{j_2}$ of the composite system of spins. Recall that for each of the Hilbert spaces $\mathcal H_{j_1}$ and $\mathcal H_{j_1}$ there exist orthonormal bases of eigenvectors of $\mathbf ...


3

Consider two bodies A and B. With respect to an inertial coordinate system with origin at point O, the coords of the particles in A are vectors $x_{a}\in V_{3}$ with $a=1,2,\ldots, N_{A}$ and similarly the coords of the particles of B are $x_{b}\in V_{3}$ with $b=1,2,\ldots,N_{B}$. The momenta wrt the inertial frame with origin at point O of the particles ...


0

The problem here was when doing the integration by parts. One has to be very careful when remembering which operators act on which expressions. In my case, it turned out that the derivative was acting on the wrong terms when doing the integration by parts, which indeed did add an extra minus sign.


0

I've been wondering a lot about this too. So far I haven't got any straight answers, and I've asked around a lot. Angular acceleration is easy - if you have either a moment of inertia (2D) or an inertia tensor (3D+) I, then the angular acceleration due to a force is simply $\alpha=I^{-1}\tau$ for torque $\tau=\vec{F}\textrm{x}\vec{r}$, with a force ...


1

Hints: Notice that if we define $\mathbf S_{123} = \mathbf S_1 + \mathbf S_2 + \mathbf S_3$ and $\mathbf S_{12} = \mathbf S_1 + \mathbf S_2$, then we have \begin{align} \mathbf S_{123}^2 = \mathbf S_{12}^2 + \mathbf S_3^2 + 2\mathbf S_{12}\cdot\mathbf S_3 \end{align} Notice that your hamiltonian can be written as follows: \begin{align} H = ...


0

It looks like you've missed a sign: \begin{align} Q_i &= \epsilon_{ijk} \int_{p,q} \sqrt{\frac{E_q}{E_p}}p^k \int d^3x (a_qe^{iq\cdot x}-a_q^\dagger e^{-iq\cdot x}) \left(a_p (-i\frac{\partial}{\partial p^j})e^{ip\cdot x} -a_p^\dagger (i\frac{\partial}{\partial p^j})e^{-ip\cdot x}\right). \end{align} Then it looks like the term that goes like ...


0

To the original poster: You appear to be operating under the "hollywood" misconception that a black hole somehow "sucks harder" than the same amount of mass in a non-black-hole form. However, this false "black holes produce an enormous sucking" misconception is one of the many, many concepts of physics that "hollywood" gets totally wrong; a black hole of a ...


0

I'm not totally understanding the question, but I think the critical aspect here is that in atoms, the energy gap between successive levels is much greater than the "thermal" energy of the electrons. So the energy levels really do have to fill up pairwise, with each pair of electrons forming a singlet state with spin zero. I think it might be otherwise in a ...


4

Such an ordering arises from the fact that are arranged chronologically, i.e., according to the dates they were "discovered". The principle quantum number $n$ entered the picture with Bohr's theory of the Hydrogen atom in 1913.Bohr introduced $n$ in his quantization of angular momentum postulate where $n$ is the allowed orbit. Mathematically, $L = n{h ...


2

Use Lagrangian mechanics method to answer this problem because it is easier than Newtonian mechanics (IMHO). Let $T$ be the kinetic energy, $V$ be the potential energy then the Lagrangian $L$ is given by $$ L=T-V $$ and the Lagrangian equation is $$ \frac{d}{dt}\left(\frac{dL}{d\dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0, $$ where it is assumed that ...


1

The center of mass is not $(0, 0)$. If you take the average of the two positions, you will get $(c, d)$. The angular momentum $\vec L$ is $\vec r \times \vec p$. With mass $m$, this will be: $$ \vec L = m \vec r \times \vec v $$ $\vec v_\pm$ is just $\dot{\vec r}_\pm = \frac{\mathrm d}{\mathrm dt} \vec r_\pm$. So you have to to a time differentiation of ...


0

To prove the conservation of quantities, you need to be able to compute the motion of the system so that you can directly compute these quantities from the time dependent coordinates and verify that they do not change with time. However I don't think that the motion of this system is integrable: it looks like a multiple oscillator and it is very prone to ...


2

I can get you thinking along the right path, but I'll leave most of the doing for you. I suspect that for much of this to make sense, you'll need to actually do it. The idea is to start at the "top" of the ladder: $$|2, 2\rangle = |3/2, 3/2\rangle_1\otimes|1/2, 1/2\rangle_2 =|3/2, 3/2\rangle_1|1/2, 1/2\rangle_2.$$ With this statement (and choice of ...



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