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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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 = ...


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 ...



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