Tag Info

Hot answers tagged

22

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 ...


5

The situation is simpler than you think. Basically you don't need to do Clebsch-Gordan at all! The massless irreducible representations of the (proper orthochronous) Lorentz group are $1$-dimensional, labelled by their helicity $h$. Since $h = \pm 1$ are related by parity we group them as the $\pm$ helicities of the photon. To compute the helicity of ...


5

Yes, angular momentum depends very much on the origin you pick. (You can see this most clearly by examining a single particle in free space with fixed velocity - the angular momentum is $0$ only if you pick the origin along the particle's line of movement.) It's true that linear momentum is independent of your choice of origin; however $\vec p$ is still ...


4

So, what is electron actually doing in s state. Is it actually not moving at all. I know that it has got spin, but if it is not moving, doesn't it violates uncertainty principle. The "s state" is just a name for the first eigenfunction $\phi_{100}(\mathbf r)$ of the Hamiltonian of hydrogen type. This function can be used to calculate probability that ...


4

You are still thinking of the bound electron in a classical sense. The electron is a quantum mechanical "entity", it obeys quantum mechanical laws and the definition os S state identifies the energy level the electron is allowed to occupy. Within the energy level it does not have a position or a trajectory, it has an orbital. The orbital is the probability ...


4

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

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 ...


2

At the level of particles, most people don't think of "spin" in terms of actual movement, but as something else entirely. Spin is often approached as a mathematical property only. This is because some things do not make a ton of sense if you think of electrons spinning about some axis. If you interpret spin as an electron (or other particle) actually ...


1

Just some links.. The electron is comprised of two spherical scalar waves, one inward and one outward. A phase shift of the inward wave occurs in the wave-center region near where $\tau=0$, and spin appears as a required rotation of the inward wave in order to become the outward wave. This requirement is a property of 3D space termed spherical rotation. ...


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

You seem to be asking for a comparison of the angular momentum carried by gravitational waves in classical general relativity with the the angular momentum carried by quantum gravitons. For the analogous problem in electromagnetism, you should read Beth's Mechanical Detection and Measurement of the Angular Momentum of Light, from early in the days of ...


1

Here's the algorithm for finding Clebsch-Gordan coefficients: Start with a maximally aligned state like $\big|S,M = 2,2\big> = \big| m_1,m_2 = 1,1\big>$ Operate repeatedly with the lowering operator to generate the remaining states with the same $S$, such as $$\big|S,M=2,1\big> = \frac{\left|1,0\right\rangle + \left|0,1\right>}{\sqrt2}$$ ...


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 = ...



Only top voted, non community-wiki answers of a minimum length are eligible