16

Yes, the "artificial gravity" can last for a very long time (the other answers address the caveats to "forever"), but we are not getting anything for free: the person at the edge of this rotating spaceship has a force continuously applied to them, but as long as they stay in place (with respect to the ship) this force does not do any work,...


11

You are missing the property ${\bf v}\times {\bf v}={\bf 0}$ of the vector product. So if ${\bf r}\to {\bf r}+ {\bf v}t$ the angular momentum ${\bf L} \propto {\bf r}\times {\bf v}$does not change.


6

This is an interesting question that could really be formulated as follows: When collapsing an object that has the ratio $a^* = cJ/(GM^2)$ larger than one, can an over-extremal black hole form? There is no rigorous proof, but the answer seems to be no, at least in reasonable physical situations. The simple reason for that is that when you have an object ...


5

This depends on which one is bigger. If $\ell$ is bigger than $s$, then $|\ell-s|=\ell-s$, so $j$ goes from $\ell-s$ to $\ell+s$ in unit steps, which means $2s+1$ different values. If $\ell$ is smaller than $s$, you get the same story with swapped symbols, so you get $2\ell+1$ different values. If they're equal, then both options apply.


5

Conservation laws I would start from the most basic question. Any operator evolves in time as: $$O(t) = e^{\frac{i}{\hbar} Ht} O(0) e^{-\frac{i}{\hbar} Ht}$$ (this is a consequence of the Hamiltonian being the generator of time translations) where $U(t) = e^{-\frac{i}{\hbar} Ht}$ is the time evolution operator. It follows straightforwardly that when we ...


5

Ach, yes & no, but... this is the most ponderous summary of the classic Jordan map construction of SU(2) matrix generators there is. Most of the difficulty and confusion here lies in the change of bases, which runaway abstraction of notation fosters. The notional spaces of the two oscillators are not the spacetime indexed by the generators of rotations, ...


5

The short answer to your question is yes - spin most certainly does have something to do with rotations. The answer to the subsequent question of precisely what it has to do with rotations is a subtle story. When we construct a quantum mechanical model, we need to decide on a Hilbert space to define the states for the system, and we need choose some ...


4

$L=\vec{r}\times \vec{v} = rv\sin\theta \times\vec{n}$ (where $r$ and $v$ are the magnitudes, $\theta$ is the angle between $\vec{v}$ and $\vec{r}$ and $\vec{n}$ is a unit vector perpendicular to the plane in which $\vec{r}$ and $\vec{v}$ lie) $v$ is constant but $r\sin\theta$ is also constant It's the distance between $O$ and where the dotted line crosses ...


4

The $l$ rotation representation is irreducible,so the rotation matrix cannot couple states of different $l$


4

It is all about conservation of angular momentum . A spinning body, a planet, in space would continue spinning for ever if it were the only body in the universe. We know that gravitational forces induce tides, in systems of planets and stars, and even though the total angular momentum is conserved individual bodies lose it. If the space ship is far away ...


3

I assume you're looking for the components of the state in the angular momentum representation? The state itself is technically the same, but you want it in a different basis. We need to think about the actions of $L^2$ and $L_z$ in the Fock space. Consider $$L_z(a^{\dagger})^n |0\rangle = \frac{n}{2} (a^{\dagger})^n |0\rangle$$ $L_z=\frac{1}{2}(N_a-N_b)$ is ...


3

If you consider $L^2$ and $L_z$ simultaneously then $|lm\rangle$ are non-degenerate (each has a different $\textbf{pair}$ of eigenvalues). Considering them together is enough to show that $Y_l^m$ are orthogonal: if $l\neq l'$ you have $\langle lm|L^2|l'm'\rangle = 0 $ and if $m\neq m'$ you have $\langle lm|L_z|l'm'\rangle = 0 $. Both imply $\langle lm|l'm'\...


2

Yes, is it possible to read off $\ell$ and $m$ from the shape of the spherical harmonic $Y_{\ell m}$. The rules are not complicated: Number of plane nodes (containing the $z$-axis): $|m|$ Number of cone-like nodes (around the $z$-axis): $\ell -|m|$ There is a nice interactive web page for Visualization of Spherical Harmonics. You can choose values $\ell$ ...


2

Both dipoles appear to experience an anticlockwise torque, around their own centers of mass. This is not the same as an anticlockwise torque around the origin, or around any other single point. Each individual dipole rotates anticlockwise, but the two dipoles also move, in a clockwise direction around the origin. The overall angular momentum is conserved.


2

So yes, the left dipole gets a downward force on its negative charge and an upwards force on its positive charge, giving an anticlockwise torque. But the upward force is bigger as the + charge is nearer the second dipole, so there's also a net upwards force. The right dipole gets a leftward force on the negative charge and an equal rightward force on the ...


2

"is it correct to say that a person sitting inside a spinning spaceship will experience "artifical gravity" forever, without any intervention or additional torque required. ?" Without any external torque, yes. But not without any intervention. The intervention needed is the centripetal force. In any spinning body, a particle at a distance ...


2

Consider the simple case where there are two spin-1/2 particles. The possible states of such a system are \begin{align} \vert ++\rangle\, ,\qquad \vert +-\rangle\, ,\qquad \vert -+ \rangle\, ,\qquad \vert --\rangle \end{align} and the total angular momentum operators are \begin{align} \hat L_{i}= {\hat L}_i^{(1)} +{\hat L}_i^{(2)} \end{align} Construct ...


2

Recall that a representation $\Pi: G \mapsto \mathrm{GL}(V)$ is irreducible if there are no non-trivial subspaces $U\subsetneq V$ such that $\Pi(G) U = U$. Your intuition appears to be that given a Lie group $G$, there is some positive integer $n$ such that representations on spaces with dimension $\gt n$ must be reducible, and representations on spaces ...


1

We do not see astronauts in the ISS adjusting their orientation by rotating their arms , because it is much simpler to rotate themselves by pushing off the walls of the compartment, they are in. Rotating themselves, using just their body and arms is much more strenuous and hence not very efficient. But it can be done, somewhat like how a cat rotates itself ...


1

For any unitary operator $U$, use the abbreviation $$ \sigma_U X\equiv U^{-1}XU. $$ Now, let $G$ be a group of unitary transformations $\sigma_U$ that act on field operators $X$ in this way. If $\sigma_U\in G$ and $\sigma_V\in G$, then $\sigma_U\sigma_V\sigma_U^{-1}\in G$. This follows just from the fact that unitary operators are invertible and can be ...


1

Sorry for being a bit late here. What I find to be the magic part is: rotation of a body on an axis can always translated to rotation on a different axis + a linear motion. Let's consider your case, viewed from above, before the split: You can see the linear velocity vectors on multiple points as well as the CM. Now, after the split, the vectors are the ...


1

I believe the only state that is a simultaneous eigenstate of ${L}^2, L_z, L_x$ and $L_y$ is $|l=0, m=0 \rangle$ (thus the ground state of the Hydrogen atom) with eigenvalue $0$ for all the concerned observables. As expected, this doesn't form an orthonormal basis by itself. It is easy enough to see why this is the only such state in existence. The states $|...


1

Note that if two observables do not commute, then there exists no common eigenbasis. However, this in general does not mean that there are no states which are eigenstates of both operators. For your specific example: You could try to express $\hat{L}_x$ and $\hat{L}_y$ in terms of the ladder operators $\hat{L}_+ $ and $\hat{L}_-$. Applying $\hat{L}_x$ and $\...


1

The system is not isolated. The frictionless pivots can exert forces on the system, and since they are not co-axial those forces will generate a net torque about any axis.


1

We have that the angular momentum is $$\vec L = m \vec r \times \vec v$$ where $\vec r = \vec r_0 + \vec v t$ (since no external force is applied, the object simply continues with its given velocity by Newton's first Law, $\vec r_0$ is some arbitrary position, $\vec v$ the velocity, and $t$ the time). Therefore, $$\begin{align} \vec L &= m\left( \vec r_0 ...


1

You can derive $L_x$ from the ladder operators $L_{\pm}$. These operators verify: \begin{equation} L_{\pm} | 1, m \rangle= \hbar \sqrt{2 - m (m \pm 1)} | 1, m \pm 1 \rangle \end{equation} With the given basis you can derive the matrix for these operators and find $L_x$ like: \begin{equation} L_x = \frac{1}{2} (L_+ + L_-) \end{equation}


1

The operator $\cal{D}$ is constructed by exponentiating the angular momentum operators $\hat L_i$’s, none of which change the $\ell$ value.


1

I would respectfully disagree with answers and comments saying that the explanation is generation of magnetic field and hence angular momentum of electromagnetic field. Magnetic field could give some contribution, but it is only partially relevant. Let me start with $\textbf{approach 1}$. There is nothing wrong with it, but it implicitly assumes that ...


1

These are term symbols, which are a compact way to encode the angular-momentum characteristics of a quantum state. The general scheme for this notation is of the form ${}^{2S+1}L_J$, where: $2S+1$ is the spin multiplicity, i.e., the number of linearly-independent spin states of the system. $L$ is the total orbital angular momentum of the system, denoted in ...


1

If the fixed axis is the axis of symmetry (a principal axis), and if the angular velocity vector is fixed in direction along that axis, then if the angular speed is changed the angular momentum vector and angular velocity vector remain parallel.


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