# Tag Info

5

simply the resistance of a body to rotate it over an axis? Gosh, I dislike the word resistance in this context since resistance is, in general, dissipative and, in particular, resistance to rotation would imply that an isolated object that is rotating would eventually stop. Think of moment of inertia (rotational inertia) about an axis as a measure of ...

4

I think that what your teacher has told you is that the angular momentum of a body can be split into two components: The spin angular momentum which is an intrinsic property of the body and is independent of the point about which you wish to find the angular momentum. $L_{\text{spin}} = I_{\text{cm}} \omega = \frac v r$ where $I_{\text{cm}}$ is the ...

3

The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows: $$\left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle$$ where the second factor is a Clebsch-Gordan coefficient and $q=-1,0,1$ indicates the type of transition. For ...

3

I could give you an answer by barking up a very different tree indeed! In phase space QM, and not, repeat not geometric quantization, you may work on flat phase spaces and forfeit spheres altogether, the way you actually do in Hilbert spaces. If you can stomach that, read on, otherwise not, lest you feel your expectations betrayed. Plain vanilla ...

2

You don't have to, but it makes the equations easier to deal with because you don't have to account for the moment of acceleration terms. See the 2nd part this this answer about deriving Newton's laws on an abitrary point not the center of mass. So finally the equations of motion of a rigid body, as described by a frame A not on the center of gravity C ...

2

Let's look at the hodograph of a constant radius & constant velocity motion. Left: trajectory of one of the masses. Right: hodograph, i.e. locus of the velocity vectors. Now, let's look closer at how the velocity changes during a small time interval $\mathrm dt$. A force is needed to rotate it (difference between the brown and red arrows). If you ...

2

Let the body rotate about the $z$-axis, then by the definition of angular momentum $$\vec{L}=\vec{\omega} I_z.$$ where $\omega$ is the angular velocity about the $z$-axis. So we could take the parallel axis theorem and multiply it by $\omega$: $$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$ Now ponder the terms in it. If I understand the ...

2

Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...

2

As you mentioned, $\vec {L_o}_⊥$ is proportional to $\omega$. So the middle term means $\vec {L_o}_⊥$ varies by varying angular velocity. You can also derive this. formula like this: $$\frac{d \vec {L_o}_⊥ }{dt}= \frac {d\vec{\omega}}{dt}A \hat{L_o}_⊥+ \vec{\omega}A \frac{d\hat{L_o}_⊥}{dt}= \frac{1}{\omega} \frac{d \omega }{dt} \vec{L_o}_⊥ + \vec{ \omega } ... 2 If you draw similar triangles, then you'll find that r_A/r_B = v_y/v_x, and so the product r_A v_x is equal to r_B v_y. Try drawing a line from the tip of your lower \vec{v} vector to the tip of your lower v_y component to see this. 2 I am certain that mathematical analysis of tidal locking has been done many times but I have failed to find such an analysis where the mechanism for the transfer of angular momentum to spin angular momentum is included which is the question which has been asked. Perhaps someone is able to produce a reference or an analysis? Having experienced on a number ... 2 Angular momentum is conserved only if there's no external forces, in this case the electron gains energy by light or by heat wich is kinetic energy. They are both external forces so the conservation of angular moment does not apply. 2 In short, you must consider the total elements of the system for conservation of momentum. In this case, nearly all of the momentum is exchanged between the electron and a photon that is absorbed or radiated away (the light). Momentum is conserved, and is largely balanced by this electron-photon interaction, although smaller amounts may be exchanged with ... 2 Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ... 2 The notation (\vec S_1\otimes 1) \cdot (1\otimes \vec S_2) is interpreted as a contraction over spatial indices. Hence, \vec S_1\cdot \vec S_2=\sum_{ij} \delta_{ij}S_{1i} \otimes S_{2j}. By contrast, (S_{1x}+S_{1y}+S_{1z})\otimes(S_{2x}+S_{2y}+S_{2z})=(\vec S_1\cdot [1,1,1])\otimes (\vec S_2\cdot [1,1,1]). 2 Remember that the variation of the angular momentum equals the external torque. If there are no external torque (as in your case), the angular momentum is conserved. 2 The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as:$$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain ... 2 In general, the change in angular momentum resulting from a change in moment of inertia depends on how the change is implemented, and to some extent your perspective. In physics, you can think of global conservation laws as constraints that feed into your interpretation of a system. Consider the simple problem of determining the change in linear momentum ... 2 To answer the generic dimension part, in case it is not self-evident from Valter's answer: In D dimensions, the rotation matrix is the exponential of an angle θ times a matrix K, a normalized generator of the corresponding rotation group SO(D) around some unit axis D-vector k, in the vector representation, so the matrix is D×D. The eigenvectors of these ... 2 It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results. Let \rho_m(\boldsymbol r) and \rho_e(\boldsymbol r) be the mass and charge densities of the electron. The g factor is given by$$ g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ ...

2

If you place the astronauts in a centrifuge, then at some point you're actually increasing the G's that they experience. Think about it. If the centrifuge spins about a horizontal axis, then when the astronaut reaches the bottom of that spin, they'll feel extra force downward, even though they might be weightless at the top of the spin. If the centrifuge ...

2

First draw a circular path. In order for an object to follow that path at a constant speed, there must be a force acting on it towards the centre of the circle. At any instant, that force has no component in the direction of motion, and so, if that's the only force acting on the object, its speed will stay constant. Now draw a spiral path (i.e. one in which ...

1

As MSha has explained, centrifuges won't work. The only way to protect astronauts from g-forces is to transfer the normal force that act on their bodies in a different way. The problem is that normal forces only act on the boundary surface. Obviously the entire volume of the astronauts will be accelerated, this means that that the force acting on each volume ...

1

The left hand: rotate the state $|JM>$ by applying a rotation $R$ on it. Right hand side: insert completeness condition $\sum_{M'} |JM'><JM'|$ $D$ is the matrix representation of rotation matrix $R$ in basis ${|JM>}$. The rotated state is expanded in terms of basis ${|JM>}$ with coefficient $D$ in terms of rotation matrix.

1

The angular momentum is conserved in central force motion (like what we have in the case of Earth-Moon system). In such a case, the force $\vec{F}$ and the radius vector $\vec{r}$ are parallel so that the resultant torque is zero $$\vec{\tau}=\vec{r}\times\vec{F}=o$$ This means the angular momentum ($\vec{L}$) of the of the Moon about the center is a ...

1

It's worth thinking about why the tidal bulges1 are not lined up with the line between the bodies (which is where you would naively expect them to) and then thinking about how that affects the gravitational interaction between them. Because the moon takes about one month to orbit and the Earth takes one day to turn, the naive location of the tidal bulges ...

1

The error lies in the theoretical confusion between forces/torques and energy. The kinetic energy is linked to the motion generated by forces and torques which are the causes of the motion itself. Understanding the energy value in the two situations results impossible without knowing the time course of the applied force, being the energy, and so the work, ...

1

The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation. When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ ...

1

So the definition of angular momentum should be this: "Angular momentum is the product of the angular velocity of the body or system and its moment of inertia with respect to the rotation axis, and that is directed along the rotation axis". That's not a useful definition at all, because (i) it does not specify what this "moment of inertia" thing is, and ...

1

I) For the record, here is the operator calculation that OP wants to avoid. The benefit of the calculation is that the operators are not sandwich with any bra/ket representation, and hence we do not have to worry about whether the bra/ket representation is faithful. Let us put $\hbar=1$ for simplicity. The starting point is the CCR  [x^i, ...

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