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I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by $$... 3 The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by:$$D_t = d_t + \omega(t)\times\tag{1}where D_t is the "total" derivative, i.e. the time derivative in the inertial frame, d_t is the time derivative in the frame fixed to the body. Since there there are no torques ... 0 I would do a hand analysis at first. You have two torques on the arrow. One is the drag torque, which is maximum when the arrow is transverse to the orbital velocity, as in your initial condition. The second is gravity gradient torque, which will be maximum when the arrow is horizontal. Compute each of these for your arrow. If one is much greater than ... -1 Oh, absolutely. With a Superman-like body you can also 1) Lift an entire mountain in two hands without the rock disintegrating under the local pressure, and without the mountain falling apart due to preexisting weaknesses in the rock, 2) Travel faster than light 3) Travel faster than light and travel backwards in time 4) Hear sounds so faint that they ... -1 I believe you could overcome gravity (in the atmosphere) in the same way as birds do - you just need enough muscles to do that. 0 By definition, both torque and angular momentum are relative to a point. So when you say \tau=r \times f and L=r \times p, the vector r originates from some fixed point. In light of these definitions, you can also easily prove that dL/dt = \tau and various other useful properties. In classical mechanics, these are always true. One very useful ... 1 Torque is the angular force (the moment) - i.e. \tau = r \times F, not r \times p, which is angular momentum. When you are dealing with a rotating or rotational system (like a disc on an axis), then when the torque and angular momentum are always aligned along that axis, it is much easier to use the magnitude of the torque and the magnitude of the ... 2 The answer from WetSavannaAnimal is much more comprehensive (and better) than my own will be. But just in case this helps you, I will have a go. No offence intended, I have a feeling you might be getting the spin axis and orbital axis of the electron mixed up. Just a quick analogy with the earth, it's tilted axis precesses, that means it describes a ... 1 Larmor precession is the steady rotation of the direction of a magnetic moment of a particle about a magnetic field I'm pretty sure the word is simply meant to be an analogy with the precession of a spinning rigid body's angular momentum when the body is acted on by a torque that is not aligned with the angular momentum. The reason for the analogy is that ... 0 If the electron is confined to the x-y-plane, it's z-position is fixed, i.e. certain, and hence the z-momentum infinitely uncertain by the uncertainty relation. That paragraph is trying to say that, if the magnitude of \vec L is larger than L_z, then \vec L is not fixed, and if angular momentum is not fixed, i.e. conserved, then the motion does ... 0 If the particle is known to be in the xy-plane, then \Delta z = 0, and so (by the uncertainty principle) \Delta p_z = \infty. Roughly speaking, this means that there's a good chance that the particle's momentum would be greater than "escape momentum", and thus that it could escape from the hydrogen atom. (I'm kind of dubious about this argument ... 2 As the other answers have said, the true reason for this result is that the angular momentum of a quantum system cannot really be thought of as a classical vector with a magnitude and a direction, and well-defined values on all three components simultaneously. What Beiser is trying to do is to go against this direction as far as possible, and see how ... 1 Honestly, it's because the ideas of "parallel" and "antiparallel" and "perpendicular" don't quite apply to quantum angular momentum. The angular momentum of a quantum particle is not really a vector in the sense you're probably used to thinking about it. What I mean is, it's not really something that points in a direction and has a magnitude. It's more ... 1 I think the statement is nonsensical. In quantum mechanics the vector \mathbf L is not a world-vector. Its components are not numbers, they are operators. However, the expectation value of \mathbf L in a given state, \langle \mathbf L \rangle is a world-vector and it can be aligned with any given axis. (The magnetic field is a red herring.) It doesn't ... 1 I think you could work it like this: X_+ ={1 \over \sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix}) =a (\begin{matrix} 1 \\ 0 \end{matrix} ) +b(\begin{matrix} 0 \\ 1 \end{matrix} ) .where X_+ is the eigenvector on the positive axon of S_x Solve and find a,b and there you are. Note also that you can write a general spinor as (\begin{matrix} cos\theta ... 1 Are you sure that's what the book is asking you to find? \hbar/2 is the eigenvalue of the S_{x} operator corresponding to spin up, but it is not part of the state vector. If the question is really asking you to express the \mid S_{x};+\rangle ket in the S_{z} basis, then you're nearly correct, just a minor sign error:\mid S_{x};+\rangle = ...

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Never seen that before, so I just tried it. Cool. I believe that the membrane between the yolk and the white is elastic, so when you first, gently, give the egg a little angular momentum, you are only spinning the white. As the yolk catches up the effective moment of inertia drops, and conservation of momentum therefor implies a higher angular velocity.

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I guess mark this question as answered, though not fully. See the comments posted on the question itself.

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If you have an observable, that is like a symmetric matrix. So its eigenvectors span the whole space so you can form a basis for the whole space. So the span of the eigenvectors really is the whole space. What about eigenspaces? You did have to use all the eigenvectors to get the whole space, not just ones that share the same eigenvalue. If you only use ...

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Since the unfractured disk is assumed (it is so assumed, right?) to be of uniform density, it has a center of mass at the geometric center of the disk. Then, since the center of mass of two distinct masses lies on a line connecting the two, the CMs of the two fragments must lie on a diameter of the disk. Not only that, if the radial distance of the CM of ...

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That's correct so long as the rotation axis passes through the centers of mass of both objects. In general, the moment of inertia about a fixed axis (the $z$-axis, say) will be something like $$I = \int_\text{object} \rho (x^2 + y^2) \,dV$$ But if we can split this integral up into two disjoint volumes (a cylinder and a sphere, say), we will have $$I = ... 1 I don't know your textbook, but I might have understood the point. Considering a \frac{1}{2}-spin particle (for example) you should know that a suitable Hilbert space to represent the system is \mathbb{C}^2. Moreover Pauli matrices \sigma_i  for i = x,y,z toghether with the identity matrix form a base for the vector space of linear operators ... 1 Both negative sign and positive sign are correct. When you make an infinitesimal rotation with angle d\phi about the z-axis, then both two following representations for transformed coordinates are true:$$ \left\{ \begin{array}{ll} x'=x-d\phi y \\ y'=y+d\phi x \end{array} \right. $$and$$ \left\{ \begin{array}{ll} x'=x+d\phi y \\ y'=y-d\phi x \end{array} ...

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The answers to all three of your questions are "there is no effect due to the rotation." A mirror moving parallel to its own surface does not cause any Doppler shift in the reflected light, or any Doppler aberration in the angle of reflection. This is an exercise in Zangwill's Modern Electrodynamics (Problem 22.21, to be precise), and can be proven (rather ...

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The spin of a quantum field is related to the representation of the Lorentz group they transform under: scalar fields transform under the trivial representation, spinors transform under the spinorial representation, gauge bosons under the vectorial representation, gravitons (if they exist) under the second-rank tensorial representation... If you restrict to ...

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Elaborating on ACruiosMind's comment, assume that the matrices $A$ and $B$ are defined the following way: $$A=\begin{pmatrix} 1 & 2 \\ 5 & 4 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$ Notice that the eigenvectors of $A$ are \begin{pmatrix} 1 \\ 5/2 \end{pmatrix} \quad \text {and} \quad ...

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Although $[ \hat{L_{x}}, \hat{L_{y}}] \phi_{l, m_{l}} = 0$, $\phi_{l, m_{l}}$ is neither $\hat{L_{x}}$'s nor $\hat{L_{y}}$'s eigenstate.

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To answer your questions: Yes, the rod will start to rotate; and if the force continues to act in the same direction and the wheel rotates for long enough, then the torque will change. However, we can take the limit where a very large force acts for a very short time, in such a way that their product is constant. (That is, instead of 120 N on the CM for ...

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In any situation, momentum (linear or rotational) is conserved. So if there are no external forces, you can use that. But energy is only conserved in certain situations (such as elastic collisions). The deformation of the door by the bullet is inelastic. You can consider it similar to energy losses from friction. In this case, the (mechanical) energy in ...

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