# Tag Info

0

This question of the causation of gyroscopic torque (and its magnitude and direction) troubled me all my life until I had a 'Eureka' moment about 20 years ago. I have written up the explanation of this phenomenon on my website, www.newtontime.com, using nothing more than Newton's Laws of Motion and without any need for 'fancy' mathematics. The analysis is ...

2

One way to understand it is to recognize that for the spherical harmonic $|l,m\rangle$ with $l=0$ (and obviously $m=0$), we have $\hat L_i|0,0\rangle=0$, where $\hat L_i$ is the angular momentum operator in the direction $i=x,y,z$. It is obvious for $\hat L_z$, which eigenvalue is $m=0$, and can be verified for the other two. Then, the rotation operator ...

1

Suppose that there existed a spherically symmetrical wavefunction $\psi({\bf r})=f(r)$ for which $l\neq0$. This cannot be, for if we calculate $\langle \psi | L^2 | \psi \rangle$ we will always get zero, as each term in $L^2$ has derivatives with respect to $\theta$ and $\phi$. Conceptually speaking, a spherically symmetric state gives the electron the ...

0

If C is the center of mass of the ball, and A is the contact point then the velocity of the ball at the contact point is $$\vec{v}_A = \vec{v}_C + \vec\omega \times (\vec{r}_A-\vec{r}_C)$$ If you have infinite friction then you have no slipping which means $\vec{v}_A =0$. If there was slipping then only the speed in vertical direction should be zero ...

0

For a system of point particles, the definition $$\vec{L}_i=\vec{r}_i\times\vec{p}_i$$ is always true; it's just a definition. I see no reason why that won't work here. The only choice you have to make is where to measure the position vectors $\vec{r}_i$ from. A particularly convenient position from which to measure $\vec{r}_i$ is the rotation axis. One ...

2

Here is the diagram you are discussing: It seems you are worried by the angular momentum carried by the W+. The W+ is a virtual particle in this reaction. In virtual paths the particle is off mass shell and its mass is unphysical, and angular momentum as a part of a four vector will be a complicated function also having unphysical measure, so ...

1

The W is massive so can be in the spin 0 state (or $s=-1, 0, 1$ in general). The photon is massless so does not have this "longitudinal" polarization. For the massive vector boson, the relevant symmetry group is the little group $SO(3)$, and for the photon it is $SO(2,1)$.

2

It is not immediately obvious, but the block has calculable angular momentum at the point just before impact. the block has velocity $v$ tangential to the disk's center of rotation which is a distance $r$ away., and so has angular velocity $\omega=v/r$. the block also has calculable moment of inertia around that center, $I=mr^2$. Then, it is simply ...

0

I suppose you know the mass and extent of the disk. Let's just ignore the mass of the stick. (We don't have to do that, but it makes everything simpler). We can then just use conservation of energy: You can calculate the kinetic energy of the block. Once it stops, this energy will be transferred completely to the rotational energy of the disk. Using the ...

3

Here's a motivation for where the inertia tensor $I=(I_{ij})$ (and by extension moments of inertia) comes from. It's a quantity that's analogous to mass for rotational motion in the sense that the kinetic energy of a rotating object is essentially proportional to the inertia tensor times the square of the body's angular velocity. More precisely ...

3

I think you are confussed between mass moment of inertia and area moment of inertia. The first is an equivalent of mass in angular direction and is defined as $\int_V{r^2\rho dV}$. An angular equivalent of $F=ma$ is: $$\tau=I\alpha$$ where $\tau$ is torque (angular equivalent of force, with units $[Nm]$), $I$ is mass moment of inertia (angular equivalent of ...

2

The angular momentum of a single particle with mass $m$ in motion about an axis, with angular speed $\omega$, a distance $r$ from the axis, is $L = r (m v) = m r^2 \omega$. When we consider an extended body, the sum up the contribution ($m r^2$) from each particle in motion inside the body, and this is the moment of inertia. More generally, \begin{align} ... 1 Your solution looks fine to me. Yes: the angular momentum is preserved in the horizontal plane (the weight is vertical and the reaction of the sphere surface is a central force) so your first relation is fine, just remember that \theta is not the vertical angle, but lies on the plane tangent to the sphere at point B. There are two kinds of rotational ... 0 First, you are right: in this problem angular momentum around the vertical axis is conserved. This is because all forces acting on the particle have no azimuthal component. Note, that even if we assume that the particle is rolling on the bowl surface without slippage, the angular momentum of its own rotation would be of negligible compared with the angular ... 0 I do not have a solution, just some steps to get there. I have parametrized the problem with spherical coordinates, \varphi is the azimuthal angle (around the hoop), \psi is the nutation angle (drop from horizontal plane) for a position vector \vec{r} = \begin{pmatrix} r \cos \varphi \cos \psi \\ -r \sin \psi \\ -r \sin \varphi \cos \psi ...

0

The angular momentum of a massive sun may cause the freely falling spaceship to start spinning in the direction of the sun's angular momentum for an effect of frame dragging. You can take a look at the Kerr metric which describes the behaviour of the spacetime near a massive spinning object. If you're not familiar with general relativity it could be ...

0

The only way for a falling object to be made to rotate and translate is if there was a separate force causing this rotation. In an atmosphere this is a net force on one side of the craft whose surface area (and therefore drag) is the highest, causing this part of the craft to rotate away from the direction the entire craft is translating. Essentially ...

1

Since the operator $L^2$ yielded $6\hbar^2$, one can deduce that $l=2$ after measurement. In general, the only possible measurements of one of the $L_i$ components are from $-l\hbar$ to $+l\hbar$ in integer steps. If you're comfortable with this for $L_Z$, note that none of the angular momentum operators $L_i$ are fundamentally more special than the other ...

3

If $L^2$ is $6\hslash^2$ this means that the quantum number $l$ is $2$, hence the quantum number $m$ goes from $-2$ to $+2$, and the possible values of $L_x$ are $-2\hslash, -\hslash, 0, \hslash, 2\hslash$ Conventionally $L_z$ is used instead than $L_x$, but what applies to $L_z$ must apply also to $L_x$ and $L_y$: there are no preferential directions in ...

0

If you know the motion of a point A on a rigid body, with linear velocity $\vec{v}_A$ and angular velocity $\vec{\omega}$ then the formulas below will give you the linear momentum of the rigid body, and the angular momentum about point A. If the body is pivoting about A then $\vec{v}_A=0$, otherwise in the general case $\vec{v}_A \neq 0$. Linear momentum ...

4

The moment of inertia tensor is not constant in the external reference frame (http://en.wikipedia.org/wiki/Precession#Torque-free )

0

Here there is an error in your angular momentum calculation.$L=mvr$ is valid only if that is a point size particle.Here it is a cylinder,so angular momentum $L$ about your point of contact should be $$L=L_{cm} +I_{cm}\omega$$ where $L_{cm}=$angular momentum of CM about your orgin $$and$$ $I_{cm}=$moment of inertia about CM So applying angular momentum ...

3

Well, the angular momentum of a rigid body is equal to the sum of the angular momentum of the body around it's center of mass, plus the angular momentum of the center of mass. Having said that, suppose the rod is rotating about one end (I imagine a pendulum motion; correct me if I'm wrong), you can calculate the angular momentum by $L = I \omega$ if you ...

0

It looks like the friction force will decelerate the cylinder until (at some time t) there is no sliding, just pure rolling on the surface. Until then the friction force will equal $\mu mg$ (the sign depends on the direction of the axis). Therefore, $\mu m g t=m(v_0-v_f)$, where $v_f$ is the final velocity of the center of mass. On the other hand, we can ...

0

Explaining conservation of angular momentum is a good idea. I would like to add another explanation that is probably at the 9-year-old level. Imagine the top from, well... the top. Let's say it is going clockwise. Suppose the mass begins to tip to the right. In a short amount of time, the rightmost part of the top would have experienced a downward ...

0

I think the best way to understand spin is to look at it from a complete abstract point of view: do not try to find classical analogs to it. What you find when you perform experiments with electrons is that they interact with a magnetic fields as if they had an intrinsic magnetic moment with two possible values: a positive one and a negative one (loosely ...

1

It is true, you have to "rotate twice" (or by $720^\circ$) to recover the original state. You can prove this in the following way. Let $$|a\rangle=|+\rangle\langle+|a\rangle+|-\rangle\langle -|a\rangle$$ be a general ket. Consider now a rotation by a finite angle $\theta$ around the $z$ axis. I remind here that if a ket of a spin $1/2$ system is ...

-1

Along time ago (1960's) I saw on TV a show where a scientist was explaining the working of a spinning top in front of an audience of children. He had a roundabout with a seat in the middle on which he sat a young boy. The boy held a metal bar with a wheel on its end (Quite heavy!) The scientist spun the wheel and rotated the roundabout. I cant remember ...

3

First, to check the decomposition of a product of representations, you may use, as noticed by user26143, the tool Form Interfact to Lie. Choose Tensor product decomposition, then choose $A_1$ for $SU(2)$, or $A_2$ for $SU(3)$,click sur "Proceed", type your representation, and click on "Start" to have the decomposition. The name of the representations in this ...

1

Because $SU(2)$ is not the same as $SU(3)$ ? The closest analog to your $SU(3)$ case would be two doublets: $\mathbf{2} \otimes \mathbf{2} = \mathbf{1} \oplus \mathbf{3}$, as you already know :) Afaik, $SU(3)$ has two independent $SU(2)$ subgroups, i.e., it has two "$L^2$" operators. You can still do Clebch-Gordan-style coefficients calculations but it ...

2

It is proven in Tong's QFT script http://www.damtp.cam.ac.uk/user/dt281/qft.html section 4.1. in a quite nice fashion.

2

I struggled with this one as well and once I found I have written it in LaTeX which I will copy here below. Do note that I am using slightly different conventions than P&S, however it should still work out the same. \begin{aligned} S^{\mu \nu} & = - \frac{i}{4}[\gamma^\mu,\gamma^\nu] \\& = - \frac{i}{4}(\gamma^\mu \gamma^\nu - ...

1

A force affects the motion of the center of mass only (call it point C). The rotational motion is defined by the total torque applied on the center of mass. If the applied force $(F_x,F_y)$ is at location $(r_x, r_y)$ relative to the center of mass then  \begin{aligned} F_x &= m \ddot{x}_C \\ F_y &= m \ddot{y}_C \\ r_x F_y - r_y F_x ...

1

When you hit anything with a force, it causes an acceleration in the direction of the force. So if you hit the square at the corner like so: It would start moving upwards. However, it also experiences a torque about its center of mass. This torque can be calculated as the force multiplied by the perpendicular distance between the center of mass and the ...

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