New answers tagged

0

It can be easily seen using the Heisenberg picture. Take a Hermitian operator $A$ that commutes with the Hamiltonian $H$. Remember that the eigenvalues of $A$ are observables. Then $A$ also commutes with the time evolution operator $U(t) = e^{-i H t}$. $[A,U(t)] = 0 \quad \rightarrow AU(t) = U(t)A \rightarrow A = U(t)AU(t)^\dagger $ So the operator $A$ ...


0

If you have an eigenstate of an operator $A$: $$A|Ψ(t=0)>=a|Ψ(t=0)>$$ Then apply $A$ to the time evolution of the wavefunction: $$A|Ψ(t)>=Ae^{-iHt}|Ψ(t=0)>=e^{-iHt}A|Ψ(t=0)>=e^{-iHt}A|Ψ(t=0)>=a|Ψ(t)>$$ Note that the second to last equality is only true if $[A, H]=0$. So the eigenstate of $A$ remains an eigenstate of $A$! This applies ...


2

A conserved quantity is one that commutes with the Hamiltonian for the simple reason that $[A,H] = 0$ implies $$ \frac{\mathrm{d}}{\mathrm{d}t} A = 0$$ in the Heisenberg picture. Another way to see that commuting with the Hamiltonian means conservation is to consider that the time evolution operator $U(t) = \exp(-\mathrm{i}Ht)$ is just the exponential of ...


-1

Do electrons in an atom always have the same 'direction'? No. They can have different 'directions'. Note the wikipedia atomic orbitals article which says an atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. Later on, the article says the electrons do not orbit the ...


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Modern high-end bowling balls are extremely non-homogeneous in construction. If you look around at vendor sites, you'll see that there's an off-center "dumbell" of different density from the body of the sphere. All sorts of eccentric motion can occur as a result.


3

Yes, quantum mechanics allows you to speak of clockwise or anti-clockwise motion, but it comes with the usual caveats of quantum mechanics. The tool that tells clockwise from anti-clockwise motion is the angular momentum. Motion is anti-clockwise around an axis, say the $z$-axis, if the component of angular momentum along that axis is positive. There is an ...


18

As you intuit, it is indeed pretty hard to define a sense of "direction" for an atomic electron within quantum mechanics when the electron doesn't have an orbit but it is instead some fuzzy ball of probability, but it is doable and in fact it is one of the central constructs in atomic physics. What ends up mattering is angular momentum, i.e. how much the ...


0

Invariably, a good measurement involves comparing something "known" with something "unknown". For example, you put an unknown length next to a ruler. In your case, you have an "unknown" thrust resulting in some acceleration. You could tie a string to your bottle car, run the string over a pulley, and hang a small weight from it. Measure the acceleration for ...


1

We know that $\psi_{l,m}$ satisfies, for each $l$ and $m$, the equations $$L^2\psi_{l,m}(r,\theta,\phi)=l(l+1)\hbar^2\psi_{l,m}(r,\theta,\phi),$$ $$L_z\psi_{l,m}(r,\theta,\phi)=m\hbar \psi_{l,m}(r,\theta,\phi).$$ But we also know that, by definition, $Y_{l, m}(\theta, \phi)$ satisfy the same equations. Then, unless the eigenvalue equations for $Y_{l,m}$ ...


0

It seems Wikepedias article "Two-body problem in general relativity" adresses the subject in detail. Obviously things are very much more complicated than I imagened as it involves GR and Einsteins field equations to which there are no closed form general solutions. To sign off on the matter I have found the following supposedly correct answers to my 7 ...


1

Pouring anything into the cup adds mass but doesn't add angular momentum. Conservation (plus friction between fluid and cinnamon) of angular momentum requires the new total mass to rotate slower than the original contents.


0

It is straightforward matrix multiplication and the operators enter through from either side. $exp(iS_zϕ/ℏ)$ would act from the left on the kets while $exp(−iS_zϕ/ℏ)$ would act from the right on the bras. You can see it more clearly if you do it explicitly. Use the basis $|+⟩$ and $|−⟩$ to represent the operators as matrices $|+⟩$ = ...


2

You could try measuring the effects of the Lense-Thirring effect. It is an example of frame dragging. Essentially, an object that is orbiting near a massive object that is also rotating will have its axis undergo a change in its orientation. There are two problems here: The rotating object must be large. The rotating object must not be rotating slowly. ...


3

One can use the Doppler effect, which will shift spectral lines to the red at the side the rotates away from us and towards the blue on the side that rotates towards us. This is being used by astronomers who measure "rotational broadening" on stars which can not be resolved in telescopes. In that case it's all about measuring the rate of rotation, of course, ...


0

To hit a rigid body such that it rotates about a specified point, you need to hit it at the instant center of percussion. If the pivot point is a distance $c$ from the center of mass, then the percussion center is located a distance $$\ell = \frac{I_{cm}}{m c} = \frac{\kappa^2}{c} $$ away from the center of mass. Here, $I_{cm}$ is the mass moment of inertia ...


1

If a ball of say radius $R$ rolls without slipping it has both linear ($p$) and rotational momentum ($L$): $$p=mv$$ $$L=I\omega$$ Where $m$ is mass of ball, $I$ is inertial moment of ball, $v$ is translational (linear) speed and $\omega$ is angular speed. For rolling without slipping the following condition also holds: $$v=\omega R$$ The ball will have ...


0

If we observe the path of the ball as radius reduces, we can see that the path followed is not circular. So between the radius and tangent, angle formed will not be 90 degrees.. This creates a component of tension along the path of the ball thus increasing its velocity


3

The picture of a black hole being like a huge vacuum is pretty misleading, not to say wrong. The gravitational field around the black hole is exactly the same as around any other object of the same mass. Like any other object, it's perfectly possible to orbit a black hole, just like we orbit the Sun and don't fall in. In fact, if you were to magically ...


0

You've the answer in your statement: ...black holes are giant vacuums that will absorb/consume the area around it. The key is that the area around it is a few times the event horizon of the black hole and we are very far from the event horizon of the supermassive black hole at the center of the Milky Way. The event horizon for Sag A* (the name of the ...


1

There is a pattern. If you fix the eigenbasis of $S_z$ you have for any up $+$ or down $-$ eigenstate in $\vec{u}$ direction: $$ \vec{u}=\sin\left(\theta\right) \cos \left(\phi\right)\vec{x}+\sin\left(\theta\right) \sin \left(\phi\right)\vec{y}+\cos\left(\theta\right)\vec{z} $$ then $$ S_u=\sin\left(\theta\right) \cos ...


0

The left hand diagram is in the plane of the couple produced by the weight of the rod and disc and the reaction force at the pivot point. The right hand diagram is my attempt at a 3D schematic. The key to understanding what is going on is first to realise that the direction of the external torque $\tau$ is at right angles to the direction of the ...


1

The impulse is the change in total momentum of the body of mass. The momentum $P$ of a rigid body is the product of the velocity of the centre of mass $v$ and total mass $M$. A free rigid body's centre of mass translates at constant velocity, while the body itself performs a rotation about the centre of mass. Thus, after the impulse $J$ has been applied, ...


1

"Say we have two spinning spheres, each spinning about an axis going through their own centre of mass. I have highlighted a very important statement in your question. If you have a body rotating about its centre of mass then if the centre of mass does not move the angular momentum of the body will be the same about any point. So the angular momentum of ...


0

Intuitively the motion of a rigid body is split between, translation of the center of mass, and rotation about the center of mass. The intrinsic inertia about those two motions are the mass and the mass moment of inertia. To rotate a body about any other point away from the center of mass means that the center of mass has to translate in addition to the ...


1

Yes you have to take the frictional force F into account so this exerts a negative impulse moment RFdt on the right hand disk with radius R as well as rFdt on the left hand disk with radius r. So the delta of the rotational moments have the ratio r/R. Assuming the left hand disk spins with rim velocity v0 before impact and after impact they have equal rim ...


1

In order for the spinning disc to set the stationary one in motion, friction forces need to act between the contacting surfaces: That requires a Normal force $F_N$ to act between them. Using the kinetic friction coefficient $\mu_k$ we can then state: $$F_F=\mu_k F_N$$ The friction force on the $m,r$ disc causes angular acceleration $\alpha=\frac{d ...


1

Define $D=\{(x,y)\in\mathbb R^2:\|(x,y)\|\leq R\}$ as the disk of interest. There are two spaces of interest here: the space of square-integrable functions on $D$, $L_2(D)$, and the space of such functions with Dirichlet boundary conditions, $\mathcal H=\{\psi\in L_2(D):\psi(p)=0\:\forall p\in \partial D\}$. You're interested in the hamiltonian ...


2

We generalize the result from the second answer you linked: $$\vec{L}=\sum_i \left(\vec{r}'+\vec{{r}_i}' \right)\times \vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\sum_i \vec{{r}_i}'\times\vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\vec L' $$ Now $\vec{r}'$ is some arbitrary vector, not just the vector to the center of mass. Thus when the system is moving ...


1

When you drop a stationary disc onto a rotating one there must be a time when there is relative motion between the discs as you cannot have an infinite acceleration. If there is no friction then nothing much happens and the spinning disc carries on spinning and the other disc just sits still on top of it. To get an interaction between the discs you need ...


1

yes i think your assumption is correct based on the fact that in equilibrium the discs shouldn't slip on each other. but then you need to change your equation for angular momentum. use parallel axis theorem to write down the moment of inertia for the stationary disc to get the answer in the equation. since you have chosen your axis to be passing through the ...


4

We only need worry about angular momentum when calculating other angular behaviors. In your case, you want to find the linear speed (velocity but without worrying about direction) of each ball. The speed of a point on a circle is the angular speed in radians/second multiplied by the radius of the circle, $2\pi r$ (to get to circumferences traveled per ...


-1

As L=m*(r^2)*w m= mass, r= radius, w is angular velocity case 1 L= 2*(5^2)*10 = 500 case 2 L= 2*(2^2)*25 = 200 now off course one with larger momentum will have larger inertia when stopping hence case 1. so no real need to go in the linear scenario.


1

Since you know about $SU(2)$ characters, this is a doable exercise. Let me remind you that the character of spin-$j$ is $$\chi_j(t) = \sum_{m=-j}^j e^{2imt} = \frac{\sin (2j+1)t}{\sin t}.$$ The crucial property is that the character of a tensor product is the product of characters, i.e. $\chi_{j \otimes j'}(t) = \chi_j(t) \chi_{j'}(t).$ In your case, ...


1

I mean, if you believe what you've just written, $$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1)$$ Then I think you have already proven it, with a little more formal calculus. First rewrite that expression above as $\Delta \theta(t)=\Delta \theta(t')$, with the prime indicating the other interval. If $$t_2-t_1=t'_2-t'_1,$$ Then just set $\Delta ...


0

There are two kinds of operators which have an intuitive meaning when you apply them to a state: Evolution operators like $e^{-i\hat H\Delta t/\hbar}$ simply take the state at time $t$ and give you the state at $t+\Delta t$ Projectors tell you what the state will be after you take a measurement. For example, supose you measure an observable $\hat A$ on ...


1

The intuitive explanation is that an eigenstate of $\hat L_x$ with eigenvalue $m_x=0$ is invariant by rotation around the $x$ axis (by definition), but not by rotations around any other axis. Thus it cannot be an eigenstate of $\hat L_z$ with eigenvalue $m_z=0$. See for example this picture of the eigenfunctions of $\hat L{}^2$ and $\hat L_x$ :


3

When you say that $L_x$ and $L_y$ vanish for a point confined to move in the plane $z=0$, you mean that the the solution $\vec{x}=\vec{x}(t)$, $\vec{p}=\vec{p}(t)$ describes a curve in the given plane with tangent vector parallel to that plane. So that, exactly along that curve, $$L_x(\vec{x}(t),\vec{p}(t))= L_y(\vec{x}(t),\vec{p}(t))=0\quad \forall t \in ...


3

Refs. 1 and 2 only discuss a generic/bulk notion of a constant of motion in phase space; not any refinement thereof restricted to a subsurface of phase space. Let us here carefully explain the generic/bulk notion of a constant of motion used in Refs. 1 and 2. Definition 1: An on-shell constant of motion $F(q,p,t)$ satisfies $$\tag{1} ...


2

The torque exerted by $\vec B$ is perpendicular to it, so the $z$ component of angular momentum is conserved.


1

This is not my answer, it's one of the answers you can find here Is there a reason why the spin of particles is integer or half integer instead of even and odd? I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers The "spin" tells us how the wavefunction ...


-4

I studied the subject for 5 years and came to the conclusion that in average the toppling movement of the earth occurs each 12,000 years in average. The mass of the meteorite hitting the earth must be equalor more then 10 12th kg See for details in www.couldthesunriseinthewest.com/- Johan Leupen



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