In reading a postulate on angular-momentum coupling, the author made this statement:     In Q.M, the angular momentum cannot be complete aligned with the $z$-axis, otherwise its $x$ and $y$-components would be zero and be known simultaneously.   I don't fully understand this statement.  Is there a better way to gain insight into this statement?
123891
This is the excerpt from my book, Arthur Beiser's Concepts of Modern Physics:      An atom with a certain value of $\displaystyle{m_l}$ will assume the corresponding orientation of its angular momentum $\mathbf{L}$ relative to an external magnetic field if it finds itself in such a field. However, we note that $\mathbf{L}$ can never be aligned exactly parallel or anti-parallel to $\mathbf{B}$ because $L_z$ is always smaller than $\sqrt{l(l + 1)\hbar}$ of the total angular momentum.   Why can't $\mathbf{L}$ be parallel to the applied magnetic field? I am not getting the reason showed by the author. Can anyone help me conceive the reason stated above?
45316
It is commonly asserted that no consistent, interacting quantum field theory can be constructed with fields that have spin greater than 2 (possibly with some allusion to renormalization). I've also seen (see Bailin and Love, Supersymmetry) that we cannot have helicity greater than 1, absenting gravity. I am yet to see an explanation as to why this is the case; so can anyone help?
51815
I am a bit confused about the kinetic energy of a rotating object when the axis of rotation does not pass through the center of mass. For example, consider a thin hoop of mass $m$ and radius $R$ that is hanging from a pivot and is free to rotate about that pivot. If the hoop were rotating about it's center of mass, the kinetic energy would just be  $$K = \frac{1}{2}I\omega^2 = \frac{1}{2} m R^2 \omega^2,$$ because the center of mass of the hoop is not moving (i.e. the translational kinetic energy is zero). However, when the hoop is rotating about the pivot, the center of mass is no longer stationary. Is this compensated for in the change in the moment of inertia, or does this translational piece need to be added? In other words, do we have $$ K_{tot}= K_{trans}+K_{rot}=\frac{1}{2} m v^2+\frac{1}{2}I\omega^2=\frac{1}{2}m(R\omega)^2+\frac{1}{2}(2mR^2)\omega^2 = \frac{3}{2} m R^2 \omega^2 $$ or $$ K_{tot}= K_{trans}+K_{rot}=0+\frac{1}{2}I\omega^2=\frac{1}{2}(2mR^2)\omega^2 = m R^2 \omega^2 $$ I am fairly certain that the latter is the correct answer but I just wanted to make sure I am thinking about this correctly.
33170
is one of those surprisingly clear results of mathematical calculations, for which I am inclined to think that some kind of intuitive understanding should or must be possible. However I don't know of any, do you?  *Independence of time $\leftrightarrow$ energy conservation. *Independence of position $\leftrightarrow$ momentum conservation. *Independence of direction $\leftrightarrow$ angular momentum conservation.  I know that the mathematics leads in the direction of Lie-algebra and such but I would like to discuss whether this theorem can be understood from a non-mathematical point of view also.
46296
In quantum mechanics, when two observables commute, it implies that the two can be measured simultaneously without perturbing each other's measurement results. Or in other words, the uncertainty in their measurements are not coupled.   But in classical mechanics, in an analogous way we have the Poisson brackets instead, where when two functions/variables of the systems have a vanishing Poisson bracket with each other, we say they are in "involution".  I understand that when a variable $j$ has a vanishing Poisson bracket with the Hamiltonian of the system, means that $j$ is a constant of motion, conserved in time. But for the general case:      If the Poisson bracket of $f$ and $g$ vanishes ($\{f,g\} = 0$), then $f$ and $g$ are said to be in involution.    What does the above mean physically? Is it in some sense related to measurement results as it is with commutators in QM, meaning two variables in involution are completely independent of each other? Finally, is there a simple classical example, where one can see 2 observables are in involution, but 2 other (same system) may not be (i.e. counter-example for involution)? I think it would help better understand what in "involution" means.
84011
I've heard it from two teachers and saw a task with a solution based on this assumption:     If the net torque is zero when measured from one frame of reference, it is equal to zero   in all other frames of reference   This actually seems clear, especially if you put it to simple words like:     If a body doesn't rotate from one point of view, it won't start rotating if you look at it    from another point of view   Still, I wanted to test it and made such an image:   I calculated the net torque measured from point A (I'll skip the designations such as $m_1, r_1, m_2$ etc. and get to the value):  $$1m\cdot1N + 5m\cdot1N - 4m\cdot1.5N = 0$$  Now, let's calculate the net torque from B:  $$-2m\cdot1N + 2m\cdot1N - 1m\cdot1.5N = -1.5Nm$$  From this it seems, that the body doesn't rotate around an axis in A, but does around the axis in B.
117040
So this Youtube video has been around for a while now  . Also this  may be related.  I just wanted to know if there is any standard hydrodynamic terms to explain it. Derek clearly states it is not about Bernoulli principle, as in this case the ball rotation is important.  Has any recent paper/videos discussed what happens if you do the same experiment with hydrophobic spheres? Can it be predicted with laminar flow theories or does it need turbulence?  Too many questions have been posed, but not so many responses on the internet yet.  Edit: I said Derek stated "it wasn't Coandă effect", yet what he says is that it is not "just Bernoulli's principle". Coandă effect is the typical hairdryer ping-pong ball levitation which is explained by Bernoulli's principle.   Edit2: rephrasing.  Additional comment: There seems to be two explanations "water adhere to the sphere and is ejected conserving horizontal momentum" and "Magnus effect" (ME). The later is not so obvious to me as the system is in equilibrium and there are two fluids, water and air, (does ME means air pushes the ball horizontally?).   Edit3: DESCRIPTION To avoid rot links and people that do not want to see the video, I'll try to explain the phenomena. You have a sphere and a water jet. The water jet is vertical (opposite to gravity), and touches the sphere almost tangentially. The ball not only rotates due to the water jet but also stays at constant height from the ground. The angular velocity can be defined in the direction of the cross product between the position of the contact point between the jet and the sphere (origin in the center of the sphere) and the direction of the jet. Some water drops fly away at the top of the sphere away from the jet.  Edit4: New question, would this phenomenon occur if there was no air?
14650
I am learning P&amp;S's Introduction of quantum field theory. My teacher said that if the mass of neutrino is exactly 0, then we should not observe any right-handed neutrinos and left-handed anti-neutrinos according to Weyl's theory. It is because $$ \left( i\gamma^\mu \partial_\mu - m \right) \psi = \left( \begin{array}{cc} -m &amp;  i\left(\partial_0 + \vec\sigma \cdot \vec\nabla\right) \\ i\left(\partial_0 - \vec\sigma \cdot \vec\nabla\right) &amp; -m \end{array} \right) \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right) =0. $$  If we set m=0, then we will have:  \begin{align} i\left(\partial_0 - \vec\sigma \cdot \vec\nabla\right) \psi_L &amp;= 0; \\ i\left(\partial_0 + \vec\sigma \cdot \vec\nabla\right) \psi_R &amp;= 0. \end{align}  And if these formulas can describe neutrinos we must conclude that there are two types of neutrinos.  My confusions are:   It is known that the mass of neutrino is not 0, therefore there are chances (depending on the mass) to find right-handed neutrinos and left-handed anti-neutrinos. But we don't. If I accept that there are only these two types of neutrinos (right-handed anti-neutrinos and left-handed neutrinos), this property of neutrino causes the parity violation, so parity violation must relate to angular momentum and therefore spinor space. Am I write? Are there any reasons why neutrino is so weird that is has mass but still has helicity + Or - 1. This goes against with what I have learned that only when a particle  moves at the speed of light can its spin aligned with its momentum.
94636
Suppose we have a disk of radius $r$ and mass $m$ travelling at velocity $v$. I want to calculate the instantaneous angular momentum with axis through the edge of the disc (on the circumference).   Angular momentum $= I \omega$. $I = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$ by the parallel axis theorem. $\omega = \frac{v}{r}$. Therefore, angular momentum $= \frac{3mrv}{2}$.  Alternatively, angular momentum $=p\times r= m r \times v = mrv$.  Why do these two methods differ? Which, if any, are correct?
57315
How do I prove the relation between the vectors of magnetic moment $\vec\mu$ and angular momentum $\vec L$, $$\vec\mu=\gamma\vec L$$  ?  Many text books and lecture notes about the principles of magnetism show the relation of $\mu$ and $L$ as scalars only and then just state that the relation holds also for the vectors. An example:   $$\mu=I\cdot A = \frac{q}{t}\pi r^2 = \frac{qv}{2\pi rm}m\pi r^2$$  ($I=q/t$: current, $A$: area of a loop, $q$ charge, $t=2\pi r/v$: time of 1 rotation, $v$: velocity of particle, $m$: mass )  The angular momentum is $\vec L = \vec r\times\vec p$ or $L=mrv$ and therefore $$\Rightarrow \mu = {\frac{q}{2m}} L = \gamma L.$$  Why is this also true for the vectors? Is there a general explication by classical physics without the need of quantum theory?
85623
I have read , and am still confused about the entire matter.  My understanding is that, using a local coordinate system, if you rotate a model about its X axis then its Y and Z axes experience gimbal lock, and become "locked" together.  However, my understanding is that if you rotate the X axis, then both the Y and Z axes rotate with it, and as such remain separate.  Wikipedia, as always, while a good resource is difficult to understand on this topic.  Could someone please endeavor to explain this to a person who does not have a Master's in physics or mathematics?
45177
I'm having trouble understanding how a  (not the  that is generated as a result but the material itself) work. The particles are aligned in a specific direction to give rise to force but I don't see how this alignment gives rise to "attraction" or force.
48870
Angular momentum of an object is a physical quantity that depends on the chosen point about which to calculate the angular momentum. It is often said that an object that has been thrown up in the air and is rotating, is physically rotating about the center of mass. I don't think that is what is physically happening. We chose the center of mass as the point of rotation because it is convenient mathematically (it makes the separation of translation and rotational energy easier). We could choose any other point (inside, outside the object, in motion or at rest relative to the object) and calculate the rotation about that arbitrary point. It is not a physical fact that the free object rotates about the center of mass. Even an object that is constrained to rotate about a fixed axis, we could describe the rotation about any point, not necessarily points on the fixed, constrained axis.
14578
The origin of spin is some what a puzzle to me, everything spin from galaxies to planets to weather to electrons.  Where has all the angular momentum come from? Why is it so natural?  I was also thinking do photons spin? we always think of the wave as a standard 2d sin wave but could this rotate in 3d? What implications would this have?   And what about spacetime how does all the spinning effect?  This has always been avoided in all lectures and classes I ever went to.
116307
Two discs mounted on different thin, lightweight rods oriented through their centres are made to rotate about their axes seperately such that the angular momentum of the two about their respective axes is same in both magnitude and direction. When both are brought in contact, they stop due to force of friction. Why is the angular momentum not conserved about their axes in this case?(Was positive before contact, but zero after the process completes, no external force present either)
46448
It is well known from Noether's Theorem how from continuous symmetries in the Lagrangian one gets a conserved charge which corresponds to linear momentum, angular momentum for translational and rotational symmetries and others.  Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians? By elementary I mean, "if this is not so, then this unreasonable thing occurs".   Of course, we can say "if we want our laws to be the same at a different point in space then linear conservation must be conserved", but can we derive mathematically the expression for the conserved quantity without using the Lagrangian?  I want to explain to a friend why they are conserved but he doesn't have the background to understand the Lagrangian formalism.
65510
I think I've a conceptual lacuna that needs to be filled, when it comes to a rigid body possessing angular velocities along more than one axes.  Here's my doubt -   Consider the following solid cone (mass $m$), with semi-vertical angle $\alpha$ and radius $r$, rotating about the Z axis with some angular velocity $\omega$. The axis of the cone, is always at a height equal to its radius, from the XY plane.     1. How is the total kinetic energy of the cone calculated?  2. How is the angular momentum of the cone calculated? (about Z axis, for example)  My doubt is not limited to the above question, which has been taken only for the sake of an example - and I believe a solution to the above question will help me understand the concept better.   While calculating the kinetic energy of a rigid body rotating about more than one axis, what exactly is done and WHY?  The translational kinetic energy is $\dfrac{mv²}{2}$, where $v$ is the velocity of the center of mass; and the rotational kinetic energy is $\dfrac{I\omega²}{2}$, where I is the moment of inertia about the center of mass (really? not sure) and $\omega$ is the angular velocity about Z axis (or is it the resultant angular velocity of the body?). Also, is the moment of inertia supposed to be about an axis through the center of mass, or about Z axis?   It's a mess, in short. Please explain the solution of the problem, and also the concept in detail. Thanks a lot!    For greater clarity as to what area I face problems in, I add a second problem too, in which the body posseses angular velocities about three axes.     You may assume the necessary moments of inertia, and dimensions for the above object. Just want to know how it's done, the approach, more than the exact answer.