I know that having a net angular momentum will contradict isotropy of the universe by preferring a specific direction. But is there any experimental data on the total angular momentum of the universe and how much it deviates from zero (perhaps according to the CM frame)?
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Suppose in the milliseconds after the big bang the cosmic egg had aquired some large angular momentum. As it expanded, keeping the momentum constant (not external forces) the rate of rotation would have slowed down, but it would never reach zero.  What implications would that have to measurments to distant supernova and CMB radiation? Do we have any experimental data that definitely rules out such as scenario? And to what confidence level?  Edit A  suggests that the universe might indeed be spinning as a whole. Anyone care to poke holes at it?  Edit 2 Even more  places limits on possible rotation.
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Are angular quantities vector?   ... It is not easy to get used to representing angular quantities as vectors. We instinctively expect that something should be moving along the direction of a vector. That is not the case here. Instead, something(the rigid body) is rotating around the direction of the vector. In the world of pure rotation,a vector defines an axis of rotation, not a direction in which something moves.   This is what my book (authored by Halliday, Walker, Resnick) says. But they mentioned with a caution that      Angular displacements cannot be treated as vectors. ....to be represented as a vector, a quantity must also obey the rules of vector addition..... Angular displacements fail this test.   Why do they fail in this test? What is the reason?
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Is there an actual definition of "direction" (that is, spatial direction) in physics, or is it just one of those terms that's left undefined? In physics textbooks it's always just taken for granted that the reader knows what it means (and it is true that just about everyone does indeed have an "intuitive" idea of what it is). But it would be more satisfying to have an concrete definition, if possible.
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I am considering the unit sphere (but an extension to one of radius $r$ would be appreciated) centered at the origin. Any coordinate system will do, though the standard angular one (with 1 radial and $n-1$ angular coordinates) would be preferable.   I know that on the 2-sphere we have $ds^2 = d\theta^2+\sin^2(\theta)d\phi^2$ (in spherical coordinates) but I'm not sure how this generalizes to $n$ dimensions.  Added note: If anything can be discovered only about the determinant of the tensor (when presented in matrix form), that would also be quite helpful.
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Problem #25 from the 2009 $F=ma$ contest:     Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs. These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii $r_1$ and $r_2$. The discs are given angular velocities of magnitudes $\omega_1$ and $\omega_2$, respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? [...] Ignore effects associated with the vertical rods.   Diagram here:     Probably conservation of angular momentum is useful here, but I am not clear how it can be applied. There are two distinct axes of rotation, one for each disc, so we cannot talk about conservation of angular momentum about one axis. If the two disks rotated about the same axes then the solution is straightforward, but this is not the case.
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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?
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In the Futurama episode "" they save the world by rotating the Earth backwards saying it shouldn't matter (which direction Earth rotates). If Earth rotated clockwise and remained in it's current counter-clockwise orbit around the Sun, would it affect anything beyond the direction the constellations move across the sky?
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It is well known that quantum mechanics and (general) relativity do not fit well. I am wondering whether it is possible to make a list of contradictions or problems between them?  E.g. relativity theory uses a space-time continuum, while quantum theory uses discrete states.  I am not merely looking for a solution or rebuttal of such opposites, more for a survey of the field out of interest.
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It is usually said that existence of discrete spacetime violates Lorentz symmetry. What quantity is used to quantify such violation? I mean could someone points a reference for a derivation that shows such analysis.  My other question is: if Lorentz symmetry is violated, does that imply space-time is discrete? or not necessarily?)
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Can we assume that most (if not all)  black holes are rotating, due to conservation of momentum?  I am excluding the micro world from this question, just thinking of the range of stars on the main sequence of the H-R diagram as they shrink over time.    Can anyone give me an example of non-rotating macro sized B.H., e.g a binary system where each star counter-rotates relative to the other and explain how they may, (if they can)  at least slow down each others's rotational velocity?
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This is really a follow up to the  question.  In the first answer to the question, Ted Bunn says:     However, the best available data seem   to indicate that the Universe is very   nearly spatially flat. This means   that, if we do live in a 3-sphere   Universe, the radius of the sphere is   very large, and the distance you'd   have to travel is much larger than the   size of the observable Universe.   My question is, can the available data be used to set a lower bound to the size of the Universe?   In other words, suppose the Universe was actually shaped like a 3 sphere. Then based on our best data, what would the "radius" of that sphere be? How much larger is that than the observable universe?
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Starting  with the Einstein-Hilbert Lagrangian  $$ L_{EH} = -\frac{1}{2}(R + 2\Lambda)$$  one can formally calculate a gravitational energy-momentum tensor   $$ T_{EH}^{\mu\nu} = -2 \frac{\delta L_{EH}}{\delta g_{\mu\nu}}$$  leading to  $$ T_{EH}^{\mu\nu} = -G_{\mu\nu} + \Lambda g_{\mu\nu} = -(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R) + \Lambda g_{\mu\nu}. $$  But then, in the paragraph below Eq. (228) on page 62 of  paper, it is said that this quantity is not a physical quantity and that it is well known that for the gravitational field no (physical) energy-momentum tensor exists.  To me personally, this fact is rather surprising than well known. So can somebody explain to me (mathematically and/or "intuitively") why there is no energy-momentum tensor for the gravitational field?
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We know that if an imaginary astronaut is in the intergalactic (no external forces) and has an initial velocity zero, then he has is no way to change the position of his center of mass. The law of momentum conservation says: $$ 0=\overrightarrow{F}_{ext}=\frac{d\overrightarrow{p}}{dt}=m\frac{d\overrightarrow{v}_{c.m.}}{dt}$$  But I don't see an immediate proof, that the astronaut can't change his orientation in the space. The proof is immediate for a rigid body (from the law of conservation of angular momentum). But the astronaut is not a rigid body.   The question is: can the astronaut after a certain sequence of motions come back to the initial position but be oriented differently (change "his angle")? If yes, then how?
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Why is orbital angular momentum not 0 when spin and linear momentum are not collinear?   Why can it be 0 when spin and linear momentum are parallel?  Like in the example of a scalar field at rest decaying into an electron and a positron via interaction $\frac{1}{2}g\phi \bar{\Psi}\Psi $   If spin is along x axis and three-momenta is along z axis, decay rate will be 0 if two spins are opposite because orbital momentum is not 0 after decaying.
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Usually, the angular frequency $\omega$ is given in $\mathrm{1/s}$. I find it more consistent to give it in $\mathrm{rad/s}$. For the angular momentum $L$ is then given in $\mathrm{rad \cdot kg \cdot m^2 / s}$.  However, the relation for torque $\tau$ says: $$ \tau \cdot t = L$$  So the torque should not be measured in $\mathrm{N \cdot m}$ but $\mathrm{rad \cdot N \cdot m}$. Would that then be completely consistent?
52873
thanks for any help.  I'm trying to show that in a 2body problem, angular momentum is conserved given that $\dfrac{dp}{dt}=\dfrac{-GMm(rv)}{r³}$, where p is momentum, t time, G gravitational constant, M mass of 1 object, m mass of the other, (rv) is the vector between them and r is the magnitude of the vector between them.  I've had a lot of attempts but I don't seem to get very far.  Where L is angular momentum, $r_1$ is position vector of mass M and $r_2$ is position vector of mass m, $p_1$ is momentum of mass M and $p_2$ is momentum of mass m.  L=$r_1Xp_1+r_2Xp_2$  If $\dfrac{dL}{dt}=0$ then dL will be conserved over time, applying product rule             $L=\dfrac{dr_1}{dt}Xp_1+r_1X\dfrac{dp_1}{dt}+\dfrac{dr_2}{dt}Xp_2+r_2X\dfrac{dp_2}{dt}$  noticing that $p_1=m\dfrac{dr_1}{dt}$ and $p_2=m\dfrac{dr_2}{dt}$ clearly $p_1$ is parallel to$ \dfrac{dr_1}{dt}$ and $p_2$ is parallel to $\dfrac{dr_2}{dt}$ and hence when cross producted with them, the result is 0.  Giving $L=r_1X\dfrac{dp_1}{dt}+r_2X\dfrac{dp_2}{dt}$  I gather I must now show that either these two terms are 0, or they are equal and     opposite. I can't see any reason why they would be 0, and (slightly guessing) it seems to    make sense to me that they are terms related to how the angular momentum of one body changes    with the other body, and I assume they must therefore be equal and opposite. However when I    do  $r_1X\dfrac{dp_1}{dt}=-r_2X\dfrac{dp_2}{dt}$      I notice $\dfrac{dp_1}{dt}=\dfrac{-GMm(rv)}{r³}$ and $\dfrac{dp_2}{dt}=\dfrac{GMm(rv)}{r³}$ (since (rv)'s direction is flipped)     Therefore $r_1X\dfrac{-GMm(rv)}{r³}=-\dfrac{r_2XGMm(rv)}{r³}$     Defining $w=\dfrac{GMm}{r³}$:  $r_1X(rv)(-w)=r_2X(rv)(-w)$     Scalar products are commutative so the (-w)'s easily cancel giving  $r_1X(rv)=r_2X(rv)$  Which then the only solutions are if all components of $r_1$ are equal to all components of      $r_2$, however angular momentum should clearly be conserved in all positions, not just when the masses are ontop of each other.  Thanks again for any help.
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As a start I assumed  which is the reciprocal of  as the frequency upper limit, but not so sure. On the other hand, is there a lower limit? Can we consider the reciprocal of our universe as a theoretical limit which is not yet limited, or just assume 0? I am just curious to know if you have any ideas with thanks.