# Tag Info

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I have received an answer but I do not understand it. It uses relativity and the Lorentz contraction. Let us assume for sake of discussion that I am standing on the equator of my intergalactic planet. Let us assume further that there is a distant galaxy which is co-planar with my intergalactic planet. At some point, when I am standing at a right angle ...

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this also true for two point mass system (each of mass m) connected by a string in which one point mass performs circular motion around the other. Then in this case is it possible to have different velocity of the point mass (which is performing circular motion around the other) in inertial and Center of mass frame of reference ? Yes, velocity will be ...

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Yes velocity of a particle is different when you shift to com frame or any other moving frame whose velocity is different from the frame initially from where you were observing the particle.

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Yes , it is possible that you get different values for inertial frame and frame of centre of mass When you solve the velocity of a point on a ball in frame of centre of mass we assume the ball is pursuing a rotational motion in the frame if reference of centre of mass because in this frame the translational motion of centre of mass and that point nullify In ...

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Other than the specified reasons in previous answers, the major reasons I choose to use the centre of mass frame of reference is because It is easier to visualize the motion of the object, as it performs pure rotation about the centre of mass. You do not need to bother about pseudo forces. You can see for yourself that torques due to all the pseudo forces ...

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Why a body does not rotate if force is applied on the centre of mass? The question is: Why does it rotate if it is not applied to the center of a mass? First we look at two rules that students of mechanical engineering (I don't know about physics) learn at university: Multiple forces applied to the same point of a solid body have the same effect as a ...

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This can be explained with a simple case - torque with zero net force. Consider a rod in free space being acted upon by two equal and opposite forces at the ends. The subsequent motion of the rod is dictated by - No net force on the COM The rod is a rigid structure What it means is individual particles of rod must move without changing their relative ...

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The rotation of a body depends upon the point at which it is hinged. So if a force applied at the center of mass (COM) then it may cause rotation about the axis (if it's not the one through COM) at which it is hinged by causing a torque to act. But what about the case when it isn't hinged at any point with the force still acting on the COM? A torque still ...

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Center of mass is a point in space, that is it doesn't have a length. Also note that a point is not a circle, it doesn't have a circumference nor area. If you replace a collection of particles with centre of mass you will end up having a point with same mass. Now just try to imagine applying a force to the point . The moment you apply a force to it , if ...

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Okay so let me give you the intuition. Now you would've heard of cases of a pendulum made of a rod instead of a string. (You can checkout this: http://hyperphysics.phy-astr.gsu.edu/hbase/penrod.html) Okay, so in this case you can clearly see that the torque is taken about the hinge. And you apply gravitational force at the centre of mass and take its torque ...

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Yes, black holes can precess. In binary consisting of two black holes, both spins and the orbital angular momentum will precess about the total angular momentum. In April, LIGO and Virgo announced the first detection of a binary black hole merger where the black hole spins were precessing: GW190412

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Let's not take into account the decrease in the pace of time caused by the gravity on (and in) Earth. As the Earth's rotation reduces with time the rotation will eventually come to a halt (even though this can be when time reaches infinity). Yes, if the earth's rotation reduces time will go faster for particles on Earth (and inside the Earth but at different ...

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For the speed at which the earth rotates, relativistic time dilation would be pretty much unnoticeable. For the satellites in the GPS system, it does have to be taken into account in order to predict accurate positions on the earth.

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Most of the iron and heavy elements in the solar system are in the Sun. They aren't all in the Sun, because the material from which the solar system formed was well-mixed and turbulent. The reason that the concentration of heavier elements is higher in the small, rocky, inner planets (the gas giants have an abundance mixture similar to the Sun), is that ...

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This question is tantamount to asking why do stellar systems, galaxies, and even planet-moon systems have a spatial extent as opposed to forming centralized blobs of decreasing density radially. The answer lies in the fact that all these objects have a rotational motion that spreads the mass outward, with the occasional accretion of heavier elements ...

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That is actually somewhat the case... The planetesimals formed in the inner Solar System were rocky planets (our terrestrial planets evolved from these), whereas the outer planets (the gas giants) in the cooler zone evolved from lighter elements such has hydrogen and helium. This is of course an incomplete answer. From exoplanets surveys we have found many ...

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Note: Your equation 2 has assumed $v_a'$ towards the right, and I will do the same for my equations. Also, I'm assuminng $d$ as the perpendicular distance between the centres of mass of the two rods (perpendicular to their lines of motion). $d=r+\frac{l}{2}cos(\alpha)$ $m_b v_a d = m_b d (v_a' - v_b') + I_b \omega_b' + I_a \omega_a'$ $m_a v_a d = m_a d (... 2 The general statement is that if$V$is rotationally symmetric about a given axis, then the component of$\vec{L}$parallel to that axis is conserved. This is easiest to demonstrate by rewriting the potential in terms of polar coordinates and showing that the torque is always perpendicular to the axis of symmetry. Alternately, this can be viewed as a ... 1 There is a simple answer to your question. First, in$\mathrm{Cl3}$(geometric algebra of 3D Euclidean vector space), the elements of the even part of the algebra are just quaternions. Second, the simplest representation of orthonormal vectors in$\mathrm{Cl3}$are the Pauli matrices. In addition, we have a geometric meaning of both the Pauli matrices and ... 1 Well, I'll take a shot at a classical explanation. Imagine a large spinning star, and a small spinning object passing close by. The principle you mention is really the "weak" equivalence principle, and it says that any free particle at a given place and time, with a given velocity, will follow the same path. In general relativity, Einstein made ... 3 The last step is incorrect. The sequence is Atoms are split into$|\uparrow \rangle$and$|\downarrow \rangle$, but only the$|\uparrow \rangle$-atoms are kept (=atoms possessing a spin which points in the$|\uparrow \rangle$direction). Atoms are split into$|\leftarrow \rangle$and$|\rightarrow \rangle$, but only the$|\leftarrow \rangle$atoms are kept.... 2 First, look at the$L=1$case: $$Y_1^1 = N(-x - iy)$$ $$Y_1^0 = \sqrt 2 Nz$$ $$Y_1^{-1} = N(x - iy)$$ where$N = \frac 1 2 \sqrt{\frac{3}{2\pi}}$. To get the eigenvalues in$x$, you need to do a coordinate change: $$(x, y, z) \rightarrow (y, z, x)$$ so in that basis (I'll call it$F): $$F_1^1 = N(-y - iz)$$ $$F_1^0 = \sqrt 2 Nx$$ F_1^{-1} = N(... 3 The problem lies in your understanding of the operator algebra and Hilbert spaces not so much on your calculus I think. Let us break it down: Can the PDE above be solved to get the general solution of the eigenfunction of Lx?(Something looks like Yml, maybe some ugly special functions) The solutions are given by linear combinations of the Y_\ell^m, which ... 1 we cannot apply it about centre of mass as external torque about centre of mass is not 0. Is my reasoning correct? Once the ball enters pure rolling friction force stops acting hence you can apply conservation of angular momentum but before that you can't. Now after starting rolling, it collides with a wall inelastically so that linear velocity becomes say ... 3 I think the physical context of your question is spin angular momentum of elementary particles (for example: an electron) and rotations of this particle in space. Let's begin with spin angular momentum \bf{S} (a vector with 3 components S_1, S_2, S_3). Physical experiments showed that the measured spin component S_j of the particle is either +\... 3 \hbar=\frac{h}{2\pi} is proportional to the Planck constant h and has the dimension of angular momentum, so the observables consist of Pauli matrices multiplied by a \hbar correspond to the angular momentum, which is not just real numbers, but have direct physical meaning. Sometimes physicists would ignore the dimension of an obervable and only ... 2 They do, for a short time. However, a spin which is precessing has higher magnetic potential energy than a spin which is completely aligned with the field, so the precessing state is unstable. The mechanism by which spins "decay" from a precessing state to a state in which they are aligned with the external field is provided by the spin-phonon ... 1 Is the author only mentioning spherical as we're concerned with conservation of ang. mom. at the moment? Yes The above mentioned Wiki article is more clear to me, but I'm still struggling with the jump from L=r\times p=r\times \mu\dot{r} is constant to \dot{r} must be perpendicular to L as well. The cross product is defined in such a way, that if ... 1 Torque, angular velocity and angular momentum are not inherently vectorial quantities. One way to demonstrate that is to consider motion in some space with more spatial dimensions than our familian three spatial dimensions. In a space with 4 spatial dimensions the following applies: to specify orientation of a state of rotation you need to specify the plane ... 1 I believe the only difference is the labelling, since due to the degeneracy introduced by the reflection symmetry you need to use a different set of commuting observables, see (Axial Symmetry bellow): https://en.wikipedia.org/wiki/Symmetry_of_diatomic_molecules#Molecular_term_symbol,_%CE%9B-doubling For the situation you describe L^2 and L_z are not a ... 0 Both of your formulas say that the total angular momentum is that associated with the rotation of the object around the center of mass plus that associated with the center of mass velocity not passing through some chosen reference point. For the latter contribution, the, r, vector goes from the reference point to the center of mass. If you choose the ... 1 In the formula \vec{L} = \vec{L'}+M\vec{r}_{cm}\times\vec{v}_{cm}$$Which represents the total angular momentum we have two parts/components: First let's look at the second term, which is M\vec{r}_{cm}\times\vec{v}_{cm}: This represents only the linear motion of the center of mass. Take the vector from the origin to the COM and call it r_{cm} and take ... 1 The formula$$\mathbf L=\mathbf L' +M\mathbf r\times\mathbf v_{\rm com}is used only for finding the value of angular momentum with respect from different locations. It is not a dynamical formula explaining any dynamics of the system. So it can't give you any relation between the torque applied and the subsequent angular momentum. So the answer of How ... 0 The vectors describing rotation are chosen along the axis of rotation because that is the only fixed direction in a rotating system. Given that choice, they do a good job of representing the magnitude and direction of rotational quantities, and as long as the axle remains fixed, they relate to each other and the various linear quantities in a ... 0 If the object precesses or not, or does something inbetween, namely nutate, depends on the ratio of Larmor frequency to the rotation frequency of the object. If you start from a situation where the magnet is tilted with respect to the field and then you let go, you obtain perfect precession only, if the rotation frequency is infinite. One may think of ... 4 Short story: it's zero, but with interesting physics to explore along the way. However, since it is in general not a product of single-electron wavefunctions the angular momenta of the two individual electrons are not defined (which makes it difficult for me to exploit symmetries for the integration of the two coordinates). In principle, yes, it's true ... 0 The magnetization in a bar magnet is due to the alignment of the spin magnetic moments of the electrons, not to currents. The precession would depend on the cross product\mu\times{\bf B}$. Since$\muis aligned along\bf B$, there is no torque and no precession. 4 The two Regge symmetries both have order$2$, but they don't both double the number of operations, because they don't commute with all of the other operations (or with each other). The Wikipedia article linked in the question actually mentions a nice way of expressing the operations that makes the full structure of the group (and the fact that it has$72$... 0 If you consider dark matter to be in the form of massive particles that have kinetic energy but only interact gravitationally, then there is a simple way to look at this. If the particles start off in a configuration whether their total energy (the sum of positive kinetic energy and negative gravitational potential energy) is zero; then they are on the cusp ... 2 Dark matter without strong self-interaction is indeed pressure-less, so it cannot form a static ball. However, since it does not interact with electromagnetism or give off any other kind of radiation, it also cannot lose energy. When normal matter clouds collapse to form stars it happens because the involved atoms and molecules can radiate away their energy ... 1 The electrons do not have zero angular momentum; they have zero orbital angular momentum, meaning that the total angular momentum of the electrons (and to a good approximation, the silver atoms) is given by the intrinsic angular momentum (spin) of the electrons, which was assumed to be uniformly distributed across all directions. Thus, the magnetic moment is ... 1 Stern and Gerlach didn't know that electrons carry spin, so they did in fact think that their intrinsic magnetic moment is 0. But through experiments they still knew that the magnetic moment of a silver atom was non-zero. They attributed this to an orbital angular momentum with quantum number$l=1$, though, so they expected three dots on the screen (one each ... 0 You are simply forgetting that the first$\vec{r}$and putting$\hat{r}$in it's place. Remenber:$\vec{r}=r\hat{r}$and$\hat{r}=\dfrac{\vec{r}}{r}$. A simple dimensional analysis shows that it`s$\dfrac{1}{r^4}$the correct proportionality. Just look the first line. Since$\hat{r}$is dimensionless and$[\vec{r}]=[r]=$meter ([X]= dimension of X) we have ... 0 There is an$\vec r$factor in the front of the first big expression and$\vec r=\color{red}{r}\hat r$. 1 By definition, a central force is spherically symmetric. From Noether's theorem, this implies that angular momentum is conserved. Equivalently, the angular momentum operator commutes with the Hamiltonian. This implies that$L^2$and$H$have simultaneous eigenvectors, so there are stationary states (i.e. eigenvectors of$H$) which have definite angular ... 1 Spin effects do not only occur in a variable magnetic field. It is certainly true that a Stern-Gerlach type experiment is a clear demonstration of the existence of spin-1/2 objects, but the existence of spin could have (eventually) been deduced without this result. Important examples of where spin comes up explicitly are the Pauli exclusion principle, the ... 1 Quantization in a narrow sense refers to discreteness of energy levels. In the context of atomic levels such discreteness is usually observed via optical experiments -e.g., via absorption, which has resonance at the frequency corresponding to the level spacing:$\hbar\omega = E_2 - E_1\$. Studying optical absorption spectrum necessarily implies the use ...

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It is perhaps best to start this out from the classical perspective. In classical electromagnetism, a spinning electrically-charged object generates a magnetic field due to the fact that the spinning of the charged object is charge in motion and thus technically is an electric current (even though it may not be what one thinks of an electric current which is ...

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