New answers tagged

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Comments to OP's post (v4): OP is trying to prove via Noether's theorem that no explicit time dependence of the Lagrangian leads to energy conservation. OP's transformation seems to be a pure horizontal infinitesimal time translation $$\tag{A} t^{\prime} - t ~=:~\delta t ~=~-\epsilon, \qquad \text{(horizontal variation)}$$ $$\tag{B} q^{\prime i}(t) - ...


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The easier way of doing this is to just consider a generic transformation, G, such that the canonical co-ordinates of the Hamiltonian are shifted as below: $$ \delta p = \frac{\partial G}{\partial q} \delta \lambda$$ and $$ \delta q = - \frac{\partial G}{\partial p} \delta \lambda\,,$$ where $\lambda$ is the transformation parameter determining how much of ...


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The compression pulse that propagates through the metal spheres of Newton's cradle are not ordinary sound waves. They are approximate solitons (a nonlinear wave form that balances dispersion against nonlinearity). It is this property of soliton pulses that is responsible for the observed behavior. More Details: Newton's cradle is a physical manifestation ...


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The simple answer is that not only momentum but also energy needs to be conserved. This puts constraints on the number of balls that can be activated in the cradle. Note that this does not always give a unique solution either. But it enforces that $ n $ balls to $ n $ balls is a "stable" solution.


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Take for example the double-slit experiment interpreted in the Copenhagen sense. The particle leaves as an object with mass, yet passes through the slits as a massless wave, only to collapse again as a particle. We can consider this example as a generalisation of the principle of anti-realism posited by Bohr. Where does the energy "go" when the ...


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Short answer: no. I'll give some context with the details of the simplest examples. In the context of conservation laws, "energy" refers to the Hamiltonian. In classical mechanics, a quantity without explicit time dependence is conserved iff its Poisson bracket with the Hamiltonian is 0. In quantum mechanics, quantities are promoted to operators on a ...


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Finding the missing equations Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the ...


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You have drawn a Feynman diagram. Feynman diagrams are iconic shorthand for integrals over the variables of the problem. The calculation gives the probability for the reaction to happen, in this case the decay of a neutron . The observables are the four vectors of the initial (neutron) and final particles. The integral is over the variables . Here is a ...


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It can be stated this way: In this particular diagram, the W boson is in a state named off-shell i.e. we say that this boson is virtual. Virtual particles are allowed to have any mass value. They can't although violate charge conservation at the vertex. This 80 GeV mass of the W boson, is for a real W boson, which is on the on-shell state. The real ...


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The situation where the equation holds is when the fluid is imcompressible. In your example where you have a fluid flowing in a tube under gravity, you can imagine two situations. One is where there is no dissipative interaction between the fluid and the tube, and one where there is. In the situation with no dissipation, then the fluid should accelerate ...


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Classically, angular momentum is only conserved in a central potential by considering the torque (correct me if I am wrong). In quantum mechanics, it is also true, isn't it? In QM, an operator is conserved iff it commutes with $H$, because $$ i\dot {\mathcal O}=[H,\mathcal O] $$ Therefore, the angular momentum is conserved iff it commutes with $H$. As ...


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Newton's third law is a statement that momentum is conserved, so it is equivalent to the law of conservation of momentum. Conservation of momentum follows from a fundamental symmetry (of the action) called space shift symmetry and as far as we know this applies to all our physical theories. So Newton's law is still valid but has to be treated with some care ...


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Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ...


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Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor $g_{\mu\nu}$ in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action $S_{\rm m}[\Phi, g]$ should be a general relativistic diffeomorphism-invariant functional. Then the Hilbert stress-energy-momentum ...


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In the Standard Model, the baryon and lepton number are accidental global symmetries. However, they are conserved only at the classical level: quantum corrections do not respect them, i.e., they are anomalous. The interesting thing is that they are violated by exactly the same amount. In terms of currents we write: $$\partial_\mu J_B^\mu=\partial_\mu ...


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Mandelstam $t = (p_1 -p_1')^2$ corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process. If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum.


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Newton't third law is just the conservation of momentum, so the question is whether conservation of momentum applies in general relativity. This turns out to be a rather complicated issue, and for the details I suggest you look at General relativity and the conservation of momentum. However you specifically ask why we feel weight due to equal but opposite ...


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This is a quite subtle problem. You have to be careful about three different situations. A ball can be thrown with velocity (relative to the ground): a) $v_0-v_e$. b) $v(t)-v_e$, where $v(t)$ is the velocity of the car just after the ball is thrown. c) $v(t)-v_e$, where $v(t)$ is the velocity of the car just before the ball is thrown. You actually stated ...


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All of your math is correct, but your result conflicts with our intuition as well as the reality of how rockets work. Here is why: You assume that "The ball will be moving at $v_\mathrm{ball}=v_0-v_e$". This approximation is valid only in the limit where $m$ is much smaller than $M$. (i.e. the limit of a continuous stream of tiny balls, like a normal ...


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What's special about $H$ is that it generates time translations. This means that operators evolve, by postulate, through Heisenberg Equations of motion $$ i\frac{\mathrm d}{\mathrm dt}\mathcal O(t)=[H,\mathcal O] $$ modulo an explicit time dependence. Therefore, if an operator commutes with the Hamiltonian, $[H,\mathcal O]=0$ we automatically conclude that ...


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I am very bad at drawingin "paint", but the process can go as antineutrino +neutrino to Z0 , Z0 to s antis (or up antiup), a gluon vertex from the quark to an up antiup (or strange antistrange quark) the parenthesis are the alternate diagram. , So it is not forbidden, it has two weak vertices and so very small cross section.


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Your first diagram is wrong, since there is no vertex in a Lorentz invariant theory where three fermions and a vector meet. However, I don't see why you say the interaction is forbidden. It would surely be insanely suppressed since amplitudes are extremely weak, but I don't see a problem with the diagram (for instance): Notice that quarks mix, so the ...


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The problem with this process is, that there is no term in the standard model which allows a coupling of a fermion a W-boson and a quark-antiquark pair. Even with the neutrino having no charge, you have to take this arrow you draw on the fermion line seriously, so the fermion flow is violated at this interaction points. Furthermore you can only couple ...


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"when it collides directly" is to be interpreted as a "head-on" collision? So the balls go off along the line of the original velocity direction. The question tells you that the direction of $A$ is reversed so the direction of $B$ must be in the opposite direction ie in the original direction of $A$'s velocity otherwise momentum cannot be conserved.


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I've noticed that you have referred to energy conservation a couple of times in your question and comments. So there is something that we should make very clear. Conservation of energy in general does not mean the conservation of any given kind of energy. Potential energy can turn into kinetic energy can turn into nuclear, chemical, thermal or acoustic ...


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Without further information it is not possible to predict if the two bodies will stick together or coalesce. This depends on the nature of the materials which come into contact, and the presence of any interlocking mechanism (such as used to couple carriages of a train). Momentum is always conserved in collisions; kinetic energy is not always conserved. ...


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Komar mass is associated to one asymptotically flat end. So a wormhole has two Komar mass, one for each side, and in principle they can even be different! Indeed the simplest wormhole you can imagine is simply two copies schwarzschild spacetimes glued together at the would-be horizon. As an aside, even time can run differently on the two sides (often ...


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Energy is not conserved during an inelastic collision so you cannot find $v$ using conservation of mechanical energy concept.


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The point is that you cannot "make the gun unmovable". Total initial momentum is $0$, so, by conservation of momentum, we will have: $$m_{(gun)} v_{(gun)} + m_{(bullet)} v_{(bullet)} = 0$$ So that $$v_{(gun)} = - \frac{m_{(bullet)}}{m_{(gun)}} v_{(bullet)}$$ If you want $v_{(gun)}$ to be exactly $0$ (with $v_{(bullet)}\neq 0$, of course) you have to ...


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When you say we make the gun unmovable what this really means is that you are fixing the gun to the Earth. So now when you fire the gun the momentum of the bullet must be equal and opposite to the momentum of the gun + the Earth. So when you fire the gun you change the velocity of the Earth very slightly. However the Earth is so much more massive than the ...


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That the bomb breaks apart due to the explosive forces which are internal to the system, has nothing to do with the trajectory of the center of mass. As user Sammy Gerbil points out correctly, the initial trajectory of the bomb was parabolic and even if the bomb exploded into million fragments of different masses, the center of mass would continue on it's ...


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UPDATE : The explosion itself conserves linear momentum, regardless of how small the fragments are. If we ignore gravity and air resistance and all other external forces, there is no change in total momentum. This is because the internal forces all occur in equal and opposite pairs (Newton's 3rd Law). If we take the external forces into account, then ...


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It will stay the same, if we neglect the variation due to gravity (every external force is going to change the momentum). If we assume a uniform distribution of the shrapnels' mass (same size for all shrapnels), the shrapnels going in the direction the bomb was originally going will have, on average, higher velocity. With a great simplification, we can say ...


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The surrounding system is in a very large very mixed state, and that washes out the distinction in angular momentum. Trivial Example Consider the trace distances between the following nearly-maximally-mixed density matrices: $$\begin{align} D \left( \frac{1}{2} \begin{bmatrix} 1& & \\ &1& \\ & &0 \end{bmatrix}, \frac{1}{2} ...


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Consider two objects, 1 and 2, colliding for some short time interval $\delta t$. During $\delta t$ let's ignore all forces except the contact force that 1 exerts on 2, $\vec{F}_{12}$ and the contact force that 2 exerts on 1, $\vec{F}_{21}$. As long as the objects touch each other, both of these forces exist, and by the principle of Newton's 3rd Law we know ...


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You are right that the magnetic field, or rather the apparatus used to generate it, absorbs some angular momentum from the quantum spin. This apparatus includes, for example, some coils of wire containing more than $10^{23}$ electrons, as well as many other macroscopic components. The upshot is that increasing the angular momentum of this apparatus by an ...


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I come to think that when a mass collides with another, both of them should always have equal velocities post-collision. To paraphrase Feynmann, no matter how beautiful you may believe your reasoning to be about this, if it doesn't agree with experiment, it is wrong. If the two objects 'stick together' after the collision (the collision is totally ...


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Often when two object collide it is often represented as an instantaneous impulse exchange. However in reality this happens continuously. Namely both objects are not completely rigid and will deform during the collision, storing energy in the elastic deformation (like a spring) and dissipating energy with any inelastic deformation. During such a collision ...


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What is conserved during a collision is not velocity, but momentum (mass times velocity). If the collision is elastic, kinetic energy is also conserved: https://en.wikipedia.org/wiki/Elastic_collision. If the collision is inleastic, only momentum will be conserved: https://en.wikipedia.org/wiki/Inelastic_collision The resulting equations (which you can ...


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Yes, there is an integral, which comes from the LSZ reduction formula, $$ \langle f|i\rangle\sim \int \mathrm dx\ \mathrm e^{ikx}\square_x G(x) $$ where $x=(x_1,x_2\cdots,x_n)$, $k=(k_1,k_2,\cdots,k_n)$ and $G$ is the $n$-point function. If you go to momentum space you'll get that integrand depends on $x$ only through exponentials, and therefore there is a ...


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Let's look at F = ma. This is not true for each force individually, for sure - each particle experiences many forces but only one acceleration. It's really Σ(all forces)F = ma. Even if we say just the force of gravity, we're looking at Σ(other objects)Fg + Σ(other forces)F = ma. So let's think about applying MOND. We have a particle that's ...


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There is no need of integrating over anything to obtain the 4-momentum conservation. Indeed, if you think in terms of perturbation theory (Feynman diagrams), each vertex conserve momentum, so that the diagram itself automatically conserves momentum.


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If we did have decay of massless particles then there'd be no reason a photon wouldn't decay from an energy of $h\nu$ into two photons of energy $h\nu/2$. Eventually all our photons would be red-shifted into oblivion and we'd have no light left in the universe.


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According to the rules of qft there are virtual photons in the vacuĆ¼m. No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way, The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing ...



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