# Tag Info

26

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 ...

8

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 ...

4

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 ...

4

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, ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

According to the rules of qft there are virtual photons in the vacuüm. 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 ...

2

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 ...

2

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 ...

2

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 ...

2

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 ...

2

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.

2

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 ...

2

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 ...

2

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 ...

2

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 ... 2 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 ... 2 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 ... 1 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. 1 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 ... 1 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. 1 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 ... 1 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} ...

1

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 ...

1

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 ...

1

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 ...

1

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible