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Noether's theorem in its usual form assumes that the system (in this case a fluid) is governed by an action principle. We assume for simplicity that the fluid consists of just one type of fluid particles. I) In the Lagrangian fluid picture, the (local) conservation of fluid particles is manifest from the onset, since the dynamical variables are the labels ...

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Newton's third law. For the force the magnet exerts on the metal, there must be an equal an opposite force on the magnet exerted by the metal. Since both form one system (metal + truck + magnet), the net force on the system is zero, and it won't move.

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Backspin! Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction. Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's ...

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The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.

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First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation! This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is ...

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The continuity equation (without sources) is usually written as follows $$\partial_t \rho + \nabla \cdot \mathbf{j} = 0$$ If you identify $\rho$ as the mass density, integrate over some volume $V$ and use the divergence theorem you get the result that you mention in your question. Namely, the change in mass in $V$ equals the amount of mass flowing through ...

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From a mathematical standpoint, any collision in which no mass is lost is described by two equations: Conservation of energy: $m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2 + E$ Conservation of momentum: $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$ You know the masses and initial velocities, so this reduces to a system of two equations in three unknowns ($... 1 To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for inelastic processes. This is described by $$\text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*}$$ This also tells you how to compute the ... 1 Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you ... 3 There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e.$E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic ... 0 It seems to me the obvious answer is that it agrees with all the experiments. I am an experimentalist after all, (the neutrino is perhaps the best "missing energy" story.) 1 You may doubt that energy is conserved, but it is a direct consequence of Noether's theorem together with the assumption of time translation invariance, and this latter assumption is perhaps a bit more palatable/fundamental. That is, it is mathematically true that if the outcome of an experiment doesn't depend on when we perform it, the quantity we call ... 3 Let's assume that an experiment is performed that seems to indicate a violation of the conservation of energy principle. Now, I suppose that it's logically possible that the experiment actually and unambiguously falsifies the principle in which case we must conclude that the principle is approximate and we must seek a deeper principle to guide our ... 4 I will try to give a short introduction into the ideas of scientific truth as I understand them. In mathematics, the world is beautifully simple. We have axioms that the set to be true, and from these we can deduce a plethora of statements to be undoubtedly true - given that the axioms are true. There may be undecidable statements about which we cannot say ... 0 It cannot be demostrated, but it can be checked. This year I've done an experiment which verifies (indirectly) this principle. It is really easy: I take an iron ball, and I shoot it to a pendulum. As a result, the pendulum goes up, and I can measure the deviation angle after the collision. With a sensor, I'm also able to measure the initial speed of the ... 0$ r^2 \dot{\theta} $is known as the specific angular momentum. Also, the correct formula for the 2 body Lagrangian is actually: $$\mathcal{L} = \frac{\mu}{2} (\dot{r}^2+r^2 \dot{\theta}^2) + \frac{GMm}{r} = \frac{Mm}{M+m} \frac{\dot{r}^2+r^2 \dot{\theta}^2}{2} + \frac{GMm}{r}$$ where$ \mu = \frac{Mm}{M+m} $is the reduced mass. 0 The collision of a rod with a point mass is the similar to the collision of two masses but with the effective mass of the rod being $$m' = m_{rod} \frac{I_{rod}}{I_{rod}+m_{rod} r^2}$$ where$r$is the distance between the point of impact and the center of mass, and$I_{rod}$is the mass moment of inertia about the center of mass. If the rod is slender ... 1 An other way to see the argument of the answer of @fuenfundachtzig , is that, concerning$SU(3)$representations, there is an equivalence between$(3*3)_\text{antisymmetrised}$representation ("red * green") and$3^*$representation ("antiblue"). Why ? Well, thanks to the completely anti-symmetric Levi_Civita symbol. Using objects upon which act the ... 1 It works if you assign colors like this: one red up, one green up, down is blue,$X$takes red and green which are equivalent to antiblue ("yellow"), thus color is conserved. I didn't take into account the last fact which explains my confusion. 4 In hydrodynamics, conservation means that what flows into the control volume is equivalent to the flow out of the control volume. With respect to momentum, we mean precisely that any change in momentum of the fluid within a control volume is due to the net flow of fluid into the volume and the action of external forces on the fluid within the volume ... 1 Here's a parallel answer to Luboš's but purely classical. Start by noting that the momentum vector of a plane wave with wavelength$\lambda$is: $$\vec{p} = \frac{2\pi}{\lambda}$$ In some elastic scattering experiment, e.g. X-ray or some other diffraction measurement, we have something like: where$\vec{p}_{in}$is the momentum of the incoming wave ... 5 Microscopically, i.e. in the quantum theory the scattering with radiation is a collision of particles with photons such as $$e^- + \gamma \to e^- + \gamma$$ The momentum vectors of the particles above are $$\vec p_1+\vec p_2= \vec p_3 + \vec p_4$$ where the identity holds due to momentum conservation. But in general$\vec p_1\neq \vec p_3$and$\vec ...

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I understand that the inner product of two 4-vectors is conserved under the Lorentz transformations Yes, $p_1.p_2$ is a Lorentz invariant So that the absolute value of the four momentum is the same in any reference frame. It is not correct to speak about the "absolute value" of a (quadri)vector. Which is conserved in a Lorentz transformation ...

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Emmy Noether proved both the theorem and its converse. Look for the book "The Noether Theorems" for a precise and discussed formulation of her statements, as well as a translation of the original paper. It seems there is a link to the pdf in the princeton math website (I don't know about copyright issues, however).

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In special relativity, if you add two velocities, you have to use the formula $$v = (v_1+v_2)\left(1+\frac{v_1v_2}{c^2}\right)^{-1} \text{ .}$$ So you cannot simply add two velocities together. Usually, velocity is not a good variable to work with in special relativity. It's much easier to use four-momentum conservation, which is simply given by $$p = p_1 ... -2 You can just verify each component and they are just momentum conservation in 3-momentum. There is no velocity conservation so you cannot add them together. 1 This is a general property of (pseudo)Riemannian geometry. I do not think there is anything specifically physical about it beyond the geometry. It holds even if \varphi in not Lorentz invariant. In (pseudo)Riemannian geometry the covariant derivative \nabla_i replaces partials \partial_i. The Laplace-Beltrami operator$$\Delta \triangleq ...

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In general, the elasticity of a collision is dependent on the properties of the colliding objects. In a perfectly elastic collision, no kinetic energy is dissipated, which means the collision creates no heat, no sound, etc. In a perfectly inelastic collision, the maximum possible amount of kinetic energy is dissipated as heat, sound, etc. This corresponds ...

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Yeah, definitely. One example is an inelastic collision, where both masses will have the same velocity after colliding. In this case, let's say a bullet of mass $m$ and speed $v_0$ hits a stationary rock of mass $M$ and they stick together and move with a final speed $v$. Intuitively, you can already tell that their velocity must be in the same direction as ...

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If the masses are the same, then as stated above, the initially moving mass stops and the other one acquires the velocity of the initially moving one. This is the mechaism behind the so called "Newton's cradle". BTW, the collision has to be head on, ie take place in one dimension.

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Normally you solve an elastic collision with just momentum and energy conservation, because you really don't know what happens at impact. The formulas are given in this answer. For equal masses in one dimension the velocities are exchanged. It turns out your interaction time and acceleration multiply to get the correct velocity change, but since you ...

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From a purely mathematical perspective, there really should not be any difference between the two forms. From a numerical simulation perspective, however, there can in fact be quite a bit of difference. In the application of the Navier-Stokes equations to Computational Fluid Dynamics (CFD) problems, it is often required to use the integral form due to the ...

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Feynman showed only that jump in the position of the particle would change mechanical angular momentum due to that particle. If the jump is due to other body or medium capable of carrying angular momentum, total angular momentum of the system could still be conserved. Continuous mass conservation is a basic idea in physics, inferred directly from experience ...

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I think the more direct way to show this is to appeal to conservation of linear momentum: I can't move or "jump" the mass without some commensurate opposite momentum. Feynman effectively mentions this later with the rocket. Linear momentum conservation really comes directly from the invariance of all known laws of physics under spatial translation (see ...

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