# Tag Info

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what the expert say that only i say

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If the 4-momentum were invariant then it would be a scalar. 4-vectors are defined by the way their components mix when we change coordinates. In particular when we apply a lorentz transformation to our coordinates the inverse transformation is applied to the vector. As a simple example consider what happens to the energy when we boost. If we start in the ...

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Conservation and invariance are fundamentally different things. Conservation means "doesn't change with respect to time". While invariance means "doesn't change with respect to Lorentz transformations". Components of four-momentum transform like vector components and are thus NOT invariant under Lorentz Transformations. But that doesn't prevent them from ...

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Momentum is really always conserved. Truly. Throwing something upward (accelerating it with your arm) causes your feet to push harder on the ground. The increased down-force causes the ground under you, and ultimately the entire Earth, to shift direction downward. Fortunately, the rock and the planet attract each other gravitationally, causing the rock to ...

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I) As a purist, I disprove of the common praxis to call the implication $$\tag{1} \{Q,H\}+\frac{\partial Q}{\partial t}~=~0 \qquad\Rightarrow\qquad \frac{dQ}{dt}~\approx~0.$$ for a 'Hamiltonian version of Noether's theorem', as explained in my Phys.SE answer here. Moreover, the implication (1) is not equivalent to the full Noether's theorem for various ...

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The batter in both cases is connected to the Earth. Momentum is conserved but you have to add the amount given to the Earth--which is impossible to detect but it is there. If you drop a ball and it lands on the Earth and stops, momentum has not disappeared, it is transferred to the Earth. The answer just above notes that the bat will give some spin to the ...

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When you hit the falling ball, friction between bat and ball will momentarily stop the side of the ball that is hit from moving down. However, since this force of friction is not applied at the center of mass of the ball, this will not result in a complete arrest of the vertical motion: instead, the ball will acquire some spin. However, it will not lose all ...

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Let's try an easier thought experiment. Get a flat bat (since it's easier to visualize, not because it changes anything), and instead of swinging it, hold it still. From a distance above it, drop the ball. Now, the key is to hold it at such an angle that the ball will travel horizontally when it hits the bat. You will see it bounce off with a forwards ...

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Provided the bat delivers exactly horizontal momentum impulsively to the second ball, it will not travel as far due to its initial downward velocity, as you say. Dissipating the downward momentum doesn't make much sense in the scenario you described.

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If there's no net external force acting on the system, the momentum is always conserved. The energy needs not be conserved for the momentum to be conserved. The only necessary condition is zero net impulse on the system from all the external forces.

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In particular, ensure no external force acts on the system. If there is an impulse on the system, there will be a net acceleration. Use Newton's Laws then.

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A quick note on the mechanics, since the question has been asked again: Force is the rate of change of momentum, so to understand the forces we need to look at how the momentum of the water is changing. Let's call the mass flow rate of the water $M$ (in kg/sec) and the upward velocity $v_u$. The momentum of the water starts off at zero because it's lying ...

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The charge associated to the $U(1)$ symmetry you mention is called weak hypercharge. The relation with the electric charge is the following $Q= T_3+Y/2$ where $T_3$ is the third generator of the $U(2)$ symmetry representing the weak isospin and $Q$ the electric charge. This relation holds for all leptons. Neutrinos have weak isospin $+1/2$ and weak ...

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You already said if you have $h$ as 1.8 m, then $\dfrac{1}{2} m v^2 = m g h$ implies $v$ is 5.9 m/s. However, you were unhappy becuase this involves energy. So just divide both sides of the equation by $m$. Now you have $\dfrac{1}{2} v^2 = g h$. The solution must still be 5.9 m/s so you get the right answer, this time using only SUVAT.

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I'll risk moderatorial opprobium with a partial answer because you have come so close. You correctly use the SUVAT equation $v^2 = u^2 + 2as$ to find that the velocity of the ball just before it strikes the ground is $v_i = -7$ m/s (using the sign convention that upwards is positive). So far so good. Now you know the ball rises back up to a height of 1.8m, ...

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This is just a property of Fourier transformations. If the correlation function is translational invariant then, by definition, the position space representation $D(x,y)$ transforms as $D(x+a,y+a) = D(x,y)$ for any constant $a$. Thus $D(x,y) = D(x-y,0)$ and so the correlator depends only on the difference $x-y$. For simplicity, we'll define $D(x-y) = ... 1 I guess what is to be noted here is the fact that, the Correlation function (operator), commutes with the momentum operator, since $$[D,\text e^{ixp}] = 0 \implies [D,p] = 0$$ Having that to be the case, one can recollect that any operator represented in its own eigenspace is diagonal should answer your question. PS : I am not completely sure about the ... 0 To expand on the answer given by @By Symmetry, let the two point function be defined as $$f(x_1,x_2)=\langle\Omega|\, \phi(x_1)\phi(x_2)\,|\Omega\rangle.$$ In order the above to be translationally invariant one must require that $$\langle\Omega|\,\left[P, \phi(x_1)\phi(x_2)\right]\,|\Omega\rangle = 0$$ where$P$is the generator of the translations as ... 0 The fact that the system is translationally invariant implies that the translation operator commutes with the Hamiltonian. This implies that they have a basis of mutual eigenstates. Since the momentum operator generates the translations, i.e. $$T =e^{-\imath x p}$$ a state is an eigenstate of the translation operator if and only if it is an eigenstate of ... 0 The law of conservation of matter (or more specifically, mass) has been disproved a long time ago. There are many ways to disprove it. For eg if you bring matter in contact with antimatter, it completely annihilates, leaving no mass behind. 3 First of all, my opinion is that the paper on your link is full of notational inconsistencies and therefore causes a great amount of confusion for someone who struggles to understand Noether’s theorem. So, allow me to formulate the field-theoretic version of Noether’s theorem in a more, according to my taste, charming way. Preliminaries 1: Lie Groups and ... 1 There is a transient tension$\Delta Tin the string that will give rise to a change in momentum of the counterweight, the pan, and the mass on the pan. At the ceiling, the pulley transfers twice this force to the support: The change in momentum of the three components is (with + direction up): $$\Delta p = m v' + m(v-v') - mv' = m(v-v')$$ This change ... 0 It is true that angular momentum is conserved in all frames, but the actual value of the angular momentum will, in general be different. If you look at the extra terms you will find that they correspond to the change in the angular momentum of the centre of mass about your chosen origin. \begin{align} \mathbf{r}_{CM}\times M\mathbf{V} & = ... 0 As several answers have stated already: A positron by itself is not known to decay at all. But if you are considering "encounters" in the course of which a given positron ceases to exist, then how about the "absorption" of a positron by a neutron, leaving a proton, accompanied by (emission of) an anti-electron-neutrino: $$\mathbf n + \mathbf e^+ \rightarrow ... 1 Yes a positron can decay without encountering an electron. But it must encounter another particle because as it is said in another answer, the positron is a stable particle (in the vacuum), so it cannot decay on its own. An example of "decay" not involving an electron:$$e^+ + \mu^- \to \bar{\nu_e} +\nu_{\mu}$this decay proceeds via the weak interaction (a ... 0 This is not correct.$\Delta t$(duration of impulse) is very short indeed, but$F_i$(force on particle$i$) is very high. It is similar to attempting to compute$\infty\ 0$, which is not zero, but undefined. However,$\int F_i\mathrm{d}t$obviously is defined, and it is not equal to zero if the velocity of your particle changes during the impulse. The sum ... 1 There is no general algorithm for doing so, and even figuring out how many conserved quantities a system has can be difficult. A famous example is the Toda lattice, a system which was originally proposed by Toda in 1967 and was believed to be chaotic, but was in fact proven to be integrable (to have too many conserved quantities to be chaotic) in 1974 by ... 0 your conclusions have nothing to do with scalar QED or quantum mechanics for that matter. Let's start from your$j^0 = -\partial_\mu F^{0 \mu}$, then you erroneously applied gauss law to find$Q$vanishes at infinity. The reason is that in your surface integral$\int_{\mathbb{R}^2}dS$need not vanish, because the field need only vanish as$1/r$for the ... 0 Whether there is a "unique" way to solve this, I'm not sure but assuming that the centre of gravity after the explosion hasn't moved with respect to before the explosion and the other conditions you imposed hold, then a triangular arrangement of the momentum vectors is the only possibility. For example, with$\alpha = 120 \text { degrees}\$, the balance of ...

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The three momentum vectors of the pieces must form the sides of a triangle so that the total momentum is conserved (all forces are internal to the system, so no momentum is added or removed). The orientation of the triangle will not be unique in space, but the lengths of the sides will uniquely determine the relative angles. If the masses or speeds vary, ...

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