# Tag Info

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At the physics 101 level, you pretty much just have to accept this as an experimental fact. At the upper division or early grad school level, you'll be introduced to Noether's Theorem, and we can talk about the invariance of physical law under displacements in time. Really this just replaces one experimental fact (energy is conserved) with another (the ...

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Warning: this is a long and boring derivation. If you are interested only in the result skip to the very last sentence. Noether's theorem can be formulated in many ways. For the purposes of your question we can comfortably use the special relativistic Lagrangian formulation of a scalar field. So, suppose we are given an action $$S[\phi] = \int {\mathcal ... 13 The symmetry you are asking about is usually called a scale transformation or dilation and it, along with Poincare transformations and conformal transformations is part of the group of conformal isometries of Minkowski space. In a large class of theories one can construct an "improved" energy-momentum tensor \theta^{\mu \nu} such that the Noether current ... 11 When the cosmonaut sneezed they would start moving, and rotating, in the opposite direction, but when the sneeze hit their faceplate (ugh!) this would stop the motion. The net result is that the velocity of the cosmonaut would not have changed, but their position and angle would have. According to Wikipedia a typical breath is 500cm^3 and a sneeze ... 10 Another way of solving such problems is to go to another reference frame, where you obviously don't have enough energy. For example you've got a 5 MeV photon, so you think that there is plenty of energy to make e^-e^+ pair. Now you make a boost along the direction of the photon momentum with v=0.99\,c and you get a 0.35 MeV photon. That is not ... 9 The stress energy tensor T^{\mu\nu} contains all the energy/momentum components of the elctromagnetic field and the conservation of these components is expressed by \partial_{\nu}T^{\mu \nu} = 0 Which states that the change in time of energy/momentum is zero. If the above is non-zero then electromagnetic field energy/momentum is transferred to charged ... 9 This is a fairly subtle question! Griffiths recently published a paper on this. Hidden momentum, field momentum, and electromagnetic impulse: Electromagnetic fields carry energy, momentum, and angular momentum. The momentum density, ϵ_{0}(E\times B), accounts (among other things) for the pressure of light. But even static fields can carry ... 9 It is possible to show that the total momentum of any static system is zero in an inertial frame where nothing is moving. This does not mean that the momenta associated with various components of that system are individually zero. As you point out, there can be finite electromagnetic momentum associated with static charge distributions. Even though there is ... 9 Here is a visualization: Momentum is mass times velocity, so draw it as the area of a rectangle: If we change the mass and velocity a little, we change the momentum: The total change in the momentum is the sum of green, blue, and purple rectangles. Their sizes are just length times width, so overall we have \Delta p = m\Delta v + v\Delta m + \Delta ... 8 If the conservation law is general, meaning that it isn't specific to one motion, but conserved in a general configuration, then the answer is yes. This follows from the theory of canonical transformations in classical mechanics. First, consider a perfectly triangular symmetric initial condition of three particles arranged on an equilateral triangle with ... 8 We can prove it in perturbative string theory but it's probably valid beyond it. In perturbative string theory, any (continuous) global symmetry has to be associated with a conserved charge which, because of the locality of the physics on the world sheet, implies the existence of a world sheet current j or \bar j or both (left movers vs right movers) ... 7 thank you for the nice question. It directly relates to the topics of conformal field theories. I found a very nice thread in another forum where I guess your question has been answered. Nevertheless, I will try to summarize the main points here and maybe add some points. Symmetries in General Relativity In general relativities, symmetries correspond to an ... 7 Anglular momentum is conserved, so any tiny initial rotation that a the original ball of gas had becomes faster as the gas collapses down into a star and disk of planets. Planets near the sun rotate slowly for the same reason that the moon always faces the same side to the Earth - tidal braking Venus probably received a hit from a some lump of rock / ... 7 I) For a mathematical precise treatment of an inverse Noether's Theorem, one should consult e.g. Olver's book (Ref. 1, Thm. 5.58), as user orbifold also writes in his answer(v2). Here we would like give a heuristic and less technical discussion, to convey the heart of the matter, and try to avoid the language of jets and prolongations as much as possible. ... 7 Yes, you can swim through space, but only if space is curved - in the vicinity of a gravitating body (which creates curvature of space-time) it is possible for an isolated body to move by only executing internal motions of parts of the body. The reason this is possible is that the center of gravity of an object is not well defined in a curved space-time. ... 6 The intuitive argument for Noether's theorem, which is also the best completely precise argument for Noether's theorem, appears in Feynman's popular book "The Character of Physical Law". I will reproduce the argument, but not the diagram. The diagram is two parallel squiggles with a line connecting them at the top and at the bottom. These represent a ... 6 The derivation in Landau and Lifschitz is making some additional implicit assumptions. They assume that all forces come from pair-interactions, and that the pair forces are rotationally invariant. With these two assumptions, the potential function in the Lagrangian is V(x1,...,xn) = \sum_{\langle i,j\rangle} V(|x_i - x_j|) And then it is easy to prove ... 6 Yes, this is the opposite of Noether's theorem. So let's call our conserved quantity A (we will consider just one conserved quantity for starters) and begin with \left \{H, A \right \} = 0 law for conservation. Because of the connection between Poisson bracket with flows on the phase space this tells you both that \mathcal{L}_{V_H} A = 0 (A is ... 6 Let me expand a bit on Manishearth's answer. There's an idea going back a long time called the principle of stationary action. See http://en.wikipedia.org/wiki/Principle_of_stationary_action for a description that isn't too mathematical. In the 18th and 19th centuries century the mathematicians Lagrange and Hamilton found ways of using this to describe ... 6 In "normal" cases, no, this is not possible. You can easily understand why by considering this process in the center-of-mass frame (which is the rest frame of the original particle). In this frame, you would start with a single particle X at rest, which has energy m_Xc^2, and wind up with 2 or more X, which will necessarily have an energy of at least ... 6 If you want to prove that \vec{L}=\vec{r}\times \vec{p} is constant with respect to time for a particle in a central force field \vec F = \phi(r) \vec r, just show that the angular momentum doesn't change with time, i.e. \frac{d}{dt}\vec{L}=0. Using the product rule we get two terms: \frac{d}{dt}\vec{L}=\frac{d}{dt}(\vec{r}\times \vec{p}) = ... 5 I agree with qftme's answer for the case of massive decay products. By energy conservation alone, \gamma \rightarrow e^+e^- should be allowed, but momentum conservation forbids it (as well as the opposite case, e^+e^- annihilation). It is only allowed if you have some other particle involved to take care of the photon momentum. In case of a massless ... 5 Jeff Harvey has of course provided you with the perfect, standardized answer: the scale invariance boils down to the tracelessness of the stress-energy tensor. But the tracelessness is not really a "conserved quantity" in the usual sense that you may have waited for. However, one may transform the problem to something that is a conserved quantity in the ... 5 Are you looking for a proof? If so, this link (which has some sign errors as pointed out in the comments) proves it as follows (without the sign errors): We start by differentiating the definition of the probability with respect to time only:$$ \frac{\partial P(x,t)}{\partial t} = \frac{\partial}{\partial t}\left (\psi^*(x,t) \psi(x,t)\right) = \left[ ...

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If by "they are associated with different masses" you mean that the flavor eigenstates have different masses then you are working from a misconception. Those states are not eigenstates of the free Hamiltonian so they don't have a mass as such. (They do have a mean expectation if you could weight a bunch of them, but it does not apply to any given neutrino.) ...

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(I only know the importance, not the mathematical treatment &c, but I doubt you want that) Significance Noether's theorem let's us obtain conservation laws. Conservation laws are pretty much the life of physics. If you want to calculate the outcome of any process, you have to see what is conserved in the process. Without these laws, you'd be left with ...

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A Stirling engine moves a fluid from a hot end to a cold end - extracting mechanical work as it does so. The power input comes from maintaining the temperature of the hot end - usually by burning some fuel externally. The 'clever' part of a Stirling engine, and the thing that gives it it's high efficiency, is that the hot end of the mechanism stays hot and ...

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$M$ is reducing. Thus, $\mathrm dM$ has a negative value. In contrast, in the above equations, you can see an $M-\mathrm dm$ term. Here, we can see that $M$ will reduce only if $\mathrm dm$ has a positive value. In other words, when time goes forwards, the mass that got thrown out ($m$) is increasing, thus $\mathrm dm$ is positive. In contrast, $M$ ...

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The two answers are physically equivalent. If you watch the Wikipedia animation you see that the fastest part of her body on the top animation is the tail. It makes a substantial contribution to the angular momentum. At any rate, the angular momentum of the tail is included in the angular momentum of the "back cylinder". The trick the cat needs to achieve ...

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