# Tag Info

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I think I have came up with the answer, which I hope is correct. Although, neither internal nor the gauge symmetry operations affect the space-time coordinates, there is a big difference: Internal symmetry is an actual symmetry of the system (field): two physically distinct field configurations (or in QM, two physically distinct states in Hilbert space) ...

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If the theory is invariant under translations in space, then linear momentum is conserved by Noether's theorem. If the theory is quantum, conservation holds only on the level of the expectation values (because that's the only meaningful level where you can talk about momentum as a number that's conserved in time), but it still holds. There is no way out. ...

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The real converse of the first theorem is the second one. Your formulation of the converse of the first theorem is too literal and thus valid as a particular case under additional conditions.

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Noether theorem tells you that if you can find a (one parameter) group of infinitesimal transformations $\alpha$ and $\beta$ such that: $$t'=t+\alpha\epsilon$$ $$q'^\mu=q^\mu+\beta^\mu\epsilon$$ and your lagrangian is invariant under this group of transformations, then the quantity ...

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It is a requirement one imposes on physical representations of the algebra of observables, known as $\alpha$-regularity, where $\alpha$ denotes the action of a certain group over the algebra. In more concrete terms, the group involved here is the Lorentz group (but more generally is the Poincaré group). Under certain hypotheses (see Wigner 1931 and Bargmann ...

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In the modern crystallography there is a notation of aperiodic crystals (or quasicrystals). They are crystals with normal basis $\mathbf a,\mathbf b,\mathbf c$ and a set of propagation (or wave) $\mathbf k$-vectors that are incommensurate with the metric $\mathbf a,\mathbf b,\mathbf c$. The atomic positions (or/and occupancies) are modulated according to ...

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the parity operation is $P: (t,x) \rightarrow (t',x') = (t,-x)$ lets see how this effects this term... $\epsilon^{\mu \nu \lambda} a_\mu \partial_\nu a_\lambda = \epsilon^{\mu 0 \lambda} a_\mu \partial_0 a_\lambda + \epsilon^{\mu j \lambda} a_\mu \partial_j a_\lambda$ $=\epsilon^{\mu 0 \lambda} a_\mu \partial'_0 a_\lambda - \epsilon^{\mu j \lambda} a_\mu ... 1 Conservation of momentum! Force × Time = Impulse = Δ Momentum Since the average force is the same going up and down, and since the momentum change is the same going up and down as well, the time during which the force is applied must also be the same. 1 Comments to the question (v1): Concerning the notion of on-shell and off-shell, see also Wikipedia and this Phys.SE post. In the context of an action formulation, on-shell means that the Euler-Lagrange (EL) equations $$\tag{1} \frac{\delta S}{\delta\phi^{\alpha}(x)}~\approx~0$$ are satisfied. It seems that the potential confusing point is the notion of an ... 1 When one throws a stone.... Your arm is capable of propelling an object at up to around 150km/h (and that's with some practice). At that speed the many factors like air resistance are negligible. Let’s load a 16-inch shell into a gun (you will find several on the ISS Iowa), aim it 45 degrees up and press the button. The shell is going to go up at ... 4 A lot of things have to hold to get that symmetry. You have to neglect air resistance. You either have to throw it straight up, or the ground over there has to be at the the same altitude as the ground over here. You have to through it slowly enough that it comes back down (watch out for escape velocity) But if you have that, then the simplest explanation ... 0 Assuming that the only forces acting on the stone are the initial force exerted by the hand on the stone AND the force of gravity, we have the following situation: Immediately after throwing, the stone has a kinetic energy in the x-direction that never changes (conservation of energy) and the stone has a kinetic energy in the y-direction that is constantly ... 6 I would consider that since acceleration is a constant vector pointing downward, that the time the projectiles downward component takes to accelerate from V(initial) to 0 would be the same as the time it takes to accelerate the object from 0 to V(final) 2 (Non-kinematics math attempt but just some principles) It is a partial observation in that It hits the ground with same speed. Angle by which it hits the ground is the same (maybe a direction change) It takes equal time to reach to the peak and then hit the ground They are equally strange coincidences. Which are more fundamental? Consider the following ... 1 Asking "Why" in physics often leads out of physics and into philosophy. Why is there light? Because God said so. Physics only answers questions about how the universe behaves and how to describe that behavior. If a why question can be answered with physics, the answer is "because it follows from a law of physics." A law is just a mathematical description of ... 3 One could say that this is an experimental observation; after one could envisage, hypothetically, where this is not the case. This is not hypothetical once you take air resistance into account. One could say that the curve that the stone describes is a parabola; and the two halves are symmetric around the perpendicular line through its apex. But ... 4 I think its because both halves of a projectile's trajectory are symmetric in every aspect. The projectile going from its apex position to the ground is just the time reversed version of the projectile going from its initial position to the apex position. 14 I would say that it is a result of time reversal symmetry. If you consider the projectile at the apex of its trajectory then all that changes under time reversal is the direction of the horizontal component of motion. This means that the trajectory of the particle to get to that point and its trajectory after that point should be identical apart from a ... 3 Comments to the question (v2): First of all, recall the notion of an (off-shell) quasi-symmetry. It means that the action$S[\phi]$changes by a boundary integral under the transformation of the fields$\phi$and spacetime point$x$, cf. e.g. this and this Phys.SE posts. Since the action$S=\int_R \!d^nx~{\cal L}$is an extensive variable, it is clear ... 1 Hint: The fusion rule Clebsch-Gordan-like coefficients$c_{ij}{}^k=c_{ijk}$are related to the 3-point function$\langle \phi_i(x) \phi_j(y)\phi_k(z)\rangle$of 3 primary fields, which in turn is totally symmetric. 1 You are right that that the symmetry breaking breaks all three symmetries of$SU(2)$. Thus the$SU(2)$generators give you three goldstone bosons in the theory with broken symmetry. However, we have not yet considered all of the symmetries of the original theory. We know that the full symmetry group has six generators and that five of them must be broken. ... 2 Suppose$g$is the generator of a certain symmetry (i.e. the generating function of an infinitesimal canonical transformation) and you are interested to know how the observable$f$changes after the "action" of$g$. In the Hamiltonian formalism the change is found to be $$\delta f \approx \epsilon\{f,g\}$$ which can be related to the time evolution of an ... 0 If your transition from high order to low order occurs slowly over a large number of lattice periods, then you can try doing a discrete Fourier transform of the area of interest taking some windowing function to focus your the computation to area of interest. Then if you get relatively sharp peaks (up to spectral leakage of the window you chose), the ... 2 Comments to the question (v5): If an action functional$S$is invariant under a Lie algebra$L$of symmetries, the corresponding Noether currents & charges do not always form a representation of the Lie algebra$L$. There could be (classical) anomalies. In some cases such (classical) anomalies appear as central extensions, cf. e.g. Ref. 1-3 and this ... 4 Comments to the question (v2): Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs$n$infinitesimal symmetries to deduce$n$conservation laws. If one is only interested in getting the$n$conservation laws one by one (and not so much interested in the fact that the$n$conservation laws often ... 1 How atomic orbitals merge into crystal band structure is, well, complicated to capture in simple models. As seen in, say, Ashcroft and Mermin (chapter 28), the energy surfaces for Si have symmetry along the <100> directions. In contrast, the surface for Ge have symmetry along the <111> directions, with the band minimum at the zone edge. A comparison ... 1 The answer is that the term symbol refers to the entire, multielectronic, molecule. You are indeed correct that a single electron in a$\Sigma$state must be symmetric under reflection about a plane that contains the internuclear axis. If you have multiple electrons, however, you can still have a global antisymmetry under such reflections, and get their ... 3 This is a very good question. First of all, you are absolutely correct that, for a single electron, invariance under$\phi\to-\phi$means invariance under$y\to-y$. This is obvious just from looking at the coordinates$x=\rho\cos(\phi)$and$y=\rho\sin(\phi)$. It is clear that$\phi\to-\phi$means the exact same thing as$y\to-y$. The answer is a little ... 1 Except that they don't have the same eigenvalue. To understand it in terms of differential operators; divide the first equation by$1/r^2$, this gives you an eigenvalue equation for the total hamiltonian operator$H = -h^2\nabla^2/2m + V$, whose eigenvalue is clearly$E$. Then we divide$\nabla^2 = \hat{R} + 1/r^2 \hat{\Omega}$in your notation (into radial ... 0 Maybe I am missing the point, but if the article says that the wave functions are "either symmetric or antisymmetric" about this axis, then for the Z=0 case they are (only ever) symmetric which is indeed one of the allowed cases. "Either... or" doesn't mean "it has to be capable of both", but "it can only be one of these". 1 JakobH's comment as an answer: The electroweak gauge group$\mathrm{SU}(2)_L \times \mathrm{U}(1)_Y$is broken into the electromagnetic$\mathrm{U}(1)_\text{em}$by the Higgs field acquiring a non-zero vacuum expectation value, granting masses to the quark and$W^\pm,Z$bosons. Thus, at the scale where up- and down-type quarks have very different masses, we ... 0 In layman's terms, it just means that the laws of physics are the same everywhere. This means that we are talking about one common set of laws. The fun part is figuring out how one common set of laws can behave the same, while they are taking place within different frames of reference. Thus we have a one, that is shared by a many. How can this be, when each ... 0 It's not that hard to see how a rotation can end up being represented by a matrix of dimension$(2j+1)\times(2j+1)$. The key concept is that this matrix acts on a subspace$V$of the Hilbert space$\mathcal H$; that is,$V$contains state vectors (kets). Generally,$V$is required to be an invariant subspace in the sense that if$v\in V$, then under a ... 0 The representation of the abstract Lie group$\mathrm{SO}(3)$on the usual space$\mathbb{R}^3$is known as the fundamental representation of the group. No other representations usually occur in classical mechanics because we mostly just have$\mathbb{R}^{3N}$as the space of spatial coordinates for$N$objects on which the rotations act. The rotations must ... 0$SO(3)$is 2-connected and it turns out that$SU(2)$is its universal (simply connected) covering group. Since there is then a covering homomorphism$\gamma:SU(2)\to SO(3)$, which is a local isomorphism, one can then consider (as a consequence of Peter-Weyl) theorem, the irreducible representations of$SU(2)\$. These can be parametrised by the points in the ...

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this phrase doesn't specify what laws are invariant It doesn't need to since it is a guiding principle, a razor. It is a statement about the nature of physical law. Put another way, on this principle, an alleged 'physical law' that isn't invariant under inertial coordinate transformations is not a genuine physical law. or even what it means to be ...

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The laws of physics are invariant means slightly different, but (almost) equivalent things depending on what formulation you are working with. Given a collection of transformations (a symmetry/transformation group) and a Lagrangian formulation, you can check whether the Lagrangian changes when you apply the transformation. If it does not change (or ...

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