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The first method is not correct by assuming a shperical symmetric geometry. For an element charge locates on the cylinder $\vec{r}' = (r', \phi', z')$. It contribution to the field point $\vec{r} = (r, \phi=0, z=0)$, $d\vec{E}$ will be along the direction $\hat{n}$ unit vector along the $\vec{r} - \vec{r}'$. The square distance $$|\vec{r} - \vec{r}'|^2 = ... 1 If you denote the external points by x,y and the internal points by 1,2,3 than the diagrams of the last line in your picture are  x1, 11,12,23,23,23,3y and x1,12,12,12,23,33,3y with integration over the internal points and what is between brackets is contracted. Yes, they are the same. 0 Forget about quantum mechanics and imagine a classical particle of mass m moving in a Mexican-hat potential (\phi^2-v^2)^2, with some friction so it doesn't roll back up the hill after rolling down. If you start it at \phi=0 then it will roll down the hill (to the right, let's say) and oscillate around \phi=v with a frequency of ω=\sqrt{8v^2/m} in ... 1$$ \frac{\exp[ik(x-y)]}{k^2-m^2+i\epsilon} = \frac{\cos(k(x-y)) + i\sin(k(x-y))}{k^2-m^2+i\epsilon} $$The k odd sin part drops out upon integration, leaving only the k even cos part. Therefore, the sign of k does not matter. 0 A simple cubic lattice is a Bravais lattice, i.e., it can be thought of as originating from the set of (infinite) translation of a cube of side a along three orthogonal axes parallel to the cube edges, according to the formula$$ {\bf R} = n_1 a {\bf \hat x} + n_2 a {\bf \hat y}+ n_3 a {\bf \hat z},~~~~~~~~~~~~(n_1,n_2,n_3) \in {\mathbb Z}^3. $$It is ... 0 By gauss law, we have that \displaystyle \int \vec{E} \cdot \mathrm{d} \vec{S}=\dfrac{Q_{\text{enc}}}{\epsilon_0}, so if for example, when a charge is kept on the face of a hemisphere, the charge is not enclosed. We want to enclose it fully in a body such that electric field still remains uniform so that gauss law can be applied. So, we complete the rest ... 1 You threw five and a half pots on the fire, intricately entangled, which are addressed by over a dozen questions and answers on this site, but I will focus on a very narrow one, which I suspect might be at the bottom of your question. That is, I am going to avoid quantization, gauge invariance, matter fields, and symmetry breaking, and focus on the globally ... 0 Sean Carroll and other "mad-dog everettians" try to move that way. He lectured about this in Harvard Foundations of Physics series. The 't Hooft's Cellular Automation Theory has similar view on symmetries also. 2 It can be attributed to the fact that the corresponding free Lagrangian density {\cal L} only contains terms with an even number of spacetime derivatives. 2 After reading the comments and searching a bit more, I think I've managed to find the solution. Considering the two displacements r_1 and r_2: applying the projection formalism should give, for r_1 or r_2,$$P r_1 = \frac{1}{4}\left(D(E)+D(C2)+D(\sigma_v)+D(\sigma'_v)\right) r_1\Leftrightarrow P r_1 = \frac{1}{2}(r_1 + r_2)$$which indicates us ... 1 welcome in the stack-community. Here are my proposed answers: \mathrm{I}. A pulsar is a neutron star that is mostly made up of neutrons. On the surface, gravitational pressure does not hinder the \beta^{-} decay of neutrons, and so charged particles such as electrons and protons can form a magnetic field due to the whirling rotation of these objects. The ... 1 When we say that an eigenstate \phi_n transforms as an irrep \rho of a group G, we mean that it belongs to a subspace of the full Hilbert space which is mapped onto itself under the action of \rho (in the present context, this subspace is an eigenspace for a particular Hamiltonian eigenvalue). That is, \phi_n belongs to a subspace V such that for ... 2 One way to see that the curvature tensor is zero is to start with the fact that there is no dependence of metric components on null coordinate v. So performing Kaluza–Klein reduction along the Killing vector \partial_v we would obtain a Newton–Cartan spacetime with two spatial coordinates (x and y), while u now becomes Galilean time coordinate. ... 0 Heuristically, the curvature being constant implies that all points are equivalent in terms of metric relations, and all directions are equivalent. More precisely, I am somewhat confused about the exact terminology used in the mathematical literature, but generally speaking a (pseudo)-Riemannian space is maximally symmetric, if it is homogenous, i.e. for ... 1 The collection of transformations that you have is indeed paramatrised by \alpha. Speaking from the math perspective, when people say that you have a continuous symmetry, they almost certainly mean that you have a representation of a Lie Group, in this case the Lie Group would be U(1) - which is the unit circle in \mathbb{C}. Note that this is ... 1 With mixed symmetry tensors (i.e. tensors that are not just one row or just one column) it is impossible to have symmetry under exchanges in the same row and antisymmetry under exchanges in the same column at the same time. One of them has to give, as you correctly noted. The difference between the two tensors is that in case (1) you do the symmetrization ... 1 The Jordan-Wigner transformation doesn't require a U(1) symmetry. All it requires is a \mathbb Z_2 symmetry. You can easily see this if you reverse engineer your way from a general fermionic Hamiltonians. Since all terms in the Hamiltonian must have an even number of fermionic creation/annihilation operators, the resulting spin model obtained from a local ... 2 The case of your model with the field along \hat{x} is known as the transverse-field XXZ chain, see e.g. Dmitriev et al., One-dimensional anisotropic Heisenberg model in the transverse magnetic field, JETP 95, 538 (2002). The combination of transverse and longitudinal (along \hat{y}) was discussed in Dmitriev and Krivnov, Phys. Rev. B 70, 144414 (2004). ... 4 There's actually no contradiction between what you say and what the two linked documents say. In fact, the two documents describe the solution of the wave equation in the form of a tight-binding solution, which is indeed a Bloch wavefunction, but it's constructed as the sum of wavefunctions with terms centered in the two sublattices. Recall that the idea of ... 3 Personally, I've always found the concept of symmetry factors more confusing than they're worth. I find it much easier and faster to simply count the number of contractions. The diagram you are looking at includes the fields$$\phi_1 \phi_2 \phi_x\phi_x\phi_x\phi_x \phi_y\phi_y\phi_y\phi_y$$\phi_1 can contract with 8 possible fields (any of the \phi_x ... 3 OP's main issue seems to be that the symmetry factor S depends on whether the external legs are distinguishable (S=4) or indistinguishable (S=8), respectively. On one hand, P&S on p. 93 consider the distinguishable situation^1 where there are not integrated over the external positions/momenta. On the other hand, OP apparently considers the ... -1 Following on the above, if we let M be a constant, it seems we can then integrate GMdm/2R^2 over dm (from 0 to M) to obtain a sum or total of gravitational forces on the surface of the sphere, obtaining GM^2/2R^2. 2 You could use the virial theorem result for a system interacting solely via a \frac{1}{r} interaction that it has an average kinetic energy equal to half the negative of the average potential energy. Since the relative positions and velocities don't change, the averages in this case are the same as the instantaneous values. There are 2 pairs a distance a ... 0 In SI units, the E-field amplitude is c=3\times 10^8 m/s larger than the B-field amplitude. But this is just a consequence of choice of units. It is perfectly possible to adopt a system of units where c=1 and E=B. Then, the diagram of the fields at a snapshot in time would be as shown in your question, with similar numbers along the axes perpendicular ... 2 The easiest way to work out your algebra, in case you haven't, and are alarmed by the missing constant term, is to rewrite your potential, eliminating the pestiferous unphysical \mu^2 that has caused grief to generations of students. Recalling that summations over repeated indices is implied, the potential is$$ \tfrac{\lambda}{4} (\phi_i \phi_i - v^2)^2 -\...

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Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\... 4 Easiest way is to do what you said. It's rather simple and quick. There are however some tricks within that method you can use to simplify the calculations. Namely, realize that the each mass will experience the same net force. This is essentially reducing the problem to a 1-body problem. The solution is exactly what you pointed out: I'll focus on the ... 1 Perhaps look at the equation of motion for the scalar length(-squared) between any two of the bodies. This is the quantity you are directly interested in but I can't promise the algebra will be significantly easier. 1 When studying a quantum field theory without spontaneous symmetry breaking, we are typically only able to access the states which perturb around the vacuum (practically speaking). These include particle states, which only make sense as an excitation from the vacuum. The same is true in the spontaneously broken case, the physical particle states only make ... -1 I guess it's because, if you look at equation 7.20, the wavefunction for the system can be separated into ψ_{system} = ψ_{CM} · ψ_{rel}. And ψ_{CM}, as shown in your figure (equation 7.30), looks like a phase factor to ψ_{rel}, i.e. e^{i·c} where c is some constant. It won't affect the calculation of probability for relative motion quantities as ... 0 Look at this example the equations of motion Newton second law and third law.$$m_1\,\vec{\ddot{r}}_1=\vec{F}_1=\frac {G\,m_1\,m_2}{\vec r_1\cdot \vec r_2}\,\vec{\hat e}_{12}m_2\,\vec{\ddot{r}}_2=-\vec{F}_1=-\frac {G\,m_1\,m_2}{\vec r_1\cdot \vec r_2}\,\vec{\hat e}_{12}$$if you add those equations you obtain.$$m_1\,\vec{\ddot{r}}_1+m_2\,\vec{\ddot{r}}...

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In the isospin formalism you treat them as identical particles. If you have two electrons, one is spin up, and one is spin down, the electrons are still identical they just have different z-projections of their spin. This same logic applies to isospin. Instead of a spin up component and spin down component like the electron, you now have $\pi^+,\pi^0,\pi^-$ ...

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The true fact is that conservation of total momentum is a consequence of translational invariace in Lagrangian mechanics. I.e. the fact that the Lagrangian is translationally invarint implies the conservation of total momentum. However in Newtonian mechanics, under the assumptions that in an inertial system and for an isolated system of material points, (a) ...

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the cubic lattice is defined by the four 3-fold rotation axes along the main diagonals of the cube, not by the presence of 4-fold ones. This is the reason why you have primitive, face-centred and body centred cubic lattices (but not a base centred cubic lattice as in the orthorhombic system that would break the 3-fold rotation symmetry). there are several ...

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$M_L$ and $M_S$ are actually still good quantum numbers for stretched states even for non-zero external fields. You have already established that the energy does not depend on $M_J$. The number of states is "conserved" as you break symmetries. Each state in the broken-LS coupling régime is "adiabatically" connected to the zero-field LS ...

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It's Wolphram jonny's answer with pictures

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Kirchhoff's junction rule: The sum of all current leaving a junction must be equivalent to the sum of all current entering it. Take the bottom-middle junction for instance. We know that $1.51$ A enters the junction from the left and that $1.51$ A leaves the junction from the right. As such, by Kirchhoff's junction rule, there is no current remaining to run ...

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Imagine voltage/potential at some point as density of electrons passing on a road. They all need space, so if there's a place near them with lower density, part of the electrons move there. Now, let's say on the bottom end of the wire in question there is a potential, so the density of electrons is higher than usual. They see they can go up to the middle ...

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You can connect a resistor (or a solid short for that matter) between any 2 equipotential points and it will have no effect on the circuit. Because both resistor terminals are at same potential no current flows. Remove your middle resistor, analyze the circuit, and you will see it clearer. p.s. Voltages are meaningless unless measured to some reference ...

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Try reversing one of the batteries or changing its voltage. This can help develop intuition. In the outer wires, both batteries try to push electrons in the same direction but on the inside they push on the opposite sides and because they are both $1.5V$, they will cancel each other's effort. If one of them was stronger you would have current in the middle.

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Another intuition: both circuits are the same but in reverse from the point of view of the center wire, so each provides the same current but in opposite directions, thus cancelling each other.

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Consider the potential. Let’s call the left upper corner A and right upper corner B. First see upper side. There are two batteries and they’re symmetric. It is easy to imagine that the potential of the mid point of the upper side is the average of A and B. Then see the bottom side. It’s also symmetric. So the potential of the mid point of the bottom side ...

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For what it's worth, it is useful to rewrite OP's scalar field $$\phi~\in~\mathbb{H}~\cong~\mathbb{R}^4\tag{1}$$ as taking values in the quaternions, $${\cal L}~=~|\partial\phi|^2+m^2|\phi|^2-\lambda(|\phi|^2)^2. \tag{2}$$ The 6-dimensional global symmetry group is $$G~=~U(1,\mathbb{H})_L\times U(1,\mathbb{H})_R.\tag{3}$$ Here \begin{align}U(1,\mathbb{H})... 4 This is something like the inverse fallacy. :) Goldstone's theorem tells you that for every broken symmetry generator, you must find one massless mode in the spectrum of the theory. It does not tell you for every unbroken symmetry generator, you must find one massive mode in the spectrum of the theory. If that were the case, then if you had chosen a symmetry-... 3 Two complex scalar fields \phi_{1} and \phi_{2} can be rewritten as four real fields, in terms of their real and imaginary parts,\Phi=\sqrt{2}\left[\begin{array}{c} \Re\{\phi_{1}\} \\ \Im\{\phi_{1}\} \\ \Re\{\phi_{2}\} \\ \Im\{\phi_{2}\} \end{array}\right].$$For the free theory, the Lagrange density is actually equal to$${\cal L}=\frac{1}{2}\partial^...

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The classical treatment is actually simpler than the quantum one, since there are no operator ordering problems. This is what Pauli relied on in his solution of the Hydrogen atom, before Schroedinger, as every decent physicist of his generation had worked out the Kepler problem like this. The idea is to find as many independent constants of the motion Q as ...

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I think the problem becomes easier upon a little re-organization. The relevant Maxwell equations may are \begin{gather} \nabla\times\vec E=-j\omega\vec B, \\ \nabla\times\vec H = j\omega\vec D, \end{gather} where we have assumed time harmonic fields and are using the $e^{j\omega t}$ time convention. The constitutive equations may be written in terms of \$...

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The article Axially symmetric spacetimes: numerical and analytical perspectives is a very nice reference and treats axially symmetric spacetimes in generality, showing how to construct the "cylindrical" slices (see also this). If your spacetime is not only "axially symmetric" but also "circular" (i.e. there are no meridional ...

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The standard way to fix this is to adopt the Belinfante-Rosenfeld tensor instead of the Noether tensor. This raises hairs. Why is the Noether tensor wrong? Not only the EM tensor must be symmetrized, also the angular momentum tensor had problems. Why is this Noether current also wrong? This problem must then be solved by ditching spin and writing it as an ...

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