# Tag Info

3

Both concepts are mathematical in character and they ultimately describe the same characteristics or situations. "Invariance" is a more technical word because it says "what has to be equal to what" for us to say that the symmetry exists. In particular, the "invariance under a symmetry transformation" means that an object, like the action $S$, has the same ...

2

Draw an arrow to represent a vector, with its length representing the vector magnitude. Draw a coordinate system and get the components of the vector. Now draw another coordinate basis, rotated with respect to the first, and get the components with respect to the new basis. The length of the arrow is the same in both systems - i.e length is invariant - ...

1

A helpful yet elementary answer may do the trick, If you are familiar with the Euler-Lagrange equation then it will be straight forward and you can skip ahead a little. If not then you have to accept that there is an equation in physics that generalises classical mechanics called the Euler-Lagrange equation. For a particle moving in one dimension under a ...

0

It is difficult to understand conservation from symmetry. But the opposite is much simpler. conservation means the invariance of equation of motion in its form under certain transformation. and the invariance of equation of motions arises as an implication of the underlying symmetry. for example i am taking one equation X2 =1 the solutions are +or-1 which ...

1

It's because first homotopy group for configuration space of one-particle states in 3D space is permutations group. In two words: scalar product $\langle \mathbf p_{1}, \mathbf p_{2}, ...| \mathbf k_{1}, \mathbf k_{2} , ...\rangle$ is invariant under simultaneous permutations $\mathbf p$ and $\mathbf k$ for different particles. For identical particles, ...

3

First, terminology: Symmetry groups are not "defined on domains". Symmetry groups exist in the abstract, and they are then represented on certain spaces. If we have a spacetime manifold $\mathcal{M}$, then the fields are functions $$f : \mathcal{M} \to V$$ where $V$ is some vector space upon which a representation $\rho : \mathrm{SO}(1,3)\to ... 2 You are approaching the question from the wrong end. The expansion of the universe is described by a particular solution to Einstein's equation called the FLRW metric. To derive this metric we have to make some assumptions, and the key assumptions are that the universe is isotropic and homogeneous i.e. that it is the same everywhere. So the universe being ... 2 Yes. You can make a model where you have coordinates$t, x,y,z$where for any$x,y,z$the universe looks the same. The metric ends up looking e.g. like $$ds^2=dt^2-(a(t))^2(dx^2+dy^2+dz^2)$$ and you can move your$x,y,z$to have any value and everything looks the same (those things do loom different for different cues of$t$). You end up with the densities ... 0 Well for the three sides touching the particle, the vector$\bf E$coming from the particle, is always contained in the surface. This means that there is no component in the direction normal to any of the surfaces. So this means$\bf E$$d \bf S and therefore the flux through them, \int \bf{E} d \bf S will result zero. 0 1) One eighth of the charge is contained in the cube because you can place eight cubes such that they have a common vertex, the one your charge is at. If your charge was on the middle of an edge, it would be shared by four cubes. If your charge was in the middle of one of the faces, it would be shared by two cubes and so on. 2) The three adjacent faces have ... 1 Suppose you have a charge in a plane. The electric field from the charge is spherically symmetric, and that means every field line from the charge that intersects the plane has an equal and opposite field line intersecting the plane. So when we integrate \mathbf{E}\cdot\text{d}\mathbf{A} the two field lines will cancel out and the net flux will be zero. ... 1 My question is how the three sides don't contribute any flux whereas we can see that a small fraction amount of flux can pass through the three sides. Not really. You see, the electrostatic field \vec{E} of the charge is always radially outwards. If the charge is situated at the exact corner of the cube, then the field is exactly coplanar with the ... 0 Hint: You can always construct imaginary cubes (7 more, in this case.) such that your charge is completely enclosed inside your multi-cube system. You can find the flux through this system via Gauss' Law. The total flux through each side (of the new configuration) would be equal via symmetry. 1 [I somewhat haphazardly pieced this answer together, so I'm not absolutely certain the conclusion is correct.] Cayley's theorem is useless here, because the group isomorphism it produces is not required to preserve any kind of topology on the groups, in particular not notions of continuity or differentiability. On the infinite symmetric group S_\infty on ... -1 The essence of the Higgs mechanism is that it allows the breaking of the (gauge) symmetry to grow a mass for the gauge (vector) bosons, which are necessarily massless in the unbroken symmetry. The Higgs scalar and the two degrees of freedom of the massless vector boson combine to form the three degrees of freedom of a massive vector boson. Goldstone's ... 0 according to me a body is homogeneous when the properties that defines its physical structure are same at all points(or space) while a body is isotropic if the value of properties,that affect some physical phenomenon,is same in all directions 0 It would be better to specify that V_0 < 0 as otherwise your problem is that of a potential barrier (no bound states) and not that of a finite potential well. The case of a finite potential well is fully developed here (Wikipedia entry), the case of a rectangular potential barrier you can find here (Wiki). 1 This question inspired me to try to write a conceptual introduction at the wikipedia article. To save you the trouble of clicking, I copied it below. (It's slightly inspired by what @Kostia wrote here) Motivating example: Position operator matrix elements for 4d→2s transition Let's say we want to calculate transition dipole moments for an electron to ... 0 I have worked recently on something related to this and I have understood that if you have a time evolution that respects a certain symmetry group, then the state space breaks down into its algebra multiplets. For instance, if you have 2 bosonic modes, you can recover the angular momentum algebra su(2) by labelling them by the total photon number and by ... 1 If your system consists of (a) a uniform electric field along direction \hat{\mathbf{n}}, or (b) a uniform magnetic field along direction \hat{\mathbf{n}}, then its symmetries are translation orthogonal to \hat{\mathbf{n}}, translation parallel to \hat{\mathbf{n}}, and rotations about \hat{\mathbf{n}}. Some of these transformations are ... 1 I edited your question (I moved one parenthesis so that your quote was not a misquote), and maybe that answered your question. In case not ... If you had a Dirac spinor \Psi then there is a transformation like \Psi\mapsto e^{\theta\gamma_{13}/2}\Psi, where \gamma_{13}=\gamma_1\gamma_3 is a product of gamma matrices (or just the unit xz plane if you ... 2 The appropriate definition of symmetry uses infinitesimal quantities, not just small quantities. Thus, in terms of your question, the Lagrangian is symmetric if dL/d\epsilon=0 at \epsilon=0. In terms of your example (rotation of a 2D harmonic oscillator), we have$$ L \to (1+\epsilon^2) L = L + \mathcal{O}(\epsilon^2)  Thus to first order in ...

Top 50 recent answers are included