# Tag Info

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 ...
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 $\rho : \mathrm{SO}(1,3)\to ... 2 Yes. You can make a model where you have coordinates$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 ... 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 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 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, ... 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 ... 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 ... 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 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 ... 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 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 ... 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. ...