Recent content by Orodruin

Of course they can, but not at all times.

I think I found my issue. I was trying ##\sin^2(\theta)##, but when mapping from SU(2) to SO(3), the ##t## here is actually ##\theta/2##? Using ##\sin^2(\theta)## I obtained that the fundamental representation should contain the trivial one once, which would be absurd. Using...

When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g)...

For things moving free or at rest, Observe what the first law does best. It defines a key frame, Inertial by name, Where the second law then is expressed. Source: https://www.physics.harvard.edu/undergrad/limericks

You’ve gotten things wrong. The basis is a choice. The typical coordinate basis depends only on the coordinates chosen - the manifold does not even need to have a metric.

Your post seemed to indicate otherwise …

This dependence is present also far away from the dipole.

There is no need to split into cases as you do. The third index is just an additional copy of the 2-index system for every possible value of the index as compared to the basic 2-index systems you mentioned. It follows directly that the symmetric case is 3x6 = 18 and the anti-symmetric 3x3 = 9.

Hint: You don’t need to compute the flux integral. Edit: with the dimensions of A, the cylinder does not completely enclose the sphere. The cylinder radius is too small.

There is actually a generalization of the cross product to arbitrary dimensions that will do the same trick asthe cross product. However, it maps two vectors anti-symmetrically to an antisymmetric N-2 tensor. The end result can be expressed using the generalized Kronecker delta.

A scalar is a function on the base manifold. A vector is a member of the tangent space of spacetime. A traceless symmetric tensor is a linear map ##T## from the tangent space to itself such that, for any orthonormal basis ##\{\vec E_\mu\}##, it holds that $$ \sum_\mu g(\vec E_\mu, T(\vec E_\mu))...

On the contrary. It is very reassuring that the components of ##\vec A## add up to ##\vec A##. You should worry if they didn’t! The more relevant conputation is showing that the orthogonal component is indeed orthogonal to ##\hat n## and that the parallel component has the same inner product...

I actually think ##\vec A - (\vec A \cdot \hat n)\hat n## is nicer than the double cross product. It also generalises to any dimension.

The answer given is a special case for three dimensions. The cross product is not defined in dimensions other than three. The decomposition is however valid in any number of dimensions.

There is no object at all in the Schwarzschild spacetime. It is a vacuum solution. The assumptions are spherical symmetry of the solution and vacuum.