Tag Info

New answers tagged

2

Electric flux is given by: Here, S (vector area is same) while the strength of the electric field E is more in A (length of arrows). $\theta=0$ in both cases. Hence flux is not same. A > B


0

After some more thought, I think I've realized what's going on. The short answer is that the above result is correct, and is just a specific case of a more general construction. Here's more explanation: the point is that the original tangent field $k^a$ is only defined on the geodesic $\gamma$, but the expression $\nabla_a k^b$ is only sensible if the ...


3

This is called Helmholtz theorem, which states that for any vector field $\vec{F}$ that is twice continuously differentiable in a bounded domain, we can perform the decomposition $$ \vec{F} = - \vec{\nabla} \Phi + \vec{\nabla}\times\vec{A} $$ See http://en.wikipedia.org/wiki/Helmholtz_decomposition for a derivation


3

Stokes' theorem needs no physical reason to be true. However, there is a nice intuitive description of the two-dimensional case. Tesselate the surface with little (infinitesimal) oriented squares and consider the integral as the sum of the curl on all these little squares: The inner sides of the squares have no contribution to this sum at all, because ...


0

Velocities and Spatial Accelerations are twists and Forces and Momenta are wrenches. Both are screws (two-vectors) with one vector free and the other a spatial field. All of them transform with the same laws and their interactions have many dual properties. NOTE: See "A treatise on the theory of screws", Stawell R Ball, ...


0

So,the integral is equal to zero because the Q that is enclosed in the Gaussian surface is zero.That does not necessarily mean that the electric field is equal to zero.The only other way for the integral to be equal to zero is if the sum of the dot products inside the integral is equal to zero.It means that you have equal negative dot products as positive ...



Top 50 recent answers are included