Anamitra Palit
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 Mar 24 comment Is it necessary to embed a 4D surface in 5D space? A tangent vector of finite length protrudes out of the manifold.The null vector is parallel to and perpendicular to itself .It is perpendicular to the manifold . It protrudes out of the surface--into what? Mar 20 awarded Popular Question Nov 5 awarded Popular Question Sep 24 awarded Autobiographer Jul 2 awarded Curious Jan 27 revised The Black Hole Problem rolled back to a previous revision Jan 27 revised The Particle-Antiparticle Problem in Relation to Special Relativity rolled back to a previous revision Jan 27 awarded Cleanup Jan 27 revised A Paradox in Special Relativity rolled back to a previous revision Jan 27 revised A Paradox in Special Relativity deleted 2 characters in body Dec 4 comment Can Parallel Transport always move a Vector Parallel to Itself? You are not allowed to rotate the tangent planes wrt each other-----they are fixed on the curved surface at the two points concerned. Dec 4 asked Can Parallel Transport always move a Vector Parallel to Itself? Dec 4 comment On Parallel Transport Does "parallel transport" move a vector parallel to itself on a curved surface even in the infinitesimal sense? You may think of two adjacent tangent planes on a curved surface.Is it always possible to have parallel vectors at the points of contact(one vector being preassigned) even if the planes are awkwardly inclined? Dec 4 comment On Parallel Transport At the point N'(referring to the original posting)you may consider a second vector tangent to the latitude-line at N'. If it(2nd vector) is parallel transported along the latitude to the point N" on the meridian NB, it is no more a tangent wrt to the latitude at N". The angle this vector makes with the tangent at N" should be equal to the angle which the first vector(moved up from the equator) makes with the meridian at N". This angle is expected to be small Dec 3 awarded Quorum Dec 3 comment On Parallel Transport Points to Observe:(1)The vector after going through a loop[by parallel transport] rotates by a negligible small angle though the enclosed area is large. You will find this in the example I have referred to in the comment after the question after Lubos Moti's comment.(2)An infinitesimally small area is not sufficient to warrant a flat space-time.The Christoffel symbols are point functions. They have non-zero value for curved spacetime. Dec 3 comment On Parallel Transport I am referring to the change between the initial and the final positions of the vector when it goes round a loop on a curved surface(by parallel transport). You may connect the point N'(in the original posting) with some point on the meridian NB by a $small{\;\;}$ curve so that the transported vector on landing on the meridian becomes tangential parallel or nearly tangential to the meridian NB. Dec 3 comment On Parallel Transport (in continuation) The above idea is embodied in the non-zero value of the Christoffel tensors in curved space. Dec 3 comment On Parallel Transport Let's's consider the vector components $A^\gamma(x^\alpha)$ and $A^\gamma(x^\alpha+d x^\alpha)$. They have an infinitesimally small separation.This is not indicative of flat space-time over the infinitesimally small spacetime region concerned .Reason:For the purpose of calculating the derivative we have to parallel transport the vector-component $A(x^\alpha +d x^\alpha)$ to the location $x^\alpha$ and this vector definitely changes its orientation wrt to its initial position even though it has moved through an infinitesimally small distance Dec 2 comment On Parallel Transport You may consider a small curved line from N' to the meridian NB so that the vector becomes parallel to the meridian NB on reaching it. At A there is no turning all due to the exclusion of a small area.