Author/Teacher from India. Interested in General Relativity and other areas of physics
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Jan
27 |
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awarded | Cleanup |
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Jan
27 |
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revised |
A Paradox in Special Relativity deleted 2 characters in body |
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Dec
15 |
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comment |
This is Quite Strange! Voters don't have to give explanations for a negative vote----it is a secret ballot.Justification is not mandatory. Deletion by software in this case is based on such negative voting which does not require any reasoning. And you call it ethics! |
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Dec
4 |
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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. |
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Dec
4 |
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asked | Can Parallel Transport always move a Vector Parallel to Itself? |
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Dec
4 |
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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? |
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Dec
4 |
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comment |
This is Quite Strange! If the software deletes an entire posting with comments without informing the concerned poster[who has an email subscription] I have to say that the software was "POORLY" developed.Proper considerations were missing while creating the software.The "software" did not notify the moderators of the deletion nor did it register the date of deletion. |
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Dec
4 |
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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 |
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Dec
3 |
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awarded | Quorum |
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Dec
3 |
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comment |
This is Quite Strange! The justification issue could have been reasonably handled at the time of closure by informing me. Keeping the original poster uninformed is quite unethical.I suddenly noticed that my posting was not there. |
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Dec
3 |
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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. |
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Dec
3 |
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This is Quite Strange! I have an email subscription by which they could have informed me of the closure in case it was closed in a legitimate manner. I have received emails in relation to comments regarding the same posting earlier.But nothing in relation to the closure! |
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Dec
3 |
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awarded | Student |
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Dec
3 |
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asked | This is Quite Strange! |
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Dec
3 |
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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. |
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Dec
3 |
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On Parallel Transport (in continuation) The above idea is embodied in the non-zero value of the Christoffel tensors in curved space. |
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Dec
3 |
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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 |
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Dec
2 |
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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. |
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Dec
2 |
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On Parallel Transport Quoting Ron Maimon:"It ends up in nearly exactly the same direction as it was when you start the parallel transport"---in such a situation if you move back to A the vector does not turn by alpha. It turns by a much smaller amount! On removing the triangle the vector in the initial and the final situation (at A) make a very small angle. If the triangle is there it turns by 90 degrees after looping round |
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Dec
2 |
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On Parallel Transport Only if you consider a geodesic the an infinitesimally think space round it and parallel to it is nearly flat--the tangent vector propagates parallel to itself |
