Recent content by the shadow

Thanks man . I understand now . And I'll try to solve the 4D sphere's equation , but it'll take a bit of time . I use a program called (TeXaide) ( you can download it from http://www.dessci.com/en/products/texaide/" is the way to use it . P.S. Using TeXaide is easier than using Latex...

Hi , wofsy look , I'll solve the equation and you tell me the meanning of the solutions : \[ \begin{gathered} X\left( {u,v} \right) = \left( {\cos u\cos v,\sin u\cos v,\sin v} \right) \hfill \\ g_{ij} = \left\langle {\frac{{\partial X}} {{\partial x^i }},\frac{{\partial X}}...

Hi , everyone I think that the length contraction is due to the Relativity of simultaneity . To measure the length of an object we must know the space-time coordinates of the point of it's beginning and the point of it's end at the same moment and form this information we can measure the...

Hi , wofsy In the the solution that I gave , I arrived at that the geodesics are straight lines . In the case of the 2 - sphere ( a sphereical surface in a 3D space ) I can't understand how dose the solutions tell us that the geodesics are great circles . Could you explain that ? Thanks .

Here is my stupid solution : \[ \begin{gathered} X\left( {a,b} \right) = \left( {\cos a,\sin a,\cos b,\sin b} \right) \hfill \\ g_{ij} = \left\langle {\frac{{\partial X}} {{\partial x^i }},\frac{{\partial X}} {{\partial x^j }}} \right\rangle \hfill \\ \frac{{\partial X}}...

Hi , wofsy The sphere parametrization that I use is : \[ x\left( {u,v} \right) = \left( {\cos u\cos v,\sin u\cos v,\sin v} \right) \] Are you sure that this is a surface ? There is four coordinates . Thanks

Hi , wosfy Let's forget the tours for now . We all know that the geodesics on a 2 - sphere are circles . When solved the geodesic equation for it I couldn't undrestand how the solutions representes circles . So could you expalin in general what the solutions of the geodesic equation means ...

Hi , wofsy Thank you very much my friend . I'd love to hear more from you . And I'd invite you to my new question ( The mystery of geodesics . ) Thanks .

Hi , everyone I have a problem with geodesic equation . I know the method of solving it , but I can't understand the solutions . When I tried to solve it for a torus I arrived at : \dot{u} = \frac{k} {{\left( {c + a\cos v} \right)^2 }} \] \[ \dot v = \pm \sqrt { - \frac{{k^2 }}...

Hello ! It's been nearly a month since I put the theard and I didn't find any replies . Is there is anything wrong in it ?

Thank you gel I understand now . Now I have a good understanding of differential geometry ( the bad thing was the connection ) thanks to you . Thanks again my friend . Thank you very much .

Thanks gel . Now I understand . The connection is a way to allow the parallel transport along a curve joining two points P and Q and the parallel transport depends on the curve that we use , so we can understand the connection as an object contains encoded information about the curvature . I...

Coordinate acceleration without a Force ! Hi GR had presented two types of motion , the geodesic motion and the non-geodesic motion . We know that the geodesic motion equation is : \[ \frac{{d^2 x^\alpha }}{{d\tau ^2 }} + \Gamma _{\beta \mu }^\alpha \frac{{dx^\beta }}{{d\tau...

Thanks gel , I understand now ( as I think ) I'll tell you what I understood and you tell me if it's wrong or right : The connection is a way to connect the tangent space at P to the tangent space at Q (as I think it makes them as one thing ) to keep a vector transports from P to Q pointing in...

Thank you gel for replying My friend , the word " connection " means that there are two things connected ( or more than two ) , in this case what are the connected things ? is the connected things are the tangent vector Y and the vectro field X ? if the answer is ( yes ) , then can you...