Recent content by PrecPoint

Let me try to give you the context again. Please pay attention to the bold words Regardless of if the space is flat or not the covariant derivative of a ##(1,0)##-tensor is given by $$X^j_{|h}=\frac{\partial X^j}{\partial x^h}+\Gamma_{l\,\,\,h}^{\,\,\,j}X^l$$ And it has nothing to do with...

I would just consider those variants of ##g_{jh}=\frac{1}{2}\frac{\partial^2F^2(x,\dot{x})}{\partial \dot{x}^j\partial \dot{x}^h}## for obvious reasons. The word pseudo is just something Riemannian people came up with to distinguish themselves from people using metrics where ##ds^2## is not...

I believe WWGD was talking about neighboring points here: Relating elements of distinct tangent spaces belonging to neighboring points is the main point of local parallelism. Your answer to WWGD makes no sense. Why would you try to relate tangent spaces of unrelated points on the manifold? A...

You mean between ##T_n(p)## and ##T_n(q)## of two neighboring points ##p,q## which are not infinitesimally close to each other on the manifold? Because otherwise the affine connection ##\Gamma_{h\,\,\,k}^{\,\,\,j}## is a unique mapping of ##T_n(p)## onto the tangent space ##T_n(q)## of a...

Quick update*; so I was not missing something. Apparently, this is a bug in Mathematica.

I'm working the integral $$\displaystyle \int_S z\,dS$$ Where S is the lateral (surface) area of the cylinder y^2+z^2=4 cut off by two planes, x=y-3 and x=6-z. Using the parametrization r(x,\theta)=(x,2\cos(\theta), 2\sin(\theta)) this is pretty straight forward $$\int...

Ok, assumed you were talking about the covariant exterior derivative. In practice, one is confronted with p-forms which are not scalars. It never occurred to me that you meant to compare ##\mathrm{d}\omega## to a covariant derivative. Sorry for the confusion guys.

Uh, ok, Let me be even more specific. By "covariant derivative" on the differentiable manifold ##X_n## I assume giulio_hep mean stuff like the covariant derivative of a contravariant vector field ##X^p(x^k)## is the (1,1) tensor field $$X^p_{|k}=\frac{\partial X^p}{\partial...

It's a contravariant 2-form and the coefficients of ##A_{hj}^k## represents the components of a type (1,2) tensor field. I have a hard time pinning down what parts of your answer actually addressed my question or how you used my concrete example. Is your point that the covariant exterior...

What do you mean when you say that the exterior derivative does not need an affine connection in the same way the ordinary covariant derivative does? Lets be concrete. The covariant exterior derivative of the 2-form ##\Pi^j= A_{hk}^j\mathrm{d}x^h\wedge \mathrm{d}x^k## is (with the usual 1-form...

Your claim is that you can construct a differentiable manifold Xn that won't allow an affine connection (affinely connected space). Please give an example of such a differentiable manifold.

On any differentiable manifold of positive dimension there are infinitely many affine connections. It should be possible to connect nearby tangent spaces, so it permits tangent vector fields to be differentiated as if they were functions on the manifold with values in a fixed vector space. A...

I don't understand, the exterior derivate still needs an affine connection?

Thank you very much TSny! Using your suggestions I first thought about "matrix A" which should be: A_{jr}=\frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^r}}} Now, using my analogy A^TA=I (the transformation is orthogonal) we note that...

I need to use some property of the relalation between the coordinate systems to prove that g_{hk} is independent of the choice of the underlying rectangular coordinate system. I will try to borrow an idea from basic linear algebra. I expect any transformation between the rectangular systems to...