Recent content by nulliusinverb

pervect, we have two variables inside the brackets, a and b. ok... but see the curvature tensor: R_{a}_{[b}_{c}_{d]}=0 it is definition equal of the tensor antisymmetric in the brackets? (where it origines ∂[aΓdb]c = ½(∂aΓdbc - ∂bΓdac) ? ) thank very much!

R_{a}_{b}_{c}^{d}ω_{d}=((-2)\partial_{[a}\Gamma^{d}_{b] }_{c}+2\Gamma^{e}_{[a]}_{c}\Gamma^{d}_{[b]}_{e})ω_{d} good, me question is about of: 1.- as appear the coefficient (-2) und the (2)? 2.- it is assumed that...

i apologize for the delay, here is the proof: F:ℝ^{n}\rightarrowℝ i have: F(\vec{x})-F(\vec{a})= \sum^{m}_{μ=1}F(t(x^{μ}-a^{μ})+a^{μ},0...,0)^{t=1}_{t=0} then: =\sum^{m}_{μ=1}(x^{μ}-a^{μ})\int^{1}_{0}\frac{\partial F}{\partial u^{μ}}((t(x^{μ}-a^{μ})+a^{μ},0...,0)dt where...

Fredrick thank you very much, the book is recommended to study this issue, although the theorem is raised from other values ​​is the same. thank! ps: "Modern Differential Geometry for Physicists" autor: Isham

hello!: my problem is about of a theorem mathematic,as I prove the following theorem? F(x)=F(a) + \sum^{n}_{i=1}(x^{i}-a^{i})H_{i}(x) good first start with the fundamental theorem of calculus: (for proof): F(x) - F(a) = \int^{x}_{a}F'(s)ds sustitution: s=t(x - a) + a...

I wanted to delete the entire message but could not, my reason was that I answered my questions, was a simple "binomial series." Anyway I upload issue again. thank you very much.

Hello!: My name is Camilo, i want answer one question about of special relativity, in Landau volumen II. the question is in doc attachments! is my first time, sorry.