@0celo7 OK.. I turned off ignore because I think you have stopped calling me stuff and other things... I don't like you because of your attitude and your language, that is why..ok?
The variation of the integral of the scalar curvature and the root of the negative metric determinant with respect to the metric gives the Ricci tensor minus one half of the scalar curvature times the metric tensor.
Nicolas Bourbaki is the collective pseudonym under which a group of (mainly French) 20th-century mathematicians, with the aim of reformulating mathematics on an extremely abstract and formal but self-contained basis, wrote a series of books beginning in 1935. With the goal of grounding all of mathematics on set theory, the group strove for rigour and generality. Their work led to the discovery of several concepts and terminologies still used, and influenced modern branches of mathematics.
While there is no Nicolas Bourbaki, the Bourbaki group, officially known as the Association des collaborateurs...
A matrix has the same relationship to a tensor as a tuple has to a vector. In both cases the latter can be represented by the former, but the latter has structure that is not uniquely captured by the former.
That kind of advice is perhaps defensible as a first introduction if you know that the students have been show the appropriate mathematical opperation on tuples and matricies, but it will teach them lessons that have to be unlearned later.
@0celo7 : no. It's just a generalization of a matrix. Woldemar Voigt introduced the word in its current meaning in 1898. He was working on stress and strain in crystals.
@dmckee Strictly speaking, only true for tensors of rank 2. For higher-rank tensors you'd need multidimensional arrays which you can't really visualize for rank higher than three.
@skullpetrol Don't know that there is blame to assign. I was ready for matricies, and not ready for abstract tensors when I first needed 2nd rank tensors.
@0celo7 : Einstein treated space as an elastic continuum. Note that people have tried to model space as a crystal. For example see epola. I'm not keen on this because it models space as an electron-positron lattice.
@0celo7: for what Einstein's got to do with this, see Wikipedia: "In the 20th century, the subject came to be known as tensor analysis, and achieved broader acceptance with the introduction of Einstein's theory of general relativity, around 1915." He treated space as an elastic continuum. Voigt introduced the word tensor for crystal stress & strain. Look up Voigt and relativity.
@Slereah No, the connected sum requires non-empty manifolds since its construction involves picking a point on each manifold where they are to be connected.
@Slereah formally, taking the connected sum of $M$ and $N$ requires two gluing maps $D^n\to M$ and $D^n \to N$, but there are no maps into $\emptyset$.
I have an embarrassingly rudimentary physic question.. I'm doing some mechanics (math course) and whilst I got the correct answer to a question in my book, their approach in the solution is quite different than mine. And it confused me about a certain aspect of static friction.
basically, i had the understanding that F_s <= u_s|F_n| is the maximum force that can be applied due to friction, and this would be applied in the opposite direction to the force being applied
@Slereah Okay. And now you need to carefully step through the definiiton of a map and convince yourself that there's exactly one map $\emptyset \to M$ and none $M \to \emptyset$ for every $M$.
It's because the map is required to assign an element of the target for every elements of the source. This is trivially so for $\emptyset\to M$, but can never be fulfilled for $M\to\emptyset$.
now, in their solution, they state that for the equilibrium condition to hold, f_s <= u_s|F_n|, and use that to derive the solution. I don't entirely get this? wouldn't this just mean that if the above holds, AND the force acting in the opposite direction is larger than F_s, THEN the equilibrium won't hold?
@JustDanyul Okay. You are correct that F_s <= u_s F_n just says the force of friction is at most u_s F_n.. It's not an equilibrium condition, the equilibrium condition should be that the net force except friction on the object is smaller than u_s F_n.