@TedShifrin the fields don't combine, but I think there's only one force (and I think there's a transform that turns electric field into magnetic field and vice versa but can Erico back me up)
Am I right in thinking that $\frac{\left<\mathbf{e}^i, \mathbf{e}'^j\right>}{||\mathbf{e}'^j||^2}$ is the correct way to convert a basis $\mathbf{e}$ to a basis $\mathbf{e}'$?
hmm but as is the case with everything else, there is dissent in the post I linked
it's so hard to get right information
@WilliamOliver only for orthogonal bases
wait I haven't thought about this clearly, ignore me for the moment
I'll call the bases $(b_1, \cdots, b_n)$ and $(c_1, \cdots, c_n)$ to avoid confusion, so given $v = v^i b_i$ we want to find $w^j$ such that $v = w^j c_j$
@LeakyNun well, specifically, I want to represent the vectors $(b_1,⋯,b_n)$ using $(c_1,⋯,c_n)$ but the more general problem is fine too.
@LeakyNun Also, in the context I am working in, it can be assumed that $(b_1, ..., b_n)$ is orthonormal (but the same cannot be assumed about $(c_1, ..., c_n)$)
we know that it must exist, so let it be that, and then $v^i b_i = w^j c_j$. take inner product with $b_k$ to get $\langle c_j, b_k \rangle w^j = \langle b_i, b_k \rangle v^i = v^k$, so $w^j = \langle c_j, b_k \rangle^{-1} v^k$?
@ÉricoMeloSilva I am kind of new to it, but it is also applied to fractions/derivatives a lot right? So $\frac{a_i}{b_i}$ is often $\sum_i \frac{a_i}{b_i}$
@LeakyNun Part of the confusion is probably that contravariant and covariant is honestly a very "physicsy" thing, and isn't very mathematically rigorous
literally all this stuff about tensor algebra is overcomplication, you can just write down the multilinear algebra using summation signs, and then suppress them, choosing components coming from the dual to go correspond to upper or lower indices in a consistent way and u have everything
einstein summation itself literally is just a convention to suppress sigmas, the multilinear is just multilinear, you just have to write things down and make a choice that's consistent
when you have a metric $g$ with componenets $g_{ij}$ then the inner product of two vectors $(v^{1}, \dots, v^{n})$ and $(w^{1}, \dots, w^{n})$ the inner product is given by $\sum_{i, j =1}^{n} g_{ij} w^{i}w^{j} = g_{ij}w^{i}w^{j}$
@LeakyNun your example sentence would be overly complicated haha but assuming the metric tensor is euclidean, it would be something like $w^j \mathbf c_j = v^j \mathbf b_j \implies w^j c_{jl} b^{kl} = v^i b_{il} b^{kl}$
yeah some of them dont know linear algebra and it makes things bad
i still truly think all this has been overcomplication, einstein is simple all it does is suppress summation signs, the confusion is all linear algebra
I think the other thing is that for a mathematician, the difference between covariant and contravariant is arbitrary. Like it depends on what you define to be the dual space. but for a physicists, the distinction is based on physical intuition and what the tensors represent in reality.
Once I understood what tensors were I was really mad haha. Its such a simple concept but there are so many different definitions which aren't equivalent (although they are isomorphic) and physicists and mathematicians talk about them in completely different ways.
Actually, this is making me think about something. I was going back through Landau and Lifshitz to try and iron out some reasonings that I skipped over the first time.
The first equation on the page is given by "neglecting terms above the first order". Why did the author feel justified in doing this? How does this make equality? Shouldn't the equation be approximate?
I was wondering when this type of thing is justified in physics reasoning. I've heard that newtonian physics is a first order approximation of QM and Relativity. So when dealing with these perturbations, do we just only consider things equal up to the first order in newtonian physics? Or is it just in this one case?
I'm working on $\begin{pmatrix}1&1&2&8\\ -1&-2&3&1\\ 3&-7&4&10\end{pmatrix}$, which the question asks to solve by Gaussian elimination, I understand that Gaussian elimination puts the matrix into row echelon (as opposed to reduced row echelon form). Should I aim to perform row operations to turn the -2 into a 1 and the 4 into a 1?
@AfronPie afron you should read the matrix from left to rigth, start with entry 1,1 and by row operations make evrything below it zero, then move to 2,1 entry make what is below it to 0
it is better to say entry than actual number, because they will change when you do row operations
this ibb.co/zQC6Yq4 is a non-convex polygon, so its one internal angle should be greater than 180 degrees. i couldn't locate that angle (its probaly on point C, but which one ACB or ACD) please help
how many distinct vectors exist, all having unit magnitude perpendicular to given line in space?
My guess is take non colinear vector take cross product of line and that vector it is perpendicular to line ? Making unit magnitude is not tough just we need normalization . Is it right? So ans to such kind of problem is infinite right
@Jacksoja the norm on $\Bbb Z[i]$ gives you integers; the "nearest integer" bit comes up when you're trying to prove that the norm induces a Euclidean domain
more generally you have a norm $N_{\Bbb Q(i)/\Bbb Q} : \Bbb Q(i) \to \Bbb Q$ which restricts to a norm $N : \Bbb Z[i] \to \Bbb Z$
If $\Gamma_1,\Gamma_2 \subset PSL(2,\mathbb{R})$ are two Fuchsian groups with $\Gamma_1 \backslash \mathbb{H}^2$ homeomorphic to $\Gamma_2 \backslash \mathbb{H}^2$, does it follow that there is an isomorphism $\Gamma_1 \xrightarrow{\cong} \Gamma_2$ which maps parabolic (resp. hyperbolic) elements to parabolic (resp. hyperbolic) elements?
Or does every isomorphism $\Gamma_1 \xrightarrow{\cong} \Gamma_2$ do so?
