Tag Info

New answers tagged

-1

That's right, there is the following equation: $$X^{\mu}=Z^{\mu}(\sigma)) (dX^{\mu}=\partial_{a}Z^{\mu}(\sigma)d\sigma^{a})$$ for 2d surface embedded into, let say, a flat finite-dimensional target space inside the Polyakov and Nambu-Goto action. The main/only "reason" why people (mostly QFTs theorists who would like to call themselves and be called by ...


0

My suggestion would be to start with the ADM formalism. The details of the Kaluza-Klein seperation will be a little different, because they're foliating on a timelike fiber, and thus, your "special" dimension is spacelike, but the idea is essentially the same -- the lapse function is your dilaton field, and the shift is your vector potential. I'm sure that ...


2

Surely there must be a better approach than this? Certainly; it's the Cartan formalism which employs differential forms. Consider your case of a sphere, with a metric tensor, $$ds^2 = r^2dr^2+r^2\sin^2 \theta \, d\theta^2$$ We can choose an orthonormal basis $e^a$ such that $ds^2 = \eta_{ab}e^{a}e^{b}$, and $e^a = e^a_\mu dx^\mu$. For our case: ...


2

Here's a way to find the Riemann tensor of the 3-sphere with a lot of intelligence but no calculations. At any point $p$ on a sphere, all directions look the same. Therefore there can be no privileged vector at a point $p$. Now consider the eigenvalue problem for the Ricci tensor, $$R^\alpha{}_\beta x^\beta = \lambda x^\alpha.$$ Since no vector is better ...


0

In 3 dimensions, all the data in the Riemman tensor is contained in the Ricci tensor as the Weyl tensor vanishes. So it suffices to compute the Ricci tensor and then using the decomposition given here.


3

The metric tensor is the only additional thing you need to define a Riemannian manifold, beyond what you'd need in order to define the manifold as a differentiable manifold anyway even if it wasn't Riemannian. In particular, topology is important. For example, just specifying that a 2-D manifold has a metric that's Euclidean everywhere is insufficient to ...


4

Yes. The electromagnetic field is nonzero, so the stress-energy tensor is equal to $\frac{1}{2}F_{ac}F^{c}{}_{b}- \frac{1}{8}g_{ab}F^{cd}F_{cd}$. And, of course, $8\pi G T_{ab} = R_{ab} -\frac{1}{2}Rg_{ab}$ EDIT: to clarify, in the riessner-Nordstrom spacetime, you have $g_{ab} = {\rm diag} (-\Delta, \frac{1}{\Delta}, {\rm sphere\; metric})$ and $F_{ab} = ...


1

Comments to the question (v2): One nice property of the 2-norm (as compared to other norms, such as, e.g. the $p$-norm) is that it gives rise to an inner product via a so-called polarization trick. E.g. in the real case the polarization formula has 2 terms: $$ \langle u, v \rangle ~:=~\frac{1}{4} || u+v ||^2 -\frac{1}{4} || u-v ||^2 .$$ There is a similar ...


0

I've since found out that where I'm going wrong is that I don't need to find the absolute derivative. I've been told on another physics forum that this problem is framed in terms of Riemann normal coordinates that makes it OK to assume $\frac{D^{2}\xi^{\mu}}{dt{}^{2}}=\frac{d^{2}\xi^{\mu}}{dt{}^{2}}$. Apparently, this is because the distance the cars travel ...


4

If we want a sense of localness (or calculus to work), we'd like to be able to obtain the length by adding up the length from pieces of the path (for example using a ruler, or counting paces as we walk along the path between two points). However, even considering just two dimensions we see something interesting for $L_p$. $$\left(|x|^p + |y|^p\right)^{1/p} ...



Top 50 recent answers are included