# Search Results

Results tagged with Search options answers only user 83398
14 results

Mathematical discipline which uses the techniques of calculus to study geometric problems. General relativity is written in this language.

The motivation goes like this. When we define things mathematically, we want to use as few separate objects as possible. We don't want to define a new object independently if it can be defined in te …
answered Aug 13 '18 by knzhou
This is an XY problem. Earlier, the OP wrote: Could you tell me what does $\nabla_{[a\Omega b]}$ represent here? How could $\Omega_b$ come down to the indices? (Is that a short notion for $\Omega … answered Mar 25 by knzhou The quote you give from Carroll about the covariant derivative is right: it quantifies the rate of change of a tensor field relative to parallel transport. The covariant derivative of a tensor at a po … answered Jul 22 '16 by knzhou I'm doubt this has anything to do with general relativity. If I'm not mistaken, the core confusion can also be found in Newton's second law, $$F = m \frac{d^2 x}{dt^2}.$$ Your question would then tran … answered Mar 5 '17 by knzhou The misconception is that everything has to transform by a single matrix multiplication. The field tensor is a rank two tensor, so it takes two (co)vectors as arguments. The components of the tensor … answered Jul 29 '16 by knzhou First a quick review: in general a metric is simply a tensor that takes in two vectors and returns a number, so the 'metric' in quantum mechanics is just the inner product, written as$\langle \psi | …
answered Sep 24 '17 by knzhou
There's a simple way to visualize differential forms that helps here. One of the most important properties of differential $p$-forms is that they are tensors that can naturally be integrated over a …
answered Feb 17 by knzhou
Yes, the method is called Young tableaux. For a rank $n$ tensor, give its $n$ slots names. Then the possible symmetries of the tensors may be classified by Young taleaux, arrays of $n$ boxes filled wi …
answered Dec 28 '18 by knzhou
These two metrics are not the same. The first metric is $$ds^2 = dr^2 + r^2 d\Omega^2$$ while the second metric is $$ds^2 = r_0^2 d\Omega^2$$ where $r_0$ is a constant, the radius of the sphere. These …
answered Jan 2 '17 by knzhou
You've just forgotten the space has to be compact. Think about the simple case $p = 0$. This is the set of harmonic functions $\nabla^2 f = 0$. Indeed, on a compact connected space the zeroth cohomo …
answered Nov 15 '18 by knzhou
This "derivation" hits a pet peeve of mine, which is that mathematical treatments of topological phases persistently confuse the phase shift resulting from a physical process with abstract, physically …
answered Aug 3 '18 by knzhou
Nature appears to be rotationally symmetric, favoring no particular direction. The Laplacian is the only translationally-invariant second-order differential operator obeying this property. Your "Lasph …
answered Apr 26 by knzhou
The divergence defined in terms of the covariant derivative, $$\nabla_i f^i$$ is indeed coordinate independent; that's the whole point of a covariant derivative. However, as you said, the divergence …
answered Jul 20 '18 by knzhou
If you remove the machinery of curved spacetime, then you've just arrived at the usual subtlety involving magnetic monopoles. That is, if you assume $F = dA$ then you automatically have $dF = 0$, whic …
answered Oct 7 '18 by knzhou