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

32
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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
4
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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
2
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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
2
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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
3
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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
1
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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
15
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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
6
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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
2
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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
0
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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
6
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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
153
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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
5
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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
2
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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