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1

Let us quickly run through the standard KK compactification. We start with a $d+1$ dimensional theory $$S = \frac{1}{16\pi G_{d+1}} \int d^{d+1}x \sqrt{G} R_{d+1}$$ More general actions on the $d+1$ dimensional space can be considered, but this will suffice for our purposes. The metric $G_{MN}$ can be decomposed as $$ds^2 = G_{MN} dx^M dx^N = e^{2\Phi} ... 4 For p=1, CTC's do not exist in Minkowski spacetime. In other 1+3 spacetimes, in principle they are admitted in the absence of further requirements (like globally hyperbolicity) on the causal structure of the spacetime. They must be present if the spacetime is compact, for instance. For p\geq 2, the answer is obviously YES. Consider a manifold M with ... 0 I read the first paragraph of the PDF recommended by Glen The Udderboat, and didn't understand it. Here's the simple method that I use: drop one of the three space dimensions and replace it with the time dimension. Use the intuitive three-dimensional visualisation that you've used all your life, and swap space dimensions in and out if you really need to ... 2 You've got two very good answers from Hunter and NowIGetToLearnWhatAHeadIs. However, it's probably useful to know that this beast O(1,3) is isomorphic or locally isomorphic (i.e. has the same Lie algebra) to a surprising number of other interesting groups, which each give you a slightly different way to think about it. First note that its identity ... 2 The way we measure length is to use the metric tensor. Any spacetime has a metric tensor associated with it, and it's the metric tensor that is responsible for the notion of distance. To make this a little less abstract consider a concrete example. In flat spacetime the metric tensor is just:$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dx^2  Suppose you want to ...

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I'm not an expert on this particular topic, but I believe I can answer your question. There are different kinds of "dimensions". The standard 3 spatial dimensions we live in are infinite in extent. However, one can also imagine dimensions that have a periodicity (like a circle). In such cases, there is a "size" to the dimension that refers to the ...

3

It's the same way you know there are three parameters in $SO(3)$. The equation $\Lambda^T \eta \, \Lambda = \eta$ has $(n^2+n)/2$ independent scalar equations. To see this, write the equation in component form: $\Lambda^{\mu\nu} \Lambda_\mu{}^\rho = \eta^{\nu\rho}$. Now we see there are $n^2$ scalar equations equations, but because $\eta$ is symmetric and ...

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From special relativity we know that a Lorentz transformation: $$x'^\mu = \Lambda^\mu {}_\nu x^\nu$$ preserves the distance: $$g^{\mu \nu} \Delta x_\mu \Delta x_\nu = g^{\mu \nu} \Delta x_\mu' \Delta x_\nu'$$ The above two equations imply: g^{\mu \nu} = g^{\rho \sigma}\Lambda_\rho ...

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No, general relativity is based on something called "intrinsic curvature", which is related to how much parallel lines deviate towards or away from each other. It doesn't require embedding space-time in a higher dimensional structure to work. You're thinking of something called "extrinsic curvature". In fact, many examples of extrinsic curvature - including ...

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Nope, spacetime curvature says nothing about the dimensionality. Your intuition here is probably wrong because human imagination needs 'some dimension to bend into' in order for something to be curved (i.e. an embedding in a higher-dimensional space). This is just our lack of imagination showing, though.

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Giving a precise answer to this is I suspect impossible, as the very notion of physical existence is quite subjective. Please therefore treat this answer as subjective - I would not expect all physicists to agree with it. But here goes… As a first stab, I’d be inclined to reason a bit like this: Your suggested philosophical definition of ‘having extension ...

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