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by messe
807 days ago
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It's not an analogy. It's literally just another basis. > In particular, for multidimensional spaces, the usual multidimensional Fourier transform only really works if you have a flat metric on that space What the hell does the metric of space-time have to do with this? When computing a fourier transform, we're not working in 3+1 dimensional space-time, we're working in either an N-dimensional (in the discrete case) or \infty-dimensional (in the continuous case) vector space; while that term contains the word "space" they DO NOT, in this context, have anything to do with Euclidean space or the Pseudo-Riemannian manifold that GR treats space-time as. |
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Which would be a bad thing, reading this kind of conversation is what makes this site worthwhile