When we try to find the geodesic of a partical at rest, in the second term of the geodesic equation we use dt/dtau = 1. Shouldn’t it be i (for imaginary number), since the time component of the 4-velocity vector is imaginary?
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1Imaginary time isn't really taught in modern GR. See, for instance, this Physics.SE Q&A or this Physics.SE Q&A – Kyle Kanos Sep 02 '22 at 21:15
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So, when we use imaginary time, vt.vt would result in a negative term. But if we use real time, vt.vt would result in a positive term. Won't that result in a wrong direction of acceleration? – Nayeem1 Sep 02 '22 at 21:24
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The signs are handled by the metric tensor. – Kyle Kanos Sep 02 '22 at 21:27
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So proper signs are handled by the Crystoffel symbols (which are derived from the metric tensor), so we don't need to worry about them in velocity components? – Nayeem1 Sep 02 '22 at 21:35
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Yes, that is the case, AFAIK. – Kyle Kanos Sep 02 '22 at 21:41
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dt/dτ for a particle at rest is not 1 in general relativity, it is √gᵗᵗ which depends on position, spin, charge, mass (and if not at rest dt/dτ depends also on the gammafactor). √gᵗᵗ is 1 in special relativity, but you tagged general relativity – Yukterez Sep 02 '22 at 21:42