Timeline for Dimensions of velocity vector in differential geometry
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2021 at 21:32 | comment | added | Leonid | Let me clarify my statement a bit: In classical mechanics we say that velocity has dimensions of length per time and interpret it as being "the rate of change of the position vector -- consisting of coordinate functions -- with respect to time", that's how we get the dimensions to be length per time in the first place. But notice that this definition depends on the choice of coordinate functions (which can look very different in different charts), that's what I mean that the definition used is coordinate dependent. Not sure if that clarified but in any case I got your point so thanks. | |
| Apr 9, 2021 at 21:19 | comment | added | J. Murray | @Leonid A vector (or vector field) can have whatever dimensions you'd like. There's no reason, for instance, that we need to interpret the curve parameter as having dimensions of time. The point of this answer is simply to demonstrate that the components of a tangent vector will generically have different dimensions than the tangent vector itself, depending on the dimensions of the basis you chose. | |
| Apr 9, 2021 at 21:17 | comment | added | J. Murray | @Leonid I'm not entirely sure I understand the question(s). What do you mean by a "coordinate-dependent description"? A tangent vector is a coordinate-independent object. If you pick some coordinates and a basis, you can expand it in components. These components will of course depend on the coordinates you choose. | |
| Apr 9, 2021 at 21:04 | comment | added | Leonid | One thing I noticed in the above example is that the dimensions of length cancelled away just like in the velocity vector case. Now that I think about it, it makes sense: If our coordinates have dimensions of length any dependence on length would inevitably lead to a coordinate dependent description yes? Because by definition it would depend on the choice of unit length that we impose on our coordinate charts. | |
| Apr 9, 2021 at 21:01 | vote | accept | Leonid | ||
| Apr 9, 2021 at 21:01 | comment | added | Leonid | Thanks, it's much clearer now. One small issue though: It seems to me then that the velocity vector as we use it in classical mechanics is actually coordinate dependent no? Also your reasoning applies to other things like Force fields for example yes? of course now this is a map from the manifold to the tangent bundle but I mean the dimensional analysis still applies: if we take our coordinates to have dimensions of length then a force vector will have dimensions of $T^-{2}M$ where M is dimensions of mass, regardless of coordinate correct? | |
| Apr 9, 2021 at 19:58 | history | answered | J. Murray | CC BY-SA 4.0 |