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Is it possible to take a time derivative of a vector given in some curvelinear coordinate system (i.e. spherical)? Mathematica would need to take into account the time dependence of the basis vectors.

$$ \frac{d}{dt}\vec{r}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}+r\sin[\theta]\dot{\phi}\hat{\phi} $$

I bet that there is something built into mathematica already to get the above result, but I can't figure it out.

Edit: Thanks for your answers, they already helped me a lot. In the end I was hoping for a quick way to get:

$$ \frac{d}{dt}\left( \begin{array}{c} f(r,\theta,\phi) \\ 0 \\ 0 \end{array} \right)=\left(\frac{d}{dt}f(r,\theta,\phi)\right) \left( \begin{array}{c} \dot{r} \\ r\dot{\theta} \\ r\sin[\theta]\dot{\phi} \end{array} \right) $$

just by typing

$$ Dt[\{f(r,\theta,\phi),0,0\},t] $$

in Mathematica. I could get this behavior by multiplying the time derivative of $f$ with the result from TransformedField, but this could quickly become tedious with higher time derivatives and more vector components. Is there a more direct way to do it?

Thanks a lot

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  • $\begingroup$ Well, if you concluded that there's something built-in why didn't you take a Documentation Centre tour ? As a starter take a look here $\endgroup$ Commented Mar 22, 2014 at 10:55
  • $\begingroup$ The documentation center tour didn't really show me any examples on what I want to achieve. But my question was not clear in that regard, for which I apologize. I have updated it (see my edit). $\endgroup$ Commented Mar 22, 2014 at 12:01

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You can do :

TransformedField["Cartesian" -> "Spherical", Dt[{x, y, z}, t], {x, y, z} -> {r, θ, ϕ}] // Simplify (* {Dt[r, t], r Dt[θ, t], r Dt[ϕ, t] Sin[θ]} *)
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  • $\begingroup$ Thanks a lot, that already started me in the right direction, but I was wondering for a more direct way of taking the time derivative of vectors (arrays) (see my edit). $\endgroup$ Commented Mar 22, 2014 at 12:00
  • $\begingroup$ I think you already get the time derivative of vectors; if you want it in a different basis, which is what your edit suggests, then this is a different question. $\endgroup$ Commented Mar 22, 2014 at 13:25

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