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I know that the following relation exists between the time derivative of a proper vector "v" in an "absolute" frame A and the time derivative of the same vector in a "relatively moving" frame B:

$$ \frac{\rm d}{{\rm d}t} \vec{v}_A = \frac{\rm d}{{\rm d}t} \vec{v}_B + \vec{\Omega} \times \vec{v} $$

Where $\vec{\Omega}$ is the angular velocity (pseudo)vector between the two frames.

This is a vectorial equation, which doesn't require the use of any coordinate system.

Now I wonder--> is there a similar equation which applies to the absolute/relative time derivative of TENSORS of generic rank? (Again involving the angular velocity between the two frames, I guess)

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  • $\begingroup$ You transformed a vector components from body system to inertial system $\left( \overrightarrow {v}\right) _{0}=R\left( \overrightarrow {v}\right) _{B}$ , take the time derivative you get $ \dfrac {d}{dt}\left( \overrightarrow {v}\right) _{0}=\dfrac {d}{dt}R\left( \overrightarrow {v}\right) _{B}+R\cdot \dfrac {d}{dt}\left( \overrightarrow {v}\right) _{B}$, hier you get derivative in inertial system and in body system $\endgroup$ – Eli Jul 28 at 19:23
  • $\begingroup$ Hi, thanks for answering. I know how to deal with vectors, my question Was about how to deal with tensors (of generic rank) time derivative expressed in differently moving frames $\endgroup$ – Federico Toso Jul 28 at 20:52
  • $\begingroup$ A vector is a $(0,1)$ tensor and @FedericoToso showed you how to calculated time derivative in two moving frames. Try looking at the inertial tensor - it's a standard calculation. $\endgroup$ – Cinaed Simson Jul 29 at 5:28
  • $\begingroup$ Hi, I can actually carry on analogous calculations with a generic tensor in place of a vector. Anyway I do require to first express my tensor in some coordinate system, transform the components and then take the derivative of the transormed components (like Eli did). I was wondering if it does exist a general tensorial equation which doesn't require to use a coordinate system (like the one I wrote in my first post) $\endgroup$ – Federico Toso Jul 29 at 7:32
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I think this can all be dealt by being explicit with your basis vectors. So let you basis vectors be $\{\mathbf{e}_i\left(\mathbf{r},\,t\right)\}$, note that I allowed for time and position dependence so this is as general is it needs to be. This basis is for normal vectors, tensors, but you can define dual basis and handle the contra-variant vectors in the same way.

So that lets say we have a tensor $\mathbf{T}=T^{i_1 i_2 \dots i_n}\left(\mathbf{r}, t\right) \:\mathbf{e}_{i_1}\otimes\mathbf{e}_{i_2}\dots\otimes\mathbf{e}_{i_n}$. Then

$\frac{d}{dt}\mathbf{T}=\frac{d}{dt}\left(T^{i_1 i_2 \dots i_n}\right)\:\mathbf{e}_{i_1}\otimes\mathbf{e}_{i_2}\dots\otimes\mathbf{e}_{i_n} + T^{i_1 i_2 \dots i_n}\:\frac{d}{dt}\left(\mathbf{e}_{i_1}\right)\otimes\mathbf{e}_{i_2}\dots\otimes\mathbf{e}_{i_n} + \dots$

I am giving this here without a proof, but I am prertty sure such proof can be established based on the universality of the tensor product.

Finally all you need is $\frac{d}{dt}\left(\mathbf{e}_i\right)$. I think it is better to evaluate this expression on case by case basis, since the general form could be ugly. For rotation, I would expect $\frac{d}{dt}\left(\mathbf{e}_i\right)=\Omega^j_{\:i}\mathbf{e}_j$ where $\Omega^j_{\:i}$ is the rotation matrix (which relates directly to your cross-product, but I cannot think of a neat notation to express it without introducing metric tensors or Hodge-duals).

I am not sure if this makes things easier, but it is coordinate independent.

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