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Unable to understand vector derivative

  1. Nov 4, 2015 #1
    Hi guys...i'm having a bad time understanding the concept of vector derivative with respect to different frames.
    Suppose i have the vector displayed in the picture, the frame A and B are in motion with respect of each other, i can understand the concept of the derivative of the vector R in the frame A...but what does it mean the derivative of the vector with respect to B since the frame's origins are not coincident?
    Thanks!
     

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  3. Nov 4, 2015 #2

    BvU

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    Vectors that are time dependent are differentiated according to the definition.
    If there is a time-dependent coordinate transformation between two frames of reference, that simply comes in:
    So if frame A origin as described in frame B is at ##\vec A_0(t) ## and moves with velocity ##\vec v_A##,
    and the derivative of ##\vec R## in frame A is ## \vec v_{\rm frame \ a} ## then $$
    {d\,\vec R_{\rm frame\ b} \over dt} = { d \left ( \vec R(t) + \vec A_0(t) \right ) \over dt } = \vec v_{\rm frame \ a} + \vec v_A $$

    (I do hope this isn't in a relativity course context ?)

    That the frame origins are not coincident does not appear in the derivative. ##\vec A_0(0) ## is a constant and drops out. You can check that by writing out the definition in full.
     
  4. Nov 4, 2015 #3
    Thanks for the reply...a bad paper misleaded me on this topic!
     
  5. Nov 4, 2015 #4

    BvU

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    How did it manage to do so ?
     
  6. Nov 4, 2015 #5
    it was using a confusing notation! :p
     
  7. Nov 4, 2015 #6
    You seem to have already solved your problem, however your question is ambiguous: what does "derivative of the vector R in the frame A" means? I know what is a "derivative of a vector with respect to time" or what is a "vector written using different systems of coordinates", but I don't know what it is what you have said.

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