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Partial Derivatives of Position Vector

  1. Nov 23, 2004 #1
    Let [itex] \vec{r} = \vec{r}(q_1,\ldots,q_n) [/itex].

    Is the following ALWAYS true?

    \frac{\partial \vec{r}}{\partial q_i} \cdot \frac{\partial \vec{r}}{\partial q_j} = \delta_{ij}

    Edit: Perhaps I should ask if it is zero when [itex] i \neq j [/itex] rather than saying that it is [itex] 1 [/itex] when [itex] i = j [/itex]

    I guess I am wondering if this statement is true ONLY IF we have an orthonormal coordinate system. Does this hold for nonorthonormal coordinate systems? Does it hold if our vectors are not normalized (I would think not)?

    Also, when we have a position vector....do we always treat it in rectangular coordinates and then express each of the coordinates in terms of our generalized coordinates. Let me clarify....for 3D, we have [itex] \vec{r} = (x,y,z) [/itex]. Do we always express x, y, and z in terms of our generalized coordinates (for example [itex] x = r \cos \theta \sin \phi [/itex])? It seems like we do. I am wondering if we ever think of [itex] \vec{r} [/itex] strictly in terms of the q---so, [itex] \vec{r} = (q_1,\ldots,q_N) [/itex]. I believe that the answer is no...since [itex] \vec{r} [/itex] must have proper units...and our generalized coordinates might not have this "feature".
    Last edited: Nov 24, 2004
  2. jcsd
  3. Nov 24, 2004 #2


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    Let's stick to simple point dynamics in the Euclidian space [tex] R^n [/tex].Writing [itex] \vec{r} = \vec{r}(q^1,\ldots,q^n) [/itex] means you have taken into account an change of coordinate.An arbitrary one.A priori
    [tex] \vec{r} = \vec{r}(x^1,x^2,\ldots,x^n) [/tex] ,where i assumed that the basis in the vector space [tex] R^n [/tex] is orthormal,which means that the metric tensor associated with these coordinates is nothing else than the Kronecker delta [tex] \delta_{ij} [/tex].If your change of coordinates is arbitrary,then the derivatives you mentioned are arbitrary functions,and more,the new basis for the same Euclidian space is not orthonormal anymore.It's just formed by "n"linear independent vectors,which can,in principle,be normalized.Since the new base is not orthogonal,this means the new metric tensor is not constant and equal to the unit matrix (as before the general change of coordinates),but has [tex] \frac{n^2-n}{2} [/tex] independent components.Written properly,your equation should have stated:
    \frac{\partial \vec{r}}{\partial q^i} \cdot \frac{\partial \vec{r}}{\partial q^j} = g_{ij}(q)

    I believe i answered your question.

    If we are in an Euclidian space,calculations are simplified by chosing the canonical basis formed by three (generally "n") normalized ans reciprocal orthogonal vectors and label the coordinates by "x","y","z" or whatever.
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