Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transformation matrix for components of acceleration

  1. Dec 31, 2011 #1
    Hello
    any body can find my mistake ?!
    TO find the component of a vector in other coordinate we can use the transformation matrix :

    http://up98.org/upload/server1/01/z/ff96m5hl2uahgjw3u2un.jpg [Broken]

    but why this does nt work for acceleration vector ?
    i mean why i cant derive the component of acceleration in spherical coordinate by use of this matrix ?
    thanks
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 31, 2011 #2
    That matrix only transforms position vectors. In spherical coordinates, the basis vectors are changing with respect to time as well as the coefficients which means that the transformation is much more complicated.

    If the basis vector in the r direction is
    [itex]\hat{r}= \sin\theta\cos\phi\hat{x}+\sin\theta\sin\phi\hat{y}+\sin\theta\hat{z}[/itex]

    then,
    [itex]\frac{d}{dt}\hat{r}=(\cos\theta\cos\phi\frac{d \theta}{dt}-\sin\theta\sin\phi\frac{d \phi}{dt})\hat{x}+(\cos\theta\sin\phi\frac{d \theta}{dt}+\sin\theta\cos\phi\frac{d \phi}{dt})\hat{y}+\cos\theta\frac{d \theta}{dt}\hat{z} [/itex]

    so, if the position of a particle is [itex] r(t)\hat{r} [/itex] then its velocity is [itex]\frac{dr(t)}{dt}\hat{r}+r(t)\frac{d \hat{r}}{dt}[/itex].
     
    Last edited: Dec 31, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Transformation matrix for components of acceleration
Loading...