# Transformation matrix for components of acceleration

1. Dec 31, 2011

### world line

Hello
any body can find my mistake ?!
TO find the component of a vector in other coordinate we can use the transformation matrix :

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. Dec 31, 2011

### DrewD

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
$\hat{r}= \sin\theta\cos\phi\hat{x}+\sin\theta\sin\phi\hat{y}+\sin\theta\hat{z}$

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

so, if the position of a particle is $r(t)\hat{r}$ then its velocity is $\frac{dr(t)}{dt}\hat{r}+r(t)\frac{d \hat{r}}{dt}$.

Last edited: Dec 31, 2011