# Coordinate transformation : translation, inclined and rotation

This article is about Coordinate transformation . In this article we will look at coordinate transformation in case of translation, inclination and rotation of S' frame of reference with respect to S frame of reference. This article is mainly for B.Sc. first year and comes under subject Mechanics.

## Coordinate transformation

If we define the position of a particle in two different frames of then in both the cases projection of the particle comes out to be different and the relation between the projections of this particle in two different frames of reference is known as their transformation equations.

If there is no relative velocity between the frames of references and they rotate through a certain degree then the transformation does not depend on time. Opposite to this when there is a certain relative velocity between them then transformation equation depends on time.

## Transformation equations for frames of reference involving translation

Consider two frames of reference S and S’ whose origins are O and O’ and observers are at the point O and O’ as shown below in figure 1. If position vector of any point P in S is $\vec{r}$ then its position vector in S’ would be $\vec{r’}$.

Figure 1

According to figure

$\overrightarrow{r’}=\vec{r}-\overrightarrow{{{r}_{0}}}$

Differentiating it w.r.t. time

$\frac{\vec{r’}}{dt}=\frac{\vec{r}}{dt}-\frac{\vec{r_0}}{dt}$

If $\vec{r_0}$ is constant then

$\frac{d\vec{r_0}}{dt}=0$

Therefore,

$\frac{d\vec{r}}{dt}=\frac{d\vec{{r}’}}{dt}$

$\vec{v}=\vec{v’}$

Again differentiating it w.r.t. time

$\frac{d\vec{v}}{dt}=\frac{d\vec{{v}’}}{dt}$

Or,

$\vec{a}=\vec{a’}$

So, in translated frame of reference S’ the position of the particle would be different but the acceleration and velocity would be same as measured in S frame of reference. Also the equations 1, 2 and 3 are time independent equations.

## Coordinate transformations in reference frames having uniform relative translational motion

In figure 1 if the frame of reference S’ has translational motion with constant velocity w.r.t. the frame of reference S than at any time their axis would be at a constant distance but the position would depend on time. If in the beginning at $t=0$ both the frame of reference has origin O and O’ at the same point. Now at time t position vector of O’ w.r.t. O would be

$\vec{r_0}=\vec{v}t$

Hence

$\vec{r’}=\vec{r}-\vec{r_0}$

$\vec{r’}=\vec{r}-\vec{v}t$

Differentiating it w.r.t. the time we get

$\vec{v’}=\frac{d\vec{r}}{dt}-\frac{d\vec{r_0}}{dt}=\vec{u}-\vec{v}$

Acceleration of the particle

$\vec{{a}’}=\frac{d\vec{u}}{dt}-\frac{{\vec{v}}}{dt}=\vec{a}$

As $\vec{v}$ is constant.

Hence position and velocity of the particle could change according to equation 4 and 5 when it is moving with uniform relative motion but acceleration remains unchanged.

Equations 4, 5 and 6 are time dependent transformation equations.

## Transformation in an inclined frame of reference

Let x, y, z, be the coordinates of a particle in a frame of reference say S as shown in figure 2.

Figure 2

So we have,

\label{eq:1}

\begin{aligned}

x=PB=OA \\

y=PA=OB

\end{aligned}

Another frame of reference S’ is inclined w.r.t. the frame of reference S such that the origins of both the frame of reference and their z-axis coincides as shown in figure 2.

Now coordinates of point P in S’ frame of reference would be

\label{eq:2}

\begin{aligned}

X’=PD=OC \\

Y’=PC=OD

\end{aligned}

Since z and z’ coincides we have

\label{eq:3}

z=z’

Now from right angle $\bigtriangleup OAE$ we have

\begin{equation*}

\end{equation*}

Again from right angle $\bigtriangleup FAP$ we have

\begin{equation*}

\end{equation*}

now,

\begin{equation*}

\begin{flalign}

&x’=PD=PF+FD \\

&\text{or,} \\

&x’=PF+OE \\

\end{flalign}

\end{equation*}

This implies that,

\label{eq:4}

x’=xcos\theta+ysin\theta

Similarly,

\begin{equation*}

\begin{flalign}

y’&=PC-EF \\

&=AF-AE

\end{flalign}

\end{equation*}

\label{eq:5}

y’=ycos\theta -xsin\theta

Equations (\ref{eq:4}) and (\ref{eq:5}) can also be written as

\begin{aligned}

\label{eq:6}

x’=xcos(X’OX)+ycos(X’OY)

\end{aligned}

\begin{aligned}

\label{eq:7}

y’=xcos(Y’OX)+ycos(Y’OY)

\end{aligned}

Equations (\ref{eq:4}) ,and (\ref{eq:5}) ,(\ref{eq:6})  and (\ref{eq:7})  are the transformation equations for an inclined frame of reference.