# Introduction

The coordinates of particle or a point is different for different frame of reference. A coordinate transformation is a transformation which established the relation between two coordinates of two different frame of reference. In this article’ Coordinate transformation and its Types ‘ we are going to discuss the basic idea of coordinate transformation. and also about the types of coordinate transformation.

## Coordinate transformation and its Types

Coordinate transformation are of two types-

- Time-independent coordinate transformation
- Time-dependent coordinate transformation

### Transformation equation for frame of reference having translation

Here, S is the stationary frame of reference. And S’ is moving frame of reference. From vector addition we have,

r₀+r’=r

Then, r’=r−r₀ (1)

differentiating with respect to time t,

dr’/dt = dr/dt – dr₀/dt

v’ = v-0

(note:- because r₀ is constant here)

v’= v (2)

And, v= The velocity of the object with respect to S frame of reference.

Also, v’ = the velocity of the particle with respect to S’ frame of reference.

Again differentiating the equation (2) with respect to time t,

dv’/dt = dv/dt

a’=a (3)

That means acceleration remains constant.

#### Transformation equations in the frame of reference having uniform relative motion of translation

At time t=0

After time t=0 or at time t,

Let us suppose that S’ frame is moving with constant velocity with respect to S frame of reference .thus at time t,

We have ,r₀=vt (3)

(that is the distance travelled in time t , by moving the frame of reference )

From the diagram we have,

r₀+r’=r

Vt+r’=r

r’=r-Vt (4)

Now differentiate this equation with respect to time t,

dr’/dt=dr/dt-V

v’=v-V (5)

And, here v’=velocity of the particle with respect to S’ frame of reference

V= the velocity of the particle with respect to S frame of reference

v= the velocity of S’ frame

Now, again differentiate with respect to time t,

dv’/dt=dv/dt-0

Because V=constant

That is a’=a

This shows that acceleration remains constant.

##### Transformation in an inclined frame of reference: for 2-D

S’ frame is inclined at angel θ, with S frame of reference. z and z’ axis are coincide with each other and their origins are same.

Now, take a point P, coordinates are (x,y) with respect to S frame of reference . And (x’,y’) with respect to S’ frame of reference.

Now, draw a perpendicular line from A to PD.

From diagram we have In ∆OAE –

∠OAE +∠EOA = 90⁰ (1)

∠OAE + ∠EAP =90⁰ (2)

After composition we get,

This shows that

∠EOA = ∠EAP = θ (3)

From ∆OAE

Cosθ = OE/OA

OE = x cosθ (4)

Similarly , sinθ = EA/OA

Sinθ = EA/x

x sinθ = AE (5)

Similarly from ∆FAP,

Cos θ = FA/AP

Cos= FA/y

FA = y Cos θ (6)

Sinθ = PF/AP

Sinθ = PF/y

PF = y Sinθ (7)

From diagram we know,

x’ = PD = PF+ FD

x’ = y Sinθ+FD

Where, FD = OE

x’ = y sinθ+x cosθ (8)

Similarly from diagram,

y’ = PC = AF-AE

y’= FA-AE

y’= y cosθ -x sinθ (9)

z’=z (10)

Equation (8),(9)and (10) are called transformation equations.

x’ = x cosθ + y sinθ

x’= x cos (x’ox) + y cos(90- x’ox)

x’ = x cos (x’ox) + y cos (x’oy) (11)

Similarly from equation (9)

(Note:- we know that -sinθ = cos (90 + θ)

θ = y’oy

= Cos (90 + y’oy)

= Cos (y’ox)

y’ = y Cos (y’ox) – x Sin (90 + x’ox)

y’ = y Cos (y’ox) + x cos (y’ox) (12)

###### Transformation in an inclined frame of reference : for 3-D

For 3D we can write

x’ = x Cos (x’ox) + y Cos( x’oy) + z Cos (x’oz) (1)

y’ = x Cos (y’ox) + y Cos (y’oy) + z Cos (y’oz) (2)

Similarly,

z’ = x Cos (z’ox) + y Cos(z’ oy) + z Cos(z’oz) (3)

We know Cos(x’ox)= iˆ.iˆ = a₁₁

Cos (x’oy) = iˆ.jˆ = a₁₂

Cos (x’oz) = iˆ.kˆ =a₁₃

Then equation will be,

x’ = x a₁₁ + y a₁₂+z a₁₃ (4)

y’ = x a₂₁ + y a₂₂+z a₂₃ (5)

z’= x a₃₁+ y a₃₂+z a₃₃ (6)

These are equation for transformation in 3-D

Now, differentiate these equations with respect to time t,

V’x = Vx a₁₁ +Vy a₁₂ + Vz a₁₃ (7)

V’y = Vx a₂₁ +V y a₂₂+Vz a₂₃ (8)

V’z= Vx a₃₁+V y a₃₂+Vz a₃₃ (9)

These are called transformation equations of velocity.

Similarly, for acceleration we Can write.

a’x = ax a₁₁+ aya₁₂ + aza₁₃ (10)

a’y = axa₂₁ + aya₂₂+aza₂₃ (11)

a’z = axa₃₁+ aya₃₂+aza₃₃ (12)

If we suppose that S is a inertial frame of reference. And there is no force applied on it then we have acceleration ax,ay,az =0

Thus from this equation (10),(11) and (12)

a’x,a’y and a’z =0

Hence we can say that S’ is also Inertial frame of reference.