- #1

Isaac Pepper

- 31

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## Homework Statement

There are three observers, all non accelerating. Observer B is moving at velocity vBA with respect to observer A. Observer C is moving at velocity vC B with respect to observer B. All three observers and all their relative velocities are directed along the same straight line. Calculate the matrix transforming the coordinates of an event from the reference frame of observer A to the reference frame of observer C. Comment of the form of the matrix

## Homework Equations

Assuming normal velocities (so we can use Galilean formulae) : $$u = v + u'$$

## The Attempt at a Solution

Hi, if anyone could just explain what it is I need to do in this question please - I have not done Matrices yet in First Year Physics, but have looked up and understood how to use them (I think). I've never seen Matrices used in Relativity before.

Any help would be greatly appreciated, thanks :)

EDIT :: So perhaps the coordinates of an event could be written as follows : $$\binom{t}{x}$$

In the reference frame of observer A, observer C would be going at a velocity of $$V = Vcb+Vba$$

Therefore in the reference frame of observer C, observer A would appear to going at the same speed in the opposite direction : $$V = -(Vcb+Vba)$$

EDIT2 :: So I'm guessing that would mean $$x' = x-vt$$

$$t'=t$$

and $$\binom{t'}{x'}=\binom{t}{x-vt}$$

And we're looking for a matrix that would help us move from ##\binom{t'}{x'}## to ##\binom{t}{x}##

So $$\binom{t}{x-vt}=\begin{pmatrix}

m&n\\

l&p

\end{pmatrix}

\binom{t}{x}$$

Therefore $$mt+nx=t \rightarrow m=1, n=0$$

$$lt+px=x-vt \rightarrow p=1, l=-v$$

Finally, $$\begin{pmatrix}

m&n\\

l&p

\end{pmatrix} = \begin{pmatrix}

1&0\\

-v&1

\end{pmatrix}$$

Is that correct?

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