# Trying to follow i beginer's proof/derivation from relativity demistified.

1. Jul 26, 2009

### Storm Butler

Trying to follow i "beginer's" proof/derivation from relativity demistified.

I'm trying to follow this proof/derivation in Relativity demystified basically the book is showing transformations and how they work according to the invariance of the speed of light. (im working on chapter 1 pg 9-13 in case any one has the book). The first question i have is, at one point in the book they say that a flash of light moving out from some origin is described by the function of C^2*t^2=x^2+y^2+z^2 (i assume a circle). then they set that equal to zero as well as another coordinate system (F), and since its in standard form (or something similar) aka only the x direction is moving the y and z and y and z cancel out so it leaves us with the equation c^2*t^2-x^2=c^2*t^2-x^2 then the author goes on to say "now we use the fact that the transformation is linear while leaving y an z unchanged. the linearity of the transformation means it must have the form x=Ax+Bc*t
c*t`=Cx+Dc*t
im confused how is it linear aren't there squared terms and how did he rearrange these equations into these two linear equations?

I have more questions but this i all i will ask for now.

2. Jul 26, 2009

### HallsofIvy

Re: Trying to follow i "beginer's" proof/derivation from relativity demistified.

Why would there be squared terms? What he is calculating is just a "change of coordinates" transformation that has nothing to do with the "c^t^2= x^2+ y^2+ z^2" (which is NOT a circle but a sphere in three-space with radius ct).

And any linear function of x and t can be written "Ax+ Bt". The "c" is extracted from B to get the units write. If x is in "meters" and t is in "seconds" Then in x'= Ax+ Bt A is dimension less since x' and x already have the same units, meters. But t has "seconds" as units so B must have units of "meter/second" and it is simplest to write that as (B/c)ct= B'ct so B' is now dimensionless.

3. Jul 26, 2009

### Storm Butler

Re: Trying to follow i "beginer's" proof/derivation from relativity demistified.

ok well if the transformation has nothing to do with the above function then where does it come from? sorry if i seem a bit slow on understanding this.

4. Jul 26, 2009

### diazona

Re: Trying to follow i "beginer's" proof/derivation from relativity demistified.

I can't say for sure without knowing what comes before this bit in the book (I don't have it myself), but the transformation doesn't really come from anywhere. Here's the deal: the book has explained why
$$c^2 t^2 - x^2=c^2 t'^2 - x'^2$$
(at least, given that you believe that the speed of light is invariant). And obviously, there must be some transformation between (t, x) and (t', x') - that is, if you're traveling at some particular speed, there must be some way for you to work out how a friend traveling at a different speed perceives the same events. The book says that this transformation is linear. Why linear? I forget what the accepted mathematical reasoning is, but you can do experiments to verify linearity of transformations. Anyway, since the transformation is linear, it can be expressed as
\begin{align}x' &= Ax + Bct\\t' &= Ct + Dx/c\end{align}
because all linear transformations look like that - it's just a definition. Hopefully you can follow things a bit better from there...

5. Jul 26, 2009

### Fredrik

Staff Emeritus
Re: Trying to follow i "beginer's" proof/derivation from relativity demistified.

It's an assumption. It's possible to make an assumption about something else and derive linearity from that, but that wouldn't be a "better" approach, just a different one. See this thread for a discussion.

I would just like to add that I strongly prefer the matrix version of this equation, with units such that c=1:

$$\begin{pmatrix}C & D\\ A & B\end{pmatrix}\begin{pmatrix}t'\\ x'\end{pmatrix}=\begin{pmatrix}t\\ x\end{pmatrix}$$

I have always found it very strange that introductory texts on SR are always using annoying units (c≠1) and never using matrices. I suppose the reason must be that instructors are assuming that their students aren't ready for matrices yet. That's what makes it so weird, because matrices are much easier than SR. It's an easy concept that makes the difficult concepts easier to understand.

Last edited: Jul 26, 2009