# Deriving Lorentz transformations using perturbation theory

1. Dec 31, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

Derive the transformations $x \rightarrow \frac{x+vt}{\sqrt{1-v^{2}}}$ and $t \rightarrow \frac{t+vx}{\sqrt{1-v^{2}}}$ in perturbation theory. Start with the Galilean transformation $x \rightarrow x+vt$. Add a transformation $t \rightarrow t + \delta t$ and solve for $\delta t$ assuming it is linear in $x$ and $t$ and preserves $t^{2}-x^{2}$ to $\mathcal{O}(v^{2})$.Repeat for $\delta t$ and $\delta x$ to second order in $v$ and show that the result agrees with the second-order expansion of the full transformations.

2. Relevant equations

3. The attempt at a solution

We need to find the third term from $x' = \frac{x+vt}{\sqrt{1-v^{2}}} = (x+vt)(1+\frac{v^{2}}{2}+\cdots)=x+vt+\frac{xv^{2}}{2}+\cdots$, and

we need to find the second and third terms from $t' = \frac{t+vx}{\sqrt{1-v^{2}}} = (t+vx)(1+\frac{v^{2}}{2}+\cdots)=t+vx+\frac{tv^{2}}{2}+\cdots$.

Using $x'=x+vt$ and $t'=t+\delta t, \delta t = rx+st+p$,

we have $t'^{2}-x'^{2}=(t+rx+st+p)^{2}-(x+vt)^{2}$

$t^{2}-x^{2}=((s+1)t+rx+p)^{2}-(x+vt)^{2}$

$t^{2}-x^{2}=(s+1)^{2}t^{2}+2(s+1)(t)(rx+p)+(rx+p)^{2}-x^{2}-2xvt-(vt)^{2}$

$0=(s^{2}+2s-v^{2})t^{2}+r^{2}x^{2}+(2r(s+1)-2v)tx+2p(s+1)t+2rpx+p^{2}$

For the constant term, $p=0$, so the terms in $x$ and $t$.

For the term in $x^2$, we have $r=0$, so we get a non-zero term in $xt$.

Where have I made the mistake?

Last edited: Dec 31, 2015
2. Dec 31, 2015

### jambaugh

your variation of t should depend on v. $\delta t = v(ax+bt)$. Then be sure you are ignoring terms of order $O(v^2)$.

3. Dec 31, 2015

### spaghetti3451

Using $x'=x+vt$ and $t'=t+\delta t, \delta t = v(ax+bt)$,

we have $t'^{2}-x'^{2}=(t+v(ax+bt))^{2}-(x+vt)^{2}$

$t^{2}-x^{2}=t^{2}+(2av)xt+(2bv)t^{2}-x^{2}-(2v)xt$, from which we deduce that $a=1,b=0$.

Using $x'=x+vt +\delta x$ and $t'=t+vx, \delta x = v^{2}(ax+bt)$,

we have $t'^{2}-x'^{2}=(t+vx)^{2}-(x+vt+v^{2}(ax+bt))^{2}$

$t^{2}-x^{2}=t^{2}+(2v)xt+v^{2}x^{2}-(x+vt)^{2}-2(x)(v^{2}(ax+bt))$

$0=(2v)xt+v^{2}x^{2}-(2v)xt-v^{2}t^{2}-2ax^{2}v^{2}-2bv^{2}xt$, from which we deduce that $a=\frac{1}{2},b=0$.

Is my approach correct?

4. Dec 31, 2015

### jambaugh

First stage, yes. That's what I got. For the second order expansion, you didn't consider additional variation of t. I'd also suggest using new constants (not a and b again).
In fact it would be good practice to keep the same definitions of $\delta x$ and $\delta t$ but just extend to higher order in v.

First stage: $\delta x = vt + O(v^2)$ and $\delta t = v(ax+bt)+O(v^2)$ .
You solved and got a=1, b=0.

Second stage: $\delta x = vt + v^2(c x + d t+O(v^3), \delta t = vx + v^2 (ex + f t)+ O(v^3)$ substitute in, solve for terms and compare with power expansion as instructed.

This format is a suggestion for clarity. You might also use subscripted constants at this stage to keep the alphabet soup in hand. $c_1, c_2, c_3, c_4$ instead of c,d,e,f.

5. Dec 31, 2015

### spaghetti3451

Alright, I've reworked my solution:

$x'=x+\delta x, t'=t+\delta t$

1st order in $v$:

$\delta x =vt+\mathcal{O}(v^{2}), \delta t = v(a_{1}x+a_{2}t)+\mathcal{O}(v^{2})$

So, $t'^{2}-x'^{2}=(t+v(a_{1}x+a_{2}t))^{2}-(x+vt)^{2}$

Using $t'^{2}-x'^{2}=t^{2}-x^{2}$ and expanding the right hand side only to first order in $v$,

$t^{2}-x^{2}= t^{2}+2tv(a_{1}x+a_{2}t)-x^{2}-2xvt$

$0=(2va_{2})t^{2}+(2va_{1}-2v)xt$

Taking the $t^2$ term, $2va_{2}=0 \implies a_{2}=0$.

Taking the $xt$ term, $2va_{1}-2v=0 \implies a_{1}=1$.

2nd order in $v$:

$\delta x =vt+v^{2}(b_{1}x+b_{2}t)+\mathcal{O}(v^{3}), \delta t = vx++v^{2}(b_{3}x+b_{4}t)+\mathcal{O}(v^{3})$

So, $t'^{2}-x'^{2}=(t+vx++v^{2}(b_{3}x+b_{4}t))^{2}-(x+vt+v^{2}(b_{1}x+b_{2}t))^{2}$

Using $t'^{2}-x'^{2}=t^{2}-x^{2}$ and expanding the right hand side only to second order in $v$,

$t^{2}-x^{2}= (t+vx)^{2}+2(t+vx)(v^{2})(b_{3}x+b_{4}t)-(x+vt)^{2}-2(x+vt)v^{2}(b_{1}x+b_{2}t)$

$t^{2}-x^{2} = t^{2}+2tvx+v^{2}x^{2}+2tv^{2}b_{3}x+2tv^{2}b_{4}t-x^{2}-2xvt-v^{2}t^{2}-2xv^{2}b_{1}x-2xv^{2}b_{2}t$

$0=(v^{2}-2v^{2}b_{1})x^{2}+(2v^{2}b_{4}-v^{2})t^{2}+(2v^{2}b_{3}-2v^{2}b_{2})tx$

Taking the $t^2$ term, $v^{2}-2v^{2}b_{1}=0 \implies b_{1}=\frac{1}{2}$.

Taking the $x^2$ term, $2v^{2}b_{4}-v^{2}=0 \implies b_{4}=\frac{1}{2}$.

Taking the $xt$ term, $2v^{2}b_{3}-2v^{2}b_{2}=0 \implies b_{3}=b_{2}$.

Therefore, $b_{3}$ and $b_{2}$ remain undeteremined.

Are my solutions correct?

6. Jan 6, 2016

### jambaugh

I think b3 and b2 should be zero. One moment while I trace your work..., yep your expansion and term cancellation is correct. I suppose we must examine the third order terms to solve for these variables. So, with your b1 and b4 values determined expand further, ignoring terms of O(v^4) and higher. (This kinda makes sense, its to second order twice since we have 2 vars each perturbed to within 2nd order.)

 PS Isn't perturbation expansion fun!!!!