Deriving Lorentz transformations using perturbation theory

[spaghetti3451]

Homework Statement

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.

Homework Equations

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?

 

[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).

 

[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?

 

[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.

 

[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?

 

[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.)

[Edit] PS Isn't perturbation expansion fun!

 

[Tianluo_Qi]

Preserving to the third order in V, you will get some conclusion like v²-2v²b₂+2v³b₄=0. But by construction, we know that all the variables are of no relevance of the velocity, thus the formula before will give the zero and the half answer at the same time.