Deriving Lorentz Transformations with 3 Postulates

[Shirish]

I'm trying my hand at deriving Lorentz transformations using 3 postulates - it's a linear transformation, the frames are equivalent, so they see the same speed of each other's origins and that the speed of light is the same. Let's say frame $S$ is moving at velocity $v$ in the $x$-direction w.r.t. $S'$. From linearity, we have
$$t'=a_1t+a_2x+a_3\\x'=b_1t+b_2x+b_3$$
Let the position of the origin of $S$ in the $S'$ frame at $t'=0$ be $x'_{O}[t'=0]$. Since $x=0$, we get $t=-a_3/a_1$ by substituting $t'=0,x=0$ in the first equation. So $x'_O[t'=0]=-\frac{b_1a_3}{a_1}+b_3$. Similarly at $t'=T'$, we have for the $S$ origin $t=\frac{T'-a_3}{a_1}$. So $x'_O[t'=T']=\frac{b_1(T'-a_3)}{a_1}+b_3$. Using $x'_O[t'=T']-x'_O[t'=0]=vT'$, we get $b_1=va_1$. Now we have
$$t'=a_1t+a_2x+a_3\\x'=va_1t+b_2x+b_3$$Now consider a photon $P$ that starts from the origin in $S$ at $t=0$. In the $S$ frame, $x_P[t=0]=0$ and $x_P[t=T]=cT$. The starting time in frame $S'$ will be $t'[t=0,x=0]=a_3$. The ending time will be $t'[t=T,x=cT]=a_1T+a_2cT+a_3$. Similarly, $x'[t=0,x=0]=b_3$ and $x'[t=T,x=cT]=va_1T+b_2cT+b_3.$ This gives
$$va_1T+b_2cT=ca_1T+a_2c^2T\\\implies va_1+b_2c=a_1c+a_2c^2$$
Finally, let the position of the origin of $S'$ in the $S$ frame at $t=0$ be $x_{O'}[t=0]$. Substituting $t=0,x'=0$ in the second equation, $x_{O'}[t=0]=-b_3/b_2$. Similarly $x_{O'}[t=T]=\frac{-b_1T-b_3}{b_2}$, and so
$$x_{O'}[t=T]-x_{O'}[t=0]=-\frac{b_1T}{b_2}=-vT\implies b_1=vb_2\implies b_2=a_1
\\\implies va_1=a_2c^2\implies a_2=\frac{va_1}{c^2}$$
The transformation can thus be restated as (on the RHS I'll replace $a_1$ by $\gamma$):
$$t'=a_1t+\frac{va_1}{c^2}x+a_3=\gamma(t+\frac{vx}{c^2})+a_3\\x'=va_1t+a_1x+b_3=\gamma(x+vt)+b_3$$
And now beyond this, I'm really not sure how to proceed. I'm lost on how to derive the value of $\gamma$. Can anyone please help? Thanks (I've looked at the Lorentz transformation derivation Wikipedia page but haven't found a resolution to my query)

 

[Dale]

it's a linear transformation

A linear transform has $a_3=b_3=0$. What you wrote is actually an affine transformation.

 

[Ibix]

And now beyond this, I'm really not sure how to proceed. I'm lost on how to derive the value of $\gamma$.

At a quick glance, you need to follow @Dale's advice then write the inverse transforms. What does the principle of relativity tell you the inverse transforms must look like?

 

[Shirish]

A linear transform has $a_3=b_3=0$. What you wrote is actually an affine transformation.

But why are Lorentz transformations linear instead of affine? The logic behind the nature of the transformation was that straight worldlines in one frame are mapped to straight worldlines in another - so doesn't that indicate an affine transformation? Maybe I'm missing an additional assumption?

 

[Ibix]

Maybe I'm missing an additional assumption?

You can choose the origins of the two frames to coincide without loss of generality.

Edit: technically you don't have to do that, but there's no reason not to do it. You can always perform a translation before/after your boost if you need to. But the frames related by what we usually call the Lorentz transformations are chosen to have coincident origins.

 

[Shirish]

At a quick glance, you need to follow @Dale's advice then write the inverse transforms. What does the principle of relativity tell you the inverse transforms must look like?

So I admit I had looked at the Wikipedia article to clear up the doubt in my post, and it seems that we can write the identity transformation as the composition of the Lorentz transformation from $(t,x)$ to $(t',x')$, and that from $(t',x')$ to $(t,x)$.

The reverse of sign of $v$ makes total sense, but why do we retain $\gamma$ as is? After all, $\gamma$ is also a parameter of the transformation, so in general shouldn't we expect it to change when we reverse the transformation?

You can choose the origins of the two frames to coincide without loss of generality.

So am I right in saying that the physical reason behind the "without loss of generality" part is that the nature of transformations wouldn't get suddenly changed if the $S'$ clock had started some time before/after, or if the origin of $S'$ were defined to be at a different point?

 

[Ibix]

The reverse of sign of $v$ makes total sense, but why do we retain $\gamma$ as is?

You don't. You know that $\gamma$ is a function of $v$ - that is, $\gamma(v)$. So your reasoning should be that the inverse transform involves $\gamma(-v)$. Deriving the inverse transforms from your forward transforms should get you an expression in terms of $\gamma(v)$, $\gamma(-v)$ and $v$, which has an obvious solution if $\gamma(v)=\gamma(-v)$.

 

[Ibix]

So am I right in saying that the physical reason behind the "without loss of generality" part is that the nature of transformations wouldn't get suddenly changed if the $S'$ clock had started some time before/after, or if the origin of $S'$ were defined to be at a different point?

Careful - the origin of $S'$ is an event not a point. I think you mean "...if the spatial origin of $S'$ were defined to be at a different point".

But basically, yes. The underlying reason is the translation symmetry of spacetime - I get the same Lorentz transforms wherever I choose my origin to be, because there's nothing fundamentally special about any event. The advantage to having coinciding origins is that the Lorentz transforms define pure boosts, and I can have my translations separately in another conceptual box and my rotations in yet a third. It turns out that the rotations and boosts form a group, the Lorentz group, which is a subgroup of the Poincare group that includes the translations - so there's a good mathematical basis to keep translations separate from boosts.

 

[Shirish]

You don't. You know that $\gamma$ is a function of $v$ - that is, $\gamma(v)$. So your reasoning should be that the inverse transform involves $\gamma(-v)$. Deriving the inverse transforms from your forward transforms should get you an expression in terms of $\gamma(v)$, $\gamma(-v)$ and $v$, which has an obvious solution if $\gamma(v)=\gamma(-v)$.

Thanks so much for the help! I originally did assume different gammas for the forward and reverse transformation, and already did what you mentioned, but I still get $\gamma\gamma'=\frac{1}{1-v^2/c^2}$ and I'm guessing there's another physical assumption I'm missing which allows me to conclude that $\gamma(v)=\gamma(-v)$. Otherwise there's no obvious mathematical reason that allows me to conclude the same (even granted that $\gamma(v)=\gamma(-v)$ is a convenient solution.

 

[Ibix]

The principle of relativity tells you that the forward and reverse transforms have the same form up to the sign of $v$. That tells you that $\gamma'=\gamma(-v)$. And then you observe that your expression is satisfied by the Lorentz gamma factor we all know and love.

 

[vanhees71]

But why are Lorentz transformations linear instead of affine? The logic behind the nature of the transformation was that straight worldlines in one frame are mapped to straight worldlines in another - so doesn't that indicate an affine transformation? Maybe I'm missing an additional assumption?

In fact you are right, because the symmetry group is not the Lorentz but the Poincare group, because Minkowski space is an affine space.

To solve your problem with the value for $\gamma$ you have to consider general spacetime events and not only the motion of a light-wave front (or a "photon" as you argue, but that I don't like, because there's no such thing as a "light particle" in any classical sense, but that's semantics at this point).

The argument however starts from this motion of a light front starting from the origin of the system $\Sigma$. You have
$$c^2 t^2-x^2=0. \qquad (1)$$
The same must now hold true in $\Sigma'$, but there the event $(ct,x)=(0,0)$ corresponds to $(ct',x')=(c a_3,b_3)$, and thus from (1) must follow (sic!)
$$c^2 (t'-a_3)^2+(x'-b_3)^2=0.$$
This implies that there must be a function $f$ such that
$$c^2 (t'-a_3)^2+(x'-b_3)^2=f(c^2 t^2-x^2).$$
Since now the inverse transformation must be of the same form it must hold as well
$$c^2 t^2-x^2=f[c^2(t'-a_3)^2 + (x'-b_3)^2].$$
So you have function $f$ such that from $y=f(z)$ follows $z=f(y)$ or $z=f[f(z)]$, i.e., the inverse function $g$ of $f$ is $f$ itself. which means $g'=1/f'=f'$ and thus $(f')^2=1$ and finally $f(z)=\pm z$. In our case $f(z)=-z$ is ruled out, because for $v=0$ you want the transformation to be $t'=t+a_3$, $x'=x+b_3$.

The upshot is that you must have
$$c^2 t^2-x^2=c^2(t'-a_3)^2 - (x'-b_3)^2$$
for all spacetime vectors $(ct,x)$. From this you should get $\gamma=(1-v^2/c^2)^{-1/2}$. Plugging in your transformation yields after some simple algebra
$$c^2 (t'-a_3)^2-(x'-b_3)^2=\frac{1}{c^2} \gamma^2 (c^2-v^2) (c^2 t^2-x^2) \stackrel{!}{=} c^2 t^2-x^2.$$
Since this must hold for all $(ct,x)$ you indeed must have $\gamma=1/\sqrt{1-v^2/c^2}$. The negative root is ruled out again, because of the case $v=0$.

 

[Shirish]

The principle of relativity tells you that the forward and reverse transforms have the same form up to the sign of $v$. That tells you that $\gamma'=\gamma(-v)$. And then you observe that that's satisfied by the Lorentz gamma factor we all know and love.

I'm sorry if I'm being really stupid (probably so) but I'm not sure how the principle of relativity would be violated if $\gamma$ were an odd function of $v$. Sure, it's absurd for it to be odd because then if $\gamma>0$, then $\gamma'<0$ and $t\to - \infty$ as $t\to \infty$, which is physically absurd. I can also understand why the magnitude of $\gamma$ needs to be the same, because then you'd see different amounts of length contraction if you switch directions, which would imply a preferred direction. But still confused whether (and if so, why) a mere change in sign of $\gamma$ would violate the principle of relativity.

 

[Ibix]

I'm sorry if I'm being really stupid (probably so) but I'm not sure how the principle of relativity would be violated if γ\gamma were an odd function of vv.

The principle of relativity just tells you that $\gamma'=\gamma(-v)$. Your expression in #9 then reads $\gamma(v)\gamma(-v)=1/(1-v^2/c^2)$. If $\gamma(v)$ is odd, then the left hand side of that is $-\gamma^2(v)$ and $\gamma$ would have to be imaginary.

 

[Shirish]

The principle of relativity just tells you that $\gamma'=\gamma(-v)$. Your expression in #9 then reads $\gamma(v)\gamma(-v)=1/(1-v^2/c^2)$. If $\gamma(v)$ is odd, then the left hand side of that is $-\gamma^2(v)$ and $\gamma$ would have to be imaginary.

You're absolutely right, sorry to bother for the trivial doubt.

 

[Ibix]

No problem. I personally favour the emoticon for those moments...

 

[vanhees71]

Well, only the part of the Poincare transformations which is continuously connected with the identity transformation is necessarily a symmetry group for any theory consistent with the special relativistic spacetime structure. This leads to the proper orthochonous Poincare transformations as the symmetry group of special relativistic theories.

Nature makes use of this "freedom" to violate the larger Poincare group: The weak interaction violates all combinations of time-reversal, space-reflection, and charge-conjugation symmetries separately. Only the "grand reflection" CPT is still a symmetry as it must be for local relativistic QFTs.

 

[Shirish]

No problem. I personally favour the emoticon for those moments...

Edited. And of course, the fact that there's no preferred direction tells us that $|\gamma|=|\gamma'|$

 

[Shirish]

Well, only the part of the Poincare transformations which is continuously connected with the identity transformation is necessarily a symmetry group for any theory consistent with the special relativistic spacetime structure. This leads to the proper orthochonous Poincare transformations as the symmetry group of special relativistic theories.

Nature makes use of this "freedom" to violate the larger Poincare group: The weak interaction violates all combinations of time-reversal, space-reflection, and charge-conjugation symmetries separately. Only the "grand reflection" CPT is still a symmetry as it must be for local relativistic QFTs.

Whatever you said went over my head Maybe in a year's time it'll start making sense to me. I'll probably have to study group theory and Lie algebra and what not

 

[Dale]

But why are Lorentz transformations linear instead of affine?

Because they are defined that way. The relevant affine transformation is called the Poincare transform. The Lorentz group is a sub group of the Poincare group, specifically the sub group that preserves the origin.

The logic behind the nature of the transformation was that straight worldlines in one frame are mapped to straight worldlines in another - so doesn't that indicate an affine transformation?

Yes, but that one is not the Lorentz transform. That one is the Poincare transform. The Lorentz transform deliberately ignores the translation of the origin.

You can certainly keep $a_3$ and $b_3$ but then you are not assuming linearity and you are not deriving the Lorentz transform. Instead you are assuming an affine transform and deriving the Poincare transform. No problem (in fact I prefer your approach), just different labels

 

[Shirish]

A few doubts related to the following part:

...and thus from (1) must follow (sic!)
$$c^2 (t'-a_3)^2+(x'-b_3)^2=0.$$
This implies that there must be a function $f$ such that
$$c^2 (t'-a_3)^2+(x'-b_3)^2=f(c^2 t^2-x^2).$$
Since now the inverse transformation must be of the same form it must hold as well
$$c^2 t^2-x^2=f[c^2(t'-a_3)^2 + (x'-b_3)^2].$$
So you have function $f$ such that from $y=f(z)$ follows $z=f(y)$ or $z=f[f(z)]$, i.e., the inverse function $g$ of $f$ is $f$ itself. which means $g'=1/f'=f'$ and thus $(f')^2=1$ and finally $f(z)=\pm z$.

* How do you go from $c^2 (t'-a_3)^2+(x'-b_3)^2=0$ to $c^2 (t'-a_3)^2+(x'-b_3)^2=f(c^2 t^2-x^2)$?

* For the inverse transformation $c^2 t^2-x^2=f[c^2(t'-a_3)^2 + (x'-b_3)^2]$, why do you assume the same form $f$ instead of some other $g$? I would've thought generally it'd be something like
$$c^2 t^2-x^2=g[c^2(t'-a_3)^2 + (x'-b_3)^2]$$
I guess the reasoning is similar to what @Ibix mentioned for $\gamma=\gamma'$, but in this case of $f$ and $f^{-1}$ it's not evident to me.

* Assuming that $f^{-1}=f$, where did $g'=1/f'=f'$ come from?

 

[vanhees71]

The Poincare transformation by definition is a one-to-one mapping between $(ct,x)$ and $(ct',x')$ and all transformations build a group (that's a somewhat hidden assumption in this discussion). This implies that the $f$ must be the same for the transformation and its inverse. I've set $g$ as the inverse of $f$ (I don't like $f^{-1}$, because in a common notation in physics this could be confused with $1/f$). Then you have
$$f[g(x)]=x \; \Rightarrow \; f'[g(x)] g'[x]=1\; \Rightarrow g'(x)=1/f'[g(x)].$$