For a future blog post I’m working on, I would like to show how the Lorentz transformations are derived. To remind you, the Lorentz transformations tell us how variables change when we go from one reference frame to another one that is moving relative to the first.
We start in a reference frame S, in which there is only one spatial dimension, $x$. Now compare S with a reference frame S’ moving with velocity $v$ in a positive $x$ direction. How do we relate $x$ and $t$ in S to their equivalents $x’$ and $t’$ in S’?
Linearity
We assume that space and time are the same everywhere we look (that space is homogeneous and time isotropic), so the relation between the variables in S and S’ are everywhere the same. In other words, we assume a linear relation: $$x’=Ax + Bt$$ $$t’=Cx + Dt$$ where A, B, C and D are constants to be found. Since $v$ is along the $x$-axis, we know that $y$ and $z$ remain unchanged.
At the origin of S’, $x’=0$, while its $x$-coordinate in S equals $vt$. If we plug that into the linear relation we get
$$0=A(vt)+Bt \rightarrow B=-Av$$
When we use that in the equation for $x’$ we started with, we find $$x’=A(x-vt)$$ for some constant A.
Constant light speed
We also know that the speed of light ($c$) is the same in all reference frames, so $x=ct$ and $x’=ct’$.
Now substitute $x=ct$ into $x’=A(x-vt)$ and also in $t’=Cx + Dt$. We arrive at $x’=A(ct-vt)$ and $t’=C(ct)+Dt=(Cc+D)t$. So requiring $x’=ct’$ yields $$ A(c-v)=c[(Cc+D)] \tag{1}$$.
For a left-moving lightpulse, $x=-ct$ and $x’=-ct’$, so we get something very similar (do try this at home!):
$$ A(c+v)=c[(-cC+D)] \tag{2}$$
If we add (1) to (2), we get $2Ac=2cD \rightarrow A=D$
If we subtract (1) from (2), we get $-2Av=2c^2 C$
$\rightarrow C=\frac{-Av}{c^2}$
So we now know: $$x’=A(x-vt)$$ $$t’=A(t-\frac{v}{c^2}x)$$
Determine A by interval invariance
Again, we make use of the fact that the speed of light is the same in all reference frames: $x=ct$ and $x’=ct’$. Squaring both sides of $x=ct$ and rearranging the terms gives us $x^2-c^2t^2=0$. The same can be done for $x’=ct’$, which gives $x’^2-c^2t’^2=0$. So we know:
$$x’^2-c^2t’^2=x^2-ct^2$$
Plug in the values for $x’$ and $t’$ from the end of the previous section, work out the squares, and out rolls:
$$A=\sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$$
We call this factor $\gamma$ – it is the factor by which time slows down in moving clocks. Let us now write down the Lorentz transformations in their full glory:
$$\boxed{x’=\gamma(x-vt)}$$
$$\boxed{t’=\gamma(t-\frac{vx}{c^2})}$$
$$\boxed{y’=y}$$
$$\boxed{z’=z}$$
The Lorentz transformations have some strange consequences. Read here about the twin paradox.
The Lorentz-transformations tell us that we live in a universe with a metric that is non-euclidean (put simply: the theorem of Pythagoras doesn’t hold) – read here about the metric which replaces it: the Minkowski metric.
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