PlanetPhysics/Spacetime Interval Is Invariant for a Lorentz Transformation

The spacetime interval between two events $E_{1}(x_{1},y_{1},z_{1},t_{1})$ and $E_{2}(x_{2},y_{2},z_{2},t_{2})$ is defined as

$(\triangle s)^{2}=c^{2}\triangle t^{2}-(\triangle x)^{2}-(\triangle y)^{2}-(\triangle z)^{2}.$

If $\triangle s$ is in reference frame $S$, then $\triangle s'$ is in reference frame $S'$ moving at a velocity $u$ along the x-axis. Therefore, to show that the spacetime interval is invariant under a Lorentz transformation we must show

$(\triangle s)^{2}=(\triangle s')^{2}$

with the reference frames related by The Lorentz transformation $x'={\frac {x-ut}{\sqrt {1-u^{2}/c^{2}}}}$ $y'=y$ $z'=z$ $t'={\frac {t-ux/c^{2}}{\sqrt {1-u^{2}/c^{2}}}}.$

The change in coordinates between events in the $S'$ frame is then given by

$\triangle x'=\left({\frac {x_{2}-ut_{2}}{\sqrt {1-u^{2}/c^{2}}}}\right)-\left({\frac {x_{1}-ut_{1}}{\sqrt {1-u^{2}/c^{2}}}}\right)={\frac {\triangle x-u\triangle t}{\sqrt {1-u^{2}/c^{2}}}}$

$\triangle y'=y_{2}-y_{1}=\triangle y$

$\triangle z'=z_{2}-z_{1}=\triangle z$

$\triangle t'=\left({\frac {t_{2}-ux_{2}/c^{2}}{\sqrt {1-u^{2}/c^{2}}}}\right)-\left({\frac {t_{1}-ux_{1}/c^{2}}{\sqrt {1-u^{2}/c^{2}}}}\right)={\frac {\triangle t-u\triangle t}{\sqrt {1-u^{2}/c^{2}}}}.$

Squaring the terms yield

$(\triangle x')^{2}=\left({\frac {\triangle x-u\triangle t}{\sqrt {1-u^{2}/c^{2}}}}\right)\left({\frac {\triangle x-u\triangle t}{\sqrt {1-u^{2}/c^{2}}}}\right)={\frac {(\triangle x)^{2}-2u\triangle x\triangle t+u^{2}(\triangle t)^{2}}{1-u^{2}/c^{2}}}$

$(\triangle y')^{2}=(\triangle y)^{2}$

$(\triangle z')^{2}=(\triangle z)^{2}$

$(\triangle t')^{2}=\left({\frac {\triangle t-u\triangle t}{\sqrt {1-u^{2}/c^{2}}}}\right)\left({\frac {\triangle t-u\triangle t}{\sqrt {1-u^{2}/c^{2}}}}\right)={\frac {(\triangle t)^{2}-2u\triangle x\triangle t/c^{2}+u^{2}(\triangle x)^{2}/c^{4}}{1-u^{2}/c^{2}}}.$

Substituting these terms into the spacetime interval gives

$(\triangle s')^{2}={\frac {c^{2}((\triangle t)^{2}-2u\triangle x\triangle t/c^{2}+u^{2}(\triangle x)^{2}/c^{4})}{1-u^{2}/c^{2}}}-{\frac {((\triangle x)^{2}-2u\triangle x\triangle t+u^{2}(\triangle t)^{2})}{1-u^{2}/c^{2}}}-(\triangle y)^{2}-(\triangle z)^{2}.$

Adding the first two terms with common denominators together yields

$(\triangle s')^{2}={\frac {c^{2}(\triangle t^{2})-(\triangle x)^{2}-u^{2}(\triangle t)^{2}+u^{2}(\triangle x)^{2}/c^{2}}{1-u^{2}/c^{2}}}-(\triangle y)^{2}-(\triangle z)^{2}.$

Pulling out a $-u^{2}/c^{2}$

$(\triangle s')^{2}={\frac {c^{2}(\triangle t^{2})-(\triangle x)^{2}-u^{2}/c^{2}(c^{2}(\triangle t)^{2}+(\triangle x)^{2})}{1-u^{2}/c^{2}}}-(\triangle y)^{2}-(\triangle z)^{2}.$

Factoring out a $c^{2}(\triangle t)^{2}-(\triangle x)^{2}$ in the numerator

$(\triangle s')^{2}={\frac {(c^{2}(\triangle t^{2})-(\triangle x)^{2})(1-u^{2}/c^{2})}{1-u^{2}/c^{2}}}-(\triangle y)^{2}-(\triangle z)^{2}.$

Finally, canceling terms gives

$(\triangle s')^{2}=c^{2}(\triangle t^{2})-(\triangle x)^{2}-(\triangle y)^{2}-(\triangle z)^{2}=(\triangle s)^{2}.$

Hence, the spacetime interval is invariant under a Lorentz transformation.