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 $$ x' = \frac{ x - u t }{ \sqrt{1 - u^2/c^2} } $$ $$ y' = y $$ $$ z' = z $$ $$t' = \frac{ t - ux/c^2 }{ \sqrt{ 1 - u^2/c^2 } }.$$

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

$$ \triangle x' = \left ( \frac{ x_2 - u t_2 }{ \sqrt{1 - u^2/c^2} } \right ) - \left ( \frac{ x_1 - u t_1 }{ \sqrt{1 - u^2/c^2} } \right ) = \frac{ \triangle x - u \triangle t } { \sqrt{1 - u^2/c^2} } $$

$$ \triangle y' = y_2 - y_1 = \triangle y $$

$$ \triangle z' = z_2 - z_1 = \triangle z $$

$$ \triangle t' = \left ( \frac{ t_2 - u x_2/c^2 }{ \sqrt{1 - u^2/c^2} } \right ) - \left ( \frac{ t_1 - u x_1 / c^2 }{ \sqrt{1 - u^2/c^2} } \right ) = \frac{ \triangle t - u \triangle t } { \sqrt{1 - u^2/c^2} }. $$

Squaring the terms yield

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

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

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

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

Substituting these terms into the spacetime interval gives

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

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.