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[parent] spacetime interval is invariant for a Lorentz transformation (Result)

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

$\displaystyle (\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

$\displaystyle (\triangle s)^2 = (\triangle s')^2 $

with the reference frames related by The Lorentz transformation

$\displaystyle x' = \frac{ x - u t }{ \sqrt{1 - u^2/c^2} } $

$\displaystyle y' = y $

$\displaystyle z' = z $

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

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

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

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

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

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

Squaring the terms yield

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

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

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

$\displaystyle (\triangle t')^2 = \left ( \frac{ \triangle t - u \triangle t } {... ...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

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

Adding the first two terms with common denominators together yields

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

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

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

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

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

Finally, canceling terms gives

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

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

Bibliography

1
Carroll, Bradley, Ostlie, Dale, An Introduction to Modern Astrophysics. Addison-Wesley Publishing Company, Reading, Massachusetts, 1996.
2
Cheng, Ta-Pei, Relativity, Gravitation and Cosmology. Oxford University Press, Oxford, 2005.
3
Einstein, Albert, Relativity: The Special and General Theory. 1916.



"spacetime interval is invariant for a Lorentz transformation" is owned by bloftin.

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Cross-references: The Lorentz transformation, Lorentz transformation, velocity, reference frame, spacetime

This is version 2 of spacetime interval is invariant for a Lorentz transformation, born on 2006-11-25, modified 2006-11-25.
Object id is 237, canonical name is SpacetimeIntervalIsInvariantForALorentzTransformation.
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Physics Classification: 03.30.+p (Special relativity)
 03. (Quantum mechanics, field theories, and special relativity )
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