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

The spacetime interval between two events E1(x1,y1,z1,t1) and E2(x2,y2,z2,t2) is defined as

     2    2   2       2        2        2
(△s ) =  c △t  − (△x )  − (△y ) −  (△z ).

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

(△s )2 = (△s ′)2

with the reference frames related by The Lorentz transformation

  ′     x − ut
x  = ∘-------2--2-
        1 − u ∕c

 ′
y = y

 ′
z = z

 ′   -t −-ux∕c2--
t =  ∘ 1 − u2∕c2.

The change in coordinates between events in the Sframe is then given by

       (            )    (             )
△x ′ =   ∘-x2 −-ut2--  −   ∘-x1 −-ut1--  =  △∘x--−-u-△t--
           1 − u2∕c2         1 − u2∕c2        1 − u2∕c2

△y ′ = y − y  = △y
        2   1

△z ′ = z2 − z1 = △z

      (             )   (             )
        t2 − ux2∕c2        t1 − ux1 ∕c2     △t −  u△t
△t ′ =  ∘-------2--2- −    ∘------2--2-  = ∘-------2--2.
           1 − u ∕c          1 − u ∕c        1 − u  ∕c

Squaring the terms yield

         (             ) (             )
    ′2      △x  − u△t       △x −  u△t       (△x )2 − 2u △x △t + u2(△t )2
(△x )  =   ∘-------2--2    ∘------2--2-  =  ---------1-−-u2∕c2----------
             1 − u  ∕c       1 − u ∕c

(△y ′)2 = (△y )2

(△z ′)2 = (△z )2

         (            ) (             )
           △t  − u△t       △t  − u△t       (△t )2 − 2u △x △t ∕c2 + u2(△x )2∕c4
(△t ′)2 =   ∘-----------   ∘------------  = ----------------------------------.
             1 − u2∕c2       1 − u2∕c2                 1 − u2 ∕c2

Substituting these terms into the spacetime interval gives

    ′2   c2((△t-)2 −-2u-△x-△t-∕c2 +-u2(△x-)2∕c4) ((△x-)2 −-2u△x-△t--+-u2(△t-)2)-      2       2
(△s )  =               1 − u2 ∕c2              −           1 − u2∕c2           − (△y ) − (△z ) .

Adding the first two terms with common denominators together yields

          2    2        2    2     2    2     2  2
(△s  ′)2 = c-(△t--) −-(△x-)-−-u-(△t-)-+-u--(△x--)-∕c-−  (△y  )2 − (△z )2.
                         1 − u2∕c2

Pulling out a u2∕c2

    ′2   c2(△t2-) −-(△x--)2 −-u2∕c2(c2(△t-)2 +-(△x-)2)       2        2
(△s )  =                  1 − u2∕c2                 − (△y )  − (△z ) .

Factoring out a c2(t)2 (x)2 in the numerator

    ′ 2   (c2(△t2) − (△x )2)(1 − u2∕c2)        2       2
(△s  ) =  ----------1 −-u2∕c2----------−  (△y ) − (△z ) .

Finally, canceling terms gives

    ′2    2   2         2        2       2        2
(△s  )  = c (△t ) − (△x ) −  (△y ) − (△z )  = (△s ) .

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

References

[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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Classification:
Physics Classification03.30.+p (Special relativity)
 03. (Quantum mechanics, field theories, and special relativity )
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