PlanetPhysics/Spacetime Interval Is Invariant for a Lorentz Transformation

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The [[../SR/|spacetime]] interval between two events E1(x1,y1,z1,t1) and E2(x2,y2,z2,t2) is defined as

(s)2=c2t2(x)2(y)2(z)2.

If s is in [[../CosmologicalConstant/|reference frame]] S, then s is in reference frame S moving at a [[../Velocity/|velocity]] u along the x-axis. Therefore, to show that the spacetime interval is invariant under a [[../CosmologicalConstant/|Lorentz transformation]] we must show

(s)2=(s)2

with the reference frames related by [[../LorentzTransformation/|The Lorentz transformation]] x=xut1u2/c2 y=y z=z t=tux/c21u2/c2.

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

x=(x2ut21u2/c2)(x1ut11u2/c2)=xut1u2/c2

y=y2y1=y

z=z2z1=z

t=(t2ux2/c21u2/c2)(t1ux1/c21u2/c2)=tut1u2/c2.

Squaring the terms yield

(x)2=(xut1u2/c2)(xut1u2/c2)=(x)22uxt+u2(t)21u2/c2

(y)2=(y)2

(z)2=(z)2

(t)2=(tut1u2/c2)(tut1u2/c2)=(t)22uxt/c2+u2(x)2/c41u2/c2.

Substituting these terms into the spacetime interval gives

(s)2=c2((t)22uxt/c2+u2(x)2/c4)1u2/c2((x)22uxt+u2(t)2)1u2/c2(y)2(z)2.

Adding the first two terms with common denominators together yields

(s)2=c2(t2)(x)2u2(t)2+u2(x)2/c21u2/c2(y)2(z)2.

Pulling out a u2/c2

(s)2=c2(t2)(x)2u2/c2(c2(t)2+(x)2)1u2/c2(y)2(z)2.

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

(s)2=(c2(t2)(x)2)(1u2/c2)1u2/c2(y)2(z)2.

Finally, canceling terms gives

(s)2=c2(t2)(x)2(y)2(z)2=(s)2.

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

All Sources

[1] [2] [3]

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, [[../SpecialTheoryOfRelativity/|Relativity: The Special and General Theory ]]. 1916.

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