Is the Horizon r=2GM a Null Surface?

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SUMMARY

The discussion centers on the concept of the event horizon in General Relativity, specifically the horizon defined by r=2GM in the Schwarzschild metric. Participants clarify that while substituting r=2GM and dr=0 into the metric yields dτ²=-(2GM)²dΩ², this does not indicate that the surface is not null. A null surface is characterized by having a null normal vector, and the normal covector to the surface defined by r-2GM=0 is indeed null, confirming that r=2GM is a null surface.

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  • Understanding of General Relativity principles
  • Familiarity with the Schwarzschild metric
  • Knowledge of null surfaces and their properties
  • Basic concepts of differential geometry
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  • Explore the Schwarzschild solution in detail
  • Learn about normal covectors and their significance in differential geometry
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quasar987
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Homework Statement


In my GR final, there was a question that asked to show that the horizon r=2GM is a null surface, i.e. [itex]d\tau^2=0[/itex]. How does that work, since just plugging r=2GM and dr=0 in the Schwarzschild metric yields

[tex]d\tau^2=-(2GM)^2d\Omega^2[/tex]

which is obviously not zero. :confused:
 
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A null surface has the property that its normal vector is null. What's a normal covector to the surface defined by r-2GM=0? Is it null?
 

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