I have a trivial question:(adsbygoogle = window.adsbygoogle || []).push({});

Let assume a world sheet of a time-like spherical shell in Minkowski space-time.

On the 2D-Minkowski diagram (R,T), where R is the radius and T is the time,

the world line is represented by a time-like curve.

Let assume that the shell collapse and its 4-velocity is

[tex]u^\alpha=(\dot{T},\dot{R},0,0) [/tex]

where [tex]\dot{T}=\mathrm{d}T/\mathrm{d}\tau >0[/tex] and [tex]\dot{R}=\mathrm{d}R/\mathrm{d}\tau <0[/tex].

Let [tex]n^\alpha[/tex] be an orthogonal vector to the 4-velocity and the surface.

With the metric diag(-1,1,1,1) we get:

[tex]n_\alpha=s (-\dot{R},\dot{T},0,0) [/tex]

[tex]n^\alpha=s (\dot{R},\dot{T},0,0) [/tex]

where s is the sign depending on the orientation.

I expect that [tex]n^\alpha[/tex] is orthogonal to [tex]u^\alpha[/tex] on the Minkowski diagram, and its space-like. For example, I expect that the outward oriented normal vector has positive coordinates. But since [tex]\dot{T}>0[/tex] and [tex]\dot{R}<0[/tex] this is not true, and for example for s=+1, [tex]n^\alpha[/tex] is a past directed vector on the diagram!?

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# Orthogonal diretion on the Minkowski diagram

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