# A Query on the Trapped Null Surface

A a compact, orientable, spacelike surface always has 2 independent forward-in-time pointing, lightlike, normal directions. For example, a (spacelike) sphere in Minkowski space has lightlike vectors pointing inward and outward along the radial direction. The inward-pointing lightlike normal vectors converge, while the outward-pointing lightlike normal vectors diverge. It can, however, happen that both inward-pointing and outward-pointing lightlike normal vectors converge. In such a case the surface is called trapped.-------- from Wikipedia: http://en.wikipedia.org/wiki/Apparent_horizon

Now a null vector is parallel to and perpendicular to itself at the same time. So the tangent plane on the concerned point on the spacelike surface should be a null surface.

What is the formal explanation for this?

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Let "a" be a spacelike vector and "b" a null vector. $b_{tt}^2-b_{xx}^2-b_{yy}^2-b_{zz}^2=0$ and $a_{tt}^2-a_{xx}^2-a_{yy}^2-a_{zz}^2<0$. Given the two stated conditions if we have,$a_{tt}b_{tt}-a_{xx}b_{xx}-a_{yy}b_{yy}-a_{zz}b_{zz}=0$ then the tangent plane at some point on the spacelike surface will be a null surface. –  Anamitra Palit May 21 '12 at 4:20
$a=(p,p,q,r)$;where p, q and r are positive quantities.Again $b=(m,m,0,0)$the first component in brackets for a and b is the time component a is a space like vector while b is a null vector a.b=0. The above possibility is a mathematically valid one. –  Anamitra Palit May 21 '12 at 4:35
The above example is with reference to Minkowski space.You may think of extending it to curved spacetime –  Anamitra Palit May 21 '12 at 4:41

The points $(k,x,y,k)$ represent a spacelike plane parallel to the x-y plane.We may take a fixed point $(k,b,c,k)$ on it.Vectors emanating from it and lying on the said plane[parallel to the x-y plane] are of the form $(0,x-b,y-c,0)$ which are again space like vectors normal to the null vector $(1,0,0,1)$.But this null vector ie,$(1,0,0,1)$ is not on the plane we are talking of.The components of the null vector are being referred from the origin and not from $(k,b,c,k)$ –  Anamitra Palit May 21 '12 at 8:55