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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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  • $\begingroup$ 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. $\endgroup$
    – Anamitra Palit
    Commented May 21, 2012 at 4:20
  • $\begingroup$ $a=(p,p,q,r)$;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. $\endgroup$
    – Anamitra Palit
    Commented May 21, 2012 at 4:35
  • $\begingroup$ The above example is with reference to Minkowski space.You may think of extending it to curved spacetime $\endgroup$
    – Anamitra Palit
    Commented May 21, 2012 at 4:41

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The surface is codimension two in spacetime, it is two dimensional inside a four dimensional space. This means that the tangent space to the surface does not consist of all vectors perpendicular to the normal to the surface, just as in three dimensions, the one-dimensional vector tangent space is not the collection of vectors perpendicular to some perpendicular line.

The tangent space to the surface does not include any null vectors, it just is what it is, some 2-dimensional space-like vector space sitting inside the manifold tangent space. To give a simple example, if the surface is the x-y plane at t=0, the future pointing perpendicular null directions are along the z axis going to the right and towards the future (1,0,0,1) and to the left and to the future (1,0,0,-1). The collection of all vectors perpendicular to (1,0,0,1) is all vectors of the form (a,b,c,a) (where a,b,c are real numbers), and includes more vectors than the tangent space of (0,b,c,0). The collection of vectors perpendicular to the other null vector, (1,0,0,-1), are (a,b,c,-a). The collection of all vectors perpendicular to both the outgoing light rays is the tangent space of the surface.

The same holds in curved space, since this is the situation locally.

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  • $\begingroup$ 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 $(k,b,c,k)$ $\endgroup$
    – Anamitra Palit
    Commented May 21, 2012 at 8:55
  • $\begingroup$ @AnamitraPalit: You are confused about where vectors are placed in GR--- a vector in GR doesn't have a beginning point and an end point, it has a beginning point and an infinitesimal displacement. The tangent planes at different places are independent things. The vector (1,0,0,1) is also defined starting at the point (k,x,y,k) (k is a fixed number, and x,y vary to sweep over the plane, you should have said it), and (1,0,0,1) is defined on the points on this plane, and it is perpendicular to the plane of points (k,x,y,k) for fixed k while varying x and y. $\endgroup$
    – Ron Maimon
    Commented May 21, 2012 at 9:06
  • $\begingroup$ You may consider my comment in relation to 4 vectors in flat spacetime $\endgroup$
    – Anamitra Palit
    Commented May 21, 2012 at 9:21
  • $\begingroup$ @AnamitraPalit: The curvature is not the issue, the issue is that vectors are located at positions--- you have to have a place for the vector. Vectors don't make sense between two points. You should always think of vectors as infinitesimal quantities attached at one point. $\endgroup$
    – Ron Maimon
    Commented May 21, 2012 at 17:47

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