Tensor question about Hypersurfaces

[hamjam9]

Homework Statement

Consider the family of hypersurfaces where each member is defined by the constancy of the function S(x^c) over that hypersurface and further require that each hypersurface be a null hypersurface in the sense that its normal vector field, n_a = S_|a be a null vector field.

Let ¡ be a member of the family of curves that pierces each such hypersurface orthogonally, meaning that the tangent vector to ¡, say k_a is everywhere collinear with the vector na at the point of piercing. Show that ¡ is a null geodesic and find the condition on the relation between n_a and k_a that allows the geodesic equation to be written in the simple form k_a||bk^b = 0.

Interpret your results in terms of waves and rays.

Where _|a denotes partial derivative with respect to a, and _||a denotes the covariant derivative with respect to a.

Homework Equations

The Attempt at a Solution

By reading through the problem it is not very hard to get the hang of what it is saying, and it seems pretty clear that must be a null surface. But I don't know where to get started in showing that it is a "null geodesic", and how to derive at the simple geodesic equation they give. I'm just very stuck here. If anyone can give me a little hint I would appreciate it. Thanks in advance.

 

[mighty2000]

Hi there, great question! Let's break down the problem step by step.

First, let's define what a null hypersurface is. A hypersurface is a mathematical object that exists in one dimension less than the space it is embedded in. For example, a 3-dimensional hypersurface would exist in a 4-dimensional space. A null hypersurface is a type of hypersurface where the normal vector field, na, is null or has a length of zero. This means that the hypersurface is tangent to the light cone, and any light rays that pass through it would not be deflected.

Next, we are given a curve, ¡, that pierces each hypersurface orthogonally. This means that the tangent vector to the curve, ka, is collinear with the normal vector, na, at the point of piercing. This makes sense since the curve is piercing the hypersurface at a right angle, similar to how light rays would pass through a null hypersurface.

Now, we need to show that this curve, ¡, is a null geodesic. A geodesic is a curve that follows the shortest path between two points. In this case, the curve is a null geodesic because it follows the path of the light rays passing through the null hypersurface. To show that it is a null geodesic, we can use the geodesic equation, which is given by ka||bkb = 0. This equation tells us that the second derivative of the curve with respect to the parameter, b, is equal to zero. This means that the curve is a straight line, which is the shortest path between two points.

Finally, we need to find the condition on the relation between na and ka that allows the geodesic equation to be written in the simple form ka||bkb = 0. This condition is that the curve, ¡, must be tangent to the hypersurface at the point of piercing, meaning that ka and na must be collinear. This makes sense since we are looking at null hypersurfaces and null geodesics, which are both related to light rays.

In terms of waves and rays, this problem is showing us how light rays would behave when passing through a null hypersurface. The curve, ¡, represents the path of the light ray, and the hypersurface represents the null surface that the light ray is passing through. The geodesic equation tells us that