## Tangent planes and normals in three and four dimensions

Posted by peeterjoot on January 4, 2013

[Click here for a PDF of this post with nicer formatting]

# Motivation

I was reviewing the method of Lagrange in my old first year calculus book [1] and found that I needed a review of some of the geometry ideas associated with the gradient (that it is normal to the surface). The approach in the text used 3D level surfaces , which is general but not the most intuitive.

If we define a surface in the simpler explicit form , then how would you show this normal property? Here we explore this in 3D and 4D, using geometric and wedge products to express the tangent planes and tangent volumes respectively.

In the 4D approach, with a vector defined by coordinates and basis so that

the reciprocal basis is defined implicitly by the dot product relations

Assuming such a basis makes the result general enough that the 4D (or a trivial generalization to N dimensions) holds for both Euclidean spaces as well as mixed metric (i.e. Minkowski) spaces, and avoids having to detail the specific metric in question.

# 3D surface

We start by considering figure 1:

We wish to determine the bivector for the tangent plane in the neighbourhood of the point

then using duality determine the normal vector to that plane at this point. Holding either of the two free parameters constant, we find the tangent vectors on that surface to be

The tangent plane is then

We can factor out the pseudoscalar 3D volume element , assuming a Euclidean space for which . That is

Multiplying through by the volume element we find that the normal to the surface at this point is

Observe that we can write this as

Let’s see how this works in 4D, so that we know how to handle the Minkowski spaces we find in special relativity.

# 4D surface

Now, let’s move up to one additional direction, with

the differential of this is

We are going to look at the 3-surface in the neighbourhood of the point

so that the tangent vectors in the neighbourhood of this point are in the span of

In particular, in each of the directions we have

Our tangent volume in this neighbourhood is

Here the shorthand has been used. Can we factor out a 4D pseudoscalar from this and end up with a coherent result. We have

This gives us

With the usual 4d gradient definition (sum implied)

we have

so we can write

so, finally, the “normal” to this surface volume element at this point is

This is just like the 3D Euclidean result, with the exception that we need to look at the dual of a 3-volume “surface” instead of our normal 2d surface.

Also note that this is not a metric free result. The metric choice is built into the definition of the gradient 1.0.17 and its associated reciprocal basis. For example with a metric where , we have and .

# References

[1] S.L. Salas, E. Hille, G.J. Etgen, and G.J. Etgen. *Calculus: one and several variables*. Wiley New York, 1990.

## Leave a Reply