# Surface Tensors

1. Jan 4, 2008

### eep

I'm reading through Pauli's "Theory of Relativity", which has a discussion of tensors in the mathematical tools section of the book.

When introducing surface tensors, he states

"Such tensors can be obtained by considering two vectors $x$, $y$ which together span a two-dimensional parallelepiped. Its projections parallel to the axes on to the six two-dimensional coordinate planes, measure in units of the six parallelepipeds of the base vectors $e_i$, are given by

$$\xi^{ik} = x^iy^k - x^ky^i$$

..."

First, since we're dealing with 4-dimensional spacetime, I'm asuuming the six two-dimensional coordinate planes would be the xy,xz,xt,yz,yt,and zt planes. Is this right?

More importantly, I don't understand what the $\xi^{ik}$ represent, can someone explain it to me in a geometrical sense?

2. Jan 6, 2008

### haushofer

Well, it looks to me that this is just a cross-product put into an antisymmetric tensor. Normally one expresses the crossproduct as

$$\xi^{i} = \epsilon_{ijk}x^{j}y^{k}$$

What your tensor is is just a generalization of this without the summation, so you obtain a second order tensor instead of a vector:
$$\xi^{ij} = x^{i} \wedge y^{j}$$
So what in the first case we called $$\xi^{3}$$ is now $$\xi^{12}$$. And, not surprisingly, this $$\xi^{3}$$ is perpendicular to the surface described by $$\xi^{12}$$.

In n dimensions $$\xi^{ij}$$ has n(n-1)/2 independent entries, which in this case is indeed 6. So in short: in the one case you choose to represent an area by a vector which is perpendicular to that area ( the crossproduct in 3 dimensions ). In the other case you choose to represent it by a rank two tensor.

3. Jan 6, 2008

### HallsofIvy

To see what they represent, think about finding the "vector differential of area" of a surface in three dimensions. If we have a surface given by z= f(x,y), then we can use x and y as parameters and write the "position vector" of the surface as $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}$. Differentiating with respect to x and then y gives the tangent vectors in the x and y directions: $\vec{r}_x= \vec{i}+ f_x \vec{k}$ and $\vec{r}_y= \vec{j}+ f_y\vec{k}$. Their cross product gives a vector perpendicular to the surface whose length is the "differential of area":
$$\left|\begin{array} {ccc}\vec{i} & \vec{j} & \vec{k} \\0 & 1 & f_y\\ 1 & 0 & f_x \end{array}\right|= f_x\vec{i}+ f_y\vec{j}- \vec{k}$$
The cross product is only defined in 3 dimensions. In higher dimensions, you need to use the generalization (with the "alternating" tensor $\epsilon_{ijk}$ as haushofer said). Actually, using the alternating tensor in 3 dimensions give a 3 by 3 "anti-symmetric" tensor: having 0s on the diagonal, the three values, given above, above the diagonal and their negatives below. That would be the $\xi^{ik} = x^iy^k - x^ky^i$ you have.

4. Jan 6, 2008

### eep

Thanks, really cleared everything up for me

5. Jan 7, 2008

### OrderOfThings

What is... a bivector?

Start with the plane and pick an origin. From two vectors $\mathbf{u}$ and $\mathbf{v}$ we can form a parallelogram which we will denote by $/\mathbf{u},\mathbf{v}/$. (Let the parallelogram have an orientation, and denote this by $-/\mathbf{u},\mathbf{v}/ = /\mathbf{v},\mathbf{u}/$.)

We will now investigate the equivalence classes of parallelograms with the same oriented area. Having not mentioned a metric or other machinery to measure area, the first question to settle is if it is actually possible to form such equivalence classes. To see that this is indeed possible, take two parallelograms $/\mathbf{u},\mathbf{v}/$ and $/\mathbf{w},\mathbf{z}/$ and split the two vectors of the first parallelogram in components of the vectors of the second parallelogram:

$$\begin{array}{l}\mathbf{u} = u_1\mathbf{w}+u_2\mathbf{z} \\ \mathbf{v} = v_1\mathbf{w}+v_2\mathbf{z}\end{array}$$

Then the two have parallelograms have the same area if

$$u_1v_2-u_2v_1=1$$.

The equivalence class of such parallelograms is called a bivector and is denoted $\mathbf{u}\wedge\mathbf{v}$. The set of bivectors in the plane is a one-dimensional vector space. Having picked one bivector $\mathbf{u}\wedge\mathbf{v}$ (generated by two vectors $\mathbf{u}$ and $\mathbf{v}$) to be a basis, any other bivector can be written

$$a\,\mathbf{u}\wedge\mathbf{v}$$

for some number $a$.

Moving up one dimension, a bivector in three dimensions is the equivalence class of parallelograms with the same area and lying in the same plane. Two bivectors $\mathbf{u}\wedge\mathbf{v}$ and $\mathbf{w}\wedge\mathbf{z}$ - not necessarily in the same plane - can be added together to produce a new bivector. First, pick new representative parallelograms $/\mathbf{a},\mathbf{b}/$ and $/\mathbf{c},\mathbf{b}/$ of the bivectors sharing a common vector $\mathbf{b}$. (This means that $\mathbf{b}$ lies in the intersection line of the bivector planes.) Then add the two other vectors together $\mathbf{d}=\mathbf{a}+\mathbf{c}$. The resulting bivector is $\mathbf{d}\wedge\mathbf{b}$.

The set of bivectors in 3-space is a three-dimensional vector space. Given a vector basis $\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z$, this also generates a bivector basis:

$$\mathbf{e}_y\wedge\mathbf{e}_z,\;\mathbf{e}_z\wedge\mathbf{e}_x,\;\mathbf{e}_x\wedge\mathbf{e}_y$$.

Any bivector, let's call it $\mathbf{\xi}$, splits into

$$\mathbf{\xi} = \xi_{yz}\,\mathbf{e}_y\wedge\mathbf{e}_z+\xi_{zx}\,\mathbf{e}_z\wedge\mathbf{e}_x+\xi_{xy}\,\mathbf{e}_x\wedge\mathbf{e}_y$$.