# Average of a tensor composed of unit vectors

1. Aug 13, 2013

### Irid

Hello, I'm studying The Theory of Polymer Dynamics by Doi and Edwards and on page 98 there is a tensor, defined as a composition of two identical unit vectors pointing from the monomer n to the monomer m:
$\hat{\textbf{r}}_{nm}\hat{\textbf{r}}_{nm}$

As far as I understood, the unit vectors have a uniform distribution in space. The authors then calculate the average of the tensor and it turns out to be 1/3 of the identity tensor:
$\text{average}(\hat{\textbf{r}}_{nm}\hat{\textbf{r}}_{nm}) = \textbf{I}/3$

No actual steps are given and I am confused by this result. I think the average should be zero. Any ideas?

2. Aug 14, 2013

### Jano L.

The rotational average is defined and can be calculated with help of rotation matrix by integration over rotation angles. But in this case, there is a simpler way.

Let the vector joining the two sites be $\mathbf n$, let us denote its $k$-th components in laboratory frame by $n_k$.

It helps to write the average of the tensor in the following way:

$$(\mathbf M)_{kl} = \langle n_k n_l \rangle.$$

There are 3x3 = 9 different terms that form the matrix 3x3. Since the expression inside the averaging bracket does not change under interchange of the two vectors(they are the same) we have $M_{kl} = M_{lk}$, so we need to consider only 6 different quantities

$$M_{11}, M_{22}, M_{33}, M_{12}, M_{13}, M_{23}.$$

Since the three axes of the coordinate system have the same relation to the axes, and since we consider all rotations equally likely, the first three have to be the same, so we really need to calculate only one of them, say $M_{11}$.

There is a nice trick to do this. What is $M_{11}$? It is an average of squares of the 1st coordinate ($x$) of the end-point of the unit vector $\mathbf n$, so the point lies on a unit sphere. For such points the sum of squares of all three coordinates is equal to 1 (3D Pythagorean theorem):

$$xx + yy +zz = 1.$$

When we average over all possible orientations of the molecule, we obtain the equation

$$\langle xx \rangle + \langle yy \rangle + \langle zz \rangle = 1.$$

Since all three average are the same, we infer that $\langle xx \rangle =\frac{1}{3}$ and similarly for $\langle yy \rangle, \langle zz \rangle$.

The off-diagonal term $M_{12}$ is the average

$$\langle xy\rangle.$$

However, for any given x there are many possible y values. Take some values $y_0$ and small interval of values of length $\Delta y$. Since all orientations are equally likely, the probability that the the y coordinate is in $(y_0, y_0+\Delta y)$ is the same as the probability that it is in $(-y_0-\Delta y_0, - y_0)$. So the expression xy is as likely positive and negative and its average turns out to be zero. Similarly for other non-diagonal terms.

So in the end, we obtain

$$\mathbf M = \left( \begin{array} ~ \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} &0\\ 0 & 0& \frac{1}{3}\\ \end{array} \right).$$

which is $\frac{1}{3} \mathbf 1$.

3. Aug 14, 2013

### Irid

Hey thanks Jano! Once you have stated the problem in a clearler mathematical notation, suddenly it starts making sense.