0:13

We're going to study another important situation, technical moment which is

necessary to understand the lectures is averaging over the directions.

Suppose we encounter the following situation that in three dimensions

we have a unit vector, which rapidly rotates in all directions.

Such that it with equal probability covers all the directions and

we want to average it over the positions of this vector.

Instead of the averaging over the time,

it is frequently used to average over the directions.

Namely, we define the following, so

the unit vector is the same as ni in tensor notations,

has the following component, it's sin theta cos phi,

sin theta sine phi, and cosine theta.

That is not hard to check that, ni times ni which is the same as n vector

squared is 1 and what does it mean that we average over all directions?

We take the following average ni

which is just d omega/4 pi,

this is solid angle, ni.

So for solid angle as you know is equal to sine theta d theta,

d phi so this is actually double integral

over 4 pi which is total solid angle of ni.

The reason why we divide by a total solid angle.

We want to have the following situation that this is just 1, so

what does it mean [COUGH] that we take this average?

We plot here the component, the first component.

Take the average.

The second component, take the average.

The third component, take the average.

This is a way we get a vector.

2:22

So, it is not hard to calculate that these average is just 0,

just by direct calculation is 0.

Well, but the explanation for that is very simple, you see,

what we should we obtain after this average.

And we should obtain invariant under rotations three-dimensional vector

invariant under rotations, three-dimensional vector.

But there is no such a vector, that is the reason we obtain 0.

It's the only vector which is invariant under rotations,

three-dimensional rotations is a vector of 0 length.

2:58

So well another way is that in this averaging for every direction we

encounter the inverse direction and then the sum, they compensate each other

because this is basically the sum of all possible directions.

They compensate each other and that's the way we get 0, so

what is the average ninj?

3:20

Well, this is by definition is d omega over 4 pi ninj.

One can directly calculate this integral, which quite tedious calculation.

But one can find the answer without even the calculation.

We should obtain a tensor with two indices which is invariant under rotations and

symmetric under exchange of indices.

The only such standard is known as delta i j where c is some

number which remains to be checked, it cannot be fixed on the symmetry grounds.

So this invariant under the rotations tensor is delta ij.

4:09

How to fix c?

Well, that is very easy, we just contract the indices i and j.

It means that we take the trace on both sides.

Trace on the left side is just ni ni,

but this is just 1 because of this and this.

On the other hand, on the left we have the trace of three by three unit matrx.

So it's c*3 = > c = 1/3,

so we have the following fact,

ni nj average is 1/3 of delta ij.

So next step is to calculate ni nj nk.

4:59

Again, by definition, this is this quantity (d omega)/4 pi ni nj nk so

how do we obtain this quantitative or maybe I should clarify.

We just plug these components here.

First components, say first component here and first component here.

That's how we obtain n 1 1 1 so if want to obtain n1,

n2, n2, we just take first component here,

first component here and second component here.

That's this way we calculate this average and

this is the way we calculated this average or this average.

5:38

So who is this guy?

This guy should invariant under rotations, a tensor with three

indices which is symmetric under exchange of all its three indices.

There is no such a tensor and as a result this is 0.

This is 0, this is straightforward.

Also one can do find this by straightforward

calculation of these integrals as this is 0.

In fact, one can show that if 1 takes

the average of any odd number, so if we have ni, ...,

6:14

ni2k+1, this is always 0 for any integer k.

This is always 0 because there is no tensor which

carries odd numbers of indices and symmetric.

There is no invariant tensor under rotations which carries odd number of

indices and which is symmetric under exchange of any of its two indices.

So next step is to calculate ni, nj, nk, nl.

Well, this is already not zero,

by definition it has similar definition to this one.

6:54

And it should be invariant under rotations.

Tensor with four indices, and

which is symmetric under exchange of any of its two indices.

So up to a constant, let me call it a, this quantity should be

according to the symmetry rules, it's delta ij, this contraction.

Delta kl, this contraction + delta ik and

delta jl, delta ik and delta jl.

And finally, this contraction and this contraction delta il delta jk.

So this quantity is invariant under rotations and

symmetric under exchange of any couple of its indices here.

So what remains to be fixed is this constant, to do that we just

contract two indices, say we contract these two indices.

It means that we equate them and sum all them, so

as a result we obtain ni nj times n squared, average,

on the left-hand side which is just ni nj.

8:14

And on the right-hand

side we obtain a (delta

ij delta ll +, which is

actually 3 delta ik delta

jk + delta il delta jl).

So this is 3, this is delta ij,

this is delta iij and what do we attain as a result?

We attain that a times 5 times delta ij

9:04

One can see using this and this,

that a should be equal to 1/15.

So as the result ni, nj,

nk, nl is (1/15) (delta ij

delta kl + delta ik delta

jl + delta il delta jk).

So we have defined and calculate the average of ni, which is just 0,

the average of ninj, which is just 1/3 in three dimensions, delta ij.

In fact, in D dimensions there, so it would be 1/ D delta ij,

where i and j are ranging from 1 to D.

And also we found that ni, nj, nk average is 0,

and found the answer for, I will not write it,

nj, nl, equals to something.

So, along the same lines, one can calculate,

find that the average of any odd number, for any odd number of n, this is 0.

And the answer for any even number of n, this is,

can be found, this is a good exercise to do to find it.

Along the same lines as we did it for this case and for this case,

now we use this knowledge to calculate.

Supposed one who needs to calculate the falling average (a,

r) scalar product b,

r scalar product and the average.

So in this average, we assume that a and

b are constant vectors not participating into the average.

And also what is fixed is the modulus of r, it's constant and

the average is done with respect to all directions of r.

11:45

And so this is equal to i,

ai bj times average of rirj.

This is equal to ai,

bj times r squared divided

by ri/r, rk/r.

This is equal to aibjr squared,

average of ninj.

This is already a unit vector of unit length and this average we know from here.

So, ai bj r-squared 1/3 times delta ij.

12:38

This together with this gives us scalar product of a and b.

So, this is r-squared/3 scalar product of (a,b).

And similarly one can calculate many other exercises in these directions,

and this will be convenient and important knowledge for

many situations we encounter in special or general theory of relativity.

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