0:13

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

Â necessary to understand the lectures is averaging over the directions.

Â Suppose we encounter the following situation that in three dimensions

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

Â Such that it with equal probability covers all the directions and

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

Â Instead of the averaging over the time,

Â it is frequently used to average over the directions.

Â Namely, we define the following, so

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

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

Â sin theta sine phi, and cosine theta.

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

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

Â We take the following average ni

Â which is just d omega/4 pi,

Â this is solid angle, ni.

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

Â d phi so this is actually double integral

Â over 4 pi which is total solid angle of ni.

Â The reason why we divide by a total solid angle.

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

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

Â We plot here the component, the first component.

Â Take the average.

Â The second component, take the average.

Â The third component, take the average.

Â This is a way we get a vector.

Â 2:22

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

Â just by direct calculation is 0.

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

Â what we should we obtain after this average.

Â And we should obtain invariant under rotations three-dimensional vector

Â invariant under rotations, three-dimensional vector.

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

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

Â three-dimensional rotations is a vector of 0 length.

Â 2:58

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

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

Â because this is basically the sum of all possible directions.

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

Â what is the average ninj?

Â 3:20

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

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

Â But one can find the answer without even the calculation.

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

Â symmetric under exchange of indices.

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

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

Â So this invariant under the rotations tensor is delta ij.

Â 4:09

How to fix c?

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

Â It means that we take the trace on both sides.

Â Trace on the left side is just ni ni,

Â but this is just 1 because of this and this.

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

Â So it's c*3 = > c = 1/3,

Â so we have the following fact,

Â ni nj average is 1/3 of delta ij.

Â So next step is to calculate ni nj nk.

Â 4:59

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

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

Â We just plug these components here.

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

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

Â n2, n2, we just take first component here,

Â first component here and second component here.

Â That's this way we calculate this average and

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

Â 5:38

So who is this guy?

Â This guy should invariant under rotations, a tensor with three

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

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

Â This is 0, this is straightforward.

Â Also one can do find this by straightforward

Â calculation of these integrals as this is 0.

Â In fact, one can show that if 1 takes

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

Â 6:14

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

Â This is always 0 because there is no tensor which

Â carries odd numbers of indices and symmetric.

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

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

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

Â Well, this is already not zero,

Â by definition it has similar definition to this one.

Â 6:54

And it should be invariant under rotations.

Â Tensor with four indices, and

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

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

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

Â Delta kl, this contraction + delta ik and

Â delta jl, delta ik and delta jl.

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

Â So this quantity is invariant under rotations and

Â symmetric under exchange of any couple of its indices here.

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

Â contract two indices, say we contract these two indices.

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

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

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

Â 8:14

And on the right-hand

Â side we obtain a (delta

Â ij delta ll +, which is

Â actually 3 delta ik delta

Â jk + delta il delta jl).

Â So this is 3, this is delta ij,

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

Â We attain that a times 5 times delta ij

Â 9:04

One can see using this and this,

Â that a should be equal to 1/15.

Â So as the result ni, nj,

Â nk, nl is (1/15) (delta ij

Â delta kl + delta ik delta

Â jl + delta il delta jk).

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

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

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

Â where i and j are ranging from 1 to D.

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

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

Â nj, nl, equals to something.

Â So, along the same lines, one can calculate,

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

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

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

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

Â now we use this knowledge to calculate.

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

Â r) scalar product b,

Â r scalar product and the average.

Â So in this average, we assume that a and

Â b are constant vectors not participating into the average.

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

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

Â 11:45

And so this is equal to i,

Â ai bj times average of rirj.

Â This is equal to ai,

Â bj times r squared divided

Â by ri/r, rk/r.

Â This is equal to aibjr squared,

Â average of ninj.

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

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

Â 12:38

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

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

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

Â and this will be convenient and important knowledge for

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

Â [MUSIC]

Â