In one of the textbooks I came across this:

"the average over the angles of the scalar product such as $(\vec{n} \cdot \vec{a}) (\vec{n} \cdot \vec{b})$ is equal to $\frac{1}{3}(\vec{a} \cdot \vec{b})$"

The author also gives the average of $(\vec{n} \cdot \vec{a}) (\vec{n} \cdot \vec{b})(\vec{n} \cdot \vec{c}) (\vec{n} \cdot \vec{d})$ which is equal to $\frac{1}{15}[(\vec{a} \cdot \vec{b})(\vec{c} \cdot \vec{d})+(\vec{a} \cdot \vec{d})(\vec{c} \cdot \vec{b})+ (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})]$.

In all of these $\vec{n} = \vec{r}/r$ and $\vec{a},\vec{b},\vec{c},\vec{d}$ are constant vectors. How to prove it? What does "average over angles" mean, just $d\theta$ and $d\phi$ or the full spherical angle $d\phi \;d\theta \sin \theta$?

Thanks in advance!

  • 1
    $\begingroup$ It's always better to be specific. What book? What dimensions? The term "angle" almost implies we are only in $2$ dimension, but that might be a translation issue. $\endgroup$ May 12, 2017 at 20:41
  • $\begingroup$ "Some problems of nuclear theory" by Akchiezer. It is angular part of integration in 3D spherical coordinates. In russian addition is it on p.115. I am not sure if translated version even exists. $\endgroup$
    – MsTais
    May 12, 2017 at 20:43
  • $\begingroup$ Maybe one can find it also in the russian multiplication. $\endgroup$
    – Jean Marie
    May 12, 2017 at 21:16
  • $\begingroup$ @JeanMarie lol, somehow with this russian multiplication they made it to work well... $\endgroup$
    – MsTais
    May 12, 2017 at 22:52
  • $\begingroup$ @Thomas Andrews: why would there be a problem speaking about an angle between with 3D vectors. It is perfectly non-ambiguous. $\endgroup$
    – Jean Marie
    May 12, 2017 at 23:24

2 Answers 2


First formula :

Here are two different proofs :

Method 1 :

Let $u \cdot v$ denote the dot product of $u$ and $v$.

Using expansion :

$$(n \cdot (a-b))^2 = (n \cdot a)^2 + (n \cdot b)^2 - 2 (n \cdot a)(n \cdot b),$$

we can express :

$$(n \cdot a)(n \cdot b) = \frac12 \left((n \cdot a)^2 + (n \cdot b)^2 - (n \cdot (a-b))^2\right)$$

which gives the equivalent identity in terms of average values :

$$\overline{(n \cdot a)(n \cdot b)} = \frac12 \left(\overline{(n \cdot a)^2 }+ \overline{(n \cdot b)^2} - \overline{(n \cdot (a-b))^2}\right)\tag{*}$$

Thus, we are brought back to compute the average value $\overline{(c \cdot n)^2} $of $(c \cdot n)^2$ for a fixed vector $c$ and a variable vector $n \in \mathbb{S}$, the unit sphere.

We are going to prove that

$$\overline{(n \cdot c)^2} = \frac13\|c\|^2 \tag{**}$$

Indeed, taking into account (**) in (*), we will get :

$$\overline{(n \cdot a)(n \cdot b)} = \frac12 \frac13 \left(\|a\|^2 + \|b\|^2 - \|a-b\|^2\right)=$$

$$= \frac12 \frac13 \left(\|a\|^2 + \|b\|^2 - \|a\|^2 - \|b\|^2+2 a \cdot b\right)=\frac13 a \cdot b$$

as desired.

It remains to prove (**). We can consider WLOG $c$ to be with coordinates $(0,0,k)$ with $k:=\|c\|$. Let $(H)$ be the hemisphere of $\mathbb{S}$ with $z>0$. If $n=(x,y,z) \in (H)$,

$$(n \cdot c)^2=k^2z^2.\tag{***}$$

(H) can be "sliced" into spherical segments of height d$z$ with area $2\pi dz$ (http://mathworld.wolfram.com/SphericalSegment.html ). Integrating all these contributions on hemisphere $(H)$, we obtain, using (***) ;

$$\overline{(n \cdot c)^2}=\color{red}{\frac{1}{2 \pi}}\int_0^1 (k^2 z^2) (2 \pi dz) = \frac13 \|c\|^2 $$

i.e., result (**). Why this front division by $\color{red}{2 \pi}$ ? Because the sum of all the spherical segments areas is equal to the value $2 \pi$ which is the area of hemisphere $(H)$. Said otherwise, dividing $2 \pi dz $ by $2 \pi$ provides a pdf (probability distribution function) whose total "mass" is 1.

Method 2 :

Identifying $a,b$ and $n$ (all assumed unit norm vectors) with the associated column vectors of coordinates with respect to a certain basis, let us consider the following symmetric matrix :


Let the dot product of $a$ and $b$ be denoted :

$$\delta:=a \cdot b \ \ \text{identified with} \ \ a^Tb.$$

Quadratic form :


is such that $Q(n)=(n^Ta)(b^Tn)+(n^Tb)(a^Tn)=2(n^Ta)(n^Tb),$

i.e., twice the expression for which we have to determine its mean.

With the hypothesis $\|n\|=1$, $Q$ is the Rayleigh quotient (https://en.wikipedia.org/wiki/Rayleigh_quotient) associated with matrix $M$. The set of values of a Rayleigh quotient is known to be expressed as the set of all weighted averages (with positive weights) of its eigenvalues.

The 3 eigenvalues of $M$ being

$$\delta-1 \leq 0 \leq \delta +1\tag{3}$$ (see appendix below), we have to compute the mean of all expressions :

$$w_1(\delta-1)+w_20+w_3(\delta+1) \ \text{such that} \ w_1+w_2+w_3=1$$

which is :

$$\frac13(\delta-1)+\frac13(\delta+1)=\frac23\delta $$

($\delta-1$ and $\delta+1$ playing symmetrical rôles).

Dividing by 2 yields the result.

Appendix Explanation of (3) ;

The eigenvalues in (3) are associated with eigenvectors $a-b$, $a \times b$, $a+b$ resp.

For example, for the first vector $a-b$ :


$$=a \delta -a\underbrace{(b^Tb)}_{=1}+b\underbrace{(a^Ta)}_{=1}-b \delta=(\delta-1)(a-b)$$

establishing that $a-b$ is an eigenvector associated with eigenvalue $\delta-1$.

2nd formula :

I have no proof for it but @ThomasTuna has given a valuable one using a tensorial method ; another very interesting answer, using almost the same tools :

Surface integral of normal components summations on a sphere

Another reference whose interest is in the question mentionning equivalent formulas in $N$ dimensions with $3$ replaced by $D$, and $15$ replaced by $(D^2+2D)$:

How to rigorously show tensor identities using symmetry arguments?

(without answer, but a valuable comment by Ted Shifrin), and finally

Eigenvalues of a rank 2 tensor defined by an integral

(without any answer...)


FIRST IMPRESSION: This problem was written in "Some problems of nuclear theory" by Akchiezer, this is a physics textbook. Classically trained physicists solve problems with tensors/indices (i.e Einstein Notation) and arguments of symmetry. Using such tools both asserted relations follow in a few lines of algebra.

GAME PLAN:I will convert the problem into tensor contraction and then I will deduce the answer by symmetry.

NOTE: I will use explicit summation, rather than Einstein's implicit summation convention.

DERIVATION: Consider $(\vec a \cdot \vec n )(\vec b \cdot \vec n)$. $$(\vec a \cdot \vec n )(\vec b \cdot \vec n) = \sum_{ij} a_i n_i b_j n_j$$ $$ = \sum_{ij} a_i b_j n_i n_j$$ Let $M_{ij} \equiv a_i b_j $ and $N_{ij} \equiv n_i n_j$. $$ = \sum_{ij} M_{ij} N_{ij} \tag{X}$$ Eq.X is how a classically trained physicist sees this problem. As the contraction of 2 different 2-tensors, here $M$ and $N$ are obviously both rank 2 tensors.

By arguments of symmetry (see Appendix), we conclude that $$N_{ij}=\frac{1}{3} \delta_{ij} \tag{XX}$$ Combine equations (*), (**), and the definition of $M$ to arrive at: $$(\vec a \cdot \vec n )(\vec b \cdot \vec n) = \sum_{ij} M_{ij} N_{ij} = \sum_{ij} (a_i b_j) (\frac{1}{3} \delta_{ij})= \frac{1}{3} \sum_{ij} a_i \delta_{ij} b_j = \frac{a^Tb}{3}$$

REVIEW AND CONNECTIONS: This is exactly, the result already provided by Jean Marie's Method 1 and 2. So this is a third method to arrive at your first relation. It has an interesting connection with Jean's method 2. In this method Jean regroups the vectors as $$(\vec{n} \cdot \vec{a}) (\vec{n} \cdot \vec{b}) = \vec{n}^T\vec{a} \vec{b}^T\vec{n}^T$$ and distinguishes the the outer product of vectors $\vec{a} \vec{b}^T$ as a matrix. Then solves the problem via properties of $M=\vec{a} \vec{b}^T +\vec{b} \vec{a}^T$.

Analogously, I have grouped $$(\vec{n} \cdot \vec{a}) (\vec{n} \cdot \vec{b}) = \vec{a}^T\vec{n} \vec{n}^T\vec{b}^T$$ and distinguished the the outer product of vectors $\vec{n} \vec{n}^T=N$ as a matrix. Then solved the problem via properties of $N$. These approaches solve the same problem because of commutativity of the inner product.

Now I will prove the second statement. GAME PLAN: Exactly as before, I will convert the problem into a tensor contraction and then I will deduce the answer by symmetry.

DERIVATION: Consider $$(\vec{n} \cdot \vec{a}) (\vec{n} \cdot \vec{b})(\vec{n} \cdot \vec{c}) (\vec{n} \cdot \vec{d}) = \sum_{ijkl} a_i n_i b_j n_j c_k n_k d_l n_l $$ Let $M_{ijkl} = a_i b_j c_k d_l$ and $N_{ijkl}=n_i n_j n_k n_l$. $$= \sum_{ijkl} M_{ijkl}N_{ijkl} \tag{*}$$ We now reduced the right hand side to a contraction of $4$-tensors

As before, by arguments of symmetry (see appendix) $$N_{ijkl}=n_i n_j n_k n_l = \frac{1}{15}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk} ) \tag{**}$$.

Combining Eq.**, Eq.*, and the definition of $M$, we arrive at the solution $$\sum_{ijkl} M_{ijkl}N_{ijkl} = \sum_{ijkl} ( a_i b_j c_k d_l ) \frac{1}{15}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk} )$$ $$ =\frac{1}{15}[(\vec{a} \cdot \vec{b})(\vec{c} \cdot \vec{d})+(\vec{a} \cdot \vec{d})(\vec{c} \cdot \vec{b})+ (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})]$$

-----------------ARGUMENTS OF SYMMETRY------------------

Consider a rotated vector $n'$ where $n'_i=\sum_{j} R_{ij} n_j = R \vec n$ (Here $R$ is defined as a rotation matrix - https://en.wikipedia.org/wiki/Rotation_matrix). Then the rotated 2-tensor $N'$ is defined as, $$N' = R \vec{n} (R \vec n)^T = R \vec{n} \vec{n}^T R^T =R N R^{-1}$$ Thus the rotation of $\vec n$ results in a basis transformation of $N$ (https://en.wikipedia.org/wiki/Matrix_similarity). NOW HERE'S THE SYMMETRY, Since the random vector $\vec n$ is sampled from spherically symmetric distribution on a sphere, then $N$ must invariant w.r.t. rotations, thus $$N'=N$$ $$N'_{ij}= R_{ik}n_k R_{jl} n_l = R_{ik} R_{jl} n_k n_l = R_{ik} R_{jl} \delta_{kj} n_j \delta_{li} n_i = R_{ij} R_{ji} n_i n_j = \delta_{ij} n_i n_j = N_{ij}$$ Thus, we conclude the only such two 2-tensors are proportional to the identity so that $N_{ij} = c_0 \delta_{ij}$.

Furthermore, $\vec n$ is normalized. Thus $\sum_i n_i n_i =1$ which is equivalently, $tr\{ N\} = \sum_{ij} N_{ij} \delta_{ij} = \sum_i c_0 \delta_{ij} \delta_{ij} = 3*c_0$, it follows that, $$N_{ij}=\frac{1}{3} \delta_{ij} \tag{XX}$$ This is result Eq.XX

Now consider the second problem, Eq.**, which is a generalization of the first. As before we still have spherical symmetry, so $$N_{ijkl}=n_i n_j n_k n_l = N'_{ijkl}$$ $$N'_{ijkl}=\sum_{wxyz} R_{iw}n_w R_{jx}n_x R_{ky}n_y R_{lz} n_z = N_{ijkl}$$ $$N'_{ijkl}=\sum_{wxyz} R_{iw}R_{jx} R_{ky} R_{lz} n_w n_x n_y n_z = N_{ijkl}$$ Now unlike above we have freedom to mix indices contractions, I could insert $n_z = \delta_{zk} n_k$ or $n_z = \delta_{zj} n_j$ or $n_z = \delta_{zi} n_i$, (from here commutativity determines all other possible combinations) and arrive at $$N_{ijkl}= c_1 (\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk} )$$ As before the normalization condition is, $$1 =\sum_i n_i n_i$$ $$= \sum_i n_i n_i \sum_k n_k n_k $$ $$= \sum_{ijkl} c_1 (\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk} ) \delta_{ij}\delta_{kl}$$ $$=9+3+3$$ Thus, we conclude that $c_1 = \frac{1}{15}$

  • $\begingroup$ Very interesting ! Happy to see a proof of the second part (and a third proof of the first part). I thought that tensor calculus could be a key by I am not enough familiar with it... $\endgroup$
    – Jean Marie
    Apr 25, 2019 at 20:18
  • $\begingroup$ Using your expression $$\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}$$ with "approach0.xyz" (a formula-based search engine for mathSE), I found an interesting reference that I have given in an "edit" to my answer. $\endgroup$
    – Jean Marie
    Apr 25, 2019 at 21:15
  • 1
    $\begingroup$ First I want to thank you for your original answer, it was incredibly useful to me when advisor first posed this problem to me. Second, the additional links you have provided in an edit to your answer are quite interesting. I think that the first link solves the second link if you swap out integration measure from a 2-sphere for the integration measure over a D-1 sphere. $\endgroup$
    – ThomasTuna
    Apr 28, 2019 at 15:57

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