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Whilst studying a book on fluid dynamics I came across a curious footnote comment which is essential in another derivation. The footnote states the following two identities:

$$ \frac{1}{|S|}\oint_S\! n_in_j\, dA = \frac{1}{3} \delta_{ij}\\ \frac{1}{|S|}\oint_S\! n_i n_j n_k n_l\, dA = \frac{1}{15} (\delta_{ij} \delta_{lm} + \delta_{ik}\delta_{jm} + \delta_{im}\delta_{jl} ) $$ where $S$ is a unit sphere, $n_i$ is a normal and $|\cdot|$ is the surface area. It looks like something pretty standard from vector calculus, but I cannot find it right now. Would anyone like to point me to a proof of these? Is there like a definitve compedium with proofs of these type of identities?

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  • $\begingroup$ Which textbook is this? $\endgroup$
    – Ian
    Feb 24, 2022 at 21:59

1 Answer 1

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Here's a unifying principle:$\,\,$ Integrating the dyadic product of radial vectors over a sphere will result in an isotropic tensor.

So assume $$\oint \hat{n}\,\hat{n}\,dA = \beta I$$ for some undetermined scale factor $\beta$.

The scale factor can be determined by contraction $$\eqalign{ I:I &= \delta_{ij}\,\delta_{ij} = 3 \cr 3\,\beta &= \oint \hat{n}\cdot\hat{n}\,dA \cr &= \oint dA = 4\pi R^2 \cr \beta &= \frac{4\pi}{3} R^2 \cr \cr }$$

Similarly $$\oint \hat{n}\,\hat{n}\,\hat{n}\,\hat{n}\,dA = \beta\,{\mathbb Y}$$ where ${\mathbb Y}$ is a 4th order isotropic tensor. There are 3 such tensors, so in general ${\mathbb Y}$ is the linear combination $${\mathbb Y}_{ijkl} = \beta_1\,\delta_{ij}\,\delta_{kl} + \beta_2\,\delta_{ik}\,\delta_{jl} + \beta_3\,\delta_{il}\,\delta_{jk} $$


We obviously need the result which is symmetric over every index, i.e. $$\beta_1=\beta_2=\beta_3=1$$ The contraction $\big\{I:{\mathbb Y}:I\big\}$ yields $$\eqalign{ \delta_{ij}\,(\delta_{ij}\,\delta_{kl} + \delta_{ik}\,\delta_{jl} + \delta_{il}\,\delta_{jk})\,\delta_{kl} &= (3\,\delta_{kl} + \delta_{kl} + \delta_{kl})\,\delta_{kl} \cr &= (5\,\delta_{kl})\,\delta_{kl} \cr &= 15 \cr }$$ and the contraction $\big\{I:(\hat{n}\,\hat{n}\,\hat{n}\,\hat{n}):I\big\}$ yields $$\eqalign{ (\hat{n}\cdot\hat{n})\,(\hat{n}\cdot\hat{n}) &= (1)\,(1) = 1 }$$ So $$\eqalign{ 15\,\beta &= \oint dA = 4\pi R^2 \cr \beta &= \frac{4\pi}{15} R^2 \cr }$$


Another approach is to use Gauss' Theorem $$\eqalign{ \oint \hat{n}\,\hat{n}\,dA &= \frac{1}{R}\oint\vec{r}\,d\vec{a} \cr &= \frac{1}{R}\int\nabla\vec{r}\,dv \cr &= \frac{1}{R}\int\,I\,dv \cr &= \frac{4\pi}{3} R^2 I \cr }$$ which matches our previous result.


On the other hand, the isotropic approach tells us immediately that $$\eqalign{ \oint \hat{n}\,dA = 0 \cr \oint \hat{n}\,\hat{n}\,\hat{n}\,dA = 0 \cr }$$ since in the first case, there is no isotropic vector.

And in the second case, there is an isotropic 3rd order tensor (i.e. the Levi-Civita permutation tensor), but it's not symmetric in all indices.

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  • $\begingroup$ That's so neat, thanks. $\endgroup$ Jan 6, 2017 at 9:43
  • $\begingroup$ Hi, thanks for this answer. If the problem is in two dimensions, can I write $\int \hat n_i \hat n_j =\frac{1}{2}\delta_{ij}$? Thanks $\endgroup$ Oct 16, 2019 at 23:59
  • $\begingroup$ Yes. Given a circle of radius $R$, integrating the quantity $({\hat n}{\hat n})$ over the circumference $(C=2\pi R)$ yields $$\frac{1}{|C|}\oint_C {\hat n}{\hat n}\,d\ell = \frac{1}{2}I$$ $\endgroup$
    – greg
    Oct 17, 2019 at 7:35

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