I encounter the following formula (for strain energy function) a lot in physics literature:

$$ W(\epsilon_{kl}) = \int_0^{\epsilon_{kl}} \sigma_{ij} \textrm{d}\epsilon_{ij} $$

where all indices ranges from 1 to 3, both $\epsilon$ and $\sigma$ are 3x3 matrices.

My question is, what exactly does it mean in non-Einstein notation? Is it an abuse of notation? Because when I see $W(\epsilon_{kl})=\cdots$, I feel it should interpreted as 9 separate equations (with $kl$ replaced by 11, 12, 13, 21, 22, 23, 31, 32, 33), which does not make sense in this case.


First of all, it is strain, and not stain, energy density function.

Second, the left hand side means that $W$ depends on all the components of the tensor $\varepsilon_{kl}$ at the same moment. So none of the sides of the equation has any free vector indices: both sides are scalars (energy is a scalar, after all)! Quite generally, if arguments are indicated in the parentheses, they don't add any free indices to an equation because the whole parenthesis with the content may be omitted.

This is true regardless of whether we use Einstein's sum rule or not.

The only difference that the Einstein sum rule makes to the equation is that we may omit $$\sum_{i=1}^3 \sum_{j=1}^3$$ at the beginning of the right hand side. This summation would have to be added if we weren't allowed to use Einstein's summation convention.

Note that the upper limit of the integral only means that we are computing a contour integral in the 9-dimensional space of tensors that starts at the point $(0,0,0,0,0,0,0,0,0)$ and ends at the point $(\epsilon_{11},\epsilon_{12},\dots , \epsilon_{33})$. The only abuse of the notation is that the dummy integration variables should be denoted by a different symbol than $\epsilon_{ij}$, e.g. by $d\epsilon'_{ij}$, but the formula makes it so clear what we mean that the prime may be omitted.

  • $\begingroup$ Ops, thanks for pointing out the typo. :) So in non-Einstein notation, this formula means: $ W(\epsilon) = \sum_i \sum_j \int_0^{\epsilon_{11}} \int_0^{\epsilon_{12}} \int_0^{\epsilon_{13}} \int_0^{\epsilon_{21}} \int_0^{\epsilon_{22}} \int_0^{\epsilon_{23}} \int_0^{\epsilon_{31}} \int_0^{\epsilon_{32}} \int_0^{\epsilon_{33}} \sigma_{ij} \textrm{d}\epsilon'_{11} \textrm{d}\epsilon'_{12} \textrm{d}\epsilon'_{13} \textrm{d}\epsilon'_{21} \textrm{d}\epsilon'_{22} \textrm{d}\epsilon'_{23} \textrm{d}\epsilon'_{31} \textrm{d}\epsilon'_{32} \textrm{d}\epsilon'_{33} $ Is it right? $\endgroup$ – James Jun 23 '13 at 1:38
  • $\begingroup$ Or it means $ W(\epsilon) = \sum_{i}\sum_{j} \int_0^{\epsilon_{ij}} \sigma_{ij} \textrm{d} \epsilon'_{ij} $ ? $\endgroup$ – James Jun 23 '13 at 1:46
  • $\begingroup$ As I wrote clearly, it's the latter, it's a one-dimensional contour integral, not a nine-dimensional integral. $\endgroup$ – Luboš Motl Jun 23 '13 at 5:09

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