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The following is from Theory of Elasticity by Landau and Lifshitz.

Landau Lifshitz theory of elasticity

Why can only two independent scalars of second order be formed from the symmetric strain tensor $u_{i k}$, which for infinitesimal strain is defined as

$$ u_{i k} = \frac12 \left( \frac{\partial u_i}{\partial x_k} + \frac{\partial u_k}{\partial x_i} \right) $$

I would assume that from a tensor that (in three dimensions) contains six independent elements, six independents elements of second order could be formed, namely just the squares of these elements. Why is this not correct?

Note that summation convention is being used, i.e.

$$ u_{ii}^2 = (u_{11} + u_{22} + u_{33})^2 $$

and

$$ u_{i k}^2 = \sum_{i j} u_{ik}u_{ik}. $$

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  • $\begingroup$ Those aren't scalars. $\endgroup$ – Javier Jan 25 '18 at 21:23
  • $\begingroup$ I edited my post the reflect that summation convention is being used. $\endgroup$ – Kappie001 Jan 25 '18 at 21:26
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Recall that a scalar isn't just any function; it needs to be a function that transforms as a scalar, i.e. one that doesn't transform at all. The general way to do this is to construct an object with all the indices contracted. We start with the general second order term $$u_{ij} u_{k\ell}$$ and need to contract indices together. The only tensors available are the Kronecker delta $\delta^i_j$ and the volume tensor $\epsilon_{ijk}$. The volume tensor doesn't give us anything: if we just use one we get an odd number of indices, and if we use two they contract together to reduce to Kronecker deltas, i.e. $$\epsilon_{ijk} \epsilon^{imn} = \delta_j^m \delta_k^n - \delta_j^n \delta_k^m.$$ Using only the Kronecker delta, we can contract $i$ with $j$, so $k$ must be contracted with $\ell$, giving the first term $u_{ii} u_{kk} = u_{ii}^2$. (Here I'm being sloppy with index placement because it doesn't matter.) Otherwise, $i$ can be contracted with $k$ or $\ell$, and it doesn't matter which by symmetry. If $i$ is contracted with $k$ then $j$ is contracted with $\ell$, giving $u_{ij} u_{ij} = u_{ij}^2$, the second term.


More generally, the fact that there are two scalars can be understood by representation theory. Your situation has $SO(3)$ symmetry, and a general symmetric rank two tensor is a six-dimensional representation that decomposes into a scalar (its trace) and a traceless part, which we write as $$6 = 1 + 5.$$ The quadratic terms are formed from a tensor product of this representation with itself, $$6 \times 6 = (1 + 5) \times (1 + 5) = 1 + 5 + 5 + 5 \times 5.$$ The first scalar is simply the trace squared, as we've seen. Now, to decompose $5 \times 5$, we use the same method you might already know from quantum mechanics. (Indeed, to translate this to spin, just subtract one from every number and divide by two.) Then $$5 \times 5 = 1 + 3 + 5 + 7 + 9.$$ In total there are two factors of $1$ and hence two scalars.

Now we come back to the first point: why are $\delta^i_j$ and $\epsilon_{ijk}$ the only tensors available? These come from the two parts of the definition of $SO(3)$. The "orthogonal" part means that the Euclidean inner product is preserved, giving the metric $\delta^i_j$. The "special" part means that the volume is preserved, giving the volume $\epsilon_{ijk}$. Nothing else is preserved, so you can't contract with any other tensors -- those would change as well under rotation.

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  • $\begingroup$ What is the volume tensor? Why are it and the Kronecker delta the only tensors available? When you say "transforms as a scalar," do you mean when we consider the quantity from another intertial frame? How can we know for sure that we have exhaustively constructed all quantities that transform as a scalar? $\endgroup$ – Kappie001 Jan 25 '18 at 21:35
  • $\begingroup$ This answer is correct, but not all that enlightening. It would be helpful to explain the nature of transformations under the rotation group and their representation by unitary matrices. $\endgroup$ – Bert Barrois Jan 25 '18 at 21:35
  • $\begingroup$ @BertBarrois I was just thinking the same thing, and finished an edit just as you posted the comment! $\endgroup$ – knzhou Jan 25 '18 at 21:36
  • $\begingroup$ @Kappie001 I edited to address both concerns, tell me if this works for you! The Levi-Civita symbol $\epsilon_{ijk}$ is defined and related to volumes here. $\endgroup$ – knzhou Jan 25 '18 at 21:40
  • $\begingroup$ Thanks a bunch. I don't really know anything about representation theory, but the explanation of which tensors are available to contract with is very insightful. $\endgroup$ – Kappie001 Jan 25 '18 at 21:44

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