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I'm working with a problem in linear elasticity, and I have to calculate the strain energy function as follows:

$$ 2W = σ_{ij}ε_{ij} $$ Where σ and ε are symmetric rank 2 tensors.

For cartesian coordinates it is really easy because the metric is just the identity matrix, hence: $$ 2W = σ_{xx}ε_{xx} + σ_{yy}ε_{yy} + σ_{zz}ε_{zz} + 2 σ_{xy}ε_{xy} + 2 σ_{xz}ε_{xz} + 2 σ_{zy}ε_{zy} $$ My question is how the expression should be for cylindrical polar coordinates $(r,θ,z)$

Many thanks!

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    $\begingroup$ (Including as a comment for now b/c I'm not sure I'm remembering right). I think all that changes is that one needs distinguish covariant/contravarient indices i.e. include the metric tensor (which is $g_{rr}=g_{zz}=1,$ $g_{\theta\theta}=r^2$ in cylindrical coords). Then one has $$2W=\sigma^{ij}\epsilon_{ij}=g_{ik}g_{jl}\sigma^{ij}\epsilon^{kl}.$$ $\endgroup$
    – Semiclassical
    Commented Jul 31, 2014 at 13:44
  • $\begingroup$ THANKS!!! Do you know how it should look in component by component? $\endgroup$
    – Jemme
    Commented Jul 31, 2014 at 13:47
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    $\begingroup$ Plug in those $g$'s, and it should pop out. For comparison, the metric tensor in Cartesian coordinates is just $g_{xx}=g_{yy}=g_{zz}=1$ (all others are zero; this was also true in the cylindrical case.) So you might see if you can reproduce your Cartesian expression. $\endgroup$
    – Semiclassical
    Commented Jul 31, 2014 at 13:57
  • $\begingroup$ Also, another check: one can get the local element of arc length by writing $$ds^2=dx_i dx^i=g^{ij}dx_i dx_j=dr^2+dz^2+r^2\,d\theta^2$$ which is correct. (Note that I'm being quite careless with upper/lower indices; an expert could state things more carefully.) $\endgroup$
    – Semiclassical
    Commented Jul 31, 2014 at 14:09
  • $\begingroup$ My problem is dealing with Einstein summation convention and rank 2 tensors! I'm not sure how to do it. I know for sure that your expression is correct. But when I try to do it component by component to actually calculate the strain energy for the bending of a thin circular plate, is when I get all confused! $\endgroup$
    – Jemme
    Commented Jul 31, 2014 at 14:14

1 Answer 1

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We'll avoid the metric tensor altogether to keep things simple. The demonstration is simpler this way, although not "proper". The polar coordinate system is rather "nice" to deal with, so you don't need the full machinery of tensors in curvilinear coordinates.

First of all, note that W is a scalar. The second fact is that you can think of polar coordinates as a set of rotated Cartesian coordinates at each point, except the angle with respect to the base / natural Cartesian coordinates is a nice function of space (which one?). Once you know that, you know there must be a change of basis matrix $\beta$ which is proper orthogonal such that

$$ \sigma_{ij} =\beta_{pi} \beta_{qj} \sigma'_{pq} \\ \epsilon_{ij} =\beta_{ri} \beta_{sj} \epsilon'_{rs} $$

Note: No need for upstairs/ downstairs indices; I've used the prime to denote the components in polar. Let's plug in to the expression for 2W. $$ 2W = \sigma_{ij} \epsilon_{ij} \\ =\beta_{pi} \beta_{qj} \beta_{ri} \beta_{sj} \sigma'_{pq} \epsilon'_{rs} \\ =\beta_{pi} \beta^T_{ir} \beta_{qj} \beta^T_{js} \sigma'_{pq} \epsilon'_{rs} \\ =\delta_{pr} \delta_{qs} \sigma'_{pq} \epsilon'_{rs} \\ =\sigma'_{rs} \epsilon'_{rs} $$

(Note that $\beta \beta^T = I$).

So you have exactly the same form as in Cartesian. Now just write the components out using the summation and naming conventions. There are 2's because the off-diagonal terms appear twice in the double sum. I'm using $r, \theta, z$ for 1,2,3.

$$ 2W = \sigma_{rr} \epsilon_{rr} + \sigma_{\theta\theta} \epsilon_{\theta\theta} + \sigma_{zz} \epsilon_{zz} + 2\sigma_{r\theta} \epsilon_{r\theta} + 2\sigma_{\theta z} \epsilon_{\theta z} + 2\sigma_{rz} \epsilon_{rz} $$

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  • $\begingroup$ Nice job. I used the metric tensor approach because I know about such things through special relativity (where knowing how to raise/lower indices is crucial) rather than from a 'curvilinear coordinates' approach. The one thing I don't like in this approach is that it sweeps the detail of getting a proper orthogonal basis away (which is important since the polar components of $\sigma,\epsilon$ aren't necessarily expressed in those terms.) $\endgroup$
    – Semiclassical
    Commented Jul 31, 2014 at 17:06
  • $\begingroup$ @user167250 You're welcome. $\endgroup$
    – user_of_math
    Commented Jul 31, 2014 at 17:18
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    $\begingroup$ @Semiclassical It's a bit tricky. I've assumed (as is usually the case in applications) that the polar components are physical polar components. You'd have to use the metric tensor if the components were not physical - there'd be extra factors of $r$ along for the ride. However, because the final expression is fully contracted (a scalar), the expression for 2W would not change even though the meaning of individual components of the tensors would change. $\endgroup$
    – user_of_math
    Commented Aug 1, 2014 at 11:28

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