## Strain tensor in spherical coordinates

Posted by peeterjoot on January 23, 2012

## Spherical tensor.

To perform the derivation in spherical coordinates we have some setup to do first, since we need explicit representations of all three unit vectors. The radial vector we can get easily by geometry and find the usual

We can get by geometrical intuition since it the plane unit vector at angle rotated by . That is

We can get by utilizing the right handedness of the coordinates since

and find

That and some Mathematica brute force can be used to calculate the differential strain element, and we find

### A manual derivation.

Doing the calculation pretty much completely with Mathematica is rather unsatisfying. To set up for it let’s first compute the unit vectors from scratch. I’ll use geometric algebra to do this calculation. Consider figure (\ref{fig:qmTwoExamReflection:continuumL2fig5})

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumL2fig5}

\caption{Composite rotations for spherical polar unit vectors.}

\end{figure}

We have two sets of rotations, the first is a rotation about the axis by . Writing for the unit bivector in the plane, we rotate

Now we rotate in the plane spanned by and by . With , our vectors in the plane rotate as

(with since ).

Now, these are all the same relations that we could find with coordinate algebra

There’s nothing special in this approach if that is as far as we go, but we can put things in a nice tidy form for computation of the differentials of the unit vectors. Introducing the unit pseudoscalar we can write these in a compact exponential form.

To summarize we have

Taking differentials we find first

Summarizing these differentials we have

A final cleanup is required. While is a vector and has a nicely compact form, we need to decompose this into components in the , and directions. Taking scalar products we have

Summarizing once again, but this time in terms of , and we have

Now we are set to take differentials. With

we have

Squaring this we get the usual spherical polar line scalar line element

With

our differential is

We can add to this and take differences

For each we have

and plugging through that calculation is really all it takes to derive the textbook result. To do this to first order in , we find

Collecting terms we have the result of the text in the braces

It should be possible to do the calculation to second order too, but to include all the quadratic terms in is again really messy. Trying that with mathematica gives the same results as above using the strictly coordinate algebra approach.

# References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

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