Change of system of coordinates for the stress matrix

I have a stress matrix in cartesian coordinates : $\begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}$. How can I convert it to spherical coordinates ?

• Google "stress tensor spherical coordinates". See section 1.7: brown.edu/Departments/Engineering/Courses/En221/Notes/… – Samuel Weir Jul 3 '18 at 19:14
• In general, $\mathbf{A}'=\mathbf{P}^{-1}\mathbf{AP}$ – Quantumness Jul 3 '18 at 19:15
• Do you know how to express the three unit vectors for Cartesian coordinates in terms of the three unit vectors for spherical coordinates (in terms of the two spherical coordinate angles)? – Chet Miller Jul 4 '18 at 12:18
• It's totally geometric, and doesn't require the answer of Joe. – Chet Miller Jul 5 '18 at 22:18
• Here is a better version of the page linked by Samuel Weir: solidmechanics.org/text/AppendixD/AppendixD.htm The equations on that page should render correctly on modern browsers. This URL was found by u2berggeist – PM 2Ring Jun 25 at 17:03

One way to conceptualize the stress matrix is to view it as a tensor. In general, your matrix

$$T = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$$

should be thought of in terms of how it relates on a displacement vector $$v^T = (\mathrm{d}x, \mathrm{d} y, \mathrm{d} z)$$. The stress tensor tells you that the energy change associated to this small displacement vector is

$$\delta E = v^T T v = a {\mathrm{d}x}^2 + b {\mathrm{d}y}^2 + c {\mathrm{d}z}^2$$

Now, let's consider what happens if we change into spherical coordinates. Recall that in spherical coordinates $$(r,\phi,\theta)$$

$$x = r \cos \phi \sin \theta \\ y = r \sin \phi \sin \theta \\ z = r \cos \theta$$

This gives the relations

$$\mathrm{d}x = \mathrm{d}r (\cos\phi \sin\theta) + \mathrm{d}\phi (- r\sin\phi \sin\theta) + \mathrm{d}\theta(r \cos\phi \cos\theta)\\ \mathrm{d}y = \mathrm{d}r (\sin\phi \sin\theta) + \mathrm{d}\phi (r \cos\phi \sin\theta) + \mathrm{d}\theta(r \sin\phi \cos\theta)\\ \mathrm{d}z = \mathrm{d}r (\cos\theta) + \mathrm{d}\theta(-r \sin\theta)$$

which can be written in matrix form as

$$\begin{bmatrix} \mathrm{d}x \\ \mathrm{d}y \\ \mathrm{d}z \end{bmatrix} = J \begin{bmatrix} \mathrm{d}r \\ \mathrm{d}\phi \\ \mathrm{d}\theta \end{bmatrix}$$

Where J is the "Jacobian Matrix" (or change of coordinates)

$$J = \begin{bmatrix} \cos\phi \sin\theta & -r \sin\phi \sin\theta & r \cos\phi \cos\theta \\ \sin\phi \sin\theta & r \cos\phi \sin\theta & r \sin \phi \cos\theta \\ \cos \theta & 0 & -r \sin \theta \end{bmatrix}$$

This means that for the displacement vector $$\tilde{v}^T = (\mathrm{d}r, \mathrm{d}\phi,\mathrm{d}\theta)$$ written in spherical coordinates, the vectors in two different coordinate systems can be related to each other by

$$v = J \tilde{v}$$

We'll end up having that the energy change can be written as

$$\delta E = (J \tilde{v})^T T (J \tilde{v}) = \tilde{v}^T (J^T T J) \tilde{v}$$

which is some pretty complicated expression in terms of the $$\mathrm{d}r, \mathrm{d}\phi, \mathrm{d}\theta$$.

So since $$(J^T T J)$$ is the matrix that generates the same energy change for the vector in different coordinates, the stress matrix $$\tilde{T}$$ in spherical coordinates is really

$$\tilde{T} = J^T T J$$

This is a pretty general lesson that will let you express the stress matrix in any coordinate system, not just spherical ones.

And note that this transformation rule is different $$T \rightarrow J^T T J$$ is different than that for a linear transformation, which is $$A \rightarrow J^{-1} A J$$. This means that T is a tensor quantity, and not a linear transformation.

The defining property of a tensor is that is defines a length, sending a vector $$v$$ to a number, $$v \rightarrow v^T T v$$. In our case is the energy change $$\delta E$$ defined above, and the goal is to keep that length the same irrespective of the coordinate system.

On the other hand, a linear transformation is defined as sending vectors to vectors, $$v \rightarrow A v$$.

So after I found the better link(http://solidmechanics.org/text/AppendixD/AppendixD.htm), I went ahead and worked the full thing out. Here are the results. Note the results assume the tensor is symmetric (ie. $$S_{ij}=S_{ji}$$).

\begin{align} S_{rr} &= \mathbf{c}_{\theta}^2S_{xx} + 2\mathbf{c}_{\theta}\mathbf{s}_{\theta}S_{xy} + \mathbf{s}_{\theta}^2S_{yy} \\ S_{r\theta} &= \mathbf{c}_{\theta}\mathbf{s}_{\theta}(S_{yy}-S_{xx}) + (\mathbf{c}_{\theta}^2- \mathbf{s}_{\theta}^2)S_{xy} \\ S_{rz} &= \mathbf{c}_{\theta}S_{xz} + \mathbf{s}_{\theta}S_{yz} \\ S_{\theta\theta} &= \mathbf{s}_{\theta}^2S_{xx} - 2\mathbf{c}_{\theta}\mathbf{s}_{\theta}S_{xy} + \mathbf{c}_{\theta}^2S_{yy} \\ S_{\theta z} &= -\mathbf{s}_{\theta}S_{xz} + \mathbf{c}_{\theta}S_{yz} \\ S_{zz} &= S_{zz} \end{align}

where

\begin{align} \cos \theta &= \mathbf{c}_{\theta} \\ \sin \theta &= \mathbf{s}_{\theta} \end{align}