# Transforming stress tensor from polar to Cartesian coordinates?

Trying to understand where my calculation goes wrong. I have the stress tensor with components $$T_{\rho \rho}, T_{\rho \theta}$$ and $$T_{\theta \theta}$$. I wish to express this in Cartesian coordinates. So, I will first transform it to Cartesian basis and then do a coordinate transformation.

I have $$r= \rho \cos \theta$$ and $$t=\rho \sin \theta$$. The jacobian, $$J$$, is $$\begin{bmatrix} \cos \theta & -\rho \sin \theta \\ \sin \theta & \rho \cos \theta \end{bmatrix}$$

Now, as I understand it, the transformation to Cartesian basis is achieved via

$$$$T^{cart}_{\mu \nu}=J T^{polar}_{\mu \nu} J^T.$$$$

The right hand side will be in terms of $$\rho$$ and $$\theta$$. So Finally I just let $$\rho \rightarrow \sqrt{r^2 + t^2}$$ and $$\theta \rightarrow \tan^{-1}\left(\frac{t}{r} \right)$$.

However, my end result becomes suspiciously messy. Is this approach correct?

## 1 Answer

In general, a coordinate change on a covariant tensor is carried out as follows:

1. $$x^\mu = (\rho, \theta) \to \bar{x}^\mu = (\sqrt{r^2 + t^2}, \tan^{-1}(t/r))$$ as you wrote.
2. Using Einstein notation for the sum the coordinates transforms as: $$\bar{T}_{\mu \nu} = T_{\alpha \beta}\frac{\partial x^\alpha}{\partial \bar{x}^\mu} \frac{\partial x^\beta}{\partial \bar{x}^\nu}.$$

In this transformation, cartesian coordinates are the $$\bar{x}^\mu$$ and the polar ones the $$x^\mu$$. With this we can see how the components of the tensor changes: $$\bar{T_{r r}} = T_{\rho \rho} \left( \frac{\partial \rho}{\partial r} \right)^2 + (T_{\theta \rho} + T_{\rho \theta})\frac{\partial \theta}{\partial r} \frac{\partial \rho}{\partial r} + T_{\theta \theta} \left(\frac{\partial \theta}{\partial r}\right)^2.$$

Computing the derivatives we obtain: $$\frac{d\rho}{dr} = \frac{r}{\sqrt{r^2 + t^2}} = \frac{r}{\rho},$$ $$\frac{d\theta}{dr} = -\frac{1}{r^2 + t^2} = -\frac{1}{\rho^2}.$$ Sustituying above leads to:

$$\bar{T_{r r}} = T_{\rho \rho} \frac{r^2}{r^2 + t^2} + (T_{\rho \theta} + T_{\theta \rho})\frac{r}{(r^2+t^2)^{3/2}} + T_{\theta \theta} \frac{1}{r^2 + t^2}.$$

I think that you tried a Jacobian matrix to go from cartesians to polars instead of polar to cartesians. The same applies for all the other components, take advantadge of the tensor symmetry to avoid extra calculations and hope this is useful to you :).

PS: For a contravariant tensor: $$\bar{T}^{\mu \nu} = T^{\alpha \beta}\frac{\partial \bar{x}^\mu}{\partial x^\alpha} \frac{\partial \bar{x}^\nu}{\partial x^\beta}.$$

• Thank you. Just to clarify. The indices on the RHS on the tensor with a bar should be $r$ and $t$ right? Commented Nov 22, 2023 at 9:42
• The truth is that the text indexes are just flags, we should write $T_{00},T_{01},T_{10},T_{11}$, but yes! corrected for clarity! Commented Nov 22, 2023 at 10:17