Integration of traceless symmetric matrices

[jouvelot]

Hi,

I stumbled upon an identity when studying tensor perturbations in cosmology. The formula states that
$$
\int d^2\hat{p} f(\hat{p}.\hat{q})\hat{p}_i \hat{p}_k e_{jk}(\hat{q}) = e_{ij}(\hat{q})/2 \int d^2\hat{p} f(\hat{p}.\hat{q})(1-(\hat{p}.\hat{q})^2),
$$ where $e_{ij}$ is a symmetric traceless matrix-valued function of $\hat{q}$ with $\hat{q}_i e_{ij}(\hat{q}) = 0$, $\hat{p}$ is the unit vector parallel to $p$ and $f$ any function of the scalar product $\hat{p}.\hat{q}$.

Is there a simpler way to get this formula than to proceed by using cumbersome spherical or euclidian coordinate expressions?

Thanks in advance.

Bye,

Pierre

 

[AndyW]

Dear Pierre,

Thank you for sharing your discovery with us. The formula you have stumbled upon is indeed quite interesting and can provide valuable insights in the study of tensor perturbations in cosmology. I have looked into it and have found a simpler way to arrive at this formula.

Instead of using spherical or euclidean coordinate expressions, we can use the fact that the unit vector $\hat{p}$ is perpendicular to $\hat{q}$ in a two-dimensional space. This means that the dot product $\hat{p}.\hat{q}$ is equal to the sine of the angle between the two vectors.

Using this, we can rewrite the formula as:
$$
\int d^2\hat{p} f(\sin\theta)\hat{p}_i \hat{p}_k e_{jk}(\cos\theta) = e_{ij}(\cos\theta)/2 \int d^2\hat{p} f(\sin\theta)(1-\cos^2\theta),
$$
where $\theta$ is the angle between $\hat{p}$ and $\hat{q}$.

This form of the formula is much simpler and can be easily understood without the need for cumbersome coordinate expressions. I hope this helps in your further study of tensor perturbations in cosmology.