Tensor Integration

1. Nov 13, 2009

tim85ruhruniv

Could someone help me out ??

I tried this integration over the surface of a sphere of unit radii,

$$$P_{mn}e_{m}\otimes e_{n}=\frac{1}{D_{pq}e_{p}\otimes e_{q}}\int e_{m}\otimes e_{n}dS_{r=1}$$$

and I always get $$$4\pi e_{m}\otimes e_{n}$$$ and the 'D' tensor as it is..

I am expecting additionally a '3' in the denominator, am I wrong ??? If i do the integration over unit volume then I get the 3 in the denominator. Sorry for sounding stupid but is there a necessity to consider the unit tensor, i just assume it as a constant under integration.

2. Nov 13, 2009

clamtrox

Homework assignments and stuff like that should be posted in the appropriate section.

I don't get your indices either; it seems they don't add up on left and right hand sides. $$e_m \otimes e_n$$ definitely needs not be constant. Consider for example usual spherical coordinates.

3. Nov 13, 2009

George Jones

Staff Emeritus
Okay, I'll bite; along with clamtrox's note about indices, I have questions. What does

$$\frac{1}{D_{pq}e_{p}\otimes e_{q}}$$

mean? How does one divide by a tensor (not the component of a tensor), which is an element of a vector space?
Yes and no. From the Physics Forums rules:

4. Nov 14, 2009

tim85ruhruniv

hey !!

Thanks guys for looking at my work.

I cant see how the indices dont add up... maybe i am missing something... but

Each component of
$$$\mathbf{P}$$$will be a function of the $$\mathbf{\mathrm{D}^{-1}}$$ tensor.

about division by the tensor..

$$x=\mathbf{D}y$$ for some 'x' and some 'y'

so I hope I can rewrite this as $$y=\mathbf{\mathrm{D}^{-1}}x$$
and probably find the Inverse at a later stage. Which for the time being I believe doesent depend on the co-ordinates of integration.

Like clamtrox said, I use spherical co-ordinates to integrate, should I worry about $$$e_{m}\otimes e_{n}$$$ should I transform the tensor basis ???