1. Oct 15, 2009

### dakg

Ok I have seen the tensor double dot scalar product defined two ways and it all boils down to how the multiplication is defined. Does anyone know which is correct? I believe the first one is correct but I keep seeing the second one in various books on finite element methods.

1. $$\nabla \vec{u} \colon \nabla \vec{v} = u_{i,j} v_{j,i}$$

or

2. $$\nabla \vec{u} \colon \nabla \vec{v} = u_{i,j} v_{i,j}$$

dakg

Last edited: Oct 15, 2009
2. Oct 15, 2009

### espen180

You mean outer multiplication between two vectors, right? The definition i have seen (using index notation) is, in $$D$$ dimensions,

$$\vec{u} \otimes \vec{v}= a_{ij}=u_i v_j\;,\;1\leq i,j \leq D$$

3. Oct 15, 2009

### dakg

sorry there is a $$\nabla$$ missing

i'll edit it

4. Oct 15, 2009

### dakg

i have it in there but it isn't printing, let me try here

$$\nabla \vec{u} \colon \nabla \vec{v}$$

5. Oct 15, 2009

### lurflurf

The first one is more common, but it is a matter of convention.

6. Oct 15, 2009

### dakg

Do you know why? I found the first one in a Lightfoot book on transport.

They make different results, so wouldn't one be correct and the other wrong?

7. Oct 16, 2009

### lurflurf

Not wrong just different.
log(e)=1
log(10)=1
3*5+2=17
3*5+2=21
Here are examples of conventions that can lead to confusion.
The convention here (using dyadic product for an example) is
1) (ab):(cd)=(a.d)(b.c) the usual rule
2) (ab):(cd)=(a.c)(b.d) the other rule
The usual rule proably is choosen because of matrix algebra
ie to be the same as matrix product