# Division with Einstein summation convention

1. Jan 5, 2015

### Powertravel

1. The problem statement, all variables and given/known data

I have the following equation

Aab= c ua ub

Where Aab is a rank 2 tensor and ua is a vector and c is a scalar and a,b = {0,1,2,3}. I know both Aab , ua and ua
I want to find c explicitly but I don't know how to interpret or calculate

c = Aab / ( ua ub )

Does anyone know how I should proceed?

2. Relevant equations
Aab= c ua ub
3. The attempt at a solution

c = Aab / ( ua ub )

2. Jan 5, 2015

### ShayanJ

Actually $A^{ab}=c u^a u^b$, in matrix form becomes the following:

$A=c\left[ \begin{array}{ccc} u^1 u^1 \quad u^1 u^2 \quad u^1 u^3 \\ u^2 u^1 \quad u^2 u^2 \quad u^2 u^3 \\ u^3 u^1 \quad u^3 u^2 \quad u^3 u^3 \end{array} \right]$.

Now if you write $A^{ab}$ in its matrix form too, you will find your answer.

3. Jan 5, 2015

### Powertravel

Thank you. I got tunnel vision on getting getting some sort of contraction going. Just to be clear, did you mean that I can get c by
c = A00 / (u0)2 = A11 / (u1)2 = ... etc?
Is it even possible to evaluate Aab / (ua ub)?

4. Jan 5, 2015

### ShayanJ

Yeah, that's correct.
But your last sentence tells me that you don't know enough about what you're doing. $A^{ab}$ is not a tensor, its only the (a,b) component of the tensor A. So $A^{ab}$ is only a number. When you work with tensors like that, you're working in the component notation. So $c=\frac{A^{ab}}{u^a u^b}$ is just the division of two real numbers and nothing more. Also it has nothing to do with Einstein summation convention. But I can tell you how it can be related. At first we should bring down one of the indices so that we have one index up and one index down to sum over. Then we should contract the indices:
$A^{ab}=cu^a u^b \Rightarrow g_{ca}A^{ab}=c g_{ca} u^a u^b \Rightarrow A_c^{\ b}=c u_c u^b \Rightarrow A_b^{\ b}=c u_b u^b \Rightarrow c=\frac{A_b^{\ b}}{u_b u^b}$
Now $A_b^{\ b}$ and $u_b u^b$ are just numbers, not components of a tensor.

5. Jan 5, 2015

### Powertravel

That's a neat trick and clears up alot of confusion I have. Thank you for the help.