# Tensor test

1. Jan 15, 2014

### Libra82

1. The problem statement, all variables and given/known data
Problem as stated: Consider a vector $A^a$. Is the four-component object $\left( \frac{1}{A^0},\frac{1}{A^1},\frac{1}{A^2},\frac{1}{A^3}\right)$ a tensor?

2. Relevant equations
Roman indices run from 0 to 3. Einstein summation convention is used.
Tensors of rank 1 (vectors) transform as
Contravariant vectors: $A^a = \frac{\partial x^a}{\partial x ' ^b}A'^b$
Covariant vectors: $A_a = \frac{\partial x'^b}{\partial x^a}A'_b$

3. The attempt at a solution
I use the above mentioned transformation rules for each of the individual components getting:
$\left( \frac{1}{A^0},\frac{1}{A^1},\frac{1}{A^2},\frac{1}{A^3}\right) = \left(\frac{1}{\frac{\partial x^0}{\partial x'^b}A'^b},\frac{1}{\frac{\partial x^1}{\partial x'^b}A'^b},\frac{1}{\frac{\partial x^2}{\partial x'^b}A'^b},\frac{1}{\frac{\partial x^3}{\partial x'^b}A'^b}\right)$

What I wanted was to transform the object to another frame and either arriving at a transformed object according to the above mentioned rules (confirming it is a tensor) or being unable to fit it into one of the rules above (disproving the object to be a tensor).

I am currently stuck with how to proceed.

2. Jan 16, 2014

### D H

Staff Emeritus
Presumably all of those Aa are non-zero, and hence 1/Aa is bounded.

Hint: Certainly you can find a transformation that sets one of the coefficients of A' to zero.