Tensor test

  • Thread starter Libra82
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Homework Statement


Problem as stated: Consider a vector [itex]A^a[/itex]. Is the four-component object [itex]\left( \frac{1}{A^0},\frac{1}{A^1},\frac{1}{A^2},\frac{1}{A^3}\right)[/itex] a tensor?


Homework Equations


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


The Attempt at a Solution


I use the above mentioned transformation rules for each of the individual components getting:
[itex]\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)[/itex]

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.
 

Answers and Replies

  • #2
D H
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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.
 

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