Transform rule for (1,2) tensor

[franznietzsche]

What is the transform rule for a (1,2) tensor? Is it:



T^a{}_{bc} = \bar T^d{}_{ef}\frac{\partial x^a}{\partial \bar x^d}\frac{\partial \bar x^e}{\partial x^b}\frac{\partial \bar x^f}{\partial x^c}



or is there an added term like in the transform rule for Christoffel symbols of the second kind?

[lethe]

Originally posted by franznietzsche

or is there an added term like in the transform rule for Christoffel symbols of the second kind?
this is correct. you only need the Christoffel symbols if you are taking a derivative.

[franznietzsche]

Thank you.

Originally posted by lethe
this is correct. you only need the Christoffel symbols if you are taking a derivative.

i know, but that transofrm is the same as the one for the christoffel symbols except the christoffel symbol's transformation rule has the added term:

\frac{\partial x^i}{\partial x^k}\frac{\partial^2 \bar x^m}{\partial x^j \partial x^k}


That is what i was referring to.

[lethe]

Originally posted by franznietzsche
Thank you.



i know, but that transofrm is the same as the one for the christoffel symbols except the christoffel symbol's transformation rule has the added term:

\frac{\partial x^i}{\partial x^k}\frac{\partial^2 \bar x^m}{\partial x^j \partial x^k}


That is what i was referring to.
yeah, the Christoffel symbols have an additional term, because they involve taking a derivative.

since you are not, then you do not need the extra term, and the equation you have in the first post is correct.

[Peterdevis]

tensors always transform always like you wrote T^a{}_{bc} = \bar T^d{}_{ef}\frac{\partial x^a}{\partial \bar x^d}\frac{\partial \bar x^e}{\partial x^b}\frac{\partial \bar x^f}{\partial x^c}
(by definition)
When you take a normal derivative from a tensor, you don't become a tensor. This is a problem for making diff equations with tensors. Therefor we define a new derivative (covariant derivative)
We becomes this by putting a second term (connection coefficients) by ten partial derivative. In general relativity we take a connection coëfficient we have derived from the metric. This is the Christoffel connection(are symbols)

[HallsofIvy]

The Christoffel symbols have that added term specifically because the Christoffel symbols are not tensors.