# Transform rule for (1,2) tensor

1. Feb 8, 2004

### 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?

2. Feb 8, 2004

### lethe

this is correct. you only need the Christoffel symbols if you are taking a derivative.

3. Feb 8, 2004

### franznietzsche

Re: Re: Transform rule for (1,2) tensor

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.

4. Feb 9, 2004

### lethe

Re: Re: Re: Transform rule for (1,2) tensor

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.

5. Feb 9, 2004

### 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)

6. Feb 10, 2004

### HallsofIvy

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