Tensor transforming law problem / metric-delta contraction

1. Aug 20, 2011

teddd

Hey guys it's me again!
I'm now asking you 2 question, I'm sure you'll be helpful as usual!!

1) changing basis- changing tensors

ok, the law of transformation of a (let's make this simple) 02 tensor is $$g_{\mu '\nu '}=\frac{\partial x^\alpha}{\partial {x^{\mu}}'}\frac{\partial x^\beta}{\partial {x^{\nu}}'} g_{\alpha\beta}$$where the prime indices indicate the new coordinate system.

I found an exercise where you're given the tensor $$S_{\mu\nu=}\left( \begin{array}{cc} 1&0\\0&x^2 \end{array} \right)$$

in coordinate $(x^1=x , x^2=y)$
and you're asked to write it down in the new primed coordinate $(x^1=x',x^2=y')$ with$$x'=\frac{2x} {y}$$ and $$y'=\frac{y}{2}$$

So that
$$\begin{array}{ccc} \frac{\partial x^1}{\partial {x^{1}}'}&=&y'\\ \frac{\partial x^ 2}{\partial {x^{2}}'}&=&x'\\ \frac{\partial x^1} {\partial {x^{2}}'}&=&0\\ \frac{\partial x^2}{\partial {x^{2}}'}&=&2\\ \end{array}$$

Now using the formula above i can write, for the first element :$$S_{1 ' 1 '}= \frac{\partial x^\alpha}{\partial {x^{1}}'}\frac{\partial x^\beta} {\partial {x^{1}}'}S_{\alpha\beta}=(\frac{\partial x^1}{\partial {x^{1}}'})^2S_{11}+2\frac{\partial x^1}{\partial {x^{1}}'}\frac{\partial x^2}{\partial {x^{1}}'}S_{21}+(\frac{\partial x^2}{\partial {x^{1}}'})^2 S_{22}=(y')^2 +(x')^4(y')^2$$

which is incorrect, becaouse it should come $S_{1'1'}=(y')^2$

And all of the rest came up to be wrong, I get$$S_{\mu '\nu '=}\left( \begin{array}{cc} (y')^2 +(x')^4(y')^2&2(y')^2(x')^3\\2(y')^2(x')^3&4(x'y')^2 \end{array} \right)$$ whil it should be $$S_{\mu '\nu '=}\left( \begin{array}{cc} (y')^2&x'y'\\x'y'&4(x'y')^2 +(x')^2 \end{array} \right)$$

what am I missing???

2) metric-delta contractions

This question is about the contraction of the metric.
I know that $$g_{\mu\nu}g^{\mu\lambda}=\delta_{\nu}^{\lambda}$$ where $\delta_{\nu}^{\lambda}$ is the Kronecker delta, equal to 1 if $\nu=\lambda$, to 0 otherwise.

But, does it works even for $\delta_{\nu\lambda}$ ?
I mean, it's still 0 if $\nu=\lambda$ and to 0 otherwise?

I am pushed to say no, becaouse if I contrac the delta with the metric I get $$g^{\kappa\nu}\delta_{\nu\lambda}=\delta^{\kappa}_{\lambda}$$ this tensor certainly isn't the kronecker delta (the identity matrix), but will contain some terms of the metric!!

isnt'it??

Thanks for the attention!!

Last edited: Aug 20, 2011
2. Aug 20, 2011

haushofer

$$\delta_{\mu\nu} = g_{\mu\lambda}\delta^{\lambda}_{\nu} = g_{\mu\nu}$$

So a delta with only lower indices IS the metric, you just give it another name.

3. Aug 20, 2011

teddd

Thanks a lot haushofer for your answes, to both of my posts!!

now I wait for the first question...:tongue2:

4. Aug 20, 2011

teddd

But (excuse me for insisting!) Gravitation (the book) and wikipedia says that the kronecker delta IS $\delta_{\mu\nu}$!!

It is becaouse actually in this form is NOT a tensor but a normal generalized function and therefore it makes no sense to contrac it with the metric?

I mean, the kronecker deta, as a tensor, is only $\delta^{\mu}_\nu$ and contracting with the metric gives the metric itself (like you rightfully say)