Tensor transforming law problem / metric-delta contraction

[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') withx'=\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 getS_{\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!!

[haushofer]

To answer your second question,


\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.

[teddd]

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

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

[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)