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 <br /> S_{\mu\nu=}\left(<br /> \begin{array}{cc}<br /> 1&amp;0\\0&amp;x^2<br /> \end{array}<br /> \right)<br />

in coordinate (x^1=x , x^2=y)
and you're asked to write it down in the new primed coordinate (x^1=x&#039;,x^2=y&#039;) withx&#039;=\frac{2x} {y} and y&#039;=\frac{y}{2}

So that
\begin{array}{ccc}<br /> \frac{\partial x^1}{\partial {x^{1}}&#039;}&amp;=&amp;y&#039;\\<br /> \frac{\partial x^ 2}{\partial {x^{2}}&#039;}&amp;=&amp;x&#039;\\<br /> \frac{\partial x^1} {\partial {x^{2}}&#039;}&amp;=&amp;0\\<br /> \frac{\partial x^2}{\partial {x^{2}}&#039;}&amp;=&amp;2\\<br /> \end{array}Now using the formula above i can write, for the first element :S_{1 &#039; 1 &#039;}= \frac{\partial x^\alpha}{\partial {x^{1}}&#039;}\frac{\partial x^\beta} {\partial {x^{1}}&#039;}S_{\alpha\beta}=(\frac{\partial x^1}{\partial {x^{1}}&#039;})^2S_{11}+2\frac{\partial x^1}{\partial {x^{1}}&#039;}\frac{\partial x^2}{\partial {x^{1}}&#039;}S_{21}+(\frac{\partial x^2}{\partial {x^{1}}&#039;})^2 S_{22}=(y&#039;)^2 +(x&#039;)^4(y&#039;)^2which is incorrect, becaouse it should come S_{1&#039;1&#039;}=(y&#039;)^2

And all of the rest came up to be wrong, I getS_{\mu &#039;\nu &#039;=}\left(<br /> \begin{array}{cc}<br /> (y&#039;)^2 +(x&#039;)^4(y&#039;)^2&amp;2(y&#039;)^2(x&#039;)^3\\2(y&#039;)^2(x&#039;)^3&amp;4(x&#039;y&#039;)^2<br /> \end{array}<br /> \right) whil it should be S_{\mu &#039;\nu &#039;=}\left(<br /> \begin{array}{cc}<br /> (y&#039;)^2&amp;x&#039;y&#039;\\x&#039;y&#039;&amp;4(x&#039;y&#039;)^2 +(x&#039;)^2<br /> \end{array}<br /> \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,

<br /> \delta_{\mu\nu} = g_{\mu\lambda}\delta^{\lambda}_{\nu} = g_{\mu\nu}<br />

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...

 

[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)