Hey guys it's me again!(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]g_{\mu '\nu '}=\frac{\partial x^\alpha}{\partial {x^{\mu}}'}\frac{\partial x^\beta}{\partial {x^{\nu}}'} g_{\alpha\beta}[/tex]where the prime indices indicate the new coordinate system.

I found an exercise where you're given the tensor [tex]

S_{\mu\nu=}\left(

\begin{array}{cc}

1&0\\0&x^2

\end{array}

\right)

[/tex]

in coordinate [itex](x^1=x , x^2=y)[/itex]

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

So that

[tex]\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}[/tex]

Now using the formula above i can write, for the first element :[tex]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[/tex]

which is incorrect, becaouse it should come [itex]S_{1'1'}=(y')^2[/itex]

And all of the rest came up to be wrong, I get[tex]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)[/tex] whil it should be [tex]S_{\mu '\nu '=}\left(

\begin{array}{cc}

(y')^2&x'y'\\x'y'&4(x'y')^2 +(x')^2

\end{array}

\right)[/tex]

what am I missing???

2) metric-delta contractions

This question is about the contraction of the metric.

I know that [tex]g_{\mu\nu}g^{\mu\lambda}=\delta_{\nu}^{\lambda}[/tex] where [itex]\delta_{\nu}^{\lambda}[/itex] is the Kronecker delta, equal to 1 if [itex]\nu=\lambda[/itex], to 0 otherwise.

But, does it works even for [itex]\delta_{\nu\lambda}[/itex] ?

I mean, it's still 0 if [itex]\nu=\lambda[/itex] and to 0 otherwise?

I am pushed to say no, becaouse if I contrac the delta with the metric I get [tex]g^{\kappa\nu}\delta_{\nu\lambda}=\delta^{\kappa}_{\lambda}[/tex] 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!!

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# Tensor transforming law problem / metric-delta contraction

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