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!!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Tensor transforming law problem / metric-delta contraction

**Physics Forums | Science Articles, Homework Help, Discussion**