- #1
teddd
- 62
- 0
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 [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!
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!
Last edited: