Finding tensor components via matrix manipulations

[spaghetti3451]

Homework Statement

Imagine we have a tensor $X^{\mu\nu}$ and a vector $V^{\mu}$, with components

$X^{\mu\nu}=\left( \begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right), \qquad V^{\mu} = (-1,2,0,-2).$

Find the components of:

(a) ${X^{\mu}}_{\nu}$
(b) ${X_{\mu}}^{\nu}$
(c) $X^{(\mu\nu)}$
(d) $X_{[\mu\nu]}$
(e) ${X^{\lambda}}_{\lambda}$
(f) $V^{\mu}V_{\mu}$
(g) $V_{\mu}X^{\mu\nu}$

Homework Equations

The Attempt at a Solution

(a) ${X^{\mu}}_{\nu}=X^{\mu\rho}\eta_{\rho\nu}=\left( \begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right) \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc} -2 & 0 & 1 & -1 \\ 1 & 0 & 3 & 2 \\ 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & -2 \end{array} \right)$,

where the rows of the left matrix are multiplied by the columns of the right matrix because the summation is over the second index of $X^{\mu\rho}$ and the first index of $\eta_{\rho\nu}$.

(b) ${X_{\mu}}^{\nu}=\eta_{\mu\rho}X^{\rho\nu}= \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right) =\left( \begin{array}{cccc} -2 & 0 & -1 & 1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right)$,

where the rows of the left matrix are multiplied by the columns of the right matrix because the summation is over the second index of $\eta_{\mu\rho}$ and the first index of $X^{\rho\nu}$.

(c) $X^{(\mu\nu)}=\frac{1}{2}(X^{\mu\nu}+X^{\nu\mu})=\frac{1}{2}\Bigg[\left( \begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right)+\left( \begin{array}{cccc} 2 & -1 & -1 & -2 \\ 0 & 0 & 1 & 1 \\ 1 & 3 & 0 & 1 \\ -1 & 2 & 0 & -2 \end{array} \right) \Bigg]=\left( \begin{array}{cccc} 2 & -0.5 & 0 & -1.5 \\ -0.5 & 0 & 2 & 1.5 \\ 0 & 2 & 0 & 0.5 \\ -1.5 & 1.5 & 0.5 & -2 \end{array} \right)$

(d) $X_{[\mu\nu]}=\frac{1}{2}(X_{\mu\nu}-X_{\nu\mu})=\frac{1}{2}(\eta_{\mu\rho}X^{\rho\sigma}\eta_{\sigma\nu}-\eta_{\nu\sigma}X^{\sigma\rho}\eta_{\rho\mu})$

Are my answers to (a), (b) and (c) correct?

With part (d), I'm not sure if I should take the original matrix to $X^{\rho\sigma}$ or the transposed matrix to $X^{\rho\sigma}$? Does it make a difference anyway?

 

[andrewkirk]

What you have written looks correct to me, including (d). If you instead used the transposed matrix in (d) you would change just the sign of the answer.

 

[spaghetti3451]

Ok!

(d) $X_{[\mu\nu]}=\frac{1}{2}(X_{\mu\nu}-X_{\nu\mu})=\frac{1}{2}(\eta_{\mu\rho}X^{\rho\sigma}\eta_{\sigma\nu}-\eta_{\nu\sigma}X^{\sigma\rho}\eta_{\rho\mu})= \frac{1}{2}\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right) \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) - \frac{1}{2}\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{cccc} 2 & -1 & -1 & -2 \\ 0 & 0 & 1 & 1 \\ 1 & 3 & 0 & 1 \\ -1 & 2 & 0 & -2 \end{array} \right) \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$
$=\frac{1}{2} \left( \begin{array}{cccc} 2 & 0 & -1 & 1 \\ 1 & 0 & 3 & 2 \\ 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & -2 \end{array} \right)-\frac{1}{2} \left( \begin{array}{cccc} 2 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 \\ -1 & 3 & 0 & 1 \\ 1 & 2 & 0 & -2 \end{array} \right) =\left( \begin{array}{cccc} 0 & -0.5 & -1 & -0.5 \\ 0.5 & 0 & 1 & 0.5 \\ 1 & -1 & 0 & -0.5 \\ 0.5 & -0.5 & 0.5 & 0 \end{array} \right)$

(e) ${X^{\lambda}}_{\lambda} =\eta_{\lambda\rho}X^{\rho\sigma}\eta_{\sigma\lambda}$

Is my answer to (d) correct?

Am I on the right track with (e)? How do I sum over $\lambda$?

 

[andrewkirk]

(d) looks OK
(e) is just the trace of a matrix you have already calculated (where?), so you don't need to do any new matrix multiplications.

 

[spaghetti3451]

(d) looks OK

Thanks!

(e) is just the trace of a matrix you have already calculated (where?), so you don't need to do any new matrix multiplications.

(e) ${X^{\lambda}}_{\lambda}={X^1}_{1}+{X^2}_{2}+{X^3}_{3}+{X^4}_{4}=-2+0+0-2=-4$, from part (a).

(f) $V^{\mu}V_{\mu}= \left( \begin{array}{cccc} -1 & 2 & 0 & -2 \end{array} \right) \left( \begin{array}{c} -1 \\ 2 \\ 0 \\ -2 \end{array} \right)=9$

(g) $V^{\mu}X^{\mu\nu}= \left( \begin{array}{cccc} -1 & 2 & 0 & -2 \end{array} \right) \left( \begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right)= \left( \begin{array}{cccc} 0 & -2 & 3 & 9 \end{array} \right)$

What do you think?

 

[Fightfish]

$V_{\mu}$ and $V^{\mu}$ have different components - don't forget that you need to apply the metric tensor to raise and lower indices!

 

[spaghetti3451]

$V_{\mu}$ and $V^{\mu}$ have different components - don't forget that you need to apply the metric tensor to raise and lower indices!

Ok!

(f) $V_{\nu}=V^{\rho}\eta_{\rho\nu}= \left( \begin{array}{cccc} -1 & 2 & 0 & -2 \end{array} \right) \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{cccc} 1 & 2 & 0 & -2 \end{array} \right)$

Therefore, $V^{\mu}V_{\mu}=V^{0}V_{0}+V^{1}V_{1}+V^{2}V_{2}+V^{3}V_{3}=(-1)(1)+(2)(2)+(0)(0)+(-2)(-2)=7$.

Is it correct now?

 

[Fightfish]

Yup, looks correct now. Same for part (g) - based on your original post, it seems to be $V_{\mu}$ instead of $V^{\mu}$. It helps to remember that in the Einstein summation convention, one index should be superscripted and the other subscripted.

 

[spaghetti3451]

Yup, looks correct now. Same for part (g) - based on your original post, it seems to be $V_{\mu}$ instead of $V^{\mu}$. It helps to remember that in the Einstein summation convention, one index should be superscripted and the other subscripted.

Ok, so using $V_{\mu}$ from part (f),

(g) $V_{\mu}X^{\mu\nu}= \left( \begin{array}{cccc} 1 & 2 & 0 & -2 \end{array} \right) \left( \begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \end{array} \right)= \left( \begin{array}{cccc} 4 & -2 & 5 & 7 \end{array} \right).$

Is it all right?

 

[Fightfish]

Yup, seems alright to me.

 

[spaghetti3451]

Thanks to both andrewkirk and Fightfish for helping me to solve the problem!