# Finding tensor components via matrix manipulations

1. Apr 29, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

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) $V_{\mu}X^{\mu\nu}$

2. Relevant equations

3. 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?

2. Apr 30, 2016

### 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.

3. Apr 30, 2016

### 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 $\lambda$?

4. Apr 30, 2016

### 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.

5. May 1, 2016

### spaghetti3451

Thanks!

(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?

6. May 1, 2016

### 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!

7. May 1, 2016

### spaghetti3451

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?

8. May 1, 2016

### 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.

9. May 1, 2016

### spaghetti3451

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?

10. May 1, 2016

### Fightfish

Yup, seems alright to me.

11. May 1, 2016

### spaghetti3451

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