Finding tensor components via matrix manipulations

  • #1
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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?
 

Answers and Replies

  • #2
andrewkirk
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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
1,344
33
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
andrewkirk
Science Advisor
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(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
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33
(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?
 
  • #6
954
117
[itex]V_{\mu}[/itex] and [itex]V^{\mu}[/itex] have different components - don't forget that you need to apply the metric tensor to raise and lower indices!
 
  • #7
1,344
33
[itex]V_{\mu}[/itex] and [itex]V^{\mu}[/itex] 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?
 
  • #8
954
117
Yup, looks correct now. Same for part (g) - based on your original post, it seems to be [itex]V_{\mu}[/itex] instead of [itex]V^{\mu}[/itex]. It helps to remember that in the Einstein summation convention, one index should be superscripted and the other subscripted.
 
  • #9
1,344
33
Yup, looks correct now. Same for part (g) - based on your original post, it seems to be [itex]V_{\mu}[/itex] instead of [itex]V^{\mu}[/itex]. 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?
 
  • #10
954
117
Yup, seems alright to me.
 
  • #11
1,344
33
Thanks to both andrewkirk and Fightfish for helping me to solve the problem!:smile:
 

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