- #1

Arman777

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- Homework Statement:
- Calculating tensor equations

- Relevant Equations:
- Tensor Identities

Let us suppose we are given two vectors ##A## and ##B##, their components ##A^{\nu}## and ##B^{\mu}##. We are also given a minkowski metric ##\eta_{\alpha \beta} = \text{diag}(-1,1,1,1)##

In this case what are the

a) ##A^{\nu}B^{\mu}##

b) ##A^{\nu}B_{\mu}##

c) ##A^{\nu}B_{\nu}##

For part (a), it seems that we are going to obtain a 4x4 matrix with components

$$A^{\nu}B^{\mu} = \begin{bmatrix}

A^0B^0 & ... & A^0B^3 \\

A^1B^0 & ... & A^1B^3\\

A^2B^0 & ... & A^2B^3 \\

A^3B^0 & ... & A^3B^3\\

\end{bmatrix}$$

For part (b) I have written something like this

$$A^{\nu}B^{\gamma}\eta_{\mu \gamma} = D^{\nu}_{\mu} =

\begin{bmatrix}

A^0B^0\eta_{00} & A^0B^1\eta_{11} & A^0B^2\eta_{22} & A^0B^3\eta_{33}\\

A^1B^0\eta_{00} & A^1B^1\eta_{11} & A^1B^2\eta_{22} & A^1B^3\eta_{33}\\

A^2B^0\eta_{00} & A^2B^1\eta_{11} & A^2B^2\eta_{22} & A^2B^3\eta_{33}\\

A^3B^0\eta_{00} & A^3B^1\eta_{11} & A^3B^2\eta_{22} & A^3B^3\eta_{33}\\

\end{bmatrix}$$

Actually the ##D^{0}_{0}## becomes ## = A^0B^0\eta_{00} + A^0B^1\eta_{01} + A^0B^2\eta_{02} + A^0B^3\eta_{03}## but that is just ##A^0B^0\eta_{00}##

For part c its just the sum I guess so I need to write

##A^{\nu}B_{\nu} = A^{\nu}B^{\gamma}\eta_{\nu \gamma} = A^0B^0\eta_{00} + A^1B^1\eta_{11} + A^2B^2\eta_{22} + A^3B^3\eta_{33}##

Are these expressions ttrue ?

If I do something like this

##B_{\mu} = \eta_{\mu \nu}B^{\nu}## and write ##B_{\mu} = (-B^0, B^1, B^2, B^3)## and just multiply this with ##A^{\nu}## I would have got the same result right ?

In this case what are the

a) ##A^{\nu}B^{\mu}##

b) ##A^{\nu}B_{\mu}##

c) ##A^{\nu}B_{\nu}##

For part (a), it seems that we are going to obtain a 4x4 matrix with components

$$A^{\nu}B^{\mu} = \begin{bmatrix}

A^0B^0 & ... & A^0B^3 \\

A^1B^0 & ... & A^1B^3\\

A^2B^0 & ... & A^2B^3 \\

A^3B^0 & ... & A^3B^3\\

\end{bmatrix}$$

For part (b) I have written something like this

$$A^{\nu}B^{\gamma}\eta_{\mu \gamma} = D^{\nu}_{\mu} =

\begin{bmatrix}

A^0B^0\eta_{00} & A^0B^1\eta_{11} & A^0B^2\eta_{22} & A^0B^3\eta_{33}\\

A^1B^0\eta_{00} & A^1B^1\eta_{11} & A^1B^2\eta_{22} & A^1B^3\eta_{33}\\

A^2B^0\eta_{00} & A^2B^1\eta_{11} & A^2B^2\eta_{22} & A^2B^3\eta_{33}\\

A^3B^0\eta_{00} & A^3B^1\eta_{11} & A^3B^2\eta_{22} & A^3B^3\eta_{33}\\

\end{bmatrix}$$

Actually the ##D^{0}_{0}## becomes ## = A^0B^0\eta_{00} + A^0B^1\eta_{01} + A^0B^2\eta_{02} + A^0B^3\eta_{03}## but that is just ##A^0B^0\eta_{00}##

For part c its just the sum I guess so I need to write

##A^{\nu}B_{\nu} = A^{\nu}B^{\gamma}\eta_{\nu \gamma} = A^0B^0\eta_{00} + A^1B^1\eta_{11} + A^2B^2\eta_{22} + A^3B^3\eta_{33}##

Are these expressions ttrue ?

If I do something like this

##B_{\mu} = \eta_{\mu \nu}B^{\nu}## and write ##B_{\mu} = (-B^0, B^1, B^2, B^3)## and just multiply this with ##A^{\nu}## I would have got the same result right ?

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