# Tensor Calculations given two vectors and a Minkowski metric

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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 ?

Last edited:
• vanhees71

Homework Helper
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In this case what are the

a) ##A^{\nu}B^{\mu}##
b) ##A^{\nu}B_{\mu}##
c) ##A^{\nu}B_{\mu}##
(c) is identical to (b). I think you must have written one of them down wrong.

I think (a) gives a coordinate-dependent scalar, not a matrix, as it simply represents the product of
##A^{\nu}## (a scalar, as it is a component of A) with ##B^{\mu}## (a scalar, as it is a component of B). I have not come across any tensor notation that allows interpretation of ##A^{\nu}B^{\mu}## as a tensor expression.

We can however interpret ##A^{\nu}B_{\mu}## as a tensor expression, and it aligns with the second matrix you show above - which you label as (a) but I think you meant (b).

For (c), later in your post you rewrite it as different to what you write at the top. I agree with the solution you give for that - a sum that returns a scalar - in this case also a 0-0 tensor (ie coordinate-independent), as we can interpret your correction of (c) as a tensor expression.

To your last question, I would say Yes, assuming you're asking about (b), and you have to use an index different from ##\nu## as the second index of the Minkowski tensor. I would write:

$$A^{\nu}B_{\mu} = A^{\nu}\left(B^{\gamma} \eta_{\mu\gamma}\right)$$

Last edited:
• vanhees71 and Arman777
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(c) is identical to (b). I think you must have written one of them down wrong.

I think (a) gives a coordinate-dependent scalar, not a matrix, as it simply represents the product of
##A^{\nu}## (a scalar, as it is a component of A) with ##B^{\mu}## (a scalar, as it is a component of B). I have not come across any tensor notation that allows interpretation of ##A^{\nu}B^{\mu}## as a tensor expression.

We can however interpret ##A^{\nu}B_{\mu}## as a tensor expression, and it aligns with the second matrix you show above - which you label as (a) but I think you meant (b).

For (c), later in your post you rewrite it as different to what you write at the top. I agree with the solution you give for that - a sum that returns a scalar - in this case also a 0-0 tensor (ie coordinate-independent), as we can interpret your correction of (c) as a tensor expression.

To your last question, I would say Yes, assuming you're asking about (b), and you have to use an index different from ##\nu## as the second index of the Minkowski tensor. I would write:

$$A^{\nu}B_{\mu} = A^{\nu}\left(B^{\gamma} \eta_{\mu\gamma}\right)$$
I have edited my post maybe you can do the same

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I think (a) gives a coordinate-dependent scalar, not a matrix, as it simply represents the product of
Aν (a scalar, as it is a component of A) with Bμ (a scalar, as it is a component of B). I have not come across any tensor notation that allows interpretation of AνBμ as a tensor expressio
This is interesting. But I cannot understand how can it give us a scalar ?
Let me write ##A^{\nu}B^{\mu} = A^{\nu}B_{\gamma}g^{\mu \gamma}##

##A^{\nu}B_{\gamma}## This is the same as ##D^{\nu}_{\gamma}## so it seems that my result is correct.

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By definition, a scalar is a quantity, which doesn't change under Lorentz transformations. I'm a bit puzzled, what the correct question c) is. If it's ##A_{\nu} B^{\nu}## it means, as was also said before ##\eta_{\mu \nu} A^{\mu} B^{\nu}## (note that you have to sum over repeated indices from 0 to 3).

Now think, how do vector components like ##A^{\mu}## and ##B^{\nu}## transform under a Lorentz transformation, and what it actually means that a transformation is a Lorentz transformation (in relation to the Minkowski "metric").

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Now think, how do vector components like ##A^{\mu}## and ##B^{\nu}## transform under a Lorentz transformation, and what it actually means that a transformation is a Lorentz transformation (in relation to the Minkowski "metric").

Its just ##A^{\alpha}\Lambda^{\bar{\beta}}_{\bar{\alpha}}= A^{\bar{\beta}}##

where

$$\Lambda^{\bar{\beta}}_{\bar{\alpha}} = \begin{bmatrix} \gamma&-v\gamma&0&0\\ -v\gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}$$

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This is interesting. But I cannot understand how can it give us a scalar ?
Let me write ##A^{\nu}B^{\mu} = A^{\nu}B_{\gamma}g^{\mu \gamma}##
What do you think either of those expressions mean? Try to express it in terms of vectors A, B and the metric ##\eta## without using subscripts - ie in coordinate-independent language. You won't be able to, which indicates that the expression is coordinate-dependent.

By contrast, I can express (b) as vector A applied to the covector of B given a metric ##\eta##, and (c) as the trace of (b). These are coordinate-independent descriptions.

$$\Lambda^{\bar{\beta}}_{\bar{\alpha}} = \begin{bmatrix} \gamma&-v\gamma&0&0\\ -v\gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}$$
$$A^{\prime \beta}={\Lambda^{\beta}}_{\alpha} A^{\alpha}.$$