Tensor Calculations given two vectors and a Minkowski metric

[Arman777]

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 ?

 

[andrewkirk]

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:

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

 

[Arman777]

(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

 

[Arman777]

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.

 

[vanhees71]

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").

 

[Arman777]

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

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

 

[andrewkirk]

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.

 

[vanhees71]

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

where

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

Good! And what follows then for $A_{\alpha} B^{\alpha}$ and $A_{\bar{\alpha}} B^{\bar{\alpha}}$? (NB I don't like to put the basis-distinguishing label to the indices of tensor components but put it on the symbol, i.e., I'd rather write
$$A^{\prime \beta}={\Lambda^{\beta}}_{\alpha} A^{\alpha}.$$
Also note that you should carefully keep both the horizontal and the vertical placement of the indices under control!