Tensor Calculations given two vectors and a Minkowski metric

In summary: The result is the same as before,Basically, the matrix you show in part (a) is just a coordinate-dependent scalar.
  • #1
Arman777
Insights Author
Gold Member
2,168
193
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:
  • Like
Likes vanhees71
Physics news on Phys.org
  • #2
Arman777 said:
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:
  • Like
Likes vanhees71 and Arman777
  • #3
andrewkirk said:
(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
 
  • #4
andrewkirk said:
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.
 
  • #5
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").
 
  • #6
vanhees71 said:
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

$$\Lambda^{\bar{\beta}}_{\bar{\alpha}} =
\begin{bmatrix}
\gamma&-v\gamma&0&0\\
-v\gamma&\gamma&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}$$
 
  • #7
Arman777 said:
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.
 
  • #8
Arman777 said:
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!
 

Related to Tensor Calculations given two vectors and a Minkowski metric

1. What is a tensor?

A tensor is a mathematical object that describes the relationship between vectors and scalars in a multi-dimensional space. It can be thought of as a generalization of vectors and matrices, and is often used in physics and engineering to describe physical quantities and their transformations.

2. What are tensor calculations used for?

Tensor calculations are used to solve problems in physics and engineering that involve multi-dimensional systems, such as relativity, electromagnetism, and fluid dynamics. They are also used in machine learning and artificial intelligence for data analysis and pattern recognition.

3. How do you perform tensor calculations given two vectors and a Minkowski metric?

To perform tensor calculations given two vectors and a Minkowski metric, you first need to represent the vectors and metric as matrices. Then, you can use matrix multiplication and other mathematical operations to compute the desired tensor quantities, such as the inner product, outer product, and contraction of the vectors with the metric.

4. What is a Minkowski metric?

A Minkowski metric is a mathematical object that describes the geometry of spacetime in special relativity. It is a generalization of the Euclidean metric, which measures distances in three-dimensional space, and is used to calculate the interval between two events in four-dimensional spacetime.

5. How are tensors related to special relativity?

Tensors are essential in special relativity as they allow us to describe the geometry of spacetime and the transformations of physical quantities between different reference frames. In particular, the Minkowski metric is a tensor that is used to calculate the spacetime interval, which is an invariant quantity in special relativity.

Similar threads

  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
15
Views
2K
  • Advanced Physics Homework Help
Replies
15
Views
3K
  • Advanced Physics Homework Help
Replies
2
Views
708
  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
1
Views
3K
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Replies
1
Views
1K
Back
Top