Special Relativity: A^μ_ν Differences Explained

[LagrangeEuler]

Can someone explain me difference between
$$A^{\mu}_{\hspace{0.2cm} \nu}$$
$$A^{\hspace{0.2cm} \mu}_{\nu}$$
and
$$A^{\mu}_{\nu}$$?

 

[vanhees71]

The first two expressions are correct and usually denote components of a 2nd-rank tensor. You can lower and raise indices with the metric components $g_{\mu \nu}$ and $g^{\mu \nu}$, respectively, i.e., you have
$${A_{\nu}}^{\mu} = g_{\nu \sigma} g^{\mu \rho} {A^\sigma}_{\rho}.$$
The last expression should be avoided, because the horizontal placement of the indices is not indicated. It's ok if the tensor $A$ is symmetric, i.e., if $A_{\mu \nu}=A_{\nu \mu}$, because (only!) then
$${A^{\mu}}_{\nu} = g^{\mu \rho} A_{\rho \nu} = g^{\mu \rho} A_{\nu \rho} ={A_{\nu}}^{\mu},$$
and the horizontal ordering is not important.

 

[martinbn]

https://en.wikipedia.org/wiki/Mixed_tensor

 

[LagrangeEuler]

The first two expressions are correct and usually denote components of a 2nd-rank tensor. You can lower and raise indices with the metric components $g_{\mu \nu}$ and $g^{\mu \nu}$, respectively, i.e., you have
$${A_{\nu}}^{\mu} = g_{\nu \sigma} g^{\mu \rho} {A^\sigma}_{\rho}.$$
The last expression should be avoided, because the horizontal placement of the indices is not indicated. It's ok if the tensor $A$ is symmetric, i.e., if $A_{\mu \nu}=A_{\nu \mu}$, because (only!) then
$${A^{\mu}}_{\nu} = g^{\mu \rho} A_{\rho \nu} = g^{\mu \rho} A_{\nu \rho} ={A_{\nu}}^{\mu},$$
and the horizontal ordering is not important.

Is there some connection with matrices? For instance, if we have two indices.
$$A^{\mu}_{\hspace{0.2cm}\nu}$$ what is the row and what is the column? And in this case
$$A^{\hspace{0.2cm}\mu}_{\nu}$$ what is the row and what is the column?

 

[Orodruin]

Tensors are not matrices and matrices are not tensors. Tensors (of rank 2) may be represented by matrices in some basis but it is then up to you to define how indices correspond to rows and columns in a consistent manner.

 

[cianfa72]

And in this case $$A^{\hspace{0.2cm}\mu}_{\nu}$$ what is the row and what is the column?

Typically when in a given basis you represent a tensor as a matrix the first index (on the left) is the row and the second the column. So in your example $\nu$ is the row and $\mu$ the column.

 

[Ibix]

I'd repeat Orodruin's caution. Any rank 2 tensor can be represented as a matrix (4×4 in relativity), and it's quite common to get metric tensors represented like this. But there isn't a way to notate upper and lower indices in that form, so it isn't clear what can be legally contracted with what and it's easy to end up writing a matrix equation that contracts over two lower indices. That's difficult to debug, so you do it at your own risk.