Raising & Lowering Indices: Q&A on Relativity

[GR191511]

TL;DR

metric tensor

I am reading《Relativity - An Introduction to Special and General Relativity》
my question:
$1=-\eta_{44}L^{n{'}}{_4}L_{n{'}}{^4}=-\eta^{n'}{^{m'}}L_{n{'}}{_4}L_{m'}{_4}$
$\eta$ is Minkowski Metric，$L$ is Lorentz transformation matrix...

1.Since $-\eta_{44}$=1,what's the usage of it here?
2.Why is $-\eta_{44}L^{n{'}}{_4}L_{n{'}}{^4}$equal to$-\eta^{n'}{^{m'}}L_{n{'}}{_4}L_{m'}{_4}$?
3.How does the $4$ of $L_{n{'}}{^4}$ get down?

 

[Ibix]

That seems a rather odd thing to be doing. Can you check that you've transcribed correctly? And is there some context to this that you can provide?

There are several conventions for notating Lorentz transforms, so check what the textbook says, but I would interpret $L^a{}_b$ and $L_a{}^b$ as forward and reverse transforms so $L^{n'}{}_aL_{n'}{}^b=\delta^b_a$ (edit: corrected index placement)and the 44 component is 1. The operation seems rather pointless, so perhaps I'm misunderstanding something.

You can lower indices on tensors by contracting with the metric and raise them by contracting with the inverse metric. I haven't seen anybody contract the metric with a Lorentz transform before, and I'm not sure it's a good idea because you need to be careful with which coordinate expression of the metric you use, but it's just a sum so it ought to be mathematically legit. And it's not a huge problem with Einstein coordinates on flat spacetime anyway because the metric doesn't change form under Lorentz transformations. So $L_{ab}=\eta_{ac}L^c{}_b$, I would presume.

 

[GR191511]

That seems a rather odd thing to be doing. Can you check that you've transcribed correctly? And is there some context to this that you can provide?

There are several conventions for notating Lorentz transforms, so check what the textbook says, but I would interpret $L^a{}_b$ and $L_a{}^b$ as forward and reverse transforms so $L^{n'}_aL^b_{n'}=\delta^b_a$ and the 44 component is 1. The operation seems rather pointless, so perhaps I'm misunderstanding something.

You can lower indices on tensors by contracting with the metric and raise them by contracting with the inverse metric. I haven't seen anybody contract the metric with a Lorentz transform before, and I'm not sure it's a good idea because you need to be careful with which coordinate expression of the metric you use, but it's just a sum so it ought to be mathematically legit. And it's not a huge problem with Einstein coordinates on flat spacetime anyway because the metric doesn't change form under Lorentz transformations. So $L_{ab}=\eta_{ac}L^c{}_b$, I would presume.

the preceding part of the text:
"Evaluating the (4,4) component of (2.5),we obtain (remember that indices are raised and lowered by means of $\eta$)"
the so called "(2.5)"are: $x^{n'}=L^{n'}{_a}x^a\qquad x_{m'}=L_{m'}{^b}x_b\qquad L_{m'}{^b}=\eta {_{m'}}{_{n'}}\eta{^a}{^b}L{^{n'}}{_a}$

 

[cianfa72]

I haven't seen anybody contract the metric with a Lorentz transform before, and I'm not sure it's a good idea because you need to be careful with which coordinate expression of the metric you use.

Indeed Lorentz transformation is not a tensor, so may be is a nosense contract it with the metric tensor (or its inverse).

 

[Orodruin]

Indeed Lorentz transformation is not a tensor, so may be is a nosense contract it with the metric tensor (or its inverse).

It most certainly is not. The Lorentz transformation coefficients are the transformation coefficients between different inertial frames.

 

[guru_4]

https://rpubs.com/lizhi1800/Einstein_Notation

Subscript and superscript are for covariant and contravariant components. They are to tell people the summation result is contravariant or covariant.