I Raising & Lowering Indices: Q&A on Relativity

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The discussion centers on the mathematical properties of Lorentz transformation matrices and their interaction with the Minkowski metric in the context of special and general relativity. Participants question the purpose of contracting the metric with Lorentz transformations, noting that while it is mathematically permissible, it may not be conceptually sound since Lorentz transformations are not tensors. The conversation highlights the importance of correctly interpreting index notation and the implications of raising and lowering indices using the metric. Clarifications are sought regarding specific components and their transformations, emphasizing the need for careful transcription and understanding of the underlying principles. Overall, the dialogue underscores the complexities involved in tensor mathematics within the framework of relativity.
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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?
 
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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.
 
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Ibix said:
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}##
 
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Ibix said:
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).
 
cianfa72 said:
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.
 
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