I Raising & Lowering Indices: Q&A on Relativity

  • I
  • Thread starter Thread starter GR191511
  • Start date Start date
  • Tags Tags
    Indices
GR191511
Messages
76
Reaction score
6
TL;DR Summary
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?
 
Physics news on Phys.org
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.
 
Last edited:
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}##
 
Last edited:
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.
 
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
Back
Top