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.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a...
Back
Top