Raising & Lowering Indices: Q&A on Relativity

  • I
  • Thread starter GR191511
  • Start date
  • Tags
    Indices
In summary: They are not needed in the notation for Lorentz transformations. They are needed when dealing with vectors and matrices, but not for Lorentz transformations.
  • #1
GR191511
76
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
  • #2
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:
  • #3
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:
  • #4
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).
 
  • #5
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.
 
  • Like
Likes vanhees71
  • #6
  • Like
Likes vanhees71

1. What is the concept of raising and lowering indices in relativity?

Raising and lowering indices is a mathematical operation used to transform tensors from one form to another in relativity. It involves multiplying a tensor by the metric tensor and its inverse to change the placement of indices.

2. Why is it important to understand raising and lowering indices in relativity?

Raising and lowering indices is crucial in relativity because it allows us to easily manipulate and compare tensors in different coordinate systems. It also helps us to express physical laws in a more compact and convenient form.

3. How does raising and lowering indices affect the properties of a tensor?

Raising and lowering indices does not change the underlying properties of a tensor, such as its rank or symmetry. However, it does change the way the tensor components transform under coordinate transformations.

4. Can raising and lowering indices be applied to any type of tensor?

Yes, raising and lowering indices can be applied to any tensor, including vectors, matrices, and higher-order tensors. However, it is most commonly used for tensors of rank 2 or higher.

5. Are there any practical applications of raising and lowering indices in relativity?

Yes, raising and lowering indices are used extensively in the calculation and analysis of physical quantities in relativity, such as energy-momentum tensors and curvature tensors. They are also important in the study of general relativity and its applications in astrophysics and cosmology.

Similar threads

  • Special and General Relativity
Replies
17
Views
1K
  • Special and General Relativity
Replies
1
Views
534
  • Special and General Relativity
Replies
5
Views
2K
  • Special and General Relativity
Replies
2
Views
1K
Replies
7
Views
1K
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
Replies
6
Views
1K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
17
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
2K
Back
Top