# Calculate Contraction 2nd & 4th Indices Riemann Tensor

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• GR191511
In summary: Yes, that's correct.You mean...##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##and##\,####g^{ea}####\,##is the linear combination...Yes, that's correct.
GR191511
How to calculate the contraction of second and fourth indices of Riemann tensor?I can only deal with other indices.Thank you！

GR191511 said:
How to calculate the contraction of second and fourth indices of Riemann tensor?I can only deal with other indices.Thank you！

Can you write out the contraction in terms of a sum?
Because of the symmetries of the Riemann tensor,
there aren't very many independent nonzero contractions.

GR191511 and topsquark
GR191511 said:
How to calculate the contraction of second and fourth indices of Riemann tensor?
$$g^{bd} R^a{}_{bcd}$$

vanhees71, GR191511 and topsquark
PeterDonis said:
$$g^{bd} R^a{}_{bcd}$$
Why it is not##\,####R^a{}_{bcb}##?What is the difference between##\,####g^{bd}R^a{}_{bcd}####\,##and##\,####R^a{}_{bcb}##?

GR191511 said:
Why it is not##\,####R^a{}_{bcb}##?What is the difference between##\,####g^{bd}R^a{}_{bcd}####\,##and##\,####R^a{}_{bcb}##?
You cannot sum over two lower or two upper indices. It must be one lower and one upper index, so ##R^a{}_{bcb}## is an illegal statement. That's why @PeterDonis raises an index with the metric before contracting.

topsquark, PeterDonis, vanhees71 and 1 other person
Ibix said:
You cannot sum over two lower or two upper indices. It must be one lower and one upper index, so ##R^a{}_{bcb}## is an illegal statement. That's why @PeterDonis raises an index with the metric before contracting.
So there are only three types to contracting a Riemann tensor?First and second;first and third;first and fourth?

GR191511 said:
So there are only three types to contracting a Riemann tensor?First and second;first and third;first and fourth?
No, you can contract any pair of indices, you just have to raise or lower them first. In a manifold with a metric you're always free to do so, so there's nothing particularly special about ##R^a{}_{bcd}## versus ##R^{ab}{}_{cd}## or ##R^a{}_b{}^c{}_d##. They contain the same information. As @robphy noted upthread, however, most contractions are either zero or a linear combination of other contractions.

topsquark, Nugatory, vanhees71 and 1 other person
Ibix said:
No, you can contract any pair of indices, you just have to raise or lower them first. In a manifold with a metric you're always free to do so, so there's nothing particularly special about ##R^a{}_{bcd}## versus ##R^{ab}{}_{cd}## or ##R^a{}_b{}^c{}_d##. They contain the same information. As @robphy noted upthread, however, most contractions are either zero or a linear combination of other contractions.
Thank you!Does ##\,####R_{abcd}=-R_{bacd}####\,##is equivalent to##\,####R^a{}_{bcd}=-R^b{}_{acd}##?I tried to prove it,but failed.

From ##R_{abcd}=-R_{bacd}## you get by contraction with ##g^{ae}##
$${R^e}_{bcd}=-{{R_b}^{e}}_{cd}.$$
Note that it is utmost important to keep the indices in the right vertical AND horizontal order!

topsquark, GR191511 and Ibix
Ibix said:
No, you can contract any pair of indices, you just have to raise or lower them first. In a manifold with a metric you're always free to do so, so there's nothing particularly special about ##R^a{}_{bcd}## versus ##R^{ab}{}_{cd}## or ##R^a{}_b{}^c{}_d##. They contain the same information. As @robphy noted upthread, however, most contractions are either zero or a linear combination of other contractions.
##g^{bd}R^a{}_{bcd}=R^{ad}{}_{cd}=-R^{ad}{}_{dc}=-R^{da}{}_{cd}=R^{da}{}_{dc}####\,##I can't get zero or a linear combination of Ricci tensor...Where am I wrong?

Why do you think it's wrong? The curvature tensor is antisymmetric in the 1st and 2nd as well as the 3rd and 4th index pairs, i.e., changing the order of the first two upper and the two lower indices just compensates and your equation thus is correct.

vanhees71 said:
Why do you think it's wrong? The curvature tensor is antisymmetric in the 1st and 2nd as well as the 3rd and 4th index pairs, i.e., changing the order of the first two upper and the two lower indices just compensates and your equation thus is correct.
But all of them are not zero or a linear combination of Ricci tensor...This is not consistent with "the Ricci tensor is essentially the only contraction of the Riemann tensor"(《A First Course in General Relativity》2nd Edition page 164)

But the Ricci tensor is
$$R_{\mu \nu}={R^{\alpha}}_{\mu \alpha \nu}.$$
The only other non-zero contraction of the curvature tensor is just ##-R_{\mu \nu}##.

topsquark and robphy
vanhees71 said:
But the Ricci tensor is
$$R_{\mu \nu}={R^{\alpha}}_{\mu \alpha \nu}.$$
The only other non-zero contraction of the curvature tensor is just ##-R_{\mu \nu}##.
I don't understand...Is##R^{da}{}_{dc}####\,##a linear combination of ##\,####R^d{}_{adc}##?Thanks

GR191511 said:
I don't understand...Is##R^{da}{}_{dc}####\,##a linear combination of ##\,####R^d{}_{adc}##?
Do you understand how indexes are raised and lowered using the metric?

PeterDonis said:
Do you understand how indexes are raised and lowered using the metric?
You mean...##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##and##\,####g^{ea}####\,##is the linear combination coefficient?

GR191511 said:
You mean...##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}##
Yes.

GR191511 said:
and##\,####g^{ea}####\,##is the linear combination coefficient?
The expression ##g^{ea}R^d{}_{edc}## is a contraction; that means it's a sum of multiple terms in which the repeated index ##e## takes all possible values. So ##g^{ea}## is not a single number.

PeterDonis said:
Yes.The expression ##g^{ea}R^d{}_{edc}## is a contraction; that means it's a sum of multiple terms in which the repeated index ##e## takes all possible values. So ##g^{ea}## is not a single number.
##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##then what?

GR191511 said:
##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##then what?
What do you want to do? If you want to contract the second and fourth indexes of ##R^{da}{}_{dc}##, how do you think you would do that?

PeterDonis said:
What do you want to do? If you want to contract the second and fourth indexes of ##R^{da}{}_{dc}##, how do you think you would do that?
I think it can be transformed into 0 or ##R_{ac}####\,##or##\,####-R_{ac}##

GR191511 said:
I think it can be transformed into 0
I believe the correct contraction here will end up being zero, yes. But you shouldn't have to guess. You should be able to work out what the correct contraction expression will look like, and then use the known symmetries of the Riemann tensor to decide whether that contraction is identically zero or not.

GR191511 said:
or ##R_{ac}####\,##or##\,####-R_{ac}##
Not the way you've written it, no. Do you know what a contraction is?

GR191511 said:
##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##then what?
Look at my post #3. Compare it with the expression on the RHS above (relabeling indexes as needed).

I'm not sure, whether the OP understands the basic notation. An expression like ##g^{ea} {R^d}_{edc}## tells you to take the product of the components of the metric and the curvature tensor and (!) sum over indices that occur twice (where always one must be an upper and the other a lower index). In this example you have to sum the expression over ##e## (from 0 to 3).

vanhees71 said:
I'm not sure, whether the OP understands the basic notation. An expression like ##g^{ea} {R^d}_{edc}## tells you to take the product of the components of the metric and the curvature tensor and (!) sum over indices that occur twice (where always one must be an upper and the other a lower index). In this example you have to sum the expression over ##e## (from 0 to 3).
I know it. It will end up being ##R^{da}{}_{dc}=R^a{}_c####\,##but##\neq####R_{ac}##

Of course, why should ##{R^a}_c## be the same as ##R_{ac}##? In fact it cannot be, because then you'd need ##g_{ab}=\delta_{ab}##, but that cannot be, because the signature of the pseudo-metric is (1,3) or (3,1), depending on whether you use the west-coast or east-coast convention of the signs.

## 1. What is the Riemann tensor?

The Riemann tensor is a mathematical object used to describe the curvature of a space. It is particularly important in the field of general relativity, where it is used to describe the curvature of spacetime caused by the presence of matter and energy.

## 2. How is the Riemann tensor calculated?

The Riemann tensor is calculated using the Christoffel symbols, which are derived from the metric tensor. The metric tensor describes the distance between points in a space, and the Christoffel symbols represent the rate of change of the metric tensor with respect to the coordinates of the space.

## 3. What do the 2nd and 4th indices represent in the Riemann tensor?

The 2nd and 4th indices in the Riemann tensor represent the two vectors that are used to calculate the curvature of a space. The 2nd index represents the vector that is being parallel transported along a geodesic, while the 4th index represents the vector that is being compared to the original vector after it has been parallel transported.

## 4. Why is it important to calculate the Riemann tensor?

The Riemann tensor allows us to understand the curvature of a space, which is crucial in understanding the behavior of matter and energy in that space. It is also used in the field of general relativity to describe the effects of gravity and the motion of objects in curved spacetime.

## 5. How is the Riemann tensor related to other mathematical concepts?

The Riemann tensor is related to other mathematical concepts such as the Ricci tensor and the scalar curvature. These concepts are used to fully describe the curvature of a space and are crucial in understanding the behavior of matter and energy in that space.

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