Calculate Contraction 2nd & 4th Indices Riemann Tensor

[GR191511]

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

 

[robphy]

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.

 

[PeterDonis]

How to calculate the contraction of second and fourth indices of Riemann tensor?

$$
g^{bd} R^a{}_{bcd}
$$

 

[GR191511]

$$
g^{bd} R^a{}_{bcd}
$$

Why it is not$\,$$R^a{}_{bcb}$?What is the difference between$\,$$g^{bd}R^a{}_{bcd}$$\,$and$\,$$R^a{}_{bcb}$?

 

[Ibix]

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.

 

[GR191511]

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?

 

[Ibix]

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.

 

[GR191511]

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.

 

[vanhees71]

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!

 

[GR191511]

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?

 

[vanhees71]

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.

 

[GR191511]

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)

 

[vanhees71]

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 $-R_{\mu \nu}$.

 

[GR191511]

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

 

[PeterDonis]

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?

 

[GR191511]

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?

 

[PeterDonis]

You mean...$R^{da}{}_{dc}=g^{ea}R^d{}_{edc}$

Yes.

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.

 

[GR191511]

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?

 

[PeterDonis]

$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?

 

[GR191511]

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}$

 

[PeterDonis]

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.

or $R_{ac}$$\,$or$\,$$-R_{ac}$

Not the way you've written it, no. Do you know what a contraction is?

 

[PeterDonis]

$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).

 

[vanhees71]

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).

 

[GR191511]

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}$

 

[vanhees71]

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.