GR191511
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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!
$$GR191511 said:How to calculate the contraction of second and fourth indices of Riemann tensor?
Why it is not##\,####R^a{}_{bcb}##?What is the difference between##\,####g^{bd}R^a{}_{bcd}####\,##and##\,####R^a{}_{bcb}##?PeterDonis said:$$
g^{bd} R^a{}_{bcd}
$$
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 said:Why it is not##\,####R^a{}_{bcb}##?What is the difference between##\,####g^{bd}R^a{}_{bcd}####\,##and##\,####R^a{}_{bcb}##?
So there are only three types to contracting a Riemann tensor?First and second;first and third;first and fourth?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.
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 said:So there are only three types to contracting a Riemann tensor?First and second;first and third;first and fourth?
Thank you!Does ##\,####R_{abcd}=-R_{bacd}####\,##is equivalent to##\,####R^a{}_{bcd}=-R^b{}_{acd}##?I tried to prove it,but failed.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?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.
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 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.
I don't understand...Is##R^{da}{}_{dc}####\,##a linear combination of ##\,####R^d{}_{adc}##?Thanksvanhees71 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}##.
Do you understand how indexes are raised and lowered using the metric?GR191511 said:I don't understand...Is##R^{da}{}_{dc}####\,##a linear combination of ##\,####R^d{}_{adc}##?
You mean...##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##and##\,####g^{ea}####\,##is the linear combination coefficient?PeterDonis said:Do you understand how indexes are raised and lowered using the metric?
Yes.GR191511 said:You mean...##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}##
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 said:and##\,####g^{ea}####\,##is the linear combination coefficient?
##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##then what?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.
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 said:##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##then what?
I think it can be transformed into 0 or ##R_{ac}####\,##or##\,####-R_{ac}##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 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:I think it can be transformed into 0
Not the way you've written it, no. Do you know what a contraction is?GR191511 said:or ##R_{ac}####\,##or##\,####-R_{ac}##
Look at my post #3. Compare it with the expression on the RHS above (relabeling indexes as needed).GR191511 said:##R^{da}{}_{dc}=g^{ea}R^d{}_{edc}####\,##then what?
I know it. It will end up being ##R^{da}{}_{dc}=R^a{}_c####\,##but##\neq####R_{ac}##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).