Is there a mistake in this tensor multiplication problem?

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The discussion revolves around the evaluation of the tensor multiplication problem involving antisymmetric tensors M and N. Participants analyze the expression ep_{ijkl} M^{ij} N^{kl} + ep_{ijkl} N^{ij} M^{kl}, concluding that it simplifies to 2ep_{ijkl} M^{ij} N^{kl}, which is not zero. They explore the conditions under which the expression could equal zero, considering the antisymmetry of M and N. The conversation highlights the possibility of a typo in the original problem, as the derived expressions do not align with the expected outcome. Ultimately, the participants seek verification of their reasoning and calculations regarding the antisymmetric properties of the tensors involved.
DuckAmuck
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Homework Statement
Show that
epsilon_{ijkl} ( M^{ij} N^{kl} + N^{ij} M^{kl}) = 0
Relevant Equations
epsilon is the 4D anti-symmetric Levi-Cevita tensor. M and N are also anti-symmetric tensors.
ep_{ijkl} M^{ij} N^{kl} + ep_{ijkl}N^{ij} M^{kl}
The second term can be rewritten with indices swapped
ep_{klij} N^{kl}M^{ij}
Shuffle indices around in epsilon
ep{klij} = ep{ijkl}
Therefore the expression becomes
2ep_{ijkl}M^{ij}N^{kl}
Not zero.
What is wrong here?
 
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I tried, but got the same result as you did. Are you sure its not supposed to be
##\epsilon_{ijkl} ( M^{ij} N^{kl} - N^{ij} M^{kl}) = 0##?
What I did was to write everything out, using all even permutations of 1,2,3,4:
{1,2,3,4}, {1,3,4,2}, {1,4,2,3}, {2,1,4,3}, {2,3,1,4}, {2,4,3,1}, {3,1,2,4}, {3,2,4,1}, {3,4,1,2}, {4,1,3,2}, {4,2,1,3}, {4,3,2,1}
and all odd ones:
{1,2,4,3}, {1,3,2,4}, {1,4,3,2}, {2,1,3,4}, {3,2,1,4}, {4,2,3,1}, {2,3,4,1}, {2,4,1,3}, {3,1,4,2}, {3,4,2,1}, {4,1,2,3}, {4,3,1,2}
and the fact that ##M## and ##N## are anti-symmetrical, i.e. ##M^{12}= - M^{21}## etc.
 
ok i think i have solid reasoning here:

Suppose ##C^{ij} = M^{ij} + N^{ij}##

From symmetry and antisymmetry we have:

##\epsilon_{ijkl} C^{ij}C^{kl} = 0##

Also if you foil the CC product in terms of M and N you get ##C^{ij}C^{kl} = M^{ij}M^{kl} + N^{ij}N^{kl} + M^{ij}N^{kl} + N^{ij}M^{kl}##

The MM and NN terms are zero for the same reason the CC product is when multiplied by epsilon.

So this demands that

##\epsilon_{ijkl} (M^{ij}N^{kl} + N^{ij}M^{kl}) = 0##

Can someone please verify?
 
DuckAmuck said:
ok i think i have solid reasoning here:

Suppose ##C^{ij} = M^{ij} + N^{ij}##

From symmetry and antisymmetry we have:

##\epsilon_{ijkl} C^{ij}C^{kl} = 0##

Also if you foil the CC product in terms of M and N you get ##C^{ij}C^{kl} = M^{ij}M^{kl} + N^{ij}N^{kl} + M^{ij}N^{kl} + N^{ij}M^{kl}##

The MM and NN terms are zero for the same reason the CC product is when multiplied by epsilon.

So this demands that

##\epsilon_{ijkl} (M^{ij}N^{kl} + N^{ij}M^{kl}) = 0##

Can someone please verify?
No, the statement as it stands seems false to me. It is not generally the case thar ##\epsilon_{ijkl} C^{ij} C^{kl} = 0##.
 
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Orodruin said:
No, the statement as it stands seems false to me. It is not generally the case thar ##\epsilon_{ijkl} C^{ij} C^{kl} = 0##.
You’re right. I am just trying to figure out *how* this could be zero at this point, as in what conditions. Otherwise I’m stumped.
 
DuckAmuck said:
I am just trying to figure out *how* this could be zero at this point, as in what conditions.

You should have been given all the conditions already, that M and N are antisymmetric rank-2 tensors.

There is always the possibility that whoever gave you this problem, is wrong / made a typo. I have been tearing my hair off several times doing excersices in general relativity books... to find out there was some typo in the problem as written.

Here is my "expanded" calculation that I did btw:

The underlined terms I will collect at the end.

## \underline{M^{12}N^{34}} + M^{13}N^{42} + M^{14}N^{23} + \underline{M^{21}N^{43}} + M^{23}N^{14} + M^{24}N^{31} + M^{31}N^{24} + M^{32}N^{41} + M^{34}N^{12} + M^{41}N^{32} + M^{42}N^{13} + M^{43}N^{21} ##
##- ( \underline{M^{12}N^{43}} + M^{13}N^{24} + M^{14}N^{32} + \underline{M^{21}N^{34}} + M^{32}N^{14} + M^{42}N^{31} + M^{23}N^{41} + M^{24}N^{13} + M^{31}N^{42} + M^{34}N^{21} + M^{41}N^{23} + M^{43}N^{12} )##
##+ N^{12}M^{34} + N^{13}M^{42} + N^{14}M^{23} + N^{21}M^{43} + N^{23}M^{14} + N^{24}M^{31} + N^{31}M^{24} + N^{32}M^{41} + \underline{N^{34}M^{12}} + N^{41}M^{32} + N^{42}M^{13} + \underline{N^{43}M^{21}} ##
##- ( N^{12}M^{43} + N^{13}M^{24} + N^{14}M^{32} + N^{21}M^{34} + N^{32}M^{14} + N^{42}M^{31} + N^{23}M^{41} + N^{24}M^{13} + N^{31}M^{42} + \underline{N^{34}M^{21}} + N^{41}M^{23} + \underline{N^{43}M^{12}} \: ) ##

The stuff I underlined:
## M^{12}N^{34} + M^{21}N^{43} - M^{12}N^{43} - M^{21}N^{34} + N^{34}M^{12} +N^{43}M^{21} -N^{34}M^{21} - N^{43}M^{12} ##

(##M^{21}= - M^{12}## and ##N^{43}= - N^{34}##)

##M^{12}N^{34} + (-1)^2 M^{12}N^{34} - (-1)M^{12}N^{34} - (-1)M^{12}N^{34} + N^{34}M^{12} +(-1)^2N^{34}M^{12} - (-1)N^{34}M^{12} - (-1)N^{34}M^{12} = 8M^{12}N^{34} ##

Well that was fun.
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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