How Can I Simplify and Solve the Einstein Summation Convention Problem?

In summary, the solution to the given problem is $$a_3 (b \cdot c) - b_3 (c \cdot a)$$ which can be obtained by using the Kronecker deltas to simplify the expression. It is helpful to use the properties of Kronecker deltas and to sum over repeated indices to simplify the problem.
  • #1
Athenian
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Homework Statement
Solve ##a_i \, b_j \, c_k \, \epsilon_{ij \ell} \, \epsilon_{3k \ell}##
Relevant Equations
See Below ##\longrightarrow##
Attempted Solution:
$$a_i \, b_j \, c_k \, \epsilon_{ij \ell} \, \epsilon_{3k \ell}$$
$$a_i\, b_j\, c_k\, (\delta_{i3} \, \delta_{jk} - \, \delta_{ik}\, \delta_{j3})$$

Beyond this, though, I am quite lost. I know I am very close to the answer, but seeing this many terms can become fairly confusing for me. Is there a way or method to better (and simply) digest the above problem and solve it?

Any help would be greatly appreciated. Thank you!
 
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  • #2
Athenian said:
Homework Statement:: Solve ##a_i \, b_j \, c_k \, \epsilon_{ij \ell} \, \epsilon_{3k \ell}##
Relevant Equations:: See Below ##\longrightarrow##

Attempted Solution:
$$a_i \, b_j \, c_k \, \epsilon_{ij \ell} \, \epsilon_{3k \ell}$$
$$a_i\, b_j\, c_k\, (\delta_{i3} \, \delta_{jk} - \, \delta_{ik}\, \delta_{j3})$$

Beyond this, though, I am quite lost. I know I am very close to the answer, but seeing this many terms can become fairly confusing for me. Is there a way or method to better (and simply) digest the above problem and solve it?

Any help would be greatly appreciated. Thank you!
That's the correct first step. Now, each Kronecker delta can be used to get rid of one of the indices that appears in it. For example, what happens if you sum over the index "i"?
 
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  • #3
@nrqed, thank you for the helpful hint and guidance! I was finally able to figure it out as seen below:

Continuing where I left off:
$$a_i \, b_j \, c_k \, \delta_{i3} \, \delta_{jk} - a_i \, b_j \, c_k \, \delta_{ik} \, \delta_{j3}$$
$$\Rightarrow a_3 \, b_k \, c_k - a_i \, b_c \, c_i$$
$$\Rightarrow a_3 (b \cdot c) - b_3 (c \cdot a)$$

Thank you for all your help!
 
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  • #4
Athenian said:
@nrqed, thank you for the helpful hint and guidance! I was finally able to figure it out as seen below:

Continuing where I left off:
$$a_i \, b_j \, c_k \, \delta_{i3} \, \delta_{jk} - a_i \, b_j \, c_k \, \delta_{ik} \, \delta_{j3}$$
$$\Rightarrow a_3 \, b_k \, c_k - a_i \, b_c \, c_i$$
$$\Rightarrow a_3 (b \cdot c) - b_3 (c \cdot a)$$

Thank you for all your help!
Good job!
 
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