- #1
Phys pilot
- 28
- 0
first of all english is not my mother tongue sorry. I want to ask if you can help me with some of the properties of the levi-civita symbol.
I am showing that
$$\epsilon_{ijkl}\epsilon_{ijmn}=2!(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
so i have this...
$$\epsilon_{ijkl}\epsilon_{ijmn}=\delta_{ii}\delta_{jj}\delta_{km}\delta_{ln}+\delta_{ij}\delta_{jm}\delta_{kn}\delta_{li}+\delta_{im}\delta_{jn}\delta_{ki}\delta_{lj}+\delta_{in}\delta_{ji}\delta_{kj}\delta_{lm}-\delta_{ii}\delta_{jn}\delta_{km}\delta_{lj}-\delta_{in}\delta_{jm}\delta_{kj}\delta_{li}-\delta_{im}\delta_{jj}\delta_{ki}\delta_{ln}-\delta_{ij}\delta_{ji}\delta_{kn}\delta_{lm}=3^2\delta_{km}\delta_{ln}+\delta_{kn}\delta_{lm}+\delta_{km}\delta_{ln}+\delta_{kn}\delta_{lm}-3\delta_{km}\delta_{ln}-\delta_{km}\delta_{ln}-3\delta_{km}\delta_{ln}-3\delta_{kn}\delta_{lm}=3\delta_{km}\delta_{ln}-3\delta_{kn}\delta_{lm}=3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
which is not equal to the supposedly correct answer. can the statement be wrong?
if not i can't see my error.
Also i need to prove that:
$$\epsilon_{ijkl}\epsilon_{ijkm}=3!\delta_{lm}$$
i know and i proved that
$$\epsilon_{ijkl}\epsilon_{ijkl}=24=4!$$
so if $l=m$ i have this
$$\epsilon_{ijkl}\epsilon_{ijkl}=3!\delta_{ll}=3!\cdot 3=18$$
which is not equal to 4!=24 so its the statement wrong again? it must be
$$\epsilon_{ijkl}\epsilon_{ijkm}=8\delta_{lm} $$
Actually i proved that:
$$\epsilon_{ijkl}\epsilon_{ijkm}=\delta_{ii}\delta_{jj}\delta_{kk}\delta_{lm}+\delta_{ij}\delta_{jk}\delta_{km}\delta_{li}+\delta_{ik}\delta_{jm}\delta_{ki}\delta_{lj}+\delta_{im}\delta_{ji}\delta_{kj}\delta_{lk}-\delta_{ii}\delta_{jm}\delta_{kk}\delta_{lj}-\delta_{im}\delta_{jk}\delta_{kj}\delta_{li}-\delta_{ik}\delta_{jj}\delta_{ki}\delta_{lm}-\delta_{ij}\delta_{ji}\delta_{km}\delta_{lk}=3^3\delta_{lm}+\delta_{ik}\delta_{km}\delta_{li}+\delta_{ii}\delta_{lj}\delta_{jm}+\delta_{jm}\delta_{kj}\delta_{lk}-3^2\delta_{lm}-\delta_{jj}\delta_{li}\delta_{im}-3\delta_{ii}\delta_{lm}-\delta_{ii}\delta_{lm}=3^3\delta_{lm}+\delta_{lk}\delta_{km}+\delta_{ii}\delta_{lm}+\delta_{lj}\delta_{jm}-3^2\delta_{lm}-3\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}=3^3\delta_{lm}+\delta_{lm}+3\delta_{lm}+\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}=8\delta_{lm} $$
Even more if i use my solution:
$$\epsilon_{ijkl}\epsilon_{ijmn}=3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
to prove
$$\epsilon_{ijkl}\epsilon_{ijkm}=3!\delta_{lm}$$
using $m=k$ and $n=m$ i have this
$$3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
$$3(\delta_{kk}\delta_{lm}-\delta_{km}\delta_{lk})$$
$$3(3\delta_{lm}-\delta_{lm})$$
$$3(2\delta_{lm})$$
$$6(\delta_{lm})=3!(\delta_{lm})$$
I am showing that
$$\epsilon_{ijkl}\epsilon_{ijmn}=2!(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
so i have this...
$$\epsilon_{ijkl}\epsilon_{ijmn}=\delta_{ii}\delta_{jj}\delta_{km}\delta_{ln}+\delta_{ij}\delta_{jm}\delta_{kn}\delta_{li}+\delta_{im}\delta_{jn}\delta_{ki}\delta_{lj}+\delta_{in}\delta_{ji}\delta_{kj}\delta_{lm}-\delta_{ii}\delta_{jn}\delta_{km}\delta_{lj}-\delta_{in}\delta_{jm}\delta_{kj}\delta_{li}-\delta_{im}\delta_{jj}\delta_{ki}\delta_{ln}-\delta_{ij}\delta_{ji}\delta_{kn}\delta_{lm}=3^2\delta_{km}\delta_{ln}+\delta_{kn}\delta_{lm}+\delta_{km}\delta_{ln}+\delta_{kn}\delta_{lm}-3\delta_{km}\delta_{ln}-\delta_{km}\delta_{ln}-3\delta_{km}\delta_{ln}-3\delta_{kn}\delta_{lm}=3\delta_{km}\delta_{ln}-3\delta_{kn}\delta_{lm}=3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
which is not equal to the supposedly correct answer. can the statement be wrong?
if not i can't see my error.
Also i need to prove that:
$$\epsilon_{ijkl}\epsilon_{ijkm}=3!\delta_{lm}$$
i know and i proved that
$$\epsilon_{ijkl}\epsilon_{ijkl}=24=4!$$
so if $l=m$ i have this
$$\epsilon_{ijkl}\epsilon_{ijkl}=3!\delta_{ll}=3!\cdot 3=18$$
which is not equal to 4!=24 so its the statement wrong again? it must be
$$\epsilon_{ijkl}\epsilon_{ijkm}=8\delta_{lm} $$
Actually i proved that:
$$\epsilon_{ijkl}\epsilon_{ijkm}=\delta_{ii}\delta_{jj}\delta_{kk}\delta_{lm}+\delta_{ij}\delta_{jk}\delta_{km}\delta_{li}+\delta_{ik}\delta_{jm}\delta_{ki}\delta_{lj}+\delta_{im}\delta_{ji}\delta_{kj}\delta_{lk}-\delta_{ii}\delta_{jm}\delta_{kk}\delta_{lj}-\delta_{im}\delta_{jk}\delta_{kj}\delta_{li}-\delta_{ik}\delta_{jj}\delta_{ki}\delta_{lm}-\delta_{ij}\delta_{ji}\delta_{km}\delta_{lk}=3^3\delta_{lm}+\delta_{ik}\delta_{km}\delta_{li}+\delta_{ii}\delta_{lj}\delta_{jm}+\delta_{jm}\delta_{kj}\delta_{lk}-3^2\delta_{lm}-\delta_{jj}\delta_{li}\delta_{im}-3\delta_{ii}\delta_{lm}-\delta_{ii}\delta_{lm}=3^3\delta_{lm}+\delta_{lk}\delta_{km}+\delta_{ii}\delta_{lm}+\delta_{lj}\delta_{jm}-3^2\delta_{lm}-3\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}=3^3\delta_{lm}+\delta_{lm}+3\delta_{lm}+\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}=8\delta_{lm} $$
Even more if i use my solution:
$$\epsilon_{ijkl}\epsilon_{ijmn}=3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
to prove
$$\epsilon_{ijkl}\epsilon_{ijkm}=3!\delta_{lm}$$
using $m=k$ and $n=m$ i have this
$$3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
$$3(\delta_{kk}\delta_{lm}-\delta_{km}\delta_{lk})$$
$$3(3\delta_{lm}-\delta_{lm})$$
$$3(2\delta_{lm})$$
$$6(\delta_{lm})=3!(\delta_{lm})$$