# epsilon delta?

• May 2nd 2009, 04:18 AM
revolution2000
epsilon delta?
The epsilon delta rule states

EijkEpqk=(delta(ip)delta(jq)-delta(iq)delta(jp))

I am constantly using this but get stuck when it is applied.

For example

EijkEpqkA(j)B(l)C(m)=(delta(ip)delta(jq)-delta(iq)delta(jp))A(j)B(l)C(m)

This then becomes

A(j)B(i)C(j)-A(j)B(j)C(i)

Can anybody please explain this result????
• May 2nd 2009, 08:26 AM
VonNemo19
I think you may have had some trouble when you did your latex, because I can't understand a word of what you wrote. My specialty is epsilon-delta proofs so I would love to help. Let me know when you fix the glitch.(Wink)
• May 2nd 2009, 10:00 AM
revolution2000
do you mean writing it in math type. I don't know how to do that.
Is there an option on the posting screen?
• May 2nd 2009, 10:09 AM
revolution2000
The epsilon delta rule states

$\displaystyle \epsilon_{ijk}\epsilon_{pqk}=\delta_{ip}\delta_{jq }-\delta_{iq}\delta_{jp}$

I am constantly using this but get stuck when it is applied.

For example

$\displaystyle \epsilon_{ijk}\epsilon_{pqk}A_{j}B_{l}C_{m}=(\delt a_{ip}\delta_{jq}-\delta_{iq}\delta_{jp})A_{j}B_{l}C_{m}$

This then becomes

$\displaystyle A_{j}B_{i}C_{j}-A_{j}B_{j}C_{i}$

Can anybody please explain this result????

Is it true that

$\displaystyle \delta_{ij}a_{i}=a_{j}$

If so does this not apply to the above