# Homework Help: Evaluate Levi-civita expression

1. Oct 3, 2009

### Meggle

1. The problem statement, all variables and given/known data
Evaluate the expression $$\epsilon_{ijk} \epsilon_{jmn} \epsilon_{nkp}$$

2. Relevant equations
$$\epsilon_{ijk} \epsilon_{ilj} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$$

3. The attempt at a solution

Let $$\epsilon_{ijk} = \epsilon_{jki}$$ by permutation of Levi-civita

$$\epsilon_{jki} \epsilon_{jmn} \epsilon_{nkp} = (\delta_{km}\delta_{in} - \delta_{kn}\delta_{im})\epsilon_{nkp}$$

$$\epsilon_{nkp}=0$$ if n=k, however $$\delta_{kn}=0$$ if $$n \neq k$$

$$(\delta_{km}\delta_{in} - \delta_{kn}\delta_{im})\epsilon_{nkp} = (\delta_{km}\delta_{in})\epsilon_{nkp}$$

At this point, can I just go through the possible values for each of the indicies and add it up?

$$(\delta_{km}\delta_{in})\epsilon_{nkp} = (\delta_{22}\delta_{11})\epsilon_{123} + (\delta_{33}\delta_{22})\epsilon_{231} + (\delta_{22}\delta_{33})\epsilon_{321} + (\delta_{11}\delta_{22})\epsilon_{213} + (\delta_{33}\delta_{11})\epsilon_{132}$$
As all other combinations result in zeros.

$$(\delta_{km}\delta_{in})\epsilon_{nkp} = 1+1+1-1-1-1 = 0 = \epsilon_{ijk} \epsilon_{jmn} \epsilon_{nkp}$$
Is that right?

I'm doing this paper extramurally and really struggling with it. My previous assignments are taking ages to come back to me, so I'm shaky about what I know and where I'm going wrong. Could someone have a look at this and tell me if it looks ok?

2. Oct 4, 2009

### tiny-tim

Welcome to PF!

Hi Meggle! Welcome to PF!

(have a delta: δ and an epsilon: ε and try using the X2 tag just above the Reply box )
No, just use the deltas to change the indices …

for example, δkmδinεnkp = εimp

3. Oct 4, 2009

### Meggle

Re: Welcome to PF!

Oh! Ok, thanks!
Now if I assign all the possible values to i, m, and p, I think I end up with 0 still?
εimp = ε123 + ε231......
εimp = 1 + 1 + 1 - 1 - 1 - 1 + (a whole bunch of zeros where i = m etc)
εimp = 0

Sorry, I know this point should be really obvious. My course readings seem to operate on the "state nothing explicitly" style of teaching, which I'm having trouble adapting to!

4. Oct 4, 2009

### tiny-tim

Noooo!

In εimp, i m and p are fixed.

εijkεjmnεnkp is a function of i m and p (after you sum all the others, i m and p are still there) …

i m and p have to be in the answer!

ok … what's the other half, δknδimεnkp ?

5. Oct 5, 2009

### Meggle

($$\delta$$km $$\delta$$in - $$\delta$$kn$$\delta$$im)$$\epsilon$$nkp = $$\delta$$km $$\delta$$in$$\epsilon$$nkp - $$\delta$$kn$$\delta$$im$$\epsilon$$nkp

$$\delta$$km $$\delta$$in $$\epsilon$$nkp - $$\delta$$kn$$\delta$$im $$\epsilon$$nkp = $$\epsilon$$imp - $$\delta$$im$$\epsilon$$nnp

That looks like I've done something wrong. What can I do with this? The indicies are different in $$\delta$$kn$$\delta$$im , so I can't use one to alter the other, but I can't use $$\delta$$im on $$\epsilon$$nnp either. If I don't do anything to it, $$\epsilon$$nnp will be zero.
It's like there's a giant penny hanging over my head, and if I just keep banging it, it'll drop... Sorry I'm being so dense. Do appreciate the help though.

6. Oct 5, 2009

### tiny-tim

Important life lesson

sometimes you have to be able to tell the difference between a problem and a solution!

In this case, yes, εnnp (btw, you could equally well have written it εkkp ) is zero.

So εnnp = 0.
don't want to worry you, but

it's actually a great sharp sword!

7. Oct 5, 2009

### Meggle

Hang on, so that's actually the solution?
$$\epsilon_{ijk}\epsilon_{jmn}\epsilon_{nkp} = \epsilon_{imp}$$
OoooOooo. I was on the wrong track! Yeepers. Thanks for all your help!!
(I'll be minding out for that sword now...)