Evaluate Levi-civita expression

[Meggle]

Homework Statement

Evaluate the expression $$\epsilon_{ijk} \epsilon_{jmn} \epsilon_{nkp}$$

Homework Equations

$$\epsilon_{ijk} \epsilon_{ilj} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$$

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?

 

[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 )

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

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

No, just use the deltas to change the indices …

for example, δ_kmδ_inε_nkp = ε_imp ​

 

[Meggle]

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, δ_kmδ_inε_nkp = ε_imp ​

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

 

[tiny-tim]

ε_imp = ε₁₂₃ + ε₂₃₁...

Noooo!

In ε_imp, i m and p are fixed.

ε_ijkε_jmnε_nkp 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, δ_knδ_imε_nkp ?

 

[Meggle]

Noooo!

In ε_imp, i m and p are fixed.

ε_ijkε_jmnε_nkp 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, δ_knδ_imε_nkp ?

Soooooo, instead of what I had done:
($$\delta$$_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.

 

[tiny-tim]

… $$\delta$$_im$$\epsilon$$_nnp
…
If I don't do anything to it, $$\epsilon$$_nnp will be zero.

Important life lesson …

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

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

So ε_nnp = 0.

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.

don't want to worry you, but …

it's actually a great sharp sword! ​

 

[Meggle]

Important life lesson …

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

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

So ε_nnp = 0.

don't want to worry you, but …

it's actually a great sharp sword! ​

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...)