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Levi-Civita Identities

  1. Sep 9, 2013 #1
    I am having trouble establishing a process to verify various identities for problems in index notation.

    Description of Problem
    Verify that [itex]\epsilon_{ijk}\epsilon_{iqr}=\delta_{jq}\delta_{kr}-\delta_{kq}\delta_{jr}[/itex]

    Attempt at Solution

    I know that the term is only positive if there is an even permutation in both terms, or an odd permutation in both terms. This means that the second indices equal and the third indices equal, creating the positive set of deltas. The term is negative if 2nd = 3rd and 3rd = 2nd in each one, respectively, creating the negative set of deltas.

    This is right. However, I do not know how to write that out mathematically. How do I expand out the Levi-Civita symbol to show that this is the case? Do I expand it out as a sum over i and write all the different possible cases? Is there a more elegant way of doing this?
     
  2. jcsd
  3. Sep 9, 2013 #2
    Not that I am aware of. Just sum it all up. But I am no mathematician. . . Proof this way is trivial, just long. Sorry, I didn't really help you did I. . .
     
  4. Sep 9, 2013 #3
    What you have written in words is very close to a proof by cases, so if you just made sure to state all the possible cases for the indices and calculate that the left hand side is equal to the right hand side in every case, then that constitutes a proof.

    You ask about a more elegant way of doing it? There are several. The first-order approximation to a more elegant method is by writing the Levi-Civita symbol in the form:[tex]\varepsilon_{klm}=\delta_{k1}(\delta_{l2}\delta_{m3}-\delta_{l3}\delta_{m2})+\delta_{k2}(\delta_{l3}\delta_{m1}-\delta_{l1}\delta_{m3})+\delta_{k3}(\delta_{l1}\delta_{m2}-\delta_{l2}\delta_{m1}).[/tex] You can prove this identity through cases, and it would be a bit easier than the above suggested proof by cases, and then you could just use that expression to write out [itex]\varepsilon_{ijk}\varepsilon_{iqr}[/itex], and then you get the answer.

    As a higher-order approximation to an elegant method, you can recognize the above expression as the determinant of a matrix of kroneckers:
    [tex]\varepsilon_{klm}=\begin{vmatrix}\delta_{k1} && \delta_{k2} && \delta_{k3} \\ \delta_{l1} && \delta_{l2} && \delta_{l3} \\ \delta_{m1} && \delta_{m2} && \delta_{m3}\end{vmatrix}
    [/tex] Then you could use the identity Det(A)Det(B)=Det(AB). This is basically the method Landau and Lifshitz suggest.
     
    Last edited: Sep 9, 2013
  5. Sep 10, 2013 #4
    Thanks for the help with this. I was told that all these solutions are only one or two lines. Both of the above methods fill the page with algebra. Is this the only way for it to be done? At this point, it seems that the paragraph method is the most elegant.
     
  6. Sep 10, 2013 #5

    vela

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    You should also show that the righthand side is 0 in cases where, say, i=j.
     
  7. Sep 10, 2013 #6
    Oops sorry for that typo that was up here for a while. It seems like you can do this in under a page if you use the expression I gave.

    [tex]\varepsilon_{ijk}=[\delta_{i1}(\delta_{j2}\delta_{k3}-\delta_{j3}\delta_{k2})+\delta_{i2}(\delta_{j3}\delta_{k1}-\delta_{j1}\delta_{k3})+\delta_{i3}(\delta_{j1}\delta_{k2}-\delta_{j2}\delta_{k1})] \cdot [\delta_{i1}(\delta_{q2}\delta_{r3}-\delta_{q3}\delta_{r2})+\delta_{i2}(\delta_{q3}\delta_{r1}-\delta_{q1}\delta_{r3})+\delta_{i3}(\delta_{q1}\delta_{r2}-\delta_{q2}\delta_{r1})][/tex]You can see that the cross terms like [itex]\delta_{i1}\delta_{i2}[/itex], immediately cancel, and the terms like [itex]\delta_{i1}\delta_{i1}[/itex] are equal to 1 since we sum over i, so you get:[tex]=(\delta_{j1}\delta_{k2}-\delta_{j2}\delta_{k1})(\delta_{q1}\delta_{r2}-\delta_{q2}\delta_{r1})+(\delta_{j2}\delta_{k3}-\delta_{j3}\delta_{k2})(\delta_{q2}\delta_{r3}-\delta_{q3}\delta_{r2})+(\delta_{j3}\delta_{k1}-\delta_{j1}\delta_{k3})(\delta_{q3}\delta_{r1}-\delta_{q1}\delta_{r3})[/tex][tex]=\delta_{ja}\delta_{qa}\delta_{kb}\delta_{rb}-\delta_{jc}\delta_{rc}\delta_{kd}\delta_{qd}=\delta_{jq}\delta_{kr}-\delta_{jr}\delta_{kq}[/tex]You might need to write out a little more of the algebra going between the last line and the line above it but it's not too tricky.
     
    Last edited: Sep 10, 2013
  8. Sep 10, 2013 #7

    WannabeNewton

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  9. Sep 10, 2013 #8
    I am not really sure what the vertical notation is. I have never been taught it, and I guess it has something to do with covariance and contravariance, but I do not know anything beyond that. That was never covered.

    Also, the professor gave me a hint that seems to have set me backwards completely. He said to let..

    [itex]\epsilon_{ijk}=e_i\cdot\left(e_j\times e_k\right)[/itex]

    Then use vector identities to show how the whole thing is true. Now, I am even more confused. I wrote the problem out using the identity, and I have no clue how to simplify it. I am currently at this:

    [itex]\epsilon_{ijk}\epsilon_{iqr}=e_i\cdot\left(e_j\times e_k\right)e_i\cdot\left(e_q\times e_r\right)[/itex]

    I know the various associative properties with dot and cross, but I do not see how any of this would lead to a simplification into kronecker deltas.
     
  10. Sep 10, 2013 #9
    If you are seeking to prove as you stated above, I believe you must go about it the long way. That way you do not require any other properties (which you cannot use unless you prove them). This is what we did in 2nd year undergraduate math physics at least.
     
  11. Sep 10, 2013 #10
    Ohhh, yes one of "those" proofs. Very physicsy. Well recall that [itex]\vec{e}_i\cdot\vec{e}_j=\delta_{ij}[/itex]. That'll be useful later. Let's write out the contracted levi-civitas using your professor's suggestion:[tex]
    \varepsilon_{ijk}\varepsilon_{iqr}=\sum_i \left(\vec{e}_i\cdot (\vec{e}_j\times\vec{e}_k) \right )\left(\vec{e}_i\cdot (\vec{e}_q\times\vec{e}_r) \right )=\sum_i (\vec{e}_j\times\vec{e}_k)_i(\vec{e}_q\times\vec{e}_r)_i[/tex] where we identify [itex]\vec{e}_i\cdot\vec{v}=v_i[/itex], the [itex]i[/itex]th component of the vector [itex]\vec{v}[/itex]. So now we recognize that this represents a certain dot product:[tex]\varepsilon_{ijk}\varepsilon_{iqr}=(\vec{e}_j\times\vec{e}_k)\cdot(\vec{e}_q\times\vec{e}_r)[/tex]
    Now we have to use the helpful vector identities: [tex]\vec{a}\cdot(\vec{b}\times\vec{c})=\vec{b}\cdot(\vec{c}\times\vec{a})[/tex][tex]\vec{a}\times(\vec{b}\times\vec{c})=\vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})[/tex]
    so applying these we get
    [tex]\varepsilon_{ijk}\varepsilon_{iqr}=\vec{e}_q\cdot(\vec{e}_r\times(\vec{e}_j\times\vec{e}_k))=\vec{e}_q\cdot(\vec{e}_j(\vec{e}_r\cdot\vec{e}_k)-\vec{e}_k(\vec{e}_r\cdot\vec{e}_j))[/tex][tex]=(\vec{e}_q\cdot\vec{e}_j)(\vec{e}_r\cdot\vec{e}_k)-(\vec{e}_q\cdot\vec{e}_k)(\vec{e}_r\cdot\vec{e}_j)=\delta_{qj}\delta_{rk}-\delta_{qk}\delta_{rj}[/tex]
     
    Last edited: Sep 10, 2013
  12. Sep 10, 2013 #11
    Well, I am definitely behind on my vector identities. Even after seeing those steps, I have no idea how you applied those vector identities. I will spend the next few hours figuring that out.

    [edit] Alright. I managed to figure it out. Thanks for the help! I would not have been able to figure out how to implement those vector identities. I guess I need to do some more memorization of these identities until they become second nature.
     
    Last edited: Sep 10, 2013
  13. Sep 10, 2013 #12
    Well I'd be happy to explain if it'll save you a few hours.

    We started with: [tex]\varepsilon_{ijk}\varepsilon_{iqr}=(\vec{e}_j\times\vec{e}_k)\cdot(\vec{e}_q\times\vec{e}_r)[/tex]
    Let us define [itex]\vec{a}_1:=(\vec{e}_j\times\vec{e}_k)[/itex]. Then we have
    [tex]\varepsilon_{ijk}\varepsilon_{iqr}=\vec{a}_1\cdot(\vec{e}_q\times\vec{e}_r)[/tex]
    This is in the form [itex]\vec{a}\cdot(\vec{b}\times\vec{c})[/itex] so we apply the rule: [itex]\vec{a}\cdot(\vec{b}\times\vec{c})=\vec{b}\cdot(\vec{c}\times\vec{a})[/itex] This yields,
    [tex]\varepsilon_{ijk}\varepsilon_{iqr}=\vec{e}_q\cdot(\vec{e}_r\times\vec{a}_1)[/tex]
    and plugging back in the definition for [itex]\vec{a}_1[/itex] brings you to the equation [tex]\vec{e}_q\cdot(\vec{e}_r\times(\vec{e}_j\times\vec{e}_k))[/tex]Now see if you can apply the second identity to finish the proof.
     
  14. Sep 10, 2013 #13

    Thanks so much for the help! I did not initially see your steps in the previous post, but I just had to stare at the identities and how to implement them some more. Eventually, I managed to get it. It was not easy, though! The vector identities were the hardest part for me. I think it was one of those things that I learned back in the day in undergrad, but I hadn't touched them in so long.
     
  15. Sep 10, 2013 #14
    Sorry guys, I have an additional higher level question regarding this stuff. Is it mathematically legal for me to work these problems out if, instead of just 1 repeated index, the epsilons have 2 or all 3 repeated indices? In the midway steps in the problems (same steps as above, but ijk, ijk, or ijk, ijr) a given index shows up more than three times. I was told that anything showing up more than 2 times makes no sense.
     
  16. Sep 12, 2013 #15
    I am solving another problem involving proving that if UV = W, then the multiplication of their respective inverses holds true as well. Does anyone know of a good site that is filled with practice problems with index notation in it? The use of the Levi-Civita symbol is blowing my mind, despite filling up 10 pages of practice notes and not understanding it whatsoever, no matter how much practice I seem to put into it. I feel like I am spinning my wheels here.
     
  17. Sep 12, 2013 #16

    WannabeNewton

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    Well to answer your previous question: in Einstein notation, dummy indices can only appear twice in an expression. For example, ##\epsilon^{ijk}\partial_j E_k = -\partial_t B^i## conforms to Einstein notation but ##\epsilon^{ijk}\partial_j E_j = -\partial_t B^i## and ##\epsilon^{ikk}\partial_j E_k = -\partial_t B^i## are nonsensical in Einstein notation.
     
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