Proof using Levi-Civita symbol

In summary: Therefore, \delta_{jk}\delta_{kj} = 1In summary, the formula for \sum_{j,k} \epsilon_{ijk} \epsilon_{ljk} = 2\delta_{il} can be simplified by replacing the Kronecker delta variables with 1 when they are the same and using the property \delta_{ij}\delta_{jk} = \delta_{ik}.
  • #1
cashmerelc
18
0

Homework Statement


Prove [tex]\sum_{j,k}[/tex] [tex]\epsilon_{ijk}[/tex] [tex]\epsilon_{ljk}[/tex] = 2[tex]\delta_{il}[/tex]

Homework Equations


[tex]\epsilon_{ijk}[/tex] [tex]\epsilon_{ljk} = [tex]\delta_{il}[/tex]([tex]\delta_{jj}[/tex][tex]\delta_{kk}[/tex] - [tex]\delta_{jk}[/tex][tex]\delta_{kj}[/tex]) + [tex]\delta_{ij}[/tex]([tex]\delta_{jk}[/tex][tex]\delta_{kl}[/tex] - [tex]\delta_{jl}[/tex][tex]\delta_{kk}[/tex]) + [tex]\delta_{ik}[/tex]([tex]\delta_{jl}[/tex][tex]\delta_{kk}[/tex] - [tex]\delta_{jj}[/tex][tex]\delta_{kl}[/tex])

The Attempt at a Solution



Okay, in cases where subscripts of the Kronecker delta are equal, then [tex]\delta_{jj}[/tex] = 1.

If the subscripts are not equal, then [tex]\delta_{il}[/tex] = 0.

So plugging those into the parenthesis of the above equation gives me:

[tex]\delta_{il}[/tex]([tex]\delta_{jj}[/tex][tex]\delta_{kk}[/tex]) ?

If that is the case, then how could the two inside the parenthesis equal 2? I know I must be missing something.
 
Last edited:
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  • #2
In your formula, replace the [tex]\delta_{jj}[/tex], [tex]\delta_{kk}[/tex] etc... where the variables are the same... with 1.

Also, [tex]\delta_{ij}\delta_{jk} = \delta_{ik}[/tex]
 
  • #3
learningphysics said:
In your formula, replace the [tex]\delta_{jj}[/tex], [tex]\delta_{kk}[/tex] etc... where the variables are the same... with 1.

Also, [tex]\delta_{ij}\delta_{jk} = \delta_{ik}[/tex]

If [tex]\delta_{ij}\delta_{jk} = \delta_{ik}[/tex] does that mean that [tex]\delta_{lk}\delta_{kj} = \delta_{lj}[/tex] and so on?
 
  • #4
cashmerelc said:
If [tex]\delta_{ij}\delta_{jk} = \delta_{ik}[/tex] does that mean that [tex]\delta_{lk}\delta_{kj} = \delta_{lj}[/tex] and so on?

Yes, exactly.
 
  • #5
Okay, I think one more question will help me get it.

[tex]\delta_{jk}[/tex][tex]\delta_{kj}[/tex] = ?
 
  • #6
cashmerelc said:
Okay, I think one more question will help me get it.

[tex]\delta_{jk}[/tex][tex]\delta_{kj}[/tex] = ?

= [tex]\delta_{jj} = 1[/tex]
 

1. What is the Levi-Civita symbol?

The Levi-Civita symbol, also known as the permutation symbol, is a mathematical object used in vector calculus and differential geometry. It is defined as a function that assigns a value of +1, -1, or 0 to every permutation of three indices (i, j, k).

2. How is the Levi-Civita symbol used in proofs?

The Levi-Civita symbol is commonly used in proofs involving cross products, determinants, and triple integrals. It allows for a compact and elegant way of representing and manipulating vector and tensor equations.

3. What are the properties of the Levi-Civita symbol?

The Levi-Civita symbol has three main properties: anti-symmetry, linearity, and the product rule. These properties allow for the manipulation and simplification of equations involving the symbol.

4. Can the Levi-Civita symbol be extended to higher dimensions?

Yes, the Levi-Civita symbol can be extended to higher dimensions. In three dimensions, it is a third-order tensor, and in n dimensions, it is an n-th order tensor. The properties and rules for manipulating the symbol remain the same in higher dimensions.

5. What are some real-world applications of the Levi-Civita symbol?

The Levi-Civita symbol has various applications in physics, engineering, and mathematics. It is used in the study of electromagnetism, fluid mechanics, and quantum mechanics. It is also used in the development of computer graphics and machine learning algorithms.

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