Help deriving Lagrange's Formula with the levi-civita symbol

In summary, the conversation discusses the use of determinants and the Levi-Civita symbol to prove a formula involving cross products and dot products. The final formula is given and the use of \cdot instead of \bullet is suggested.
  • #1
radonballoon
21
0
Ok, so I'm really at a loss as to how to do this. I can prove the formula by just using determinants, but I don't really know how to do manipulations with the levi-civita symbol.
Here's what I have so far:
[tex]
(\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})\vec{e_{i}}
[/tex]

And I'm trying to get to:
[tex]
\vec{A} \times (\vec{B} \times \vec{C}) = B(A \bullet C) - C(A \bullet B)
[/tex]

Does anyone have any suggestions?
Thanks
 
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  • #2
Ok So I figured it out, I'll just post the answer for the sake of completeness:

[tex]
(\vec{B} \times \vec{C})_{k} = \epsilon_{kmn} (B_{m} C_{n})
[/tex]

[tex]
let (\vec{B} \times \vec{C}) = \vec{N}
[/tex]

[tex]
\vec{A} \times (\vec{B} \times \vec{C}) = \vec{A} \times \vec{N}
[/tex]
[tex]
(\vec{A} \times \vec{N})_{i} = \epsilon_{ijk} A_{j} N_{k}
[/tex]
[tex]
= \epsilon_{ijk} A_{j} (\epsilon_{kmn} B_{m} C_{n}) [/tex]
[tex]
= \epsilon_{ijk} \epsilon_{kmn} (A_{j} B_{m} C_{n}) [/tex]

[tex]
\epsilon_{ijk} \epsilon_{mnk} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}
[/tex]

[tex]
\vec{A} \times (\vec{B} \times \vec{C}) = (\delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}) A_{j} B_{m} C_{n}
[/tex]
[tex]
= B_{i} A_{j} C_{j} - A_{j} B_{j} C_{i}
[/tex]
[tex]
= \vec{B}(\vec{A} \bullet \vec{C}) - \vec{C}(\vec{A} \bullet \vec{B}) [/tex]
 
  • #3
It's threads like these that seem to be causing PF to accumulate helpful Google searches. :)
 
  • #4
[tex]

(\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})\vec{e_{i}}

[/tex]

Should be:

[tex]

(\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})

[/tex]

And start using \cdot instead of that big black ball :-D
 
  • #5
Haha wow this seems like so long ago. I couldn't find the dot for dot product, so thanks for that :D
 

Related to Help deriving Lagrange's Formula with the levi-civita symbol

What is Lagrange's Formula?

Lagrange's Formula is a mathematical equation used in vector calculus that expresses the derivative of a vector function with respect to a given variable as a linear combination of the partial derivatives of the components of the vector function.

What is the Levi-Civita symbol?

The Levi-Civita symbol, also known as the permutation symbol, is a mathematical symbol used to represent the sign of a permutation of a set of indices. It is commonly used in vector calculus and differential geometry.

How is the Levi-Civita symbol related to Lagrange's Formula?

The Levi-Civita symbol is used in Lagrange's Formula to express the derivative of a vector function in terms of the components of the vector function and their partial derivatives. It helps to simplify the calculation and makes the formula more concise.

What are the steps for deriving Lagrange's Formula with the Levi-Civita symbol?

The steps for deriving Lagrange's Formula with the Levi-Civita symbol involve setting up the formula using the vector function and its components, using the properties of the Levi-Civita symbol to simplify the calculation, and then rearranging the terms to get the final result.

What are some applications of Lagrange's Formula with the Levi-Civita symbol?

Lagrange's Formula with the Levi-Civita symbol is commonly used in physics and engineering to solve problems related to vector fields, such as calculating the curl and divergence of a vector field. It is also used in the study of fluid dynamics, electromagnetism, and general relativity.

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