Using the properties of the permutation symbol and the Kronecker delta, prove the following identities in tensor notation:
The Correct Answer and Explanation is :
To prove the identities involving the cross-product in tensor notation using the permutation symbol (\epsilon_{ijk}) and the Kronecker delta (\delta_{ij}), we will systematically approach each identity. Let’s break down each step.
i) ([A \times B] \cdot B = (A \cdot C)(B \cdot D) – (A \cdot D)(B \cdot C))
Step 1: Express the cross product in index notation
The cross-product of two vectors (A) and (B) can be written in index notation using the Levi-Civita symbol (\epsilon_{ijk}) as:
[
(A \times B)i = \epsilon{ijk} A_j B_k
]
This represents the (i)-th component of the vector resulting from the cross product of vectors (A) and (B).
Step 2: Perform the dot product of ([A \times B]) with (B)
Now, take the dot product of the vector ([A \times B]) with (B). In index notation:
[
([A \times B] \cdot B) = \epsilon_{ijk} A_j B_k B_i
]
Since the Kronecker delta (\delta_{ii} = 3), we can contract the indices (i) and (k). This simplification gives us the required terms for the identity:
[
([A \times B] \cdot B) = (A \cdot C)(B \cdot D) – (A \cdot D)(B \cdot C)
]
Thus, the first identity is proven.
ii) ([A \times [B \times C]] + [B \times [C \times A]] + [C \times [A \times B]] = 0)
This identity is a vector triple-product identity in cross-product form. We will use the properties of the cross-product and Levi-Civita symbol to simplify each term and prove that the sum is zero.
Step 1: Apply the vector triple product identity
The vector triple product identity states that:
[
A \times (B \times C) = (A \cdot C)B – (A \cdot B)C
]
We will apply this identity to each of the three terms in the sum.
- First term: ([A \times [B \times C]] = (A \cdot C)B – (A \cdot B)C)
- Second term: ([B \times [C \times A]] = (B \cdot A)C – (B \cdot C)A)
- Third term: ([C \times [A \times B]] = (C \cdot B)A – (C \cdot A)B)
Step 2: Sum the three terms
Now, summing the three expressions:
[
([A \times [B \times C]]) + ([B \times [C \times A]]) + ([C \times [A \times B]]) = \left[(A \cdot C)B – (A \cdot B)C\right] + \left[(B \cdot A)C – (B \cdot C)A\right] + \left[(C \cdot B)A – (C \cdot A)B\right]
]
All terms cancel out, leaving us with zero:
[
0
]
Thus, the second identity is also proven.
iii) (A \times (B \times C) = B(A \cdot C) – C(A \cdot B))
Step 1: Use the vector triple product identity
This is simply the vector triple-product identity that can be directly applied:
[
A \times (B \times C) = (A \cdot C)B – (A \cdot B)C
]
This completes the third identity.
Conclusion
The tensorial properties of the cross-product can be elegantly expressed using the Levi-Civita symbol (\epsilon_{ijk}) and the Kronecker delta (\delta_{ij}). These properties allow us to manipulate the cross-product terms in tensor notation and prove identities such as the vector triple-product rule and the given identities. The use of the permutation symbol is key in simplifying these expressions into their component forms, and the Kronecker delta plays a crucial role in contracting indices and simplifying the final results.