use the Levi-Civita tensor to prove the following vector identities

Levi-Civita practice; use the Levi-Civita tensor to prove the following vector identities: $[ (A \times B) \times (C \times D) ] = [ (A \times B) \cdot D ] C – [ (A \times B) \cdot C ] D

The Correct Answer and Explanation is:

To prove the vector identity[(A×B)×(C×D)]=[(A×B)⋅D]C−[(A×B)⋅C]D,[(A \times B) \times (C \times D)] = [(A \times B) \cdot D] C – [(A \times B) \cdot C] D,[(A×B)×(C×D)]=[(A×B)⋅D]C−[(A×B)⋅C]D,

we will use the Levi-Civita tensor ϵijk\epsilon_{ijk}ϵijk​, which is a completely antisymmetric tensor used to express cross products and determinants.

Step 1: Express the Cross Products Using Levi-Civita

Recall that the cross product of two vectors can be written in terms of the Levi-Civita symbol as follows:(A×B)i=ϵijkAjBk,(A \times B)_i = \epsilon_{ijk} A_j B_k,(A×B)i​=ϵijk​Aj​Bk​,

where the summation convention is used over repeated indices.

Step 2: Express the Left-Hand Side

Now consider the left-hand side of the equation, (A×B)×(C×D)(A \times B) \times (C \times D)(A×B)×(C×D). Using the Levi-Civita symbol, we write this cross product as:[(A×B)×(C×D)]i=ϵilm(A×B)l(C×D)m.[(A \times B) \times (C \times D)]_i = \epsilon_{ilm} (A \times B)_l (C \times D)_m.[(A×B)×(C×D)]i​=ϵilm​(A×B)l​(C×D)m​.

Substitute the expressions for the cross products:[(A×B)×(C×D)]i=ϵilm(ϵlqkAqBk)(ϵmpnCpDn).[(A \times B) \times (C \times D)]_i = \epsilon_{ilm} \left( \epsilon_{lqk} A_q B_k \right) \left( \epsilon_{mpn} C_p D_n \right).[(A×B)×(C×D)]i​=ϵilm​(ϵlqk​Aq​Bk​)(ϵmpn​Cp​Dn​).

Step 3: Simplify Using the Identity for the Product of Two Levi-Civita Symbols

We now use the identity for the product of two Levi-Civita symbols:ϵlqkϵmpn=δlmδqn−δlnδqm.\epsilon_{lqk} \epsilon_{mpn} = \delta_{lm} \delta_{qn} – \delta_{ln} \delta_{qm}.ϵlqk​ϵmpn​=δlm​δqn​−δln​δqm​.

Substitute this identity into the equation:[(A×B)×(C×D)]i=(δlmδqn−δlnδqm)AqBkCpDn.[(A \times B) \times (C \times D)]_i = \left( \delta_{lm} \delta_{qn} – \delta_{ln} \delta_{qm} \right) A_q B_k C_p D_n.[(A×B)×(C×D)]i​=(δlm​δqn​−δln​δqm​)Aq​Bk​Cp​Dn​.

This expression can be simplified into two terms. Let’s break it down.

Step 4: Expand and Interpret the Terms

The first term is:δlmδqnAqBkCpDn=AqBkCpDk,\delta_{lm} \delta_{qn} A_q B_k C_p D_n = A_q B_k C_p D_k,δlm​δqn​Aq​Bk​Cp​Dn​=Aq​Bk​Cp​Dk​,

which corresponds to (A×B)⋅D(A \times B) \cdot D(A×B)⋅D times CCC.

The second term is:−δlnδqmAqBkCpDn=−AqBkCqDk,- \delta_{ln} \delta_{qm} A_q B_k C_p D_n = – A_q B_k C_q D_k,−δln​δqm​Aq​Bk​Cp​Dn​=−Aq​Bk​Cq​Dk​,

which corresponds to (A×B)⋅C(A \times B) \cdot C(A×B)⋅C times DDD.

Thus, we have:[(A×B)×(C×D)]i=[(A×B)⋅D]Ci−[(A×B)⋅C]Di.[(A \times B) \times (C \times D)]_i = [(A \times B) \cdot D] C_i – [(A \times B) \cdot C] D_i.[(A×B)×(C×D)]i​=[(A×B)⋅D]Ci​−[(A×B)⋅C]Di​.

Step 5: Conclusion

We have proven that[(A×B)×(C×D)]=[(A×B)⋅D]C−[(A×B)⋅C]D,[(A \times B) \times (C \times D)] = [(A \times B) \cdot D] C – [(A \times B) \cdot C] D,[(A×B)×(C×D)]=[(A×B)⋅D]C−[(A×B)⋅C]D,

using the Levi-Civita tensor and the properties of the cross product and the delta function. This result is consistent with vector identities in classical mechanics and electromagnetism.