Classical Dynamics of Particles and Systems · Problem 1-22

Newtonian Mechanics—Single Particle — Marion & Thornton

Physics — Mechanics · levi-civita-symbol

Problem

Evaluate the sum ∑kεijk εlmk\displaystyle\sum_k \varepsilon_{ijk}\,\varepsilon_{lmk} (which contains 3 terms) by considering the result for all possible combinations of ii, jj, ll, mm; that is, ((a)) i=ji = j ((b)) i=li = l ((c)) i=mi = m ((d)) j=lj = l ((e)) j=mj = m ((f)) l=ml = m ((g)) i≠ji \neq j or mm ((h)) j≠lj \neq l or mm Show that ∑kεijk εlmk=δilδjm−δimδjl\sum_k \varepsilon_{ijk}\,\varepsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} and then use this result to prove A×(B×C)=(A⋅C)B−(A⋅B)C\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = (\mathbf{A} \cdot \mathbf{C})\mathbf{B} - (\mathbf{A} \cdot \mathbf{B})\mathbf{C}
Intuitive explanation

The Levi-Civita symbol εijk\varepsilon_{ijk} encodes orientation and vanishes whenever indices repeat—the key is systematically checking when both εijk\varepsilon_{ijk} and εlmk\varepsilon_{lmk} are nonzero for the same kk, then using the Kronecker delta to express the result. Once you derive the contraction formula, the vector triple product follows directly by converting the cross product into index notation and applying your identity to simplify.

Intuitive explanation

The Levi-Civita symbol εijk\varepsilon_{ijk} encodes orientation and vanishes whenever indices repeat—the key is systematically checking when both εijk\varepsilon_{ijk} and εlmk\varepsilon_{lmk} are nonzero for the same kk, then using the Kronecker delta to express the result. Once you derive the contraction formula, the vector triple product follows directly by converting the cross product into index notation and applying your identity to simplify.

Related theory

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