Why does the first term in the final expression for the Right-Hand Side vanish?

The first term in the final expression for the Right-Hand Side (RHS), ϵijkωjωkxmxm\epsilon_{ijk}\omega_j\omega_k x_m x_m, vanishes due to a fundamental property in tensor algebra concerning the contraction of antisymmetric and symmetric tensors.

The term is:

Vρ[ϵijkωjωkxmxm]dV \int_V \rho \left[ \epsilon_{ijk} \omega_j \omega_k x_m x_m \right] dV

The integral and the scalar term (ρxmxm\rho x_m x_m) are non-zero, but the term inside the brackets, ϵijkωjωk\epsilon_{ijk} \omega_j \omega_k, is zero, causing the entire expression to vanish.

Reason for Vanishing

The term ϵijkωjωk\epsilon_{ijk} \omega_j \omega_k involves summation over the indices jj and kk.

  1. Levi-Civita Symbol (ϵijk\epsilon_{ijk}): This is an antisymmetric tensor with respect to the indices $j$ and $k$.

    • ϵijk=ϵikj\epsilon_{ijk} = -\epsilon_{ikj}.

    • For example, ϵi12=ϵi21\epsilon_{i12} = -\epsilon_{i21}.

  2. Angular Velocity Product ( ωjωk\omega_j \omega_k ): This product is symmetric with respect to the indices jj and kk.

    • ωjωk=ωkωj\omega_j \omega_k = \omega_k \omega_j.

    • For example, ω1ω2=ω2ω1\omega_1 \omega_2 = \omega_2 \omega_1.

When you contract (sum over) the indices jj and kk:

ϵijkωjωk \epsilon_{ijk} \omega_j \omega_k

Consider an example pair of terms in the summation:

  • The term with j=1j=1 and k=2k=2: ϵi12ω1ω2\epsilon_{i12} \omega_1 \omega_2

  • The term with j=2j=2 and k=1k=1: ϵi21ω2ω1\epsilon_{i21} \omega_2 \omega_1

Since the angular velocity terms are symmetric (ω1ω2=ω2ω1\omega_1 \omega_2 = \omega_2 \omega_1) and the Levi-Civita symbol is antisymmetric (ϵi12=ϵi21\epsilon_{i12} = -\epsilon_{i21}):

ϵi12ω1ω2+ϵi21ω2ω1=ϵi12ω1ω2+(ϵi12)(ω1ω2)=0 \epsilon_{i12} \omega_1 \omega_2 + \epsilon_{i21} \omega_2 \omega_1 = \epsilon_{i12} \omega_1 \omega_2 + (-\epsilon_{i12}) (\omega_1 \omega_2) = 0

Every term in the summation is exactly cancelled out by a corresponding term with the indices jj and kk swapped.

General Rule: The contraction of an antisymmetric tensor with a symmetric tensor over the contracted indices is always zero.

Thus, ϵijkωjωk=0\epsilon_{ijk} \omega_j \omega_k = 0, and the entire first term Vρ[ϵijkωjωkxmxm]dV\int_V \rho \left[ \epsilon_{ijk} \omega_j \omega_k x_m x_m \right] dV vanishes. This leaves only the second, non-vanishing term to complete the identity.

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