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

The term is:

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

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

Reason for Vanishing

The term $\epsilon_{ijk} \omega_j \omega_k$ involves summation over the indices $j$ and $k$.

1. Levi-Civita Symbol ($\epsilon_{ijk}$): This is an antisymmetric tensor with respect to the indices $j$ and $k$.
   • $\epsilon_{ijk} = -\epsilon_{ikj}$.
   • For example, $\epsilon_{i12} = -\epsilon_{i21}$.
2. Angular Velocity Product ( $\omega_j \omega_k$ ): This product is symmetric with respect to the indices $j$ and $k$.
   • $\omega_j \omega_k = \omega_k \omega_j$.
   • For example, $\omega_1 \omega_2 = \omega_2 \omega_1$.

When you contract (sum over) the indices $j$ and $k$:

$$ \epsilon_{ijk} \omega_j \omega_k $$

Consider an example pair of terms in the summation:

• The term with $j=1$ and $k=2$: $\epsilon_{i12} \omega_1 \omega_2$
• The term with $j=2$ and $k=1$: $\epsilon_{i21} \omega_2 \omega_1$

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

$$ \epsilon_{i12} \omega_1 \omega_2 + \epsilon_{i21} \omega_2 \omega_1 = \epsilon_{i12} \omega_1 \omega_2 + (-\epsilon_{i12}) (\omega_1 \omega_2) = 0 $$

Every term in the summation is exactly cancelled out by a corresponding term with the indices $j$ and $k$ swapped.

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

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

Brief audio

Why does the first term in the final expression for the Right-Hand Side vanish-L.mp4