How is the Moment of Inertia Tensor Derived for Rigid Body Rotation?

[etotheipi]

Homework Statement

Show that the moment of inertia tensor satisfies $$\frac{dI_{ij}}{dt} \omega_j = \varepsilon_{ijk} \omega_j I_{kl} \omega_l$$for a body undergoing arbitrary rotation about the origin.

Relevant Equations

N/A

We start from the definition$$I_{ij} = \int_V \rho(x_k x_k \delta_{ij} - x_i x_j) dV \iff \dot{I}_{ij} = \int_V \rho (2 x_k \dot{x}_k \delta_{ij} - \dot{x}_i x_j - x_i \dot{x}_j ) dV$$Now since the rigid body rotation satisfies $\dot{\vec{x}} = \vec{\omega} \times \vec{x} \iff \dot{x}_i = \varepsilon_{ilm} w_l x_m$, and also since $\vec{\omega}$ is the same at all points within $V$ at a given time $t$, we have$$
\begin{align*}
\dot{I}_{ij} &= \int_V \rho(2\delta_{ij} \varepsilon_{klm} x_k x_m w_l - \varepsilon_{ilm} x_j x_m w_l - \varepsilon_{jlm} x_i x_m w_l) dV \\

\dot{I}_{ij} w_j &= w_l w_j \int_V \rho (2 \delta_{ij} \varepsilon_{klm} x_k x_m - \varepsilon_{ilm} x_j x_m - \varepsilon_{jlm} x_i x_m) dV\end{align*}$$I don't know what to do next! We need to show that the integrand reduces to$$\rho (2 \delta_{ij} \varepsilon_{klm} x_k x_m - \varepsilon_{ilm} x_j x_m - \varepsilon_{jlm} x_i x_m) = \rho \varepsilon_{ijk}(x_m x_m \delta_{kl} - x_k x_l)$$Any help is appreciated, thanks.

 

[Antarres]

It looks to me like you lack some experience manipulating expressions with Einstein summation convention(aka manipulating indices).
Okay, so let's start with the expression which I think you have correctly:
$$ \dot{I}_{ij} = \int_V \rho(2\delta_{ij}\varepsilon_{klm}x_kx_m\omega_l - \varepsilon_{ilm}x_jx_m\omega_l -\varepsilon_{jlm}x_ix_m\omega_l)dV$$
If we look at the first term, we have the expression $\varepsilon_{klm}x_kx_m$, but from antisymmetry of indices in $\varepsilon_{klm}$, we have that this sum amounts to zero. The proof is rather simple:
$$\varepsilon_{klm}x_kx_m = \varepsilon_{mlk}x_mx_k = - \varepsilon_{klm}x_kx_m \Leftrightarrow \varepsilon_{klm}x_kx_m = 0$$
In the first equality we just relabeled the dummy indices, swapping m and k as a label, and then in second equality, we used the antisymmetry of $\varepsilon$ symbol. Hence we can conclude that the first term vanishes.
Now we have:
$$\dot{I}_{ij} = \int_V \rho(- \varepsilon_{ilm}x_jx_m\omega_l -\varepsilon_{jlm}x_ix_m\omega_l)dV$$
If we multiply the above expression by $\omega_j$, and sum over $j$, then the last term will vanish along the same lines(check it). We're left with:
$$\dot{I}_{ij}\omega_j = -\int_V dV \rho\varepsilon_{ilm}x_j\omega_j x_m\omega_l$$
Now we pull out everything that we're not integrating over:
$$\dot{I}_{ij}\omega_j = -\varepsilon_{ilm}\omega_j\omega_l \int_V dV \rho x_jx_m$$
Maybe I have given you too much in this explanation, but I think these tricks are something you will always use, so the purpose is to get you accustomed to the way of thinking. Now that we have this expression that looks like it has too little terms, we notice that it kind of looks like the term we would arrive at from the second term in the moment inertia, but the first term is missing. Maybe this happened, because the first term actually gives zero(due to antisymmetric sum over symmetric indices, like we had two times so far), so we should see what the first term would actually look like.
It would have the form of $\int_V \rho \delta_{jm}x_kx_k$. Putting the rest of our expression in, we find:
$$\varepsilon_{ilm}\omega_j\omega_l\int_V dV\rho \delta_{jm}x_kx_k = \varepsilon_{ilm}\omega_m\omega_l\int_V dV\rho x_kx_k = 0$$
since the sum in front of the integral vanishes.
So we conclude that our expression above is correct and equivalent to the expression we want to prove, and the rest of our job is just relabeling dummy indices to fit the expression and adding this term(which is equal to zero) inconsequentially. I'd leave that to you, so that I don't give you the full solution. If something here confuses you, I'll be happy to help out.

 

[etotheipi]

Thanks, that's great! I think I learned quite a few useful techniques there, and I was completely forgetting to consider any antisymmetry properties of the levi civita tensor. So after the considerations that you explained, we will have$$\begin{align*}

\dot{I}_{ij} w_j &= w_l w_j \int_V \rho (2 \delta_{ij} \varepsilon_{klm} x_k x_m - \varepsilon_{ilm} x_j x_m - \varepsilon_{jlm} x_i x_m) dV \\

&= - \int_V dV \rho \varepsilon_{ilm} \omega_j \omega_l x_j x_m
\end{align*}$$
Now we can relabel the dummy indices $l \leftrightarrow j$, and also $m \leftrightarrow k$, so that we get$$\dot{I}_{ij} w_j = -\varepsilon_{ijk} \omega_j \omega_l \int_V dV \rho x_l x_k$$ and then as you also explained the following integral$$\varepsilon_{ijl} \omega_j \omega_l \int_V dV \rho x_m x_m = \int_V \rho \varepsilon_{ijl} \omega_j \omega_l x_m x_m = 0$$is equal to zero because interchanging the dummy indices $j$ and $l$ will give$$\varepsilon_{ijl} \omega_j \omega_l = \varepsilon_{ilj} \omega_{l} \omega_{j} = - \varepsilon_{ijl} \omega_l \omega_j \iff \varepsilon_{ijl} \omega_j \omega_l = 0$$and finally, since we can re-write $$0 = \varepsilon_{ijl} \omega_j \omega_l \int_V dV \rho x_m x_m = \varepsilon_{ijk} \delta_{kl} \omega_j \omega_l \int_V dV \rho x_m x_m = \varepsilon_{ijk} \omega_j \omega_l \int_V dV \rho \delta_{kl} x_m x_m$$we can say$$\dot{I}_{ij} \omega_j = 0 -\varepsilon_{ijk} \omega_j \omega_l \int_V dV \rho x_l x_k = \varepsilon_{ijk} \omega_j \omega_l \int_V dV \rho \delta_{kl} x_m x_m -\varepsilon_{ijk} \omega_j \omega_l \int_V dV \rho x_l x_k$$or in other words$$\dot{I}_{ij} \omega_j = \varepsilon_{ijk} \omega_j I_{kl} \omega_l$$Thanks for the guidance!