Time derivative of the moment of inertia tensor

[George Keeling]

TL;DR Summary

I am completely stuck on problem 2.45 of Blennow's book Mathematical Models for Physics and Engineering.

I am completely stuck on problem 2.45 of Blennow's book Mathematical Models for Physics and Engineering. @Orodruin It says
"We just stated that the moment of inertia tensor $I_{ij}$ satisfies the relation$${\dot{I}}_{ij}\omega_j=\varepsilon_{ijk}\omega_jI_{kl}\omega_l$$Show that this relation is true by starting from Eq. 2.146 and using the fact that $\vec{v}=\vec{\omega}\times\vec{x}$."

$\vec{\omega}$ is the angular velocity. The overdot signifies a time derivative. We're in Cartesian coordinates. Eq. 2.146 was a formula for moment of inertia about the origin for a solid object of constant density $\rho$ as a volume integral. The $x_i$ are the coordinates (or components of the position vector $\vec{x}$) of the volume element $dV$.$$I_{ij}=\int_{V}^{\ }\rho\left(x_kx_k\delta_{ij}-x_ix_j\right)dV$$I have studied all the examples about angular stuff and learned that angular velocity $\vec{\omega}$, angular momentum $\vec{L}$ and moment of inertia $I$ are always relative to a point. (I always thought they were relative to an axis).

I tried to differentiate the integral with respect to time, but I can't get it to make sense and I'm not sure it's even possible.

I tried writing the infinitesimal version of the formula as$$dI_{ij}=\rho\left(x_kx_k\delta_{ij}-x_ix_j\right)dV$$then$${\dot{I}}_{ij}=\frac{dI_{ij}}{dt}=\rho\left(x_kx_k\delta_{ij}-x_ix_j\right)\frac{dV}{dt}=\left(x_kx_k\delta_{ij}-x_ix_j\right)\frac{dM}{dt}$$where $dM$ is the mass of the volume element. But $\frac{dM}{dt}$ is zero except when the boundary of the solid is at the coordinates and then it is very large.

I tried fiddling with the RHS of the relation in question and got $$\varepsilon_{ijk}\omega_jI_{kl}\omega_l=\omega_j\rho\int_{V}^{\ }\left[\varepsilon_{ijk}\omega_kx_mx_m-\varepsilon_{ijk}\omega_lx_kx_l\right]dV$$$$=\omega_j\rho\int_{V}^{\ }{\varepsilon_{ijk}\omega_kx_mx_mdV}-\rho\int_{V}^{\ }{\varepsilon_{ijk}\omega_jx_k\omega_lx_ldV}=\omega_j\rho\int_{V}^{\ }{\varepsilon_{ijk}\omega_kx_mx_mdV}-\rho\int_{V}^{\ }{v_i\omega_lx_ldV}$$so I managed to use $\vec{v}=\vec{\omega}\times\vec{x}$ in the last part but am not much wiser.

I tried a few other things and nothing helped.

Is there another way to get going?

 

[Orodruin]

If the integral is confusing you, I suggest starting out with a set of $N$ particles of masses $m^{(n)}$ and positions $x_i^{(n)}$. The moment of inertia tensor is then given by
$$
I_{ij} = \sum_{n=1}^N m^{(n)} [x_k^{(n)}x_k^{(n)} \delta_{ij} - x_i^{(n)} x_j^{(n)}].
$$
Consider what happens when you differentiate this with respect to time.

(I always thought they were relative to an axis).

A freely rotating object will generally not rotate about a single axis (unless its angular velocity is parallel to one of the eigenvectors of the moment of inertia tensor). If you force the object to rotate around a singe axis, then you have constrained the motion to be essentially a one-dimensional rotation around that axis and the moment of inertia will be a single number given by $I = n_i I_{ij} n_j$, where $\vec n$ is a unit vector parallel to the axis. However, if the angular velocity is not in one of the eigendirections and you let the object rotate freely, then the angular acceleration will generally not be aligned with the angular velocity and thus the angular velocity will change direction as time progresses.

 

[George Keeling]

Thanks for both of those! So the integral is just a sum of little boxes and we can sort of differentiate each one and add them up again so $${\dot{I}}_{ij}=\int_{V}^{\ }\left(2\delta_{ij}{\dot{x}}_kx_k-{\dot{x}}_ix_j-x_i{\dot{x}}_j\right)\rho dV$$then all the $\dot{x}$ turn into $v$ and we can use $v_l=\varepsilon_{lmn}\omega_mx_n$ to get rid of them. I have now got as far as $${\dot{I}}_{ij}\omega_j=\omega_j\omega_m\int_{V}^{\ }\left(2\delta_{ij}\varepsilon_{kmn}x_nx_k-\varepsilon_{imn}x_nx_j\right)\rho dV$$and$$\varepsilon_{ijk}\omega_jI_{kl}\omega_l=\omega_j\omega_m\int_{V}^{\ }\left(\delta_{km}\varepsilon_{ijk}x_nx_n-\varepsilon_{ijk}x_kx_m\right)\rho dV$$I am allowing $\omega_i$'s to come in and out of the integral freely (for some reason I think the solid body has constant angular momentum). The expression for ${\dot{I}}_{ij}\omega_j$ had three terms but one vanished because it contained an $\varepsilon_{jmn}$. So I've either got to prove that the expressions in brackets are the same or that they are the same when contracted with one or both of $\omega_j,\omega_m$. Am I on the right track?

 

[Orodruin]

I have now got as far as $${\dot{I}}_{ij}\omega_j=\omega_j\omega_m\int_{V}^{\ }\left(2\delta_{ij}\varepsilon_{kmn}x_nx_k-\varepsilon_{imn}x_nx_j\right)\rho dV$$

You may want to examine the first term a bit.

and$$\varepsilon_{ijk}\omega_jI_{kl}\omega_l=\omega_j\omega_m\int_{V}^{\ }\left(\delta_{km}\varepsilon_{ijk}x_nx_n-\varepsilon_{ijk}x_kx_m\right)\rho dV$$

In the first term, you may want to use some property of $\delta_{km}$ and then examine the symmetries.

I am allowing $\omega_i$'s to come in and out of the integral freely (for some reason I think the solid body has constant angular momentum).

Angular velocity ($\vec\omega$) is indeed a property of the rigid body as a whole and therefore independent of the spatial position. It is however not (necessarily) a constant with respect to time, but here the integral is a spatial integral so they can go in and out as they please.

Looking back at the construction of this problem, I may have been a bit clearer. You can either construct the moment of inertia tensor at time $t$ as
$$
I_{ij}(t) = \int \rho(\vec x, t) [\delta_{ij} \vec x^2 - x_i x_j] dV
$$
where $x_i$ are the integration variables and $\rho(\vec x, t)$ is the density at position $\vec x$ at time $t$ or as
$$
I_{ij}(t) = \int \rho(\vec y, 0) [\delta_{ij} \vec x(\vec y, t)^2 - x_i(\vec y, t) x_j(\vec y,t)] dV_y,
$$
where $\vec y$ labels the position of the volume element at $t = 0$ and $\vec x(\vec y, t)$ is where that volume element has traveled to at an arbitrary time $t$. Of course $\vec x(\vec y, 0) = \vec y$ so at time $t = 0$ the expressions are obviously equivalent. They are also equivalent at any later time $t$, but this is not as obvious - the difference is whether you use the position at time 0 or at time $t$ as the integration variable. I implicitly used the second definition here although it should be possible to arrive at the same result by using the first as well.

 

[George Keeling]

Thanks again. I should have spotted those. On to problem 2.46 only five to go!

 

[Orodruin]

On to problem 2.46 only five to go!

I am pretty sure there are more problems in the following chapters ...