Index Juggling: Angular Momentum Tensor & Inertia Tensor in 3D-Space

[sergiokapone]

Lets consider the angular momentum tensor (here $m=1$)
\begin{equation}
L^{ij} = x^iv^j - x^jv^i
\end{equation}
and rortational velocity of particle (expressed via angular momentum tensor)
\begin{equation}
v^j = \omega^{jm}x_m.
\end{equation}

Then

\begin{equation}
L^{ij} = x^ix_m\omega^{jm} - x^jx_m\omega^{im}
\end{equation}

Now we can lower indices near $\omega$ with metric tensor:
\begin{align}
\omega^{jm} = g^{jn}g^{mr}\omega_{nr} \\
\omega^{im} = g^{in}g^{mr}\omega_{nr}
\end{align}

So, we get

\begin{equation}
L^{ij} = \left( x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}\right) \omega_{nr}
\end{equation}

So, we can conclude $L^{ij} = I^{ijnr} \omega_{nr}$, the inertia tensor is

\begin{equation}
I^{ijnr} = x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}
\end{equation}Common form of inertia tensor
\begin{equation}
I^i_j = \delta_j^i x^2 - x^ix_j
\end{equation}

So, my question, how can I get common form of inertia tensor $I^i_j$ in case 3D-space based on my $I^{ijnr}$ (if it correct, of course)?

 

[haushofer]

What's the relation between L, I and omega for your "common form of I"? The indices don't match between that common form and your I with 4 indices.

Also, be aware of solving for tensors in contractions if these contractions involve (anti)symmetric tensors.

 

[sergiokapone]

What's the relation between L, I and omega for your "common form of I"? The indices don't match between that common form and your I with 4 indices.

Definition of angular momentum vector ($m = 1$)
\begin{equation}
L_l = \epsilon_{lij}x^iv^j
\end{equation}
($L_l$ dual to $L^{ij} = x^iv^j - x^jv^i$ tensor)

Rortational velocity of particle
\begin{equation}
v^j = \epsilon^{jrk}\omega_rx_k.
\end{equation}
($\omega_r$ dual to $\omega^{jk}$ antisymmetric tensor of angular velocity $\epsilon^{jrk}\omega_r = \omega^{jk}$)

Substitute in the angular momentum definition
\begin{equation}
L_l = \epsilon_{lij}x^i\epsilon^{jrk}\omega_rx_k.
\end{equation}

Let us use the property of the Levi-Civita tensor:
\begin{equation}
\epsilon^{lij} = -\epsilon^{jli}
\end{equation}

then

\begin{equation}
L_l = - \epsilon_{jli}\epsilon^{jkr}x^ix_k\omega_r.
\end{equation}

Let us use another property of the Levi-Civita tensor:
\begin{equation}
\epsilon_{jli}\epsilon^{jkr} = \delta_l^k\delta_i^r - \delta_l^r\delta_i^k.
\end{equation}

\begin{equation}
L_l = \left( \delta_l^r\delta_i^k - \delta_l^k\delta_i^r \right) x^ix_k\omega_r.
\end{equation}

Expand the brackets and take into account that $\delta_i^r x^i = x^r$, $\delta_l^k x_k = x_l$ and $\delta_i^k x^i = x^k$, we get
\begin{equation}
L_l = \left( \delta_l^r x^kx_k - x^rx_l\right) \omega_r,
\end{equation}

or

\begin{equation}
L_l = \left( \delta_l^r x^2 - x^rx_l\right) \omega_r,
\end{equation}

where inertia tensor
\begin{equation}
I_l^r = \delta_l^r x^2 - x^rx_l
\end{equation}

 

[samalkhaiat]

\begin{equation}
I^{ijnr} = x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}
\end{equation}

Common form of inertia tensor
\begin{equation}
I^i_j = \delta_j^i x^2 - x^ix_j
\end{equation}

So, my question, how can I get common form of inertia tensor $I^i_j$ in case 3D-space based on my $I^{ijnr}$ (if it correct, of course)?

In 3D there is no difference between upper and lower indices, and the metric is: $$g_{ij} = g^{ij} = \delta^{ij} = \delta_{ij} = \delta^{i}_{j}.$$ But even if you keep using upper and lower indices, "your tensor" is just $$I^{ijmn} = x^{i}x^{j} g^{mn} - x^{m}x^{j}g^{in}.$$ To obtain the inertia tensor from that, you just need to contract with $g_{ij}$: $$g_{ij}I^{ijmn} = x^{2}g^{mn} – x^{m}x^{n} ,$$ or $$g_{ln}g_{ij}I^{ijmn} \equiv I^{jm}_{j}{}_{l} = x^{2} \delta^{m}{}_{l} - x^{m}x_{l} .$$