# Index juggling by example of angular momentum tensor

• A
Lets consider the angular momentum tensor (here ##m=1##)

L^{ij} = x^iv^j - x^jv^i

and rortational velocity of particle (expressed via angular momentum tensor)

v^j = \omega^{jm}x_m.

Then

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

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

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

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

I^{ijnr} = x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}

Common form of inertia tensor

I^i_j = \delta_j^i x^2 - x^ix_j

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)?

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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.

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##)

L_l = \epsilon_{lij}x^iv^j

(##L_l## dual to ##L^{ij} = x^iv^j - x^jv^i## tensor)

Rortational velocity of particle

v^j = \epsilon^{jrk}\omega_rx_k.

(##\omega_r## dual to ##\omega^{jk}## antisymmetric tensor of angular velocity ##\epsilon^{jrk}\omega_r = \omega^{jk}##)

Substitute in the angular momentum definition

L_l = \epsilon_{lij}x^i\epsilon^{jrk}\omega_rx_k.

Let us use the property of the Levi-Civita tensor:

\epsilon^{lij} = -\epsilon^{jli}

then

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

Let us use another property of the Levi-Civita tensor:

\epsilon_{jli}\epsilon^{jkr} = \delta_l^k\delta_i^r - \delta_l^r\delta_i^k.

L_l = \left( \delta_l^r\delta_i^k - \delta_l^k\delta_i^r \right) x^ix_k\omega_r.

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

L_l = \left( \delta_l^r x^kx_k - x^rx_l\right) \omega_r,

or

L_l = \left( \delta_l^r x^2 - x^rx_l\right) \omega_r,

where inertia tensor

I_l^r = \delta_l^r x^2 - x^rx_l

samalkhaiat

I^{ijnr} = x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}

Common form of inertia tensor

I^i_j = \delta_j^i x^2 - x^ix_j

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 $$g_{ln}g_{ij}I^{ijmn} \equiv I^{jm}_{j}{}_{l} = x^{2} \delta^{m}{}_{l} - x^{m}x_{l} .$$

Last edited:
sergiokapone