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In 3 dimensions we have the well known relation (summation convention is being used)

$$ L_i = I_{ij} \omega_j $$

However, as is well known the angular momentum and angular velocity are not vectors but tensors and their Hodge duals are what are used in the above expression. So using the actual 2-forms $$ \tilde L_{ij} = \epsilon_{ijk} L_k~, $$

and likewise for $\tilde \omega$, we get the above relation as

$$ \tilde L_{ij} = \frac{1}{2} ~I_{kl} ~\epsilon_{kij}~\epsilon_{lmn} ~\tilde \omega_{mn} $$

My question is how do I generalize this to higher dimensions? The angular momentum, moment of inertia and angular velocity will remain second order tensors. However, in higher dimensions the Levi-Civita tensor will be of higher order and I cannot seem to find an unambiguous way to get a $3^rd$ order Levi-Civita tensor from a higher order one.

Borun Chowdhury
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    Related: http://physics.stackexchange.com/q/9864/2451 and links therein. – Qmechanic Jan 05 '17 at 12:28
  • I think you need to consider that angular momentum in an arbitrary location then you use a cross product in 3D to do the parallel axis theorem. But in higher dimensions you would need to transition to the wedge product. Then you can decompose it into components using tensor notations. – John Alexiou Jan 05 '17 at 13:49
  • @Qmechanic while the link you mentioned is related in that it talks about angular momentum in higher dimensions, I don't think deals with relation between angular momentum and angular velocity. If you think there is some connection to angular velocity could you please point it out? Thanks. – Borun Chowdhury Jan 06 '17 at 10:35
  • @ja72 You are talking about the definition of angular momentum $\tilde L_{ij}=x_i p_j - x_j p_i$. That is included in my question when I mention the 3-d angular momentum 'vector' is a Hodge dual of $\tilde L_{ij}$. My question is not about this topic. My question is how to express angular momentum in terms of angular velocity. – Borun Chowdhury Jan 06 '17 at 10:37

3 Answers3

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It turns out the answer is rather simpler in higher dimensions. It gets complicated when specializing to 3-dimensions.

If a rigid body has angular velocity $\omega_{ij}$ then given the location of a point $r_i$ its velocity is given by

$$ v_i = -\omega_{ij} r_j $$

and its angular momentum is given by

$$ L_{ij} =\sum m(r) ( -r_i v_j +r_j v_i )\\ = \sum m(r)( r_i r_m \omega_{mj} -r_j r_m \omega_{mi}) \\ $$

Specializing to 3D we get $$ \tilde L_k = \frac{1}{2} \epsilon_{kij} L_{ij}\\ =\epsilon_{mjt}\epsilon_{ijk} [\sum m(r) r_i r_m] \tilde \omega_t \\ = -\delta^{tm}_{ik} [\sum m(r) r_i r_m ] \tilde \omega_t \\ = [\sum m(r) (-r_k r_i +r^2 \delta_{ki})] \tilde \omega_i \\ = I_{ki} \tilde \omega_i $$

Thus we see that in arbitrary dimensions the relation between angular momenta, moment of inertia and angular velocity is given by

$$ L_{ij} = 2 \tilde I_{im} \omega_{mj} $$

where $\tilde I_{ij} = \sum m(r) r_i r_j$.

Borun Chowdhury
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    $L_{ij} \neq 2 [\sum m(r) r_i r_m] \omega_{mj} = 2 \tilde I_{im} \omega_{mj}$. The rhs isn't generally antisymmetric. I think $L_{ij} = \tilde I_{im} \omega_{mj} - \tilde I_{jm} \omega_{mi}$ is correct, or $L = \tilde I \omega + \omega \tilde I$ in matrix notation. Another problem with this approach is it isn't clear that principle moments of inertia exist, or what they are. The rank-4 tensor in Sergio's answer is the real generalization of the inertia tensor, but has the disadvantage of being rank 4 of course. – benrg Nov 21 '21 at 19:26
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3D common way

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}

Arbitrary number of dimensions

For Higher dimensions we can use directly angular momentum tensor \begin{equation} L^{ij} = x^iv^j - x^jv^i \end{equation} and Rortational velocity of particle we can express 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}

Sergio
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    This doesn't agree with equation (A2) in https://arxiv.org/abs/2207.03560. – tparker Jan 15 '23 at 00:45
  • Oh, I see, you haven't antisymmetrized over the pairs $i,j$ and $n,r$. I think this tensor is probably physically equivalent to the one in the source above, since you only plug in an antisymmetric 2-form $\omega$, as long as you remember to then only use the antisymmetric part of the output. Also, you forgot to weight the sum by the particle masses. – tparker Jan 15 '23 at 23:20
  • I prefer the form in the source above though, because it makes it clear which are the physically significant degrees of freedom. – tparker Jan 15 '23 at 23:24
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According to https://arxiv.org/abs/2207.03560, the appropriate generalization is $$L_{ij} = \frac{1}{2} I_{ijkl} \omega_{kl},$$ where $L$ is the angular momentum two-form (which in three dimensions is the Hodge dual to the angular momentum vector), $\omega$ is the angular velocity two-form (likewise), and $I$ is the rank-four tensor of inertia defined by $$I_{ijkl} := 4 \int dm \left( x_{[j} \delta_{i][k} x_{l]} \right).$$

Interestingly, the tensor of inertia has the same symmetries and number of degrees of freedom as the Riemann curvature tensor. (This paper (PDF) tries to explain the reason behind this surprising parallel, but it uses some obscure math involving the geometric algebra, and I personally don't find it particularly illuminating.)

tparker
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  • You don't need to cite a paper for it. I derived the same result in the answer above. It's a three line derivation. – Borun Chowdhury Jan 15 '23 at 15:08
  • @BorunChowdhury Your derivation is incorrect and gets a totally different result from the one in my answer. The moment of inertia tensor in general dimensions is rank-4, not rank-2 as in your answer. – tparker Jan 15 '23 at 18:12