Reading this. (page 5)

It gives the impulse $J$ in 3 dimensions using cross products etc. for a pair of colliding bodies. Since the cross product does not generalise to higher dimensions, is there an equivalent tensor equation that is true for $n$ spatial dimensions?

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    – ACuriousMind
    Nov 12, 2021 at 12:05

1 Answer 1


Interesting question.

The cross product does generalize to higher dimensions. Especially, for 1+3d spacetime, i.e., Minkowski, we can either consider $(A^\mu,B^\nu)\mapsto u^\mu\varepsilon_{\mu\nu\rho\sigma}A^\rho B^\sigma$ with $u^\mu$ a timelike unit vector (e.g., the observer’s 4-velocity) or $(F^{\mu\nu},G^{\mu\nu})\mapsto\Delta^{\kappa\lambda}{}_{\mu\nu\rho\sigma}\frac{1}{2}F^{\mu\nu}\frac{1}{2}G^{\rho\sigma}$ as an antisymmetric “product” between two bivectors(=$\mathfrak{so}(1,3)$ Lie algebra elements), where $\Delta^{\kappa\lambda}{}_{\mu\nu\rho\sigma}:=-8\delta^{[\kappa}{}_{[\mu}\eta_{\nu][\rho}\delta^{\nu]}{}_{\sigma]}$ is the structure constant of the Lorentz Lie algebra $\mathfrak{so}(1,3)$. Working with bivectors, which are the elements $\mathfrak{so}(1,3)$, turns out to be fruitful in special-relativistic mechanics of “rotation”:

  • infinitesimal angle $\Theta^\mu{}_\nu = (dg\hspace{0.07em}g^{-1})^\mu{}_\nu$ (the Maurer-Cartan form of $\mathrm{SO}(1,3)$ group manifold),
  • angular velocity $\Omega^\mu{}_\nu = \iota_{d/d\tau}\Theta^\mu{}_\nu$,
  • the orbital and spin angular momenta (elements of $\mathfrak{so}(1,3)^*$, while $T^*\mathrm{SO}(1,3)\cong\mathrm{SO}(1,3)\times\mathfrak{so}(1,3)^*$),

are all bivectors. Here, $g$ denotes an element of the Lorentz group $\mathrm{SO}(1,3)$ that describes the orientation of the object’s “body-attached orthonormal frame” (“body frame” in short). This is standard in the modern relativistic theory of rotating compact bodies—the Hanson-Regge spherical top.

Now let us turn to your question, starting with some preliminaries about special-relativistic orbital angular momentum (that you may already know). Recall that, in special relativity, momentum and energy gets packaged in a single variable called the 4-momentum $p^\mu = (E,p^i)$. This kind of “putting different Newton-like quantities into a single spacetime quantity” happens also for the angular momentum. Its spacetime version is the bivector \begin{align} L_{\mu\nu} := x_\mu p_\nu - x_\nu p_\mu \,, \end{align} which encodes the familiar three-dimensional angular momentum vector along with the “initial CM position”: \begin{align} L_{0i} &= -tp_i + x_iE = E(x_i - v_it)\,,\\ L_{ij} &= x_ip_j-x_jp_i = \varepsilon^k{}_{ij}L_k \,. \end{align} The conservation of $L_{0i}$ is interpreted as “conservation of linearity” of a CM worldline. We can confirm that $L_{\mu\nu}$ indeed “angular momentum” in, for instance, the Hamiltonian setting as follows: \begin{align} \{x^\mu,p_\nu\} = \delta^\mu{}_\nu \quad\implies\quad \{L_{\mu\nu},L_{\rho\sigma}\} = \frac{1}{2}\Delta^{\kappa\lambda}{}_{\mu\nu\rho\sigma} L_{\kappa\lambda} \,. \end{align} Note that, in the modern presentation, angular momentum is “by definition” generators of rotation (or Noether charges of rotation). The above calculation confirms that $L_{\mu\nu}=x_\mu p_\nu - x_\nu p_\mu$ correctly represents the Lorentz Lie algebra by Poisson brackets so that it is indeed the generator of rotations and Lorentz boosts. Also an additional exercise: check that the above equation reproduces the familiar $\{L_i,L_j\} = \varepsilon^k{}_{ij} L_k$.

Finally, consider the special-relativistic impulse formula: \begin{align} \frac{dp_\mu}{d\tau} = f_\mu \quad\implies\quad \Delta p_\mu = \int d\tau \,f_\mu(\tau) \,. \end{align} $p_\mu$ and $f_\mu$ are the 4-momentum and the 4-force, respectly. $\tau$ is the proper time. Now generalizing to rotational mechanics is straightforward: \begin{align} \frac{dL_{\mu\nu}}{d\tau} = \tau_{\mu\nu} \quad\implies\quad \Delta L_{\mu\nu} = \int d\tau \, \tau_{\mu\nu}(\tau) \,. \end{align} $\tau_{\mu\nu}$ is the “spacetime version” of torque. For the point particle, it is given by $\tau_{\mu\nu} = x_\mu f_\nu - x_\nu f_\mu$. To mimic the equations in p.5 of your reference, consider two balls “colliding,” i.e., “locally” exchanging an impulse $\pm q_{\mu}$. Or, think of the “scattering S-matrix” at the asymptotic infinity… Let $r^{(1)}{}^\mu$ and $r^{(2)}{}^\mu$ be the separation four-vector to the collision point from the center of each ball, as defined in your reference. Then, the difference in the first ball’s spin angular momentum $S_{\mu\nu}$ reads \begin{align} &S^{(1)}_{\mu\nu}(+\infty) - S^{(1)}_{\mu\nu}(-\infty) = 2r^{(1)}_{[\mu} I_{\nu]} \\ &\implies\quad I^{(1)}(+\infty) \, \Omega ^{(1)}_{\mu\nu}(+\infty) - I^{(1)}(-\infty) \, \Omega ^{(1)}_{\mu\nu}(-\infty) = 2r^{(1)}_{[\mu} q_{\nu]} \,, \end{align} and that’s it. I assumed that the ball retains its spherical symmetry so that the rotational moment of inertia $I^{(1)}$ (the inverse of the Regge slope in the HR spherical top theory) remains to be a Lorentz scalar. Here the “cross product” is directly implemented in the equation as a bivector. In 3d space, we had to “peel off” the Hodge dual by taking the cross product because angular velocity was defined as a vector. But, in fact, it has a bivector nature, and by defining it as a bivector we don’t have Hodge dual in the formula here.

There are also nontrivial cases. In such cases, we can derive Newton-like formulae by using the tips I have mentioned: 1) contract with a timelike unit vector that defines a frame, 2) take Hodge dual, 3) use the “bivector antisymmetric product” wisely. For instance, in the above equation, we could have contracted it with $q^\mu$. Then you get $r^{(1)}{}^\mu q^2 - (r^{(1)}\cdot q)\, q^\mu = q^2 b^{(1)}{}^\mu$ with $b^{(1)}{}^\mu$ being the “impact parameter four-vector” orthogonal to the impulse $q^\mu$.

In fact in the literature of applying modern QFT scattering amplitudes to Post-Minkowskian gravity you can find Lorentz-covariant formulae on impulses and impact parameters. Also you may want to take a look at the “Hanson-Regge spherical top” for more details on special-relativistic rotational mechanics. While reading this answer you may wondered, “there’s no such thing as rigid body in relativity!” But, it turns out that we can describe rotating compact bodies with rotating point particle within the “effective field theory” philosophy, for instance https://arxiv.org/abs/1501.04956. The “spin angular momentum” $S_{\mu\nu}$ used in this answer should be understood as a classical (i.e., not quantum) one in the effective theory language. In the effective description, a point particle, actually not a point but representing a rotating compact body, has spin angular momentum: addition of the orbital angular momenta of each part of the compact body.

I wonder if your question was about extending the formula to 4d “space.” But, the answer does not change. The bivector formula/formalism here applies for arbitrary spatial dimensions. It is just because Lie algebra elements of $\mathfrak{so}(p,q)$ are bivectors in the $(p,q)$-signature Euclidean space.


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