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In dimension $3$, we have the angular momentum $\omega = q \times v$, and since $$\frac{d}{dt} \omega = v \times v + q \times (F/m)$$ from the fact the force is central (so $F$ is parallel to $q$) we obtain the conservation of $\omega$, so $q$ always lies in the plane perpendicular to $\omega$.

With 4 spatial dimensions there's no such thing as a cross product so none of this makes sense. Can motion under a central force law not lie within a $2d$ plane in this case?

More specifically: is there some smooth $F : \mathbb{R}^4 - \{0\} \to \mathbb{R}^4$ such that $F(q)$ is always parallel to $q$, and some smooth $q : \mathbb{R} \to \mathbb{R}^4 - \{0\}$ satisfying $F(q(t)) = m q''(t)$ for some $m > 0$ such $q(\mathbb{R})$ does not lie in any affine 2-dimensional subspace of $\mathbb{R}^4$?

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  • $\begingroup$ Are you thinking of 4 spatial dimensions and time? I think so, since this is newtonian and not relativistic. The correct generalization of the cross product in n-dim is the wedge product acting on diff-forms, or dual vectors. To fully specify the question you would need to employ that machinery. You could also set up the lagrangian in 4-d using spherical coords (r, theta, phi, psi) and look at the equations of motion. $\endgroup$ – ggcg Dec 6 '19 at 22:41
  • $\begingroup$ This is explained for arbitrary dimensions in my Phys.SE answer here. Related: physics.stackexchange.com/q/9864/2451 $\endgroup$ – Qmechanic Dec 7 '19 at 6:11
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In general, the angular momentum is defined as $\mathbf{L}=\mathbf{r}\wedge\mathbf{p}$. In our problem, switching to index notation, we have, in the CM frame, $$L_{\mu\nu}=r_\mu p_\nu-r_\nu p_\mu=\mu(r_\mu\dot{r}_\nu-r_\nu\dot{r}_\mu)$$ Now, since there is no external torque, angular momentum is constant. If we define a Cartesian coordinate system $(w,x,y,z)$ such that the initial position and velocity vectors are coplanar with the $wx$-plane associated with, then our angular momentum tensor would look like such: $$L_{\mu\nu}=\begin{pmatrix}0 & \mu(r_{w}(0)\dot{r}_x(0)-r_x(0)\dot{r}_w(0)) & 0 & 0\\ \mu(r_x(0)\dot{r}_w(0)-r_{w}(0)\dot{r}_x(0)) & 0 & 0& 0 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \end{pmatrix}$$ because the $y$ and $z$ components of both vectors are zero. However, because angular momentum is conserved, each element of this tensor is conserved. Therefore, the vectors never leave this plane. This argument can be extended to any number of dimensions.

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    $\begingroup$ Ah! I think it's clear now. So essentially, differentiating $L$ works out the same way as differentiating $\omega$, so you get the conservation of $L$, which is the 2-plane spanned by $r, p$, so movement is confined to that plane. $\endgroup$ – Pedro Dec 7 '19 at 3:11
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Tesseract's answer is correct. More conceptually, in general dimensions the angular momentum is what is known as a "two-form", which you can think of a plane with a magnitude and orientation of in-plane rotation. This works in any dimension, and in any dimension the conservation of this plane means that motion under a central force is confined to that plane.

In three spatial (Euclidean) dimensions only, there exists a natural duality mapping from such a two-form to an ordinary "arrow-type" vector (technically a pseudovector), given by the unique vector that is perpendicular to the plane, with an orientation given by the right-hand rule and a magnitude equal to the magnitude of the two-form. This mapping, known as the Hodge dual, only works in three spatial dimensions (although there is a generalization that works in arbitrary dimensions). We are implicitly using it when we think of the angular momentum as an "arrow-type" pseudovector in three spatial dimensions.

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  • $\begingroup$ Wouldn't you mean a 2-vector rather than a 2-form? $\endgroup$ – Pedro Dec 7 '19 at 3:07
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    $\begingroup$ @Pedro Nope, I mean a two-form. You can take a wedge product of forms but not vectors. $\endgroup$ – tparker Dec 7 '19 at 3:11
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As answered elsewhere, yes, the motion is confined to a plane.

For me, analogy to cross products is less intuitive in higher dimensions. What makes it planar? Force dictates the second derivative. You always have a second order differential equation. All solutions to "nice" second order differential equations are defined by two functions. Below is a kinda-rigorous proof in $n$ dimensions:

Suppose you have the functions $F(\vec r)$ and $\vec q(t)$ that you described:

$F(\vec r)$ is parallel to $\vec r$. That means $F(\vec r) = C(\vec r) \vec r$ for a scalar function $C(\vec r)$.

So $F(\vec q(t)) = C(\vec q(t))\vec q(t)$ where $C(\vec q(t))$ is really just a function of $t$.

So $C(t)\vec q(t) = \vec q ''(t)$. (From the momentum eq. Ignore $m$.)

Since $C(t)$ is a scalar, you have the same differential equation for all four dimensions. Also, the differential equation is second order and homogeneous and "nice". That means the solution will be defined by at most two functions, weighted by constants. That is, it will have the form $q_k(t) = A_kf_1(t) + B_kf_2(t)$ in each dimension $k$. The $A$'s and $B$'s may change across the dimensions, but $f_1$ and $f_2$ are scalar functions - the same in all dimensions.

Put those equations together and make vectors $\vec A$ and $\vec B$:

$\vec q(t) = f_1(t)\vec A + f_2(t)\vec B$

$\vec A$ and $\vec B$ define the plane, and QED!

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A more mathematical way to think of it is that there are two relevant vectors: the initial velocity $\vec v$ and the initial acceleration $\vec a$. These span a two dimensional subspace (regardless of the number of dimensions in the full space). The solution will stay within this subspace.

There is the degenerate case when $\vec v$ and $\vec a$ are not independent and the subspace will be only one dimensional. This is the case when the object just falls directly into the centre.

Of course, there is the even more degenerate case in which both $\vec v$ and $\vec a$ are zero and the object just sits stationary at the centre.

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