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Let's say $r \times v = const$ and $r \times a = 0$. We should prove that the particle moves in a plane.

I thought to approach it from a differential equation perspective, where we know from the second equation that $a= r'' = kr$, which means that $r(t) = c_1e^{kt}+c_2e^{-kt}$.

If $k$ would not be a scalar, we would get that $v$ is a scalar, so it seems wrong, and the only solution is of the form $r(t) = (t, c_1e^{kt}+c_2e^{-kt})$.

Is there a way to prove that with differential equations, or should I stick instead to a different method (e.g. like here)?

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    $\begingroup$ Maybe you can try to compute the torsion of the curve described by the particle, (a curve is contained in the plane if and only if its torsion is 0) en.wikipedia.org/wiki/Torsion_of_a_curve $\endgroup$
    – 196884e
    Dec 11, 2023 at 8:29
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    $\begingroup$ The second assumption only tells that $k$ is some scalar function, one can not conclude that $k$ is constant. $\endgroup$ Dec 11, 2023 at 10:20
  • $\begingroup$ @LutzLehmann doesn't scalar $k$ imply scalar $r$ in that case? $\endgroup$
    – S11n
    Dec 11, 2023 at 12:42
  • $\begingroup$ Why? You can multiply vectors with scalars from the basis field. $\endgroup$ Dec 11, 2023 at 14:35
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    $\begingroup$ No, I mean $k$. You only know that $a=r''$ is a multiple of $r$, but nothing forces that to be a constant multiply, that the factor is a constant. One example is the gravity equation $r''=-\frac{r}{|r|^3}$. $\endgroup$ Dec 11, 2023 at 16:34

2 Answers 2

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Let $ r \times v =b$ ( constant vector).

Note

$ \left( r \cdot b \right)' = \left( r \cdot ( r \times v ) \right)' = r' \cdot ( r \times v ) + r \cdot ( r' \times v ) + r \cdot ( r \times v' ) = v \cdot ( r \times v ) + r \cdot ( v \times v ) + r \cdot ( r \times a ) = v \cdot ( r \times v ) + O + O =O. $

Hence $r \cdot b $ is a constant vector. Let $r_0 =r(0)$ and we have $r \cdot b = r_0 \cdot b$.

i.e.
$(r-r_0) \cdot b = 0$. This equation means that the particle moves in a plane which contains $r_0$ with a normal vector $b$.

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    $\begingroup$ You can get this faster and more precise, $r\cdot b=r\cdot(r\times v)=\det((r,r,v))=(r\times r)\cdot v=0$, so the plane is actually the subspace orthogonal to $b$. Additionally this first equation already implies with its derivative $0=v\times v+r\times a$, so that the second assumption is a consequence of the first. $\endgroup$ Dec 11, 2023 at 10:19
  • $\begingroup$ Thanks for the answer! Isn't $r \cdot (r \times v) = 0$ always? I think we should actually have 2 different $r$ there, and strictly speaking it is incorrect to expand the $b$ the way you did, because if we do have only one $r$, we don't get just the derivative of 0, but also the scalar product itself. However, the idea is the same. $\endgroup$
    – S11n
    Dec 14, 2023 at 7:40
  • $\begingroup$ $r$ depends on $t$ but $b$ does not. $\endgroup$
    – Basics
    Dec 14, 2023 at 10:05
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If $r = (x,y,z)$ is contained in a plane then

$$ \cases{a x + b y + c z = 0\\ a \dot x +b \dot y+ c\dot z = 0\\ a^2+b^2+c^2=1} $$

now solving for $a,b,c$ we have

$$ \cases{ f(t)a = z \dot y-y \dot z\\ f(t)b = x\dot z-z\dot x\\ f(t)c = y\dot x-x\dot y } $$

which is equivalent to the condition

$$ r\times \dot r = C f(t) $$

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  • $\begingroup$ Thanks! However, I don't see why the first set of equations is sufficient condition for the curve to be in plane. $\endgroup$
    – S11n
    Dec 12, 2023 at 14:25

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