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Just before proving Kepler's laws, my Professor claimed that if $\vec{F}$ is a central force with center $O$ and it is the only force acting on a point $P$, then the trajectory of $P$ is a curve plane. The proof of this lemma started from the fact that the angular momentum $\vec{L^O}$ is a constant of motion for central forces, so $\vec{r}\times m\vec{v} = \vec{r_0} \times m\vec{v_0} $. Then he stated that

therefore the motion must take place on a plane since $\vec{L^O} = \vec{L_0^O}$

and also that if $\vec{r_0} \parallel \vec{v_0}$ then the motion actually takes place on a straight line. Why are these statements true? Thanks in advance.

EDIT: I got the explanation of the first statement, but the second claim is still not clear to me. I would like to see that $\vec{r} \parallel \vec{r_0}$, so that clearly the motion takes place on the straight line parallel to $ \vec{r_0}$. In principle, $\vec{r_0} \parallel \vec{v_0}$ and $\vec{r} \parallel \vec{v}$, so that $\vec{L^O_0} = \vec{L^O} = 0$, say nothing about the relation between $\vec{r}$ and $\vec{r_0}$.

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    $\begingroup$ A central force, by definition, only varies with distance. So the force is not affecting any angular momentum and the angular momentum is therefore conserved. $\endgroup$ Commented Jul 11, 2024 at 21:05
  • $\begingroup$ @DavideMasi FYI, This is the correct explanation. The complete derivation which I wrote out the whole derivation on your math stackexchange post follows exactly this logic! Hopefully, you can use that to fill in the details for why that comment was correct. $\endgroup$ Commented Jul 12, 2024 at 21:11

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The definition of a plane can be written as $\vec{a}\cdot \vec{r} = 0$, where $\vec{a}$ is any vector perpendicular to the plane.

In this case, you have a vector quantity $\vec{L}$, which is from its definition, a vector perpendicular to a plane, since $(\vec{r}\times \vec{v})\cdot \vec{r} = 0$.

Then, since angular momentum is conserved, the vector $\vec{L}$ does not change, even though the position does, and so the body must still be in the plane defined by $\vec{L}\cdot \vec{r}=0$.

If $\vec{r}_0$ and $\vec{v}_0$ are parallel, then the angular momentum is zero. If it cannot change then $\vec{r}$ and $\vec{v}$ must always be parallel. The only way this can be arranged is straight line motion towards or away from the origin since any deviation from that will result in a non-zero $\vec{L}$.

I see your edit. But $\vec{v} = d\vec{r}/dt$ so any change in $\vec{r}$ can only be in the same direction as $\vec{v}$ (and hence $\vec{r}$) - i.e. along a radial line. In other words $$ \vec{r}(t) = \vec{r}_0 + \int \vec{v}\ dt = \vec{r_0} + \alpha(t)\vec{r_0} $$ where $\alpha$ is a scalar that depends on the central force considered, the mass of the object and the initial velocity. $\vec{v}$ can never have a non-radial component if its initial value is radial, since a central force imparts no non-radial acceleration.

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  • $\begingroup$ You explained the first part very well, thank you. Could you please clarify the second one too? I don't get it intuitively nor does it look formal enough to me. If you wish, edit your answer and I will accept it $\endgroup$ Commented Jul 12, 2024 at 7:39
  • $\begingroup$ Please check my edit $\endgroup$ Commented Jul 12, 2024 at 10:22
  • $\begingroup$ Thanks for your patience, but as an aspiring mathematician I must admit that I am not convinced yet - I'd like a proof with symbols and not words since I find it hard to follow your logic reasoning. What I mean is that it seems quite obvious that the motion must take place on a straight line, but I'm still not sure about this fact $\endgroup$ Commented Jul 12, 2024 at 13:29
  • $\begingroup$ Could you please explain the tricky part $\int \vec{v}\ dt = \alpha(t)\vec{r_0}$? $\endgroup$ Commented Jul 12, 2024 at 14:15
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    $\begingroup$ $\vec{v} = d\vec{r}/dt$ but since $\vec{v}$ is always parallel $\vec{r}$ then we can say $\vec{v} = \beta \vec{r_0}$ where $\beta$ is a time-dependent scalar and its integral is $\alpha(t)$. $\vec{v}$ can never have a non-radial component if its initial value is radial, since a central force imparts no non-radial acceleration. @DavideMasi $\endgroup$ Commented Jul 12, 2024 at 17:29
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The angular momentum ${\bf L}$ of a point-like particle is proportional to the vector product of position ${\bf r}$ and velocity ${\bf v}$. It is a vector orthogonal to the plane containing ${\bf r}$ and ${\bf v}$. Therefore, the constancy of ${\bf L}$ implies that ${\bf r}$ and ${\bf v}$ are always in the same plane (if it wouldn't be so, the direction of ${\bf L}$ should change).

The special case ${\bf L}=0$ corresponds to the case of the same direction for ${\bf r}$ and ${\bf v}$. I.e., the motion is rectilinear.

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  • $\begingroup$ I'm still not sure why the trajectory should be on the plane of $\vec{r}$ and $\vec{v}$. I know it seems very intuitive, but I don't see how it should be proved formally. $\endgroup$ Commented Jul 11, 2024 at 21:33
  • $\begingroup$ Probably my problem is that I don't have a clear idea of what the definition or plane/rectilinear motion is and how this is related to the direction of $\vec{r}$ and $\vec{v}$. In one exercise we solved in class, which required to prove that a trajectory was an ellipse, we were able to find the equation of the trajectory and observe it was the equation of an ellipse, but here I don't have any equations so I don't get how it is possible to say that the trajectory is plane or rectilinear. Hope I was clear enough $\endgroup$ Commented Jul 11, 2024 at 21:38
  • $\begingroup$ Note that if $L\neq 0$, the plane of motion is simply the plane perpendicular to $\vec{L}$. $\endgroup$ Commented Jul 11, 2024 at 22:19
  • $\begingroup$ @DavideMasi, as the answer stated, if the motion is not fixed to a plane, that implies that angular momentum is changing (cross product produce a vector that is perpendicular to the plane two vectors form), which violates the conservation of angular momentum, meaning the motion must take place in a fix plane that is not changing, so angular momentum does not change direction. $\endgroup$ Commented Jul 12, 2024 at 0:31
  • $\begingroup$ @DavideMasi Equations are not a substitute for concepts. However, if you feel more comfortable with equations, a "constructive" formal proof goes as follows: in the case of ${\bf L} \neq 0$, the orthogonality of angular momentum and position (easily proved algebraically) implies the constraint on the trajectory ${\bf L}\cdot {\bf r}= constant$, which is the equation of a plane. $\endgroup$ Commented Jul 12, 2024 at 3:38

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