# What's the angular momentum of a two point system of unit mass, velocity and separation when the points are orthogonal and after time t?

Question

Let the universe comprise two points travelling orthogonally with separation distance $$1$$, speed $$1$$. In the absence of gravity, what is the angular momentum of one around the other at time $$0$$ and after time $$t$$?

Discussion

As time goes on, the line between them moves away from being orthogonal to the motion, thereby reducing the angular velocity. But counter to this effect, their separation increases which increases the radius of rotation, thereby increasing the moment of intertia. Does either effect dominate or are they balanced? As I understand it, the claim in the comments here requires that they be perfectly balanced. Is this correct?

Attempt

About 35 years since I did this but I think angular momentum as they travel orthogonally is $$I\omega=\dfrac1{2\pi}$$

Then after time $$t$$ I have $$I\omega=mr^2\dfrac{d\theta}{dt}$$

$$\dfrac{d\theta}{dt}=\dfrac{v\sin\theta}{2\pi}$$

and $$\sin\theta=\dfrac1{\sqrt{1+t^2}}$$

So that gives $$I\omega=\dfrac{\sqrt{1+t^2}}{2\pi}$$

And that all appears to check out nicely because it generalises the original expression for $$t=0$$. (Boom I've still got it).

Is that all correct?

• Shouldn't the answer be independent of $t$? Dec 1, 2022 at 12:21
• Can you be more specific about the initial locations and positions? Perhaps by giving initial $(x,y)$ coordinates and velocity vectors? I'm having trouble envisioning your setup. Dec 1, 2022 at 12:28
• @garyp well that's what the linked coment claimed but I did the maths and got a different result, so I'm hoping that asking this will shake out the truth. HINT: assuming (and that's a big assume) I've not made a mistake, one obvious resolution is that okay, the two must then be rotating RELATIVE TO SOMETHING ELSE (e.g. the rest of the universe) but this highlights the fact that when you start talking about the "whole universe" there isn't some dominantly-massive "heavens" to put to one side as a point of reference. Dec 1, 2022 at 12:29
• @MichaelSeifert assume 2d for now then $(0,0)$ and $(0,1)$ are the locations and velocity vectors $(\theta,s)$ are $(0,1)$ and $(0,0)$ respectively Dec 1, 2022 at 12:32
• @garyp yes, I think in classical mechanics it must sum to zero. But when you look at this two body universe that is immediately problematic because it then implies the universe is rotating, so the obvious question will be - "with respect to what"? So this question is a simplified state of the earlier question. The earlier question will probably add light to this one because by adding the third body it is designed to enforce a non-rotating universe. Dec 1, 2022 at 12:38

I'm a little confused by your notation in the comments, but I'll put up an answer showing how it's supposed to work, and I can try to edit it if it's not quite the situation you're considering.

Suppose at $$t = 0$$ we have particle 1 at rest at $$(0,0)$$ and particle 2 at $$(1,0)$$ moving with $$\vec{v} = (0,1)$$. This means that at a later time $$t$$, the vector pointing from 1 to 2 is $$\vec{r} = (1,t)$$.

The angle between $$\vec{v}$$ and $$\vec{r}$$ is given by $$\tan \theta = t$$. To see this, draw a right triangle with vertices at particle 1, particle 2, and $$(1,0)$$; the legs have lengths $$1$$ and $$t$$ and the hypotenuse has length $$r = \sqrt{1 + t^2}$$. If we differentiate both sides of our equation for $$\theta$$ with respect to $$t$$, we have $$\frac{1}{\cos^2 \theta} \frac{d\theta}{dt} = 1.$$ But $$\cos \theta = 1/r$$ from that same triangle, and so we have $$\omega = \frac{d\theta}{dt} = \cos^2 \theta = \frac{1}{r^2}.$$ This means that the angular momentum is $$L = m r^2 \omega = m r^2 (1/r^2) = m$$, which is constant.

• Hmmmn... looks like I haven't still got the old skills after all! I'm sure you have understood my specification so I must have made a mistake so I'll check through for it. Dec 1, 2022 at 13:04
• ok I had speed of $1$ can be expressed as the sum of two vectors - $\frac{1}{\sqrt{1+2^2}}$ orthogonal speed and $\dfrac{t}{\sqrt{1+t^2}}$ radial speed. Then the orthogonal part divided by the circumference $2\pi$ is the angular velocity, that's $\frac{1}{2\pi\sqrt{1+t^2}}$, right? Dec 1, 2022 at 13:22
• Ah yes, first error found. Circumference at that time is $2\pi\sqrt{1+t^2}$, not $2\pi$ Dec 1, 2022 at 13:24

In the absence of gravity place a coordinate system in the middle of the line that separates the two paths. Each line (path of object) is a fixed distance $$h$$ from the coordinate system At least I think this is the setup you are describing.

Now angular momentum for one particle summed at the origin is

$$\boldsymbol{L}_0 = \boldsymbol{r} \times m \boldsymbol{v}$$

which results in the magnitude

$$L_0 = h\,m v$$

regardless of where along the path $$\boldsymbol{r} = \boldsymbol{h} + \boldsymbol{v}\,t$$ the particle is, since any part of $$\boldsymbol{r}$$ that is parallel to the velocity $$\boldsymbol{v}$$ is ignored by the cross product.

Now the same angular momentum can be used in $$L_0 = I_0 \omega$$ to find the apparent rotational speed that each particle has. Here $$I_0 = m r^2 = m ( h^2 +(v\,t)^2 )$$ and so you have

$$h\,m\,v = m ( h^2 +(v\,t)^2 ) \omega$$

which is solved for $$\omega = \frac{h^2}{h^2 + (v\,t)^2}$$