# (Why) Is orbital angular momentum conserved for point masses?

The introduction of the angular momentum as $$\vec l = \vec r \times \vec p$$ is also true for point particles. So $$\vec l$$ must refer to the orbital angular momentum (and not the "spin") in such a case.

If I have a particle moving straight with constant velocity and then I apply a force $$\vec F$$ perpendicular to the moving direction, I created a torque $$\vec M = \vec{r} \times \vec{F}$$ that creates the orbital angular momentum $$\vec l \neq 0$$. Now conservation of angular momentum says that $$\vec l$$ stays the same when the torque vanishes.

But wouldn't this mean that the point mass has to fly an endless circle without any force? This is in total contradiction to Newton's laws that a particle moves a straight line without force. Why isn't angular momentum conserved in this case? What actually is angular momentum, then?

• Have you tried calculating the angular momentum for a point particle moving at constant velocity along a straight line, which does not pass through the origin of your coordinate system? Doing so may allow you to answer your own question. May 3 at 15:11
• Hint: angular momentum may also be calculated and conserved for straight line motion of particles. Once your torque vanishes, so does the curvilinear path. May 3 at 16:53
• @KshitijKumar But why is it "conserved" then? It's obviously not conserved if it vanishes. If I turn off a force the momentum stays like it is at the momemt of turning off the force. This is called conservation of momentum. So conservation of angular momentum should mean "if I turn off the torque, then the angular momentum should stay like it is in the moment of turning off" - doesn't this mean the particle must fly on a curve? That's exactly my question. May 3 at 18:16

Imagine you had a particle traveling at constant $$\mathbf{v_0} = \dot{z_0}\mathbf{\hat{k}}$$ upwards. At time $$t = 0$$, your particle is at $$\mathbf{r} = z_0 \mathbf{\hat{k}}$$ and you apply an horizontal force $$\mathbf{F} = -F\mathbf{\hat{i}}$$, constant $$F$$, for time $$\Delta t$$.

The particle was subect to a force, so its velocity was at every instant of application of the force: $$\mathbf{v} = \mathbf{v}_0 + \Delta \mathbf{v} = \mathbf{v}_0 + \int_0^t\frac{d \mathbf{v}}{dt} dt = \mathbf{v}_0 + \int_0^t \mathbf{a} dt = \mathbf{v}_0 + \int_0^t \frac{\mathbf{F}}{m} dt$$ but the last integral really is just Newton's second law and: $$\int_0^t \frac{\mathbf{F}}{m} dt = -\frac{F}{m} t \mathbf{\hat{i}}$$ and the final speed: $$\mathbf{v}(t) = \dot{z_0}\mathbf{\hat{k}} -\frac{F}{m} t \mathbf{\hat{i}} \Longrightarrow \mathbf{v}(\Delta t) = \dot{z_0}\mathbf{\hat{k}} -\frac{F}{m} \Delta t \mathbf{\hat{i}} = \mathbf{v}'.$$ Its position is also given by integrating the last equation: $$\mathbf{r}(t) = \mathbf{r_0} + \Delta \mathbf{ r} = (z_0 + \dot{z}_0 t)\mathbf{\hat{k}} - \frac{1}{2}\frac{F}{m}t^2 \mathbf{\hat{i}} \Longrightarrow \mathbf{r}(\Delta t) = (z_0 + \dot{z}_0 \Delta t)\mathbf{\hat{k}} - \frac{1}{2}\frac{F}{m}(\Delta t)^2 \mathbf{\hat{i}} = \mathbf{r'}$$ After the torque/force ends, it stops accelerating and therefore just travels rectilinearly through space with $$\mathbf{v}'$$. In the same coordinate system, you can check that angular momentum is, after calculating speed and position (reset the clock, so we can start at $$t = 0$$ again, that's licit, we "stored" everything that happened before on the $$\Delta t$$s, $$\mathbf{r}'$$ and $$\mathbf{v}'$$): $$\mathbf{r}(t) = \mathbf{r}' + \mathbf{v}' t$$ $$\mathbf{v}(t) = \mathbf{v}'$$ $$\mathbf{L}(t) = \mathbf{r}(t) \times \mathbf{p}(t) = m \mathbf{r}(t) \times \mathbf{v}(t) = m(\mathbf{r}' + \mathbf{v}' t) \times \mathbf{v}' = m\mathbf{r}' \times \mathbf{v}'$$ notice that the primed vectors are constant, so this new angular momentum also is. So you really get rectilinear motion and constant angular momentum in the end.

Angular momentum(and torque) does not necessarily mean that there are rotations in place. you can define angular momentum for straight lines, however, that isn't very useful to solve any problem.

If you have a particle moving in a straight line and apply a force perpendicular to it, it won't rotate unless there is a centripetal force somewhere. In your exemple, if you only apply a perpendicular force, the particle will simply move in both x and y directions in a parabola.

What will happen is that, say, you have a particle moving in a straight line in the x direction. The angular momentum is: $$r \times p_x$$. Then you apply the perpendicular force, adding a y component into the velocity. Since the force is perpendicular only, then the linear momentum is conserved in the x direction at all times, thus

$$\frac{dp_x}{dt} = 0$$

The angular momentum while applying the force is, thus: $$r \times (p_x + p_y)$$. Where $$p_x$$ is constant and $$p_y$$ is varying due to the force, the torque is: $$r \times \frac{dp_y}{dt})$$

Once the torque vanishes, no forces act in the y direction, thus

$$\frac{dp_y}{dt} = 0$$

And then the particle continues in a straight line.Take the time derivative of angular momentum after the torque vanishes: $$\frac{dL}{dt} = v \times (p_x + p_y) + r \times (\frac{dp_y}{dt} + \frac{dp_y}{dt})$$

The first term vanishes because the velocity vector is always parallel to the linear momentum, then you are left with the second, but both time derivatives are zero, thus:

$$\frac{dL}{dt} = 0$$

What does this mean? Nothing, because there is no rotations, however it's still a true result to say that angular momentum is conserved in absence of torques.

For a particle of mass $$m$$ with velocity $$\vec v$$, with respect to a point $$O$$ the angular momentum is $$(1) \vec l = \vec r \times m\vec v$$ where $$\vec r$$ is the vector distance from $$O$$ to the particle. The torque due to a force $$\vec F$$ on the particle is $$(2) \vec N = \vec r \times \vec F$$ The change in the angular momentum is $$(3) {d\vec l\over dt} = \vec N$$ The angular momentum and torque depend on the point taken as $$O$$. For a particle with no applied torque, the angular momentum is constant.

Using (2), the magnitude of the torque is $$(4)rFsin(\theta)$$ where $$\theta$$ is the angle between $$\vec r$$ and $$\vec F$$. With no force there is no torque; for example a particle with constant $$\vec v$$. A force parallel to $$\vec r$$ produces no torque; for example a particle in uniform circular motion under a centripetal force.

For a particle moving in a straight line under no force, the velocity $$\vec v$$ is constant. There is also no torque on the particle, so the angular momentum is also constant. Specifically, from any point $$O$$, the angular momentum is constant, since $$\vec l = \vec r \times m\vec v$$ has constant magnitude $$r_{perpen}mv$$ where $$r_{perpen}$$ is the constant component of $$\vec r$$ perpendicular to $$\vec v$$.

For a particle moving at constant speed in a circle, there is a force, the centripetal force (provided by gravity, or tension in a string, etc.) The direction of the centripetal force is radially inward towards $$O$$ at the center of the circle. The velocity $$\vec v$$ is always tangential to the circle. Although there is a force, there is no torque about $$O$$ using (4), since $$\vec v$$ and $$\vec r$$ are always anti-parallel and $$sin(\theta) = 0$$. The angular momentum is constant, equal to $$rmv = mr^2\omega$$ where $$\omega$$ is the constant angular speed.