# Can soneome help me understand this part of the Feynman's Lost Lecture?

Feynman's lost lecture

So at 17:38, 3Blue1Brown states:

We know that once the planet has turned an angle $$\theta$$ off the horizontal with respect to the sun, that corresponds to walking $$\theta$$ degrees around our circle in our velocity diagram, since the acceleration vector rotates just as much as the radial vector.

I'm not sure I quite follow the logic here.

Yes, I can see that the acceleration vector rotate just as much as the radius vector since the sun exerts its force along the radius vector. But what I don't understand is how that translate into saying that the when the position of the planet has turned an angle of $$\theta$$ with respect to the sun, that means that the "that corresponds to walking $$\theta$$ around our circle in our velocity diagram", and certainly not how that can be derived from the fact that the acceleration vector rotates just as much as the radius vector.

I think specifically, I'm not sure the meaning of this sentence "walking $$\theta$$ degrees around our circle in our velocity diagram".

Acceleration is the change in velocity (per unit time). Since the velocity vectors trace out a circle in velocity space, the acceleration vectors are tangent to this circle. Therefore, they are always perpendicular to the 'radii' of the circle, so if the acceleration changes an angle $$\theta$$, the 'radius' also changes by the same angle.

In this post, I discuss Feynman's geometric proof. You can see that if the changes in velocity ||$$\Delta \boldsymbol{v}$$|| are kept constant, they trace out a polygon, which in the limit becomes a circle.

So if ||$$\Delta \boldsymbol{v}$$|| changes by a constant angle $$\Delta\theta$$, the radius in yellow also changes by $$\Delta\theta$$ (see the post for more detail).

In another post I use a similar idea to solve Kepler's problem with ordinary calculus. It is easy to see that

$$\boldsymbol{a} = -\frac{k}{r^2}\boldsymbol{e}_r = -\frac{k}{h}\dot{\theta}\boldsymbol{e}_r = \frac{k}{h}\dot{\boldsymbol{e}}_\theta = \dot{\boldsymbol{v}}_\text{c},$$ where $$h = r^2\dot{\theta}$$ is the magnitude of angular momentum per unit mass, $$\boldsymbol{e}_r$$ and $$\boldsymbol{e}_\theta$$ are the orthonormal basis vectors in polar coordinates, and $$\boldsymbol{v}_\text{c} = \frac{k}{h}\boldsymbol{e}_\theta$$ is a 'circular velocity', corresponding with the radius of the circle in velocity space. Again, since $$\boldsymbol{e}_r \perp \boldsymbol{e}_\theta$$, $$\boldsymbol{a} \perp \boldsymbol{v}_\text{c}.$$

• Thanks for you very detailed response. In hindsight, I think I confused "the change in velocity" with "the change in the direction of the velocity":) As a sidenote, thanks for showing me the Hall&Higson lecture notes. I would really want to do the exercises there but I don't do exercises that don't have answer keys. Do you happen to have the answer keys so that I can see whether I got the right answers? Commented Jun 9, 2021 at 14:20

On the right diagram, there are 4 sections, each section is an equal angle of 15 degrees, so 60 degrees around the orbit.

On the left velocity diagram, it said earlier that the tips of the velocity vectors are equally spaced, so 4 of those sections is a fraction of 4 out of 24 'steps' around the circle or 60 degrees again.

True, the mention of 'acceleration vector' was not helpful on the video.

• check out time 18:14. I respectfully dieagree. What you said may be true of a circular motion, but probably not an eclipse. For if that is the case, at 18:14 theta and theta? would be the same. Commented Jun 8, 2021 at 9:47
• The link didn't connect, if it does connect I'll have a look and try and improve the answer or delete if it doesn't help Commented Jun 8, 2021 at 9:49
• Would you check it again. Hopefully I fixed it already Commented Jun 8, 2021 at 10:40
• video now seen and the answer has been edited Commented Jun 8, 2021 at 12:10