Geometric Algebra Applications - Kepler Problem (Part 3)
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- Опубликовано: 5 фев 2025
- In this video, we will finish our discussion of the Kepler problem by deriving Kepler's 3rd Law, total energy, the Vis-Viva equation, and the Virial Theorem. Along the way, we'll review some facts about the ellipse and observe the total energies for the different conic sections.
References / Further reading:
References / Further Reading:
1. adsabs.harvard....
2. Lasenby and Doran's "Geometric Algebra for Physicists"
3. en.wikipedia.o...
4. en.wikipedia.o...
5. en.wikipedia.o...
6. en.wikipedia.o...
This treatment of the Kepler problem is beautiful, but we can also use something called eigenspinors in geometric algebra to reduce the Kepler problem to a simple "harmonic oscillator" equation. Powefull!
Wow, nice!
Wow. That was excellent. I love how the conic sections featured. Thanks!!! Got a quiet introduction to the Geometric Calculus too!!
Thank you!
Please continue the series.
Looking forward to 3D Geometric Algebra!
+Jo Reven
Exporting now...
At 23:45 it is mentioned a hyperbola is an orbit that it is not bound. But I'm really struggling to grasp how could it be. It doesn't seem physical to me, does it take one of the paths only? Or what is going on when the orbit goes to infinity and then just comes back from the other side? Doesn't make any sense to me.
I don't know a rigorous mathematical proof that could help off the top of my head, but since this is physics allow to leverage some intuition from other cases and the physical context to give a sense of why I don't intuitively think that the hyperbolic orbit is particularly unphysical.
Think about how an elliptical orbit works physically. Setting aside that there is an an actual maximum speed IRL (because that relies on special relatively which is distinct from Euclidean 2d geometric algebra), my intuition is that the time needed to complete an elliptical orbit increases with the size of the orbit. In fact, that's what he proved earlier isn't it when he showed that the time is proportional to the 3/2 power of the semimajor axis. As the semimajor axis a goes to infinity, the time T also goes to infinity. At a parabola it "reaches" infinity, and a hyperbola is even further past that.
Therefore, parabolas and hyperbolas could be seen as an "orbit" whose period is "infinity". Might sound a bit like a semantic game, but both statements actually contain relevant information. It is an orbit because it is a path that is determined by the gravitational interaction, and the period is infinity because it will not loop back around (in this model due to gravitational effects).
We also don't run into any issues with Energy, as since 2E=(μ^2/h^2)(1-e^2), and since μ, h, and e are all well defined we run into no mathematical issues. Furthermore, if the energy equation actually works when describing IRL situations which are modeled by hyperbolic orbits (I suspect it corresponds to when planets are flung out of their elliptical orbits), then that would be further proof that we should be thinking about circular, elliptical, parabolic, and hyperbolic paths similarly.
On the topic of which path the physical object actually takes, well it can only be on one point on an orbit, let's call that point p1. The path it takes will be determined by where it is on the hyperbola and the velocity it is traveling at (flip the sign to go the other way around).
Now, I am gonna get more intuitive here so bear with me and grain of salt. I'm gonna think about it sort of like projective geometry for a second to address your question about whether you can get from one side of the orbit to the other with a hyperbolic orbit.
Imagine we have a point at infinity that we call p' which connects the two paths of the hyperbola, meaning we sort of have a "closed path" in some sense. So then, what path can we take to get from p1 to a point "on the other side of the orbit" p2? Hard to say, but I can make an guess.
To get from a point on an elliptical orbit which is more than halfway around the ellipse the long way, passing through the point which intersects with the negative side of the x axis, requires a time T greater than or equal to T/2. This is because the time to get from one point on an ellipse one way (call it t1) and the opposite way (call it t2) should together cover the whole ellipse (t1+t2=T), and we are taking the long way around so t1>t2. Well, as we take the limit as the semimajor axis a goes to infinity, T/2 is also going to go to infinity, therefore the time it takes to get from a point on one side of the hyperbola to the other can be said to be infinite. Thus, for parabolic and hyperbolic orbits we can never see an object get from one "side" of the orbit to the other.
If that helped let me know, and if I am mistaken in my reasoning or conclusions here I totally welcome corrections.
Wow what a video , what a channel you're awesome.
Very nicely done. What's interesting to me is that there is nothing special about m1 or m2 so that from the point of view of the earth, the sun moves in an identical elliptical orbit around the earth, (just backwards), is that right? What would the derivation look like if the origin was the center of mass of the two objects? The direction of r would be the same but for two identical masses I would imagine the radius would be half of that described here. Perhaps I will try my own derivation using these videos as hints.
+Marco Dalla Gasperina
Yes, taking one to be stationary or the other to be stationary both result in the same interplanetary vector up to a sign flip. That also means the same conic section must be swept out taking either one to be the stationary planet. The barycentric (center of mass) approach is something you should look into. As a hint, the barycenter has to be on the line joining the two masses and that fact alone tells you a lot already but you can work out where the barycenter has to be as a function of m1 and m2.
Taking the center of mass as the origin effectively yields the same derivation with a different mu. My intuition would tell me that 'e' would be the same and 'p' would be scaled by the ratio of the gravitational parameters. The period would need to be the same.as well. I will try and prove that to myself.
I was right about the 'e'. It doesn't change based on changing the origin to the center of mass. I was wrong about the SLR though. Using the center of mass, it shrinks by a factor of m1/(m1 + m2). This makes perfect sense now that I think about it.
So, am I correct in assuming that the hyperbolic angle is related to eccentricity? is it exp(HypAng * ε) where ε²=0 ?
Example: consider two concentric circles, a big circle C and a little circle c, and an ellipse that is tangent to both of them call it "z".
If a circle is a plane that slices the cone perpendicular to the cone's axis of revolution, then the two concentric circles correspond to two parallel planes intersecting the cone at different 'heights'.
And if the ellipse "z" is formed by the plane at an angle that intersects the parallel planes at the same points those planes contact the cone, well heck, that's the ellipse tangent to both circles.
In orbital mechanics, this is called a Hohmann Transfer orbit. How to extract the change in velocities of the object by the change in the ε-angle?
Great videos. But, hey, every time you multiply something by a negative number without putting the minus sign inside some parentheses, it makes me cringe. In this video, it happens at 29:29. Honestly, this is a minor detail, but it produces a written equation with the wrong mathematical meaning. If I show the equation as it is written at 29:34 to anyone, I bet it will be interpreted as a subtraction, not as a multiplication. I'm sorry, your videos are excellent, but this detail is distracting.
+Doutsoldome
A fair criticism. I'll watch out for that in the future.
Thanks for the reply.
But don't get me wrong - I really like your videos. Very clelar and precise.
I discovered geometric algebra and calculus recently, and I think it's fascinating. (I'm a physicist.) Perhaps in the future it will become more mainstream.
Keep up the good work!
+Doutsoldome
Well make sure you point out the flaws where they exist, since it'll confuse someone eventually if it's not noticed. I'm merely an amateur with this stuff (in the true sense of the word 'amateur') so I don't really have any stake in whether geometric algebra/calculus becomes mainstream. Even if it doesn't, it's still knowledge.
It does, however, seem to expedite algebraic manipulations without having to use cross products, which are rather artificial, mathematically speaking. In fact, this Kepler problem video is one of my favorites because of the relative ease of getting to the equation for a conic section from the inverse square law, which I wasn't taught in college physics, even though it's an important result in the history of science.
@@Math_oma it is becoming mainstream. I'm, as a mechanical engineer, very keen on learning it fully. Mainly because it simplifies all of classical mechanics.
thank you