Geometric Algebra Applications - Kepler Problem (Part 1)

Transcript

what I'd like to do in this video is introduce you guys to a neat little application of geometric algebra called the Kepler problem now the Kepler problem asks the question what is the motion of two bodies because of gravity now what we're going to do first is set up the problem make it mathematically precise and discover the equation that we're going to have to solve then we're going to talk about two very important conserved quantities of motion first angular momentum and secondly energy let's get right into it so let's say I have two masses M1 and M2 what I'm going to do is first sub some vectors let me pick an arbitrary Point let's say right there going to draw a vector from that point to mass one let me call that displacement Vector R1 and I'm going to do the same thing for M2 I'm going to define the second displacement Vector as R2 and what I'm going to do is join the tips of these two vectors like so so this is going to be a vector from Mass one to Mass 2 pointing to Mass 2 I'm going to call this Vector R and hopefully you can you can convince yourself that this relative displacement Vector R is equal to R2 minus R1 just through simple vector addition to make this situation a little bit less abstract perhaps you can think of M1 as the sun and M2 as the Earth and R is just this relative displacement Vector pointing from the Sun to the Earth and where we're going to do is attempt to describe the evolution of this Vector R now the next thing I'd like to do is use new newon second law on M1 and on M2 to describe the accelerations of the vectors R1 and R2 so Newton Second Law is going to say that the acceleration of R1 which I'll notate as R1 double dot is equal to the force acting on mass one due to Mass 2 divid by mass one and we have a similar equation for R2 double dot we have the R2 dot is going to be the force acting on Mass 2 due to mass 1 / Mass 2 Now by Newton's third law we have a special relation between F12 and F21 we know that they're going to be equal and opposite so mathematically what that says is that the force acting on one due to two is equal to the negative of the force acting on two due to mass one what I'm going to do is substitute that right into this equation immediately and I get that's minus F12 over M2 by the way these dots up here are indicating time derivers as they usually do now what I'm going to do is I'm going to write down what F12 is so F12 this is a vector quantity so I'm going to have to tell you what the magnitude is I'm going to have to tell you what the direction is now F12 is the is given by the universal gravitational law and the direction for the force of gravity is going to be in the line joining the two masses so what I'm going to do is I'm going to define a unit Vector a vector of length one in the same direction as R I'm going to call R hat so the gravitational law says that the force acts in the same direction as our hat and the scaler out in front is going to be G M1 M2 over R 2 and R is going to be the length of this Vector R is going to be the distance separating the two bodies squared and G is universal gravitational constant since I know what F12 is and I have equations relating F12 and both R1 Dot and R2 dot I'm all set so let me substitute those right in so I have that R1 dble dot is this quantity divid by M1 and what that means is that that M1 right there is going to cancel so what I'm left with is that R1 dot is equal to G M2 because the M1 cancels divide r 2times r hat so what that's telling me is that the acceleration of the vector R1 is in that direction it's in the same direction as R hat it's going to be a scaled up version of R hat and let me do a similar thing with R2 double dot and I have the R2 double dot since I'm dividing by M2 this time the M2 is going to disappear here and I'm going to have to throw in a minus sign so what I'm left with is minus G M1 over R 2 time R hat now here notice that I have minus r hat so that means the acceleration of this Vector is going to be that way it's going to be in the opposite direction as our hat which makes sense right M1 is going to accelerate that way and M2 is going to accelerate this way next what I'm going to do is I'm going to return to R which is that relative displacement Vector R2 minus R1 I'm going to take two time derivatives and what I'll find is that taking two derivatives from both sides is that r do is equal to R2 dot minus R1 double dot so what that's saying is that the acceleration of this relative displacement Vector R is equal to the difference of these two accelerations here and since I know what these two are let me just subtract the two so I'm going to have to take this thing and subtract away this and what I'll find upon doing that subtraction and doing a little bit of factoring is that I'll get minus G * the sum M1 + M2 over R 2 * R hat and for my last little simplification what I'm going to do is just Define up a new constant which is going to be G * M1 M2 so what I'm going to do is I'm going to take out this G M1 M2 and I'm going to call that mu which is I guess called by people the gravitational parameter it's actually the sum of gravitational parameters of M1 and M2 but this will just be a new constant mu which is going to allow me to write this in a more notationally compact form as minus mu over r^ 2 * R hat and this is really the equation that we're going to be dealing with in the next few videos and we're going to be philosophizing about let's make sure we understand what this equation is saying about the acceleration of R it's saying that R double dot is some negative number times our hat and what that means is that this Vector R accelerates in a direction opposite to R hat so it accelerates that way which makes sense because from the perspective of M1 the force acting upon M2 is always in a direction toward M1 so if the forces act toward M1 by Newton Second Law the acceleration should also be in that direction as I said this is the main equation that we're going to be dealing with but let me put this equation on the back burner and talk about angular momentum for a little while let me first talk about how you normally talk about angular momentum let's say I had some mass m and I had some arbitrary point over here maybe some origin and I draw the displacement from the origin to the mass and I call that displacement Vector r as I have been doing let's say the mass is in motion and let's give it some velocity and I'll write that as r dot the first time derivative of R now what you normally do is you say let's have the angular momentum which I'll call L be equal to R crossed with the momentum which is the mass times r dot so you're going to take the cross product of this Vector with M times this vector and using the right hand rule what you on the beginning in this case is a vector pointing out of the page so this would be the vector L it's actually a pseudo Vector but we'll leave that aside and you might also write this as R cross with P where p is just m * r dot the linear momentum that's all well and good talking about this angular momentum pseudo Vector the only problem is is that if we want to talk about motion at a plane to start invoking this angular momentum concept because we're using the crossr it seems that we have to invoke a third dimension where it's not obvious that you need a third dimension to describe this angular momentum because remember the cross product only works in three dimensions it makes no sense to talk about a cross prodct in two Dimensions now what I'm going to do is talk about angular momentum from the point of view of geometric algebra what I'm going to do is I'm going to define the angular momentum which I'm going to call H because it's actually it's going to be the angular momentum per unit mass as you'll soon see so H is going to be defined as the wedge product between R and the velocity r dot so the wedge product between this vector and this Vector so remember what the uh pictorial interpretation of this is taking the wedge product of this Vector with this vector we draw in that parallelogram using those two vectors as the sides this is an oriented area the magnitude is given by the amount of area here and the orientation is going to be counterclockwise in this case if you check the direction if you check the order r and r dot but you can see this also describes this angular momentum concept just as well and there's no need to invoke any Third Dimension there's no you don't need to invoke some dimensional orthogonal to the plane and hopefully you can see that because we got rid of that M there that this is actually that this H thing is the angular momentum per unit Mass let's return to our gravitational problem and we'll start using this angular momentum concept and see what we get out of it remember that R is just that relative displacement Vector from M1 to R2 for example from the Sun to the Earth and r dot is just the velocity of that Vector R and let's uh let's explore a little let's take the derivative of H that by Vector so it's going to be DDT of R wedged with r dot and we see we have a product here so if we're going to take the derivative of the product let's just go ahead and use the product rule be a little bit bold here so I'm going to take derivative of the first Factor so it's going to be r dot leave the second Factor alone plus leave the first Factor alone take the derivative of the second so I get R double dot we have a vector wed with itself that's always zero so we're actually we're left with this uh R wedged with r double dot now remember what we were saying about this R double dot here if we consider that unit Vector R hat this equation is saying that R double dot is in the exact opposite direction as our hat so our double dot might look something like that so what happens when you take this R and wedge it with this Vector what's the area of the parallelogram swept out by those two vectors well that's zero so actually should we find that DDT of this quantity this angular momentum per unit Mass R wedged with r dot the derivative is equal to zero which is saying that this quantity is conserved throughout motion so R and the velocity can be changing over time but if you wedge these two together it's always going to give you the same B Vector so actually we have a conserved B vector quantity this is a famous physics physics concept called the conservation of angular momentum what I'd like to do now is tie in that conservation of angular momentum with a very famous physics fact called Kepler's second law which is that equal areas and equal times law so here's what I'm going to do we're going to consider that M2 and we're going to consider that relative displacement Vector R and what I'm going to do is I'm going to suppose this Mass has some velocity let's say the velocity is in that direction so that's r dot and what I'm going to do is I'm going to consider a an infinitesimally small amount of time I'm going to multiply that by r dot so what I get is actually a tiny Vector from here to there so this is going to be r dot time DT and this Vector describes how this Mass moves over the small time segment so it's the mass is going to move from here to approximately there and I'm going to join those two points and what I'm going to ask is what is the area swept out in that little time interval what's the area of that triangle well what I know is that if I were to take R and wedge it with r. DT I would get the area of the parallelogram which is going to look something like this so let me say that again I take the wedge product of r with r do DT and that gives me the area of the parallelogram so what that means is that the area of this triangle is actually 1/2 this quantity and I'm going to call this small change in area swept out over time da so this is a small amount of area given by 1/2 of this wge product and what I'm going to do is divide both sides by DT so I have the time rate of change of the area which I'm also going to call a DOT is given by 12 R wedge with r dot now what we just found is that this quantity R wedge with r dot was a conserved quantity it's just a constant which we'll just call H so we have that a DOT is equal to a constant so another way to visualize this is that if I were to make a graph of the area swept out in an orbit over time let's say this is time and this is a let's say at time equals zero you have zero area the plot of this thing since a DOT is a constant that means I have constant slope this plot is just a line and what this applies is Cap Second Law over equal times you sweep out equal areas no matter what the orbit actually looks like and I encourage you to think about this conservation of angular momentum at capar second law in the context of probably what you already know that the gravitational law generates circles and ellipses as examples of orbits so think about what this means for the way the velocity changes as you move closer to the Sun or further away from the Sun or you don't change your distance with respect to the Sun so we've seen that this angular momentum per unit Mass R wedged with r dot is a pretty useful Bor concept and what I'd like to do is do a little manipulation since we're going to need this little manipulation for future videos what I'm going to do is I'm going to remember that the unit Vector R hat is generated by taking the vector R and scaling down by its magnitude which I just called R I'm going to rearrange this and just say the vector R is equal to some scalar R time R hat I'm going to take this expression here and substitute it into that first factor in the wedge product so I have H is equal to our R hat wedged with r dot so I'm gonna take the derivative of this thing over here so it's a product so I'm going to have to use a product rule in this thing so what I have is r dot time R hat plus r times R hat dot now if you think about this expression this is just saying that the the way you can change this displacement Vector is move either in the radial direction or tangent ually now I'm going to take the wedge product here going to distribute this thing to both terms now notice here that I have our hat wedged with our hat and that's going to go to zero which means the only term that's going to survive after this wedging is going to be this thing wedged with this thing so what I'll have is R our hat wedged with r our hat dot what I'm going to do next is I'm going to move this scaler to the front so I have our s r hat wedged with r hat dot now I'm going to remember that the geometric product between R hat and R hat dot so let me write that over here R hat R hat dot is equal to the dot product between those two vectors plus the wedge product between those two vectors so I see the wedge product over here so if I just subtract both sides by this thing I can rewrite the wedge product as the geometric product minus the dot product R hat dot R hat dot now what I claim is that actually this second term is going to disappear these two things are going to dot to zero and we're going to be left with just this first term but let me show you why these two vectors are orthogonal why they dot to zero and this is going to be a little trick that we're going to to discover once again when we talk about energy so let me move that over here R hat dot with r hat Dot and here's the little trick I'm going to interpret this quantity as being a result of taking a derivative of another product so here's the Insight that this thing is 1/2 DDT of R hat dot with r hat that is I'm interpreting this thing as coming from a product rle actually one half of that now why is that if you take a look taking the derivative of this thing you get a DOT there and then in the second term you get a DOT there remember the dot product is commutative so actually those two sum to two times this quantity so that's why you have to scale down by 1/2 to set these two equal but what is r hat. r hat that's just equal to the magnitude squared of our hat now over time that magnitude doesn't change so that's always equal to one it's equal to one because it was that unit Vector so this is actually DDT of a constant namely one and that goes to zero so that implies that R hat do R hat dot is indeed zero those two vectors are orthogonal so that means I can discard that term and what I find is that H is equal to R 2times the geometric product R hat rhead dot so I think that's enough talk about angular momentum so I'd like you to remember that conservation of angular momentum concept and next we're going to talk about another conserved quantity namely the conservation of energy so let's return to this gravitational equation r dot is equal to minus mu / R 2 * R hat and here's what I'm going to do I'm going to take this equation and Dot both sides with r dot the velocity Vector so I'm going to take have R double dot dotted with r dot is equal to see scaler sitting out here times R hat dotted with r dot now what I'm going to do is I'm going to recycle that little trick that I employed uh for the other dot product I'm going to interpret this as2 the derivative of another product now what is this going to have to be it's actually going to be r dot dotted with r dot very similar logic as as before when you do the product r on these two you're going to get a DOT there in the first term second term you're going to get a DOT there but you can always flip that those two factors around in the dot product so you're actually going to get two times r dot dot with r dot so to set this equal to this I'm going to have to scale down by2 so it's a very similar little trick so I'm going to do that for the left hand side what I'm going to do for the right hand side I'm going to have to calculate this thing r dot. with r dot so let me do that over here so I have R hat dotted with remember what r dot was that was r dot time R hat just using that product rule plus r r hat dot now notice that this thing when it gets distributed to this is going to cancel out it's going to going to dot to zero because we found that R hat is orthogonal to R hat dot so I'll just take care of this first term so what I get is R hat times r hat. r hat remember that's just a magnitude squared of R hat R hat was a unit Vector so the magnitude squar is one so what we get out of that little dot product is just simply r dot so I'll write that in there the right hand side is equal to minus mu over r^ 2 time r dot now what I'm going to do let me return to this left hand side here again I have a vector dotted with itself that's going to be the magnitude squar of the Velocity Vector so just the magnitude squar of the Velocity Vector I'll call that whatever one else calls it which is v^2 and then what I'm going to do on the right hand side is I'm going to integrate it so I'm going to ask taking a derivative of what gives me this and what I'm going to have to do is integrate this 1/ R 2 thing and if you check this for yourself that this thing is equal to D over DT of mu over R now why is that you could just move that constant out ignore it for a second the derivative of 1 / R is equal to minus 1 r^ 2 but you're going have to use a chain rule so that our DOT is going to come out front so I just integrate that thing and I set the constant of integration equal to zero just for convenience so I have this equality here for my next trick what I'm going to do is move this thing to the left hand side and I'm going to write DDT actually I'm going to move that constant inside right there so DDT of 12 V ^2 minus DDT of mu / R is equal to zero and I'm going to use the fact that the derivative is a linear operator so I have a derivative of a sum or really a difference whatever that's equal to the derivative of 12 V ^2 minus mu / R is equal to zero and again we notice that we're saying that the time rate of change of this quantity here is equal to zero which is to say that this quantity right here is a conserved quantity and this time we have a conserved scalar quantity which means that over time V and R might change as they like but if you come back and compute this quantity 12 V ^ 2us mu over R it's the same at every point in the evolution of this system here I just move this equation to the top again we have this very special conserved quantity 12 V ^2 minus mu / R and being a nice conserved scalar quantity let me give it a special name called the energy which I'll denote as e and actually this is the total energy per unit Mass if you just check the units on this thing you'll see that we've actually divided out the mass here another way to see that too is if you've already seen what kinetic energy looks like you're probably used to seen as 12 mv^2 so we've actually divided out the mass here so this is total energy per unit mass and actually let's define up the kinetic energy and the gravitational potential energy per unit Mass so we're going to look at this first term and Define that to be the kinetic energy per unit Mass which is 12 v^2 and the second term is going to be the gravitational potential energy per unit Mass which I'll to know as U which is going to be minus mu over R so if I make these two definitions what I can say is that the total energy is the sum of kinetic and potential energy and again I emphasize that the kinetic energy in the potential energy might independently be changing over time but the sum is a conserved quantity now if we think of our orbits let's say that the we have an elliptical orbit and at some point in the orbit the Earth moves closer to the Sun and what does that mean for the gravitational potential energy well R decreases so that means this whole quantity gets more negative It's a larger negative quantity which means it decreases it moves in a negative Direction so in order to keep the total energy constant the kinetic energ is going to increase which is probably something you already know as the Earth moves closer to the Sun in the in the elliptical orbit it moves faster V has got to increase and you can do the reverse argument as the Earth moves farther away from the Sun R increases which means U moves in the positive direction which means that the kinetic energy has to decrease the Earth has to slow down and its orbit to keep the total energy constant so I think that'll do it for this video now I realize this video is a little bit difficult so feel free to leave any questions you may have in the comment section below and hopefully you stay tuned for future videos on both geometric algebra and the keplar problem as we dive further into this problem and actually show that the orbits produced by the law of gravitation arconic sections which will be the topic of the next video so as always if you enjoy the content feel free to subscribe leave your angry comments below and thank you for watching