Two-Body Problem in Classical Mechanics: Lagrangian Approach | Physics Tutorial

[00:00:00]So we will now turn to a classic application classic example of classical mechanics namely the twobody problem. We start with a general two-body problem and then later on we will focus on a particular um kind of force between the two bodies namely the Kepler force. Um but let us first start with sort of the general setup and the general setup is two bodies two masses m1 and m_sub_2. We have a some reference frame in space and with respect to that reference frame the first body has is at position R1. The second body is at position R2.

[00:01:06]And these two bodies they exert forces on each other. So the um second body exerts on the first body a force f12 and the first body exerts on the second body a force f_sub_21 and by Newton's law action equals reaction.

[00:01:36]This force f_sub_21 must have the same magnitude but opposite direction of the force f_sub_12. So we have f_sub_21 is equal minus f_sub_12 and we will treat that problem in the lrange formalism. So the starting point is that we write down the lrange function in cartisian coordinates.

[00:02:11]The lrange function depends on the positions and the velocities of the two bodies. So it's a function of r1 r2 r1 dot and r2 dot It's kinetic energy minus potential energy.

[00:02:38]The kinetic energy is mass over two velocity squared of the first body plus the kinetic energy of the second body. That's mass 2 /2 and then velocity of the second body squared. And then we have minus the potential.

[00:03:04]And um we had seen earlier when we discussed um Galileain invariance that if we um require for this um for the system we require Galileain invariance then the forces and hence also the potential may depend on the distance between the two bodies only. So we have a potential that depends on the distance. So the absolute value of the distance vector R1 - R2.

[00:03:45]Yeah. So this form comes from Galileain invariance.

[00:03:58]This is the starting point and the basic setup for the two body problem.

[00:04:04]And we'll treat that two-body problem in five steps. And let me outline these steps on the next page.

[00:04:15]In the first step, we will reduce the twobody problem to an effective onebody problem. That's the first step. by which we simplify the problem before us. So we reduce the two body to an effective onebody problem.

[00:04:43]At this point it's a one it's a one body problem in 3D but we will actually succeed in simplifying that further and we will reduce that to an effective problem in one dimension only.

[00:04:59]So this is the next simplification that we reduce this problem to one dimension.

[00:05:08]In the next step, we will make use of energy conservation.

[00:05:17]So we will look at energy conservation and the consequences that that has for the solution of our problem. It will bring an even further simplification.

[00:05:31]Then we are ready to look at the solution.

[00:05:35]So we look at the general solution and finally we will discuss physically what that situ solution means and what it looks like.

[00:05:51]These are the basic steps in which we are going to proceed.

[00:05:56]As always or as very often it's a matter of finding suitable convenient coordinates. So we will change coordinates and a first change of coordinates will be from the positions R1 and R2 in our fixed uh reference frame in space. we will move to center of mass coordinates and relative coordinates.

[00:06:29]This means the following. Let us recall the two bodies m_sub_1 and m_sub_2.

[00:06:42]our reference frame in space with the original coordinates R1 and R2.

[00:06:55]Now somewhere between the two bodies lies the center of mass of the system.

[00:07:09]This is the center of mass and the center of mass has position capital R.

[00:07:23]Now the formula for the center of mass position that is something that we had looked at before that is It's the weighted average of the positions of the two bodies and the the weight of each body uh corresponds to its share in the total mass. So it's um m1 over the total mass m1 + m_sub_2 * r1 plus m_sub_2 divided by the total mass time r2.

[00:08:10]That's the center of mass.

[00:08:12]And then we can introduce the relative position of the two of the two bodies.

[00:08:25]And this is a vector little r. This relative position little r this is defined as r1 minus r2.

[00:08:39]That's the relative position.

[00:08:42]So what we are doing is that we change coordinates from the original R1 and R2.

[00:08:54]We go to center of mass position capital R and relative position little R.

[00:09:08]It's an alternative way of specifying the configuration of these two masks.

[00:09:14]And then we can express the lrange function as a function of these new coordinates.

[00:09:24]So we have the lrange function now as a function of relative position relative velocity center of mass position and center of mass velocity.

[00:09:48]You obtain this lag range function simply by taking the original lrange function that we had on the previous page and expressing the old coordinates in terms of the new coordinates. So you insert this transformation law. It's a timeindependent coordinate transformation. So it's a rather straightforward substitution.

[00:10:13]And then you find the lagr function in terms of the new coordinates.

[00:10:18]It turns out this lag range function is the total mass over two where I have introduced a capital m for the total mass m_sub_1 + m_sub_2 times the velocity of the center of mass squared. So that looks like the kinetic energy. But now it's the kinetic energy of the combi of the of the entire system or the kinetic energy of the center of mass. So we cons we can consider the system composed of two bodies now as a single system with total mass capital m uh that moves with the velocity of the center of mass and the associated kinetic energy that would be this term m* r dot squ capital r dot squ.

[00:11:22]But then of course the system is not just a single body.

[00:11:27]There are two bodies and they can also move relative to each other. So there is another contribution to the kinetic energy that comes from the relative movement of the two bodies.

[00:11:42]So there's a term proportional to little r dot squared. So the relative velocity squared and the factor in front is mu / 2 where mu is defined as m_sub_1 * m_sub_2 divided by m_sub_1 + m_sub_2 and this is the so-called reduced mass.

[00:12:16]And finally we still have the um potential term which is u and that was a function of the distance between the two bodies. Now the distance is simply the absolute value of the relative position little r.

[00:12:40]Now the new form of the lagrosian the starting point is this transformation law that I gave on the previous page that expresses the new coordinates in terms of the old coordinates.

[00:12:59]What we need in order to obtain the new form of the lrange function we need the reverse. We need to express the original coordinates in terms of the new coordinates.

[00:13:10]So we have to invert these two invert these two equations and write R1 and R2 in terms of the new coordinates capital R and little R.

[00:13:32]Let me give you the results of this uh inversion.

[00:13:37]So the position R1 of the first mass that's the position of the center of mass plus well then we have to go some distance from the center of mass along the distance vector uh towards the mass number one. And when you do the calculation, you find that it's that this is just the ratio m_sub_2 over m_sub_1 + m_sub_2, the total mass times r. This gives you when you invert the two equations, this gives you the position of the first mass.

[00:14:30]The position of the second mass is the position of the first mass minus the distance vector little r. So you simply subtract from this expression above. You subtract little r and you find that this is the position of the center of mass minus m_sub_1 over m_sub_1 + m_sub_2 time little r.

[00:15:05]Is that fine with you?

[00:15:07]I mean here I skipped some details but this is a really think a simple inversion of the two equations.

[00:15:15]You can also plug in uh these two equations in the left hand side and verify that they are consistent.

[00:15:29]Now as a consequence of that we find the velocities.

[00:15:36]We can write the velocity of the first mass R1 dot.

[00:15:45]This is capital R dot plus M2 over the total mass. But the let me write for the total mass now the capital m that's a bit simpler times distance vector time derivative and for the velocity of the second mass you find velocity of center of mass minus m_sub_1 over total mass time distance vector time derivative Now we we go back to the original form of the lron function. Remember that as a function of the original coordinates r1 r2 r1 dot r2 dot.

[00:16:47]This was um m1 / 2 and then r1 dot.

[00:16:58]But let's now replace r1 dot by the term that we have above capital r dot plus m_sub_2 over total mass little r dot squared.

[00:17:14]And now that we do the that we express this immediately in terms of the of the new coordinates actually what we have what we obtain is the lrange function already in terms of the new coordinates. So little r little r dot capital r and capital r dot.

[00:17:36]So there's the kinetic energy of the first mass.

[00:17:43]Then we have the kinetic energy of the second mass that was the second mass over 2 times the velocity of the second mass. The velocity of the second mass we found was capital R dot minus M1 / capital M little R dot squared minus the potential energy which is simply given as U of the distance between the two masses.

[00:18:25]All right. Then we have to evaluate these squares.

[00:18:34]From the first bracket squared, you see we get an R dot squared. We get an M1 / 2 R dot 2. from the second bracket squared we get an m2 over two capital r dot squared. So summing these summing up these two we get m1 + m2 which is the total mass.

[00:18:59]So total we get total mass over two capital R dot squared that was one of the terms that we had encountered in the new form of the lash right then we have these um mixed terms so from the first bracket we get um 2 * capital R dot scalar product with little r dot time m_sub_2 over total mass and then multiplying this in front with m_sub_1 / 2.

[00:19:41]So we really have m_sub_1 * m_sub_2 divided by total mass time velocity of the center of mass scala product with relative velocity.

[00:19:53]That's a mix term that we get from the first bracket squared. Now from the second from the kinetic energy of the second mass this is the second bracket squared we get exactly the same mixed term but with but with the opposite sign because we have a minus inside the bracket. So these mixed terms they cancel each other out.

[00:20:18]And then finally we have the remaining squares.

[00:20:25]So these are um from the kinetic energy of the first mass we have a m1 / 2 and then we have an m_sub_2 squared over total mass and then we have an um relative velocity squared.

[00:20:47]And from the kinetic energy of the second mass we get also a term proportional to the relative velocity squared.

[00:20:58]So I can take this outside a bracket and the factor in front is in that case m2 time m1 squared over total mass squared.

[00:21:14]And I forgot the square here.

[00:21:18]All right. And then the potential energy.

[00:21:23]Now, but what we have here is we can take um we have two terms that look very similar.

[00:21:30]Um only that in the first case we have m1 * m2 2 and in the second case we have m_sub_2 * m1 2. We can factor out an m1 * m2 and then we have inside a bracket we have m_sub_2 + m1 which is a total mass which cancels against one of the capital m's in the denominator and what is left is 12 m1 m2 divided by capital M. And this is precisely what we defined as the reduced mass on the previous page.

[00:22:22]That's how you obtain the formula on the previous page.

[00:22:27]In these in these new coordinates, it's it's easy to see that the motion of the center of mass and the relative motion of the two bodies actually decouple from each other.

[00:22:41]To see that, let us copy the new form of the lrange function.

[00:22:48]In this new form of the lrange function, you see that the lagrange function is the sum of two terms.

[00:22:56]There's one term this one here which depends only on center of mass coordinates or their velocities.

[00:23:08]So this is you can view that as the lag function L1 of the center of mass and then you have terms that depend only on the relative coordinates and their time derivatives.

[00:23:34]So you can view that as a second contribution to the lagrange function which depends only on the relative positions and velocities of the two bodies.

[00:23:48]Now the fact that that you can write the lag range function for the entire system as the sum of two functions one of which depends only on the center of mass degrees of freedom and the other one depends only on the relative degrees of freedom.

[00:24:05]This implies that the equations of motion for the center of mass and for the relative um degrees of freedom they decouple.

[00:24:22]So this implies that the equations of motion for the center of mass coordinates and the relative coordinates they decouple.

[00:24:38]So in other words, the um the position of the center of mass does not enter in any way in the equation of motion for the relative position of the two bodies and vice versa.

[00:24:57]Let's quickly see what we can say about the dynamics of the center of mass.

[00:25:05]That is actually very simple because we immediately see that the lrange function that describes the center of mass motion this L1 does not depend on the center of mass position. It doesn't depend on capital R. It only depends on the velocity capital R dot.

[00:25:29]So the position of the center of mass is a cyclic coordinate.

[00:25:39]This immediately implies that the associated momentum is conserved is a constant of the motion. So this implies that the center of mass momentum capital P which is defined as the partial of the LRO function with respect to velocity. Let me write it in vector notation this way.

[00:26:08]This is when you uh do the differentiation. This is simply the total mass times the velocity of the center of mass that this is constant constant of the motion and this immediately implies you can a very simple equation of motion for the center of mass and the general solution is a uniform motion.

[00:26:39]So the position of the center of mass as a function of time has the general form. It's the position at time zero plus the velocity capital v of the center of mass time t. And this velocity is constant in time.

[00:27:03]Momentum is constant in time. They are proportional to each other via the total mass. So the velocity is also constant in time and this is simply a uniform motion.

[00:27:20]The center of mass of a twobody system simply performs a uniform motion. It's a very simple dynamics.

[00:27:29]So all the interesting stuff is in the relative motion.

[00:27:34]So from here on we can really focus on the relative motion which means the time dependence of the little r that's where the interesting physics is.

[00:27:51]Yeah. So we we focus on the relative motion from here on and the relative motion is described is governed by this second term in the lrange function the L2 which depends on the relative coordinates only. So this L2 will govern will determine the relative motion of the two bodies.

[00:28:26]So now we focus on the on the relative motion and the second part of the lag function which depends only on the relative coordinates and the relative velocities.

[00:28:41]And with this we have achieved our first goal. The first goal that I mentioned in the procedure in the outline of of our steps namely that we've effectively reduced the problem to a one body problem.

[00:29:01]If you look at the lron function that we want to focus on now this L2 and I omit the subscript now which has the form reduced mass over 2 * relative velocity squared minus the potential energy.

[00:29:26]This is precisely the lounge function of a single body of mass mu moving in a central potential.

[00:29:36]So effectively we have reduced our problem to a one body problem.

[00:29:55]namely a single body moving in a central potential.

[00:30:00]So from now on at least for a good while we can forget that we are dealing with two bodies. So we can simply now focus on the problem of a single body moving in a central potential.

[00:30:17]a central potential or a central force. That is a situation that we had already considered in the framework of Newtonian mechanics and we had already looked at consequences when you have central forces and one important consequence was that the angular momentum is conserved.

[00:30:46]So we can immediately take over these results from our discussion in Newtonian mechanics that a central force implies conservation of angular momentum.

[00:31:03]Now conservation of angular momentum uh you remember had two implications. One implication was that the and that comes from the fact that the direction of the angular momentum in particular the direction of the angular momentum is conserved that implies that the motion is effectively constrained to a two-dimensional plane.

[00:31:34]So you have the initial position and the initial velocity of the particle. They span a two-dimensional plane and the body will never leave that two-dimensional plane.

[00:31:49]So that was the first um consequence that we had discussed earlier of angular momentum conservation that the motion is constrained to a two-dimensional plane.

[00:32:07]And the second consequence was the area law. And we'll come back to that on the on the next page.

[00:32:16]If motion is constrained to a two-dimensional plane, then we can change coordinates yet again. Let me sketch that plane here, that two-dimensional plane.

[00:32:32]And let's say that the um our body moves in that two-dimensional plane. One thing we can do is we can introduce cylindrical coordinates.

[00:32:49]And the way we do that is that we choose the zaxis of the c cylindrical coordinates to be orthogonal to this two-dimensional plane in which the body moves. And in in the two-dimensional plane we introduce coordinates row and phi where row is the distance to the origin in the two dimensional plane and phi is the angle with respect to the x axis.

[00:33:25]So the idea here is that we um go to cylindrical coordinates.

[00:33:35]The two dimensional plane as I said earlier corresponds to z equal to zero. So we change the z-axis in such a way that the motion is in the plane where z is equal to zero.

[00:33:51]Now when we do that we go from the relative position little r now to cylindrical coordinates then we have to again express the lrange function in terms of these new coordinates in terms of the cylindrical coordinates.

[00:34:11]The transformation law between cartisian and cylindrical coordinates is um x is given by row * cossine phi, y = row * sin and z = z.

[00:34:33]There's the transformation law.

[00:34:36]Now the kinetic energy had the form in the original cartisian coordinates uh mu / 2 times relative velocity. So little r dot squared.

[00:34:56]Now express x doty dot and z dot in terms of the new coordinates.

[00:35:05]Then you see that this becomes row dot squared plus row ^ 2 dot squared.

[00:35:18]And in principle you also have a z dot squared.

[00:35:23]But since we chose the orientation of our coordinates in such a way that the motion is always constrained to the plane corresponding to Z equal to zero.

[00:35:38]The Z is always zero. The Z dot is always zero. And therefore the Z dot squared um we don't have to carry the Z dot squared because the problem is effectively two-dimensional. and the zcoordinate doesn't play a role. Yeah.

[00:35:57]So I close the bracket here and as a reminder I write Z and Z dot are always zero.

[00:36:10]The potential energy that was a function of the distance between the two bodies.

[00:36:19]Again, we know the distance between the two the the the distance vector is confined to that two-dimensional plane.

[00:36:32]So again, there's no Z component to worry about.

[00:36:36]And the the length of the vector R is simply in cylindrical coordinates. That's simply our row.

[00:36:52]So again using the fact that Z is zero we have eliminated the third coordinate.

[00:37:02]This has allows us to write now the lrange function in terms of the cylindrical coordinates.

[00:37:11]So in terms of row row dot phi and phi dot It's kinetic energy minus potential energy.

[00:37:22]So we just take the two terms that we have for the kinetic energy and the potential energy.

[00:37:38]We immediately see that this lrange function actually does not depend on phi.

[00:37:46]Phi is cyclic which implies that the associated momentum is a constant of the motion. The associated momentum that's the partial of the lag function with respect to fi dot.

[00:38:06]This is mu * row ^ 2 dot and we had seen earlier that that is the z component of the angular momentum.

[00:38:29]When we had earlier when we had talked about angular momentum angular momentum conservation in the context of Newtonian mechanics, we had discussed about also about the area law and we found that um this is in fact the z component of angular momentum which in our case since the motion is constrained to that two-dimensional plane perpendicular to the z-axis this equals the absolute value of the the magnitude of the angular momentum vector. So this is actually equal to the absolute value of the angular momentum which we also simply write as a little l. So this is a constant of the motion and this constant magnitude of angular momentum as we had discussed earlier corresponds to the area law.

[00:39:33]the area law. Just as a reminder, when you have here motion in a central potential around the center of the potential, then the area covered by the ray connecting the center of the potential with the body.

[00:40:02]uh the area covered in a certain time interval delta t is always the same. So when the body moves from here to here in a certain time delta t the ray connecting the two the body to the center of the potential covers a certain area this green shaded area if the body is here and again after some time deltat t the ray covers that area and these two areas are identical in the ray always in a given time interval always covers the same a constant area.

[00:40:48]The area law in particular gives you immediately an expression in a differential equation for phi. Now you have this um uh this equation mu * row ^ 2 * fi dot equals this constant angular momentum l.

[00:41:07]you can solve that for phi dot and then you have a differential equation of the form phi dot equals something.

[00:41:15]Um on the right hand side you have row of t.

[00:41:20]Um so once you know the time dependence of row you can plug it plug it in there and then you have a differential equation for phi and you can solve for ph of t.

[00:41:34]So our next task must be to find the solution for row of t.

[00:41:41]To do that we have to consider the equation of motion for row. That equation for row is also known as the radial equation. We obtain the equation of motion for row in the usual manner from the lrange function. It's an oiler lounge equation of the form d / dt of the partial of l with respect to row dot. This must be equal the partial of l with respect to row.

[00:42:19]When you consider the form of the lrange function in particular the kinetic energy you see that the partial of L with respect to row dot that's mu * row dot and the time derivative of that is mu * row double dot.

[00:42:45]And on the right hand side when you take the partial of L with respect to row then of course you get the partial derivative of the potential with respect to row but you also get the contribution from the kinetic energy because in cylindrical coordinates the kinetic energy also had a dependence on row. So you get two terms. The contribution from the kinetic energy has to form row * mu * row * 5 dot squared.

[00:43:18]And the contribution from the potential is the partial of u with respect to row.

[00:43:31]At this point you can exploit the area law that we had found on the previous page.

[00:43:39]You use the area law to express fi dot in terms of the conserved angular momentum the reduced mass and row.

[00:43:56]Using that you find mu row double dot equals and then you have for the first term you get l 2 over mu * row cube minus the partial of the potential with respect to row. And this equation of motion inside the box is that's the radial equation. It's the equation of motion for row. And the interesting aspect of this equation is that it's not simply mass times acceleration is minus the gradient of the potential, but there's this extra term there. In this extra term, you can uh so furistically you can regard that as an effect as as a centrifugal force.

[00:44:51]uh it's an it's an extra acceleration that comes from the angular momentum the finite angular momentum of the system if the angular momentum is zero so if l is equal to zero this term does not contribute but as soon as the body is moving around the center and has a finite angular momentum l is different from zero and then you will have that contribution So let me write that in quotation marks.

[00:45:26]You can view that as a kind of centrifugal force.

[00:45:31]And this is now an equation of motion for a single degree of freedom for the row. So you've effectively after changing coordinates twice, you've reduced the problem to an effectively one-dimensional problem.

[00:45:49]That was step two of our procedure that I had outlined at the beginning.

[00:45:56]Reducing it to a one-dimensional problem.

[00:45:59]The next step in the procedure that I had outlined was to exploit the conservation of energy.

[00:46:07]Keep in mind that all the coordinate transformations that we had done were time independent.

[00:46:16]Note also that the lagrange function does not depend explicitly on time. So the partial derivative of L with respect to time is equal to zero.

[00:46:31]And as we know from our general considerations, this implies that the Hamilton function actually equals the total mechanical energy and is a constant of the motion. So the Hamilton function under these circumstances is the total physical energy. So it's t plus u k kinetic plus potential energy and is conserved. So it has a constant value e that is the total energy and remember we are considering here the the relative part of the motion only so not uh the motion of the center of mass.

[00:47:26]So it's the total energy of our relative motion. Now let's let's um evaluate the Hamilton function. Let's insert for the kinetic energy and the potential energy the um formulas that we had obtained earlier. The kinetic energy had two contributions. One was mu half row dot squared.

[00:47:58]The other contribution was included a phi dot squared which we express using the area law in terms of row. So that's l 2 over 2 mu row 2.

[00:48:16]That's the kinetic part.

[00:48:19]And then we have the the potential energy u of row.

[00:48:26]Now even though the second term in this uh expression in this formula for the Hamilton function the second term comes originally from the kinetic energy.

[00:48:42]It's a function that does not depend on velocities but it depends on row depends on position.

[00:48:51]And therefore what we will do is that rather than considering it as a part of the kinetic energy we group it together with a potential and call that an effective potential as a function of row.

[00:49:13]And then we have written in this way we have for the Hamilton function for the total energy we have we have a form that we know from one-dimensional problems where we have a kinetic term which is the velocity squared plus some potential which depends only on position.

[00:49:42]So this is um we will now treat this as a as a problem in one dimension with effective potential u effective as I had indicated here in in blue.

[00:49:56]Now the the sum of kinetic and potential energy is this constant E.

[00:50:02]We can solve that for row dot. So we can first write mu / 2 row dot squared equals this constant e minus the effective potential as a function of row or we solve it for row dot. This is the square t of 2 over mu * e minus the effective potential of row.

[00:50:38]And this is now a a first order differential equation which at least in principle, not always analytically, sometimes only numerically, um but in one way or the other, we can solve for row of t.

[00:51:02]So here we are back to the procedure that we had at the very beginning of the semester when we started out with one-dimensional problems as the very simplest examples.

[00:51:16]We considered problems exactly of that form. We used energy conservation uh in order to find the solution of our problem.

[00:51:26]Before we come to general form of the solution and the discussion of the solution, let's uh pause for a minute and uh take counts of the degrees of freedom and the integration constants.

[00:51:44]As for the integration constants, what would you expect?

[00:51:48]We have two masses.

[00:51:52]Each mass performs a three-dimensional motion.

[00:51:58]So if we want to specify the initial state of our system, we would expect that we need three parameters to specify the position of the first mass. Three parameters to specify the velocity of the first mass makes six parameters for the first body.

[00:52:20]and then another six parameters for the second body. So we would expect that we we we need in total 12 parameters to specify the initial state of our two body system that should correspond to the total number of integration constants that we have in our general solution of the problem.

[00:52:45]So let's check whether this is indeed the case or whether perhaps we have lost integration constants along the way.

[00:52:56]In our general solution we have first of all we had separated the motion of the center of mass from the relative motion.

[00:53:08]The motion of the center of mass had the simple form of a uniform motion r0 plus the constant velocity of the center of mass time t.

[00:53:21]In this general solution for the center of mass there are already six integration constants namely the components of r0 and v. Next we had um introduced cylindrical coordinates and we had oriented our coordinate system in such a way that the motion is confined to the two-dimensional plane corresponding to z equals z.

[00:53:54]So we had um now the question is how many parameters do we need to specify the orientation of the plane? To specify the orientation of a two-dimensional plane in a three-dimensional space, you need two parameters because you need one way you can do that is that you specify the direction perpendicular to the plane. So you have to specify a vector perpendicular to the plane. The length of the vector is irrelevant. Only the direction of the vector is relevant. And the direction of a three-dimensional vector is specified by two parameters.

[00:54:40]What we did was that we oriented the cylindrical coordinates such that Z is equal to zero.

[00:54:52]Now in this choice we have two parameters that specify the orientation of the plane. Then we had our radial equation. Then we had reduced that to a one-dimensional problem. We had our radial equation. Um so the radial equation followed from energy conservation and from that form mu/ row do 2 that must be equal to the constant energy minus this effective potential which depends on amongst others depends on this conserved angular angular momentum.

[00:55:46]Now this equation this expression we we solve for row of t.

[00:55:52]Now in the solution for row of t there will be another three parameters namely the conserved energy E the conserved angular momentum L and an initial condition for row. So the row at some initial time t0 once we have row as a function of t for the complete solution we still need phi the angular coordinate phi as a function of t.

[00:56:36]In order to obtain phi as a function of t, we must insert the solution for row of t that we had just obtained. We have to insert that solution into the area law.

[00:57:00]That gives us an equation for fi dot.

[00:57:06]This is this conserved angular momentum L divided by mu * row squared and row is the solution that we had just obtained row of t.

[00:57:21]So this is a first order differential equation for ph of t.

[00:57:31]And the solution five of t will contain one last integration constant in addition to all the others that we already have. namely the initial condition for phi the value of phi at some initial time t0.

[00:57:55]So this is a complete list of all the integration constants that we encounter along the way.

[00:58:03]Let's add them up and they are in total indeed 12 as we had expected. So we haven't lost any integration constants along the way.