These are interesting comments. This note is based entirely on standard equations of the Lagrange and Hamilton dynamics applied to vectors, notably Eq. (3), which gives the correct momentum p from the Lagrangian (2), these are all contained in Marion and Thornton. The vector Euler Lagrange equation is Eq. (11), and leads correctly to the well known equations (20) and (21), the Leibniz equation and the equation of constraint (21). The kinetic energy is p dot p / (2m). The primary purpose of the note is to show that the correct momentum must be defined by p bold = partial lagrangian / partial r dot bold (see for example Marion and Thornton). Eqs. (22) – (24) work correctly for classical dynamics, but no longer work correctly for fluid dynamics. The correct momentum of fluid dynamics must be calculated from eq. (3) using the lagrangian (35). The correct momentum is p bold = m v bold, where v bold is given by Eq. (26). This is the same as the momentum used in UFT363, and leads to Eqs. (33) and (34). The spin connection partial R sub r / partial r must be regarded in the same way as the potential energy U(r). Neither is a Lagrange variable. The key point is that the momentum p bold can be obtained correctly from the lagrangian (2) if and only if Eq. (3) is used. This is checked from the fact that p bold is r bold dot in Eq.(2). Then use the rules of differentiation with r dot bold. For classical dynamics, Eqns (22) to (24) happen to work fortuitously, and these are of course the equations used by Marion and Thornton in their chapter seven. However, for fluid dynamics they no longer work, because the complete momentum is now:

p bold = x r dot e sub r bold + r theta dot e sub theta bold

where

x = (1 + partial R sub r / partial r)

Using this in Eqs. (2) and (3) gives the correct momentum from the correct lagrangian, containing the correct kinetic energy. The correct momentum is Eq. (28) multiplied by m. When used in Eq. (29) it leads to Eqs. (33) and (34). Eq. (33) is different from that found in UFT363, because in UFT363, the correct factor x in Eq. (33) turned out to be x squared, as in Eq. (39) of this note. Therefore the lagrangian (35) cannot be used with Eq. (38). This result is by no means obvious. It shows that there is a certain amount of subjectivity in the Lagrange method as is well known. It is by no means obvious how to choose the Lagrange variables, and the choice of lagrangian is also subjective to some degree. These things emerge in for example quantum field theory. Fortunately the answer is simple, use Eq. (13), in which there is only one Lagrange variable, vector r bold. This leads to Eqs. (33) and (34). I suggest putting Eqs. (43) to (45) through Maxima to see how the orbital precession behaves. I do not think that the replacement of x squared of UFT363 by the correct x of this note will make any qualitative difference to the precession that you have already inferred numerically. It might affect the details of the precession, but the precession will remain.

To: EMyrone@aol.com

Sent: 28/03/2017 14:40:47 GMT Daylight Time

Subj: Re: 374(2): Complete Analysis of UFT363It is difficult for me to understand this note for principal reasons. My interpretation is the following:

The Lagrangian method is based on the kinetic energy and generalized coordinates. The Euler-Lagrange equations are based on the kinetic energy of the generalized coordinates. These coordinates are found by coordinate transformations. In our case the radial coordinate is transformed by

r –> r + R_r(r)

where R_r(r) is a “distortion” of radial motion of a particle inferred by fluid dynamics. For the Lagrange mechanism this function has to be known a priori, it cannot result from the Euler-Lagrange equations. If we assume that the R_r function is to be determined dynamically by the dynamics, we need an additional equation of motion or state or whatever. In Lagrange theory, energy conservation is fulfilled. This is not necessarily the case if a “free floating” function is introduced. I guess that you had this in mind when saying that a Hamiltonian formulation is needed in addition to the Lagrangian formulation to determined the dynamics consistently.

So the question is where to take the conditions for R_r that must appear as a constraint in the Lagrange mechanism. The generalized coordinates should be r and theta, but what is the kinetic energy? Let’s assmume that the velocity, eqs.(26,27) of the note, is that derived from the coordinate transformation. Then the Euler-Lagrange equations (33,34) are correct, although they contain an unspecified function R_r (which is not time dependent).

I do not understand the part of the note after eqs.(33,34). Why do you introduce the Lagrangian (35)? Obviously this belongs to a different problem to be solved. And why should it be re-expressed to (36)? The momentum in Lagrange theory is a generalized momentum and needs not have the form (37).

On page 6 of the manuscript I cannot decipher the sentence “It is not possible to choose … as Lagrange varibles”. Which variables do you mean?

Eqs. (44) and (45) are derived from the same Euler-Lagrange equation and are not independent. It is true that (45) is a constant of motion but this is not suited for solving the equations because it is only of first order. What about usingH = 1/2 m v^2 + U(r) = const.

instead? Then we can determine partial R_r/partial r , and replace it in (43,44) so that we have only derivatives of time and the equation system could be solved by Maxima for example. In general, combination of Lagrange theory (which is for mass points primarily) and fluid dynamics (which is for distributed fields) may be a bit tricky.

Sorry for having written such a long sermon today.

HorstAm 28.03.2017 um 10:44 schrieb EMyrone:

This note shows that the complete Lagrangian and Hamiltonian formulations are needed to describe fluid dynamics self consistently. When this is done UFT363 is slighly corrected to Eqs. (43) to (45), which can be solved simultaneously using Maxima to give the orbit and spin connection.