402(2) : Rigorously Self Consistent Lagrangian Analysis

This note reviews the lagrangian analysis and shows that it is rigorously self consistent. The formal Euler Lagrange equation (2) is valid in any coordinate system. In the Cartesian system it is equivalent to Eq. (6), and in the next note it will be expressed in plane polars. These are the fundamentals, and it is always important to keep on checking the fundamentals. These fundamentals can now be augmented by vacuum interaction in many interesting ways.

a402ndpapernotes2.pdf

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402(1): Origin of Retrograde Precession

In general, if one sets up an equation F bold = gamma cubed m g bold, then it follows that moduli or magnitudes can be taken both sides, giving F = gamma cubed mg. This is because gamma cubed is a premultiplying scalar.

Date: Sat, Feb 17, 2018 at 6:53 PM
Subject: Re: 402(1): Origin of Retrograde Precession
To: Myron Evans <myronevans123>

I think that constructing a vector equation from a scalar equation is not unique, see my other emails from today.

Horst

Am 16.02.2018 um 13:46 schrieb Myron Evans:

It is shown that the lagrangian that gave retrograde precession in UFT377, Eq.(1), must be interpreted as Eq. (4), which splits into Eqs. (10) and (11). In retrograde precession, the relativistic Newtonian force is used as in Eqs. (12) and (13). Forward precession is given by Eqs. (28) and (29), which are found from the same Eqs. (10) and (11) as retrograde precession, but without the specific use of the relativistic force. The vector form of the relativistic force, Eq. (24), is found from its magnitude, Eq. (15) in the development from Egs (15) to (24). So the retrograde precession is given by the relativistic Newtonian force, which accounts for relativistic conservation of momentum (Marion and Thornton, chapter 14) because the relativistic velocity is deduced from conservation of momentum in special relativity. ECE2 relativity is essentially special relativity in a space with finite torsion and curvature. TRhe latter are missing completely from ordinary special relativity. The retograde precession equations (25) and (26) therefore take account of conservation bothf of relativistic energy and linear momentum but the forward precession equations (28) and (29) consider only the lagrangian and conservation of relativistic energy (relativistic hamiltonian). So the retrograde precession theory is more complete. Both retrograde and froward precession obey the laws of the Hamilton Lagrange dynamics. By changing initial conditions, retrograde precession may become forward precession, and vice versa.

402(1): Origin of Retrograde Precession

It is shown that the lagrangian that gave retrograde precession in UFT377, Eq.(1), must be interpreted as Eq. (4), which splits into Eqs. (10) and (11). In retrograde precession, the relativistic Newtonian force is used as in Eqs. (12) and (13). Forward precession is given by Eqs. (28) and (29), which are found from the same Eqs. (10) and (11) as retrograde precession, but without the specific use of the relativistic force. The vector form of the relativistic force, Eq. (24), is found from its magnitude, Eq. (15) in the development from Egs (15) to (24). So the retrograde precession is given by the relativistic Newtonian force, which accounts for relativistic conservation of momentum (Marion and Thornton, chapter 14) because the relativistic velocity is deduced from conservation of momentum in special relativity. ECE2 relativity is essentially special relativity in a space with finite torsion and curvature. TRhe latter are missing completely from ordinary special relativity. The retograde precession equations (25) and (26) therefore take account of conservation bothf of relativistic energy and linear momentum but the forward precession equations (28) and (29) consider only the lagrangian and conservation of relativistic energy (relativistic hamiltonian). So the retrograde precession theory is more complete. Both retrograde and froward precession obey the laws of the Hamilton Lagrange dynamics. By changing initial conditions, retrograde precession may become forward precession, and vice versa.

a402ndpapernotes1.pdf

Definition of Derivative with Respect to a Vector

I will have a look at this in detail tomorrow.

Pleasure, the discussion of basics is always important. The derivation by Marion and Thornton is the one that is always used, but that does not mean that it is correct. On going from their equation (4.55) to their equation (14.56) their bold u dot bold u is the product of velocity along the direction of the force, so they get equation (14.56). The Lorentz factor is defined as gamma = (1 – (u / c) squared) power minus half, where u squared is a scalar. I will go through this derivation in due course but today I will write up UFT401. In problem 14-38 Marion and Thornton give the magnitude of the relativistic Newtonian force as F = gamma cubed m du / dt . Since gamma cubed is a scalar it is clear that F bold = gamma cubed m d u bold / dt.

Date: Wed, Feb 14, 2018 at 7:47 PM
Subject: Re: Definition of Derivative with Respect to a Vector
To: Myron Evans <myronevans123>

OK, many thanks for clarification, with these definitions the Lagrange equations can be written in this vector form.
Meanwhile I came across another problem in M&T and note 377(1). The time derivatve of gamma is computed via the substitution

d gamma/dt = d gamma / dv0 * dv0 /dt.

Here v0 is the modulus of the velocity bold v0. I think it is not admissible to replace the term dv0/dt by a vector term in the result

F = m gamma^3 dv0/dt

in order to obtain the threedimensional result. Instead one had to do a different substitution for each component, for example

d gamma/dt = d gamma / dv0_x * dv0_x /dt

etc. Because of

v0^2 = v0_x^2 + v0_y^2 + v0_z^2

this gives a different result with cross-terms of velocities.

Horst

Am 14.02.2018 um 14:30 schrieb Myron Evans:

After some discussions with Horst the derivative of a scalar with respect to a vector is defined as the well known vector gradient (Google "derivative with respect to a vector") so that is how the Euler Lagrange equation (4) of the note should be defined. I exemplify this definition by deriving Eq. (16) of UFT377. I also give an example of a lagrangian differentiated with respect to bold r dot from Appendix 21 of Atkins, "Molecular Quantum Mechanics", third edition and also show how this equation can be interpreted. So in vector algebra
del phi : = partial phi / partial bold r

That is what I meant in UFT377. The important result is that the relativistic Newton equation gives retrograde precession. Discussions such as this are very important in order to check fundamentals. The definition (1) of this note can be extended to n dimensions, notably four dimensions, and the Lagrange method is important throughout modern physics.

Definition of Derivative with Respect to a Vector

After some discussions with Horst the derivative of a scalar with respect to a vector is defined as the well known vector gradient (Google "derivative with respect to a vector") so that is how the Euler Lagrange equation (4) of the note should be defined. I exemplify this definition by deriving Eq. (16) of UFT377. I also give an example of a lagrangian differentiated with respect to bold r dot from Appendix 21 of Atkins, "Molecular Quantum Mechanics", third edition and also show how this equation can be interpreted. So in vector algebra
del phi : = partial phi / partial bold r

That is what I meant in UFT377. The important result is that the relativistic Newton equation gives retrograde precession. Discussions such as this are very important in order to check fundamentals. The definition (1) of this note can be extended to n dimensions, notably four dimensions, and the Lagrange method is important throughout modern physics.

anoteonderivativewithrespecttoavector.pdf

equations for forward and backward precession

In reply to the second remark, Note 377(1) uses exactly the same method as Marion and Thornton, third edition,Eqs. (14.53) to (14.58), the derivation of the relativistic kinetic energy from the work integral. The relativistic Newtonian force is used in Eq. (14.54). Eq. (3) of the note is Eq. (14.56) of Marion and Thornton. It is seen that dt cancels out in Eq. (2), giving Eq. (3). nothing else is assumed. Integrating Eq. (3) by parts gives the well known relativistic energy T = (gamma -1) m c squared.

OK thanks. I will reply to these two interesting remarks in two postings. In point (1) I used the expression for the Newtonian force given by Marion and Thornton, third edition, problem (14.38). As you know, the relativistic Newtonian force is the differential of the relativistic momentum with respect to the time t in the lab frame. The Minkowski force is the differential of the relativistic momentum with respect to the proper time tau. In component form the relativistic Newton force is given by Eqs. (21) and (22). These two simultaneous equations when integrated numerically give retrograde precession, as you showed. The retrograde precession comes directly from the relativistic Newton equation. The Lagrangian (11) with proper Lagrange variable bold r gives the relativistic Newton equation directly. As you know there is freedom to chose the proper lagrange variable, the choice bold r leads directly to the well known relativistic Newton equation, so it is a valid choice. The lagrangian (11) is also correct. The rules of the Hamilton / Lagrange dynamics are also obeyed, the kinetic energy is a function only of dr(bold) / dt, and the potential energy is a function only of r bold. The forward precession is obtained by rewriting the lagrangian (13) as the lagrangian (1). The latter is used in the Euler Lagrange equations (5) and (6) with two proper variables X and Y. The key point is that choice of proper variables is different for forward and retrograde precession. The choice for forward precession also obeys the rules of the Hamilton / Lagrange dynamics because the kinetic energy is a function only of X dot and Y dot, and the potential energy is a function only of X and Y. It is well known that different choices of proper Lagrange variables leads to different dynamics. They must be chosen by experience or "by inspection" as they say in mathematics. The conservation of relativistic angular momentum can be demonstrated y by a choice of Lagrange proper variables r and phi of the plane polar system. As shown in Note 401(1), Relativistic angular momentum is always conserved and the relativistic hamiltonian is always conserved

Equations for forward and backward precession
To: Myron Evans <myronevans123>

I came across the following two problems.

1.
In UFT 377, eq,(16), the relativistic Newtonian force is derived from the expression

d bold p / dt = gamma m d bold v / dt. (1)

This gives

gamma^3 m r dot dot (3)

by evaluating the above expression (1). The result is equated to the result from the Euler-Lagrange equations by definition but both results are probably not the same. This then is the reason for giving forward and backward precession.

2.
When evaluating (1) in note 377(1), a substitution is made by the

d gamma/dt = d gamma/dv0 * dv0/dt.

However in the gamma factor there is not v0^2 but the modulus of v in form of v_x ^2 + v_y ^2. When inserting this term, the differentiation gives a tensor expression, see protocol. I am not sure if this can be further simplified, but it is quite different from the simple result (3).

Horst

405-1.pdf

equations for forward and backward precession

OK thanks. I will reply to these two interesting remarks in two postings. In point (1) I used the expression for the Newtonian force given by Marion and Thornton, third edition, problem (14.38). As you know, the relativistic Newtonian force is the differential of the relativistic momentum with respect to the time t in the lab frame. The Minkowski force is the differential of the relativistic momentum with respect to the proper time tau. In component form the relativistic Newton force is given by Eqs. (21) and (22). These two simultaneous equations when integrated numerically give retrograde precession, as you showed. The retrograde precession comes directly from the relativistic Newton equation. The Lagrangian (11) with proper Lagrange variable bold r gives the relativistic Newton equation directly. As you know there is freedom to chose the proper lagrange variable, the choice bold r leads directly to the well known relativistic Newton equation, so it is a valid choice. The lagrangian (11) is also correct. The rules of the Hamilton / Lagrange dynamics are also obeyed, the kinetic energy is a function only of dr(bold) / dt, and the potential energy is a function only of r bold. The forward precession is obtained by rewriting the lagrangian (13) as the lagrangian (1). The latter is used in the Euler Lagrange equations (5) and (6) with two proper variables X and Y. The key point is that choice of proper variables is different for forward and retrograde precession. The choice for forward precession also obeys the rules of the Hamilton / Lagrange dynamics because the kinetic energy is a function only of X dot and Y dot, and the potential energy is a function only of X and Y. It is well known that different choices of proper Lagrange variables leads to different dynamics. They must be chosen by experience or "by inspection" as they say in mathematics. The conservation of relativistic angular momentum can be demonstrated y by a choice of Lagrange proper variables r and phi of the plane polar system. As shown in Note 401(1), Relativistic angular momentum is always conserved and the relativistic hamiltonian is always conserved

Equations for forward and backward precession
To: Myron Evans <myronevans123>

I came across the following two problems.

1.
In UFT 377, eq,(16), the relativistic Newtonian force is derived from the expression

d bold p / dt = gamma m d bold v / dt. (1)

This gives

gamma^3 m r dot dot (3)

by evaluating the above expression (1). The result is equated to the result from the Euler-Lagrange equations by definition but both results are probably not the same. This then is the reason for giving forward and backward precession.

2.
When evaluating (1) in note 377(1), a substitution is made by the

d gamma/dt = d gamma/dv0 * dv0/dt.

However in the gamma factor there is not v0^2 but the modulus of v in form of v_x ^2 + v_y ^2. When inserting this term, the differentiation gives a tensor expression, see protocol. I am not sure if this can be further simplified, but it is quite different from the simple result (3).

Horst

405-1.pdf

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