I need to give some explanation.

1. The task can be considered as a two body problem in the gravity.

In the lowest order of expansion (v/c)^2 the Lagrangian has a form (attached file). It is a copy of a page from the book of Fok, 'Theory of the space, time and gravity', Sec. 81 (in Russian but I am sure there is English translation).

This Lagrangian follows from Einstein-Infeld-Hoffman's derivation of the Einstein equations. The derivation of E-I-H is too complicated; Fok's derivation is much more transparent.

In the blue box, special relativity correction to the masses.

In the purple box, the term that gives the Newtonian gravitational force.

In the red box, the term which gives (v/c)^2 correction to the gravitational force.

In this system, this term is a consequence of retardation effect in the gravity. This effect has been considered by Steven Carlip to disprove Van Flandern's argument on the absence of the gravitational aberration [S. Carlip, Phys. Lett.. A267 (2000), 81 - a copy can be found at Arxiv.org].

So I don't need in solving the original Einstein equations because the task has been reduced to two body equations with some correction terms.

2. There is some analogue between the gravity and electromagnetism. Carlip and Ibison in their discussion with Van Flandern used this analogue. The system I propose is like the analogue of one 'thought device' proposed by Georgy Ivanov. The author stated his device should have one-direction motion. Kirk McDonald resolved the problem:

http://www.hep.princeton.edu/~mcdonald/examples/onoochin.pdf

(I must say I am not the author of the paradox).

But McDonald didn't lead the solution to the end. The system I propose is certain improvement of the device we discussed at one Russian forum. In 'thought operation' of Ivanov's device some unsolved points still exist. But in electromagnetism, it is difficult to calculate some points, for example, the effect of radiation. In the gravity, there is no radiation up to (v/c)^5 so I am able to state that the energy should conserve up to (v/c)^2.

Even forgetting the energy violation, there is a violation of momentum conservation. Now you refer to a Lagrangian derived by Fock. Fair enough: is this Lagrangian translation invariant?It surely seems to be. If it is, momentum is conserved and, for a reason which should readily be traceable if you look in detail at what happens to momentum, your conclusions are ill-founded.

To Leyvraz

"Lagrangian translation invariant?It surely seems to be. If it is, momentum is conserved....your conclusions are ill-founded."

It is no problem to derive the force from Fok's Lagrangian and to see that the gravitational forces between two bodies are different. The same can be made for Robertson's solution who used original E-I-H equations.

If the forces are not compensated why my conclusion should be ill-founded? Do you mean Robertson and Fok made errors in derivation of their Lagrangians that lead to non-compensated forces?

No, I believe the Lagrangian to be fine, and attribute no errors whatever to Robertson or Fock. I mean that translational invariance implies that the sum of momenta: partial L wrt r_1 dot and partial L wrt r2 dot is conserved, and that that contradicts your claim concerning the system's global velocity increasing.

You say ``It is no problem to derive the force from Fok's Lagrangian and to see that the gravitational forces between two bodies are different.'' You should beware: what do you mean by force? From Lagrange's equations, you have an acceleration multiplied by a velocity-dependent prefactor (which I would not call mass) being equal to the partial derivative of L wrt r1 or r2. But the latter, which you may call force, contains a large number of terms (many velocity-dependent) you have not mentioned. Besides, the whole computation must be performed vectorially. Thus the computation is by no means straightforward.

Happily the computation is quite unnecessary: the conservation of momentum is quite sufficient to rule out the kind of motion you have in mind. Simply do compute the total momentum (defined as above, not, of course, m1 dot r1 + m2 dot r2), and you will see that the perpetuum mobile, as all its brethren, fails to move.

Why do you decide that the theorem from the classical mechanics (translation invariance) can be applied to this system? Fok's Lagrangian is some approximation of order of (v/c)^2.

So happily I don't need to seek why my statement contradicts to the translation invariance. Mooreover, you cannot apply this theorem to the gravitational systems if you accept the basic points of the general relativity.

Vladimir: you cannot have it both ways! Five posts ago, you said.

``So I don't need in solving the original Einstein equations because the task has been reduced to two body equations with some correction terms.''

So you say: all I need is to look at the Lagrangian. But any dynamics derived from a Lagrangian that is translation invariant, must satisfy Noether's theorem. As long as you solve the Lagrangian you showed us, it is easy to show explicitly that the total momentum, that is, the sum of all momenta associated to to dot r_i, must be a constant.

I agree that applying Noether to Einstein's equations themselves might be tricky, especially because of radiation. But you have justified your computations by an ordinary N-particle Lagrangian, for which such theorems do apply unrestrictedly. It does not depend where the Lagrangian comes from.

Dear Vladimir,

from Newton’s gravitational law I can calculate the force exerted by a pound of butter sitting on a scale. I also know that this force satisfies actio = reactio. General relativity claims that Newton’s law has to be corrected, but I am not aware of a formula which modifies the weight of a pound of butter. A clear cut answer did not emerge from the discussion so far. It makes me suspicious, if I read in Cornwall’s comment that mass is not involved in the equation. Maybe GRT cannot be used at all to calculate gravitational forces between masses moving or not moving. The “equivalence principle” can definitely not be applied to the force field inside a rotating cylinder: There does not exist any mass distribution which could produce the centripetal inertial force by gravitation.

Moreover, in the limit of weak gravitational fields GRT tends towards SRT, but SRT is wrong as it violates “Einstein’s Third Postulate”. This was demonstrated by Einstein himself (https://www.researchgate.net/publication/279530691_Einstein%27s_Third_Postulate?ev=prf_pub) and is obvious from first order stellar aberration (https://www.researchgate.net/publication/2616265_Relativistic_Doppler_Eect?ev=prf_pub) or from the Sagnac effect (https://www.researchgate.net/publication/286091936_Classical_and_Relativistic_Derivation_of_the_Sagnac_Effect?ev=prf_pub). So, you should not put too much weight on any predictions of GRT.

As far as classical electrodynamics is concerned there exist similar unsolved problems with the Lorentz force: It violates both energy (see Appendix of this paper: https://www.researchgate.net/publication/2174184_Is_a_Plasma_Diamagnetic?ev=prf_pub) and momentum (see this note: https://www.researchgate.net/publication/282134734_The_Lorentz_Rocket?ev=prf_pub)

Best wishes for your research!

Wolfgang

Research Einstein's Third Postulate

Article Relativistic Doppler Eect

Data Classical and Relativistic Derivation of the Sagnac Effect

Article Is a Plasma Diamagnetic?

Research The Lorentz Rocket

To Dr. Leyvraz,

The Noeter theorem is valid for the classical mechanics where the retarded quantites are absent. Both the electrodynamics and the gravity have a deal with the retarded quantities which means that the observer is able to collect information coming to her/him with the speed of light/gravity. So one observer located at point A 'sees' one state of the system, the other observer located at point B 'sees' different state of the system. So it is difficult to determine the translation invariance for the systems in the electrodynamics and the gravity.

I have one more example. McDonald in his analysis of Ivanov's system calculated the Lorentz force acting on the moving charges. He found that sums of the forces (magnetic + retarded electric) leads to non-compensated forceacting on the system. Following Page and Adams paper, he stated that the total momentum conserves

$$

\frac{dP_{mech}}{dt}+\frac{dP_{el-mag}}{dt}=0

$$

But the problem is that change of the mechanical momentum gives acceleration of the system (the Lorentz force acts on the charges) but change of the EM momentum doesn't act on the EM fields - the EM fields instantaneously follow the charges.

In Ivanov's system the center-of-mass oscillates and the only reason why this system has no one-directional motion is that it is impossible to arrange such a motion - because of the form of the Darwin potentials.

I can calculate (actually I did it) the gravitational forces acting on the bodies using the 1PN approximation (as Steven Carlip and Michael Ibison did) and find the non-compensated force. In 1PN approximation the parameter m is the mass + small addition ~ 1/c^2. It should be so because for c -> infinity the general relativing should give Newtonian mechanics.

In this way, I don't need in the Lagrangian.

The Lagrangian is some abstract subject but the forces are observable quantities.

Good! So *you* are the one now claiming that Fock and Robertson made an error in the derivation of their Lagrangian. Because the derivation I sketched follows immediately from the Lagrangian and is not at all difficult.

In GRT, I surely do not know what you mean by saying that forces are observable quantities: is it not a basic hypothesis of GRT that forces can be transformed away by using a change of coordinates. So forces are not fundamental in GRT, and calculating them is nontrivial. 

In any case, I cannot agree to keep on talking on a subject that changes at every post. The problem you posed was clear enough when it dealt with the Lagrangian. I do not believe the 1PN approximation is any different (in fact, in previous posts, you said that the Lagrangian was a systematic approximation). Your computations, lacking any of the velocity-dependent forces arising from the Lagrangian, and not being even vectorial, I cannot believe to be the result of a full 1PN computation.

"Good! So *you* are the one now claiming that Fock and Robertson made an error in the derivation of their Lagrangian. Because the derivation I sketched follows immediately from the Lagrangian and is not at all difficult. "

Very good! A day or two ago I wrote:

"Do you mean Robertson and Fok made errors in derivation of their Lagrangians that lead to non-compensated forces?"

and now you charge me that I claim 'that Fock and Robertson made an error in the derivation of their Lagrangian'.

Conservation of the total momentum requires

$$

\sum_i \frac{\partial L}{\partial r_i}=0

$$

which means that the sum of internal forces is equal to zero. For this system it isn't so - the sum of all forces gives non-compensated coponent. I hope you will be able to check it by yourself.

Moreover, this system contains the mechanical linkages (the rods). For example the system of binary stars contains no such a linkage. Consideration of this linkage should give correct expression for the acceleration

$$

a=\frac{\sum_i F_i}{\sum m_i} instead of a=\frac{\sum_i F_i}{M} where M is the mass f the lower body.

$$

Presence of \sum m_i indirectly shows that the coordinate of the center-of--mass should be involved into calculations.

Just now I am not interested in motion of the system. It has some more other interesting properties.

Finally we agree. However, the expresssion you give, is indeed equal to zero. You can either verify this by explicit computation, or try to understand how it follows from translational invariance. In your case, the former might be recommendable. That the linkage forces add up to zero is, of course, a basic assumption in mechanics.

We absolutely disagree.

The expression has non-compensated addition. It is easily to see by computing the partial derivative of the Lagrangian.

Another way is to compute the force from the scalar potential (the GR vector potential is equal to zero so one doesn't need to compute this term).

To Leyvraz,

The essential property of the Lagrangian is that it depends on the 'present time' variables (Darwin especially mentioned this aspect of the Lagrangian). So to calculate the total force in the system one should compute the partial derivatives of the Lagrangian with respect to the coordinates.

Some days ago I downloaded a file with Fok's Lagrangian. Computation of $\partial L/\partial r_i$ (with accuracy to \gamma^2, r_2 = 0, (v_1*(r_1 - r_2) = 0) gives

$$

\sum_i \frac{\partial L}{\partial r_i}=\gamma\frac{3m_1 m_2 v_1^2}{|r_1 - r_2|^2}

$$

I hope you will be responsible for your words and show how to make this term to be equal to zero.

One more comment. Fok writes (the file in Russian but I translate)

"By making these approximations in Eq. (81.07),we can rewrie it in the form....

One can conclude from this equation that in case of finite motion, The Newtonian center-of-mass oscillates aroung its average point of location"

It means that the center-of-mass of the closed system without external forces moves. Or in other words, the 3rd Newtonian law isn't fulfilled for such systems.

Your computations are simply manifestly wrong. Since the only position dependent terms in the Lagrangian are of the form r_1-r_2, it is clear that all the terms arising in the term partial L/partial r_1 are the same as those arising from partial L/partial r_2, apart from a sign, so that the sum vanishes. This has strictly nothing to do with keeping any order of gamma: it is true exactly within the approximation defined by Fok and Robertson.

As for the Newtonian center of mass, I do not deny that it may move. What is conserved is the total momentum, which is the sum of all momenta

partial L/partial dot r_i

which is, of course, not the total Newtonian momentum, and in particular not the time derivative of the position of the Newtonian center of mass. . What you call Newton's third law may indeed not be valid. But Noether's theorem does, of course, apply to any Lagrangian.

Note in particular that an oscillatory motion of the Newtonian center of mass goes altogether against your own claim, that the system starts moving coherently iin one direction.

$$

\frac{\partial}{\partial r_1} \gamma\frac{m_1m_1}{|r_1 - r_2|}( 1- 3v_1^2/c^2)}=

\gamma\frac{m_1m_1(r_1-r_2)}{|r_1 - r_2|^3}( 1- 3v_1^2/c^2)}

$$

$$

\frac{\partial}{\partial r_2 }\gamma\frac{m_1m_1}{|r_1 - r_2|}( 1- 3v_1^2/c^2)}=-

\gamma\frac{m_1m_1(r_1-r_2)}{|r_1 - r_2|^3}( 1- 3v_1^2/c^2)}

$$

So it seems that $\sum {\bf n}*\frac{\partial L}{\partial r_i}=0$ where n is the unit vector directed along the axis of symmetry.

But it primitive approach. Calculation of the accelerations gives

$$

\frac{d}{dt}\frac{\partial L}{\partial v_1}=a_1 \left(m_1 + \gamma\frac{3m_1m_1 }{|r_1 - r_2|}\right)

$$

$$

\frac{d}{dt}\frac{\partial L}{\partial v_2}=a_2 m_2

$$

The accelerations cannot be different - the bodies are linked by the rigid rods. So additional term in the acceleration is equal to addition of the force. It is just the non-compensated force.

This additional force term can be derived in simpler way directly from the potentials. Actually, the Lagrangian in 1PN order has been derived in the same way as the Darwin Lagrangian has been derived in the electrodynamics, i.e. from the retarded potentials. In the electrodynamics, non-compensated force appears too. For example in the Borh model of the hydrogen atom. I hope you derive the non-compensated force in this case by yourself.

So your arguments based on translation symmetry etc. don't work.

I do not know about ``primitive approach'' You have clearly convinced yourself that the sum of partial derivatives of L wrt r1 and r2 vanishes. If you do everything vectorially, you will see that it is not only the normal component of this which vanishes, but everything. Now, for such complicated systems, you should stick to Lagrange's equations. In fact, an easy way to incorporate the constraints is simply to introduce angle coordinates in which the constraints are automatically satisfied. You should not arbitrarily change Lagrange's equations. Note that, in the presence of constraints, Lagrange's equations read

d/dt (partial L/ partila v_i) = partial L/partial r_i + Q_i

where Q_i is the generalized force. Under any circunstances these will sum to zero as well, so there is no violation of momentum conservation.

In fact, I do believe the top you have described does move, in an oscillatory way. But that does not affect conservation of total momentum and energy, since these do not have their Newtonian forms.

I don't understand why should I rewrite the Lagrange equations. The Fok Lagrangian is written in Cartersian coordinates. I explicitly showed that the accelerations of the upper bodies and the lower body are different. This difference leads to additional force (inverse transformation of the Lagrangian). My aim is to show that the additional force appears but not to study the Fok Lagrangian + constrains.

I can believe in anything but except the belief I have the equations. In opposite you appeal to common principles like 'it cannot be because it can never be' and offer me to check my calculations. Let's check your calculations first.

Simpler explanation of the appearance of non-compensated force.

Despite the effect responsible for appearance of non-compensated force belongs to the general relativity, to explain the appearance of this force the only fact is needed.

The flat metrics of the system should be accepted because the total mass of the system is too small to create curvature of the space-time. Even in calculation of the gravitational field of the Earth, nobody uses the tool of the general relativity because possible correction terms are too small. So in consideration of the system, we should use the Newtonian potential. But according to the special relativity, the speed of any interaction cannot exceed the speed of light. So for the static case the gravitational potential of the body of mass M1 is \Phi1 = \gamma \frac{M1}{R}.

But if the body moves uniformly with the velocity V , its gravitational potential should be

$$

\Phi2=\gamma \frac{M1}{\sqrt{X^2+[1 - (V/c)^2][Y^2+Z^2]}}.

$$

(X = x - Vt)

For the velocity V

You are making everything much too complex.

You have a Lagrangian: let us assume that Fok and Robertson knew what they were doing.

Then the equations of the system you look at are simply

d/dt (partial L/partial dot r_i) = partial L/partial r_i

Now an elementary calculation shows that the right hand sides add up to zero. This simply follows from the fact that the Lagrangian only depends on terms of the form r_i-r_j, so that to any term derived with respect to r_i, there corresponds another term, identical but with opposite sign, corresponding to the same part of the Lagrangian but derived with respect to another r_j. That is easy, and you should be able to check it for yourself. In fact, I see that you have actually done so, at least partly.

Now if the sum of the right-hand sides adds up to zero, do does that of the left-hand sides. This simply means that

P = sum_i partial L/partial dot r_i

is time independent. But P is simply a sum of terms involving the velocities. It is easy to check that P = 0 in your initial configuration, so that P will remain zero forever. This clearly contradicts your idea of perpetual motion.

Of course, you then argue that this is ``naive''. Why? The Lagrangian is the correct description for this complicated system. You keep talking of forces and accelerations, as if these were meaningful in that context. They are not: for example,  the left-hand side of the i-th Lagrange equation as written above contains a linear combination of all bodies' acceleration. Thus it is not reasonable to call the right-hand-side the ``force'' either, since it contributes to all bodies' acceleration.  Calculating with ``forces'' is working in an unduly simplified approximation. Retarded forces cannot be computed by ad hoc calculations such as the one you present: we have a Lagrangian, let us see what the Lagrange equations give.

That, at least, is my suggestion to the perpetuum mobile you suggest.

I explained for some times that the system contain the linkage. The Lagrange equations give different accelerations for the upper and lower bodies. The presence of the linkage doesn't allow for the bodies to have different accelerations. So one should transform the additional acceleration term to provide equal accelerations. It gives additional force.

The Fock Lagrangian obtains from the equations of motion. In Newtonian limit these equations are the same as the 2nd Newton law. Yes, Fock doesn't consider the linkage because for the cosmic bodies there are no rods etc. But Fock writes that the Newtonian center-of-mass oscillates. It doesn't lead to any difficulties since this center-of-mass doesn't moving in one direction.

By using two 2nd Newton laws for the bodies I showed how non-compensated force appears. My aim isn't investigation of symmetry of the Lagrangian of the system. My aim is to show that this force is a result of retardation of the gravitational interaction.

'Retarded forces cannot be computed by ad hoc calculations'.

The retarded forces are not computed by ad hoc calculations. Retardation type of the gravitational potential, as existence of the gravitational waves, follows from the postulate of the special relativity.

If 2nd Newton law cannot be applied to describe this system? No. For small velocities d(P)/dt = md(v)/dt = F

If propagation of the gravitational interaction isn't the wave process? No. Therefore, the gravitational potential is retarded. In all papers on the GR, the authors use not the instantaneous but retarded potentials.

So what is wrong in my equations written above?

What is wrong is that you simplify the problem unduly. Fok and Robertson have made a systematic derivation of a Lagrangian up to order v^2/c^2. Stick to it! There are many more effects in that Lagrangian than mere retardation. In fact, retardation is not explicitly present in the Lagrangian, but taken into account in an appropriate way using the instantaneous coordinates.

No, Newton's laws do not apply: the Lagrange equations combine several accelerations and equal them to a function of velocity and position: this is a very considerable generalization of ``Newton's laws''. You cannot leave the secure foundation of Lagrange's equations to guess ``force laws'': they do not exist in te usual sense: ``forces'', (meaning partial L/partial r_i) in this Lagrangian are velocity dependent,  and cause accelerations to all bodies at once: this is not a good setup in which to discuss any of Newton's three laws.

You say one has to take the constraints into account. Two remarks to this: foirst, physically, it is questionable to use rigid constraints in relativity, since, as we know, rigid bodies have infinite sound velocity, whereas the latter can only go up to c. There are all sorts of paradoxes with relativistic rigid bodies.

Second remark, mathematical: whenever you have a Lagrangian, there is a way to implement constraints, and that will work independently of whether the constraints mke physical sense or not. So we can do this here. We need not worry about whether this is meaningful, and we may even forget about relativity: Lagrangians are ``all the same''. 

The easiest way to implement a holonomic constraint is to go to coordinates, in which the constraint is automatically fulfilled. For example, take the origin at the center of the propeller, define ts position to be z, and the angle of the propeller to be phi. If you then fix the distance between the propeller centre and the lower point to be h, the length between the center and the 4 bodies to be l, then you can find the position and velocities of all 5 bodies in terms of dot z and dot phi. The Lagrangian L then reduces to a two variable Lagrangian, which can be solved exactly. I have done the computations, and the body's velocity is conserved. In fact, I must even retract an earlier statement of mine: it just stays put, without oscillating.

This is the easiest way to take the constraints into account, without leaving the Lagrangian formalism. As I said, any computations involving Newton's equations involve of necessity complex uncontrolled approximations. It is much simpler to stick to a well understood recipe, as the Fok-Robertson Lagrangian.

"Fok and Robertson have made a systematic derivation of a Lagrangian up to order v^2/c^2. Stick to it!"

I don't need to derive the Lagrangian. I uploaded the expression of the Fock Lagrangian to show that the non-compensated force can be obtained from it too. I explained before that Fock doesn't need in any constrain because there are no linkages between the cosmic bodies.

Moreover, I explained for some times that in the Newtonian approximation it is enough to use the potential to calculate the force.

"No, Newton's laws do not apply"

It is something new in the physics. The magnetic field is the relativistic effect but the equation

$$

m\frac{dv}{dt} = F_{Lor}= q(E + [v \times B])

$$

is widely used in the Newtonian approximation.

"Two remarks to this: foirst, physically, it is questionable to use rigid constraints in relativity, since, as we know, rigid bodies have infinite sound velocity, whereas the latter can only go up to c. There are all sorts of paradoxes with relativistic rigid bodies."

Do you seriously think so? Rotation of the system is the stationary process. When Lorentz suggested the hypothesis in 1892 on contraction of the rigid bodies on molecular level to explain the experiment of Michelson, if he didn't know that the speed of light is much greater that speed of sound? According to Lorentz the interferometer arms are being in new state of equilibrium due to the contracted EM forces because it is stationary process and the time of relaxation is enough large for the sound waves (phonons) to provide this state of equilibrium.

Also I hope you will present your calculations with the holonomic constrain.

``$$

m\frac{dv}{dt} = F_{Lor}= q(E + [v \times B])

$$

is widely used in the Newtonian approximation.''

Indeed, but  you are looking at a *post-Newtonian* approximation. I agree you do not need to derive the Lagrangian: Fock and Robertson did that. What you should do is *take it seriously*. Write down Lagrange's equations *IN DETAIL*! You will see they bear little resemblance to Newton's equations: they have complicated velocity dependent forces on the one hand. On the other, the left-hand side (the one with the accelerations) in the i-th equation does NOT only contain the acceleration of the i-th body: this coupling across accelerations is an essential feature of these equations, quite absent in Newton's. That is why I prefer to call them Lagrange's equations, and not to talk carelessly of force and acceleration: that just is not the structure of the equations you have.

Physically, if you take into account the retardation time t_r=L/c, you should also take into account the acoustic time t_a=L/c_S, where c_S is the speed of sound, since that time is much larger. So making a rigid body approximation (t_a=0), but at the same time considering relativistic retardation is inconsistent.

On the other hand, it can in fact be done. The meaning is doubtful, but there is no difficulty in describing the coordinates of all particles by a height Z of the center of the propeller (say) and the propeller's rotation angle phi. Once this is done, it is not hard to substitute all this in the Lagrangian, from which you get a Lagrangian depending only on dot Z and dot phi, which cannot lead to any motion. I leave the detailed calculations to you. In any case, this, and not your somewhat bizarre ways of computing accelerations using generalisations of Newton's laws, which simply are not the correct equations of motion for the system. This simply follows from Fock and Robertson's approach, who showed that their Lagrangian describes the motion correctly.

An incomplete sentence in the previous message: it should read:

In any case, this is the correct way to build consytraints within an existing Lagrangian framework, and not your somewhat bizarre ways of computing accelerations using generalisations of Newton's laws, which simply are not the correct equations of motion for the system.

My apologies.

What acoustic time should I use? Have you read about the Michelson experiment? Lorentz correctly suggested that if the interferometer moves through the absolute ether the distance between the atoms contracts. The velocity of the inteferomenter is much larger the speed of sound in the metal but if it is the obstacle for the atoms to displace to new position of equilibrium?

Other example. International space station moves 8 times faster the speed of sound. But nobody of the crew reported about the sonic effects caused the retardation of the sound. The sound propagates as usual.

Moreover, the vertical rod of the system, where the non-conpensated force is essential, doesn't move. Can you explain what relativistic effects can be found in non-moving rod independently of is it absolutely rigid or not.

On Fock and Robertson derivation of the Lagrangian. I could follow this derivation but I used the 2nd Newton law instead. Do you know that to derive this Lagrangian one should use the 2nd Newton law?

I repeat that I can work in 1PN approximation. But it doesn't mean I should use the Fock Lagrangian with necessity. To derive the gravitational force in 1PN approximation, both Carlip and Ibison use the expression for the scalar potential but not the Lagrangian.

Moreover, I use the only fact: the speed of gravity is finite. The difference in the gravitational potentials follows from this fact with necessity. It is correct in the general relativity, in the special relativity, in the classical mechanics.

One more question: you uploaded some messages and no one equation. Where are your calculations on the absence of non-compensated force?

``On Fock and Robertson derivation of the Lagrangian. I could follow this derivation but I used the 2nd Newton law instead. Do you know that to derive this Lagrangian one should use the 2nd Newton law?''

No, and No. But once more, I do not think you should follow the derivation. You should WRITE DOWN Lagrange's equations for this Lagrangian. Have you ever done it? If you have, you know that the equations do not look at all like Newton's second law. If, as I suspect, you have not, then do it!

As to saying ``to derive this Lagrangian, one should use Newton's second law'': of course not, unless you use Newton's second law in a very broad sense. This Lagrangian concerns relativistic corrections, and so describes---approximately---a much more general theory than Newton's gravity, namely GRT. What you do, on the other hand, is unsystematic: you somehow put retardation in Newton's laws: that is not what you should do: Fock and Robertson gave a systematic derivation of a Lagrangian, and that is why you should use this Lagrangian and nothing else.

As to ``uncompensated force'': for the (hopefully last) time let me reply: what do you call force? I make no calculations on Newton's force, retarded or otherwise, because that is not what should be done. On the other hand, I do compute the right-hand sides of Lagrange's equations, given by

partial L/partial r_i

This sums up to zero, is therefore ``compensated'' in your terminology, simply by looking at which variables appear in which combinations in the Fock-Robertson Lagrangian, so without need to compute anything.

Fianlly: concenring rigidity. You say ``Moreover, the vertical rod of the system, where the non-conpensated force is essential, doesn't move.'' Oops: maybe I did not understand anything after all: are you not saying that a motion will arise in the direction of the vertical rod? And for it to arise, is it not necessary that it should be communicated via the constraint---or equivalently, elastic---forces?

If you push one end of a steel bar of length L, the other end will move after a time L/c_S, where c_S is the speed of sound. For steel, it is about 5 km/sec. So if the steel bar is 1 meter, acoustic retardation is 200 microseconds. Relativistic retardation, on the other hand, is L/c, so roughly 3.3 ns, hence 60'000 times faster. But if you use a constraint, you are assuming that, say, if the propeller moves forward, the lower particle follows instantly. However, if the connection is steel, it only follows after 200 microseconds. This can be a good approximation: I neglect such short times, because I do not care. But then it is inconsistent to say that you do care about relativistic retardation, which is much shorter.

I do not see any connection with Michelson's experiment. Lorentz was thinking about elastic contraction, but we know that there is no such and that we are dealing with relativistic contraction. On the other hand, in your machine, once a part gets moving, the other parts should move as well, carried by the constraint force. But, just as gravity is not instantaneous, neither are constraints. In fact, they are usually *much slower* than gravity.

"As to saying ``to derive this Lagrangian, one should use Newton's second law'': of course not, unless you use Newton's second law in a very broad sense. This Lagrangian concerns relativistic corrections, and so describes---approximately---a much more general theory than Newton's gravity, namely GRT."

I see that you don't know how Fock derived his Lagrangian. The 2nd Newton law is used by Fock before derivation of the Lagrangian (his Eq. 76.01). I don't understand how can you speak about the derivation without looking at it.

I should say too that after your words about the retardation times I realize that you don't understand how this system works. To force the upper bodies to rotate with some speed (for example, 30000 rpm) some hours will be needed. This time is much greater any 'retarded times'.

Also I see that you don't know how Lorentz came to his hypothesis about contraction of the bodies.

' but we know that there is no such and that we are dealing with relativistic contraction.'

Nobody knows it. Within the classical electrodynamics, Lorentz's theory of absolute ether explains the results of all experiments that the SR does.

' On the other hand, in your machine, once a part gets moving, the other parts should move as well, carried by the constraint force. But, just as gravity is not instantaneous'

You don't understand that times of propagation of the gravity play no any role in work of the system. The retardation give 'deformation' of the gravitational potential but this potential is constant with time.

I repeat my question: where are your calculations?

Once more and for the last time: the derivation of Fock's Lagrangian is irrelevant. Let us assume it to be correct. From that assumption, and from a treatment of constraint forces as is usual in Lagrangian mechanics, one gets, essentially without computations, the fact that it does not move. This is due to the fact that, having a Lagrangian, we must use Lagrange's equations, not any modified Newton's equations. In the Lagrangian, the ``forces'', in other words, the right-hand sides of the equations, trivially add up to zero. The constraint forces do not affect this.

As to the issue of retardation: I think we are clearly talking past each other, and I do not see how to make sense of you remarks.

The derivation of the Fock is relevant. For the last time or not, but first you had to look at this derivation. It has been made by means of the 2nd Newton law, or more correctly, via

dP/dt = F

that is equal to

mdv/dt = F

So I can use the 2nd Newton law instead of the Lagrange equation - which you couldn't present.

Regarding your words on the retarded times. If you say so it means that you don't understand why the effect of retardation is esssential but the retardation times aren't.

All your arguments are only the words. I see no one equation written by you.

Work the equations out for yourself! Write down Lagrange's equations and add  them all up. There is no need to do any of the complicated calculations you refer to. But do use Lagrange's equations: they are surely simpler than any Newton's equations, and they must, by construction, be equivalent. In any case, Lagrange's equations are comparatively simple. If you get different results with ``Newton's equations'' than with Lagrange, there must be an error.

In fact I have looked at a derivation of the Lagrangian, namely Landau-Lifshitz, chapter 106. They do not start from Newton's equations,but from Einstein's equations. In any case, the difficulty is how to compute F, or generally speaking the right hand side of the equations. For this, the Lagrange equations are easy, whereas I do not have a clear ide how to derive the ``force''.

"Work the equations out for yourself! "

Why? Why should I write the equations which I uploaded before? I don't need in your directives what should I do. Either you explain by means of the caculations why - to your point of view - the non-compensated force is equal to zero or you admit that you cannot show it.

"But do use Lagrange's equations: they are surely simpler than any Newton's equations".

What is simpler than the 2nd Newton law? This law is studied at the school if I remember.

"If you get different results with ``Newton's equations'' than with Lagrange, there must be an error."

I repeated some times that I am able to obtain the presence of the non-compensated force by two ways.

"In fact I have looked at a derivation of the Lagrangian, namely Landau-Lifshitz, chapter 106. They do not start from Newton's equations,but from Einstein's equations."

It is written in Landau-Lifshitz, chapter 106 after Eq. (106.16) that 'it is that easy to form...', which means that L-L reconstruct the Lagrangian from the Lagrange equations, namely from the 2nd Newton law.

You have asked for comments on the puzzle. I have offered all I am willing to do. From what I have said, you should easily, if you so wished, understand why your calculations are wrong. Let me, as an aside, point out that you never  made a full derivation of what you call Newton's equations. As for Newton's equations being ``simpler'' than Lagrange's, this is obviously wrong: Lagrange's equations were developed, by no less a one than Lagrange himself, one of the greatest exponents in Celestial Mechanics, to simplify problems which are impossibly complex using Newton's equations. In particular, changing coordinates is very easy in the Lagrange formalism. That is why I have told you repeatedly, that you need do no calculations if you attack the problem correctly, by using the Lagrangian and incorporate the constraints by describing the 5 particles using two variables: the height of the whole system and the angle of the propeller. This can be done trivially, and I leave it to you as an exercise. It takes 5 minutes, if you are but willing to reflect a little.

Of course, I am not telling you what to do. I am merely showing you how your problem is solved. You can still go on denying and post complicated calculations. If a very simple reflection, needing no calculations, gives a different result from an intrinsically horrendous computation, it is clear that the horrendous computation should be distrusted.

Recall the beginning of the post: everybody told you they did not believe your expressions for the force. Then you went and gave the Lagrangian as an indication of what should be done. You should, then, be consistent and use that Lagrangian: it does greatly simplify things.

'You have asked for comments on the puzzle.'

I haven't asked for the comments. I asked if anyone can explain why one-directional motion cannot be in this system.

'As for Newton's equations being ``simpler'' than Lagrange's, this is obviously wrong:'

Have you looked at Fock's derivation of his Lagrangian? Fock starts from the equations of motion which are actually the 2nd Newton law (because there are no other equations of motion for the bodies). He obtains the force are then reconstructs the Lagrangian. I repeat again that I can don't follow Fock but use the 2nd Newton law with his expression for the force. It is absolutely corrrect operation.

'I am merely showing you how your problem is solved.'

I don't need in showing by words and saying that you made calculations. If you made calculations you should show them. I am sure you haven't such calculations.

'Recall the beginning of the post: everybody told you they did not believe your expressions for the force. '

To believe or not to believe, it is only emotions. I need in math calculations.

All (correct) ways of solving a problem are identical. You say you do not follow Fock, but some obscure Newton law, which you have never further described. I know that the Lagrangian, which I have reason to believe is correct, gives equations which are not of Newton form. Why, since you posted the Fock Lagrangian, do you so much resist using it?

``If you made calculations you should show them. '' Why? I tell you how to make them, and believe you will learn more, and possibly understand where your own calculations are wrong, if you follow the very explicit scheme I outlined above, you will see why your system stays put. But if you are too arrogant to try and do things in the simple way I sketched, I have nothing to say.

I hoped it is easy to understand that solution based on the 2nd Newton law is simpler. I wrote that if I use the Fock Lagrangian, I should make two additional operations. Fock constructs the Lagrangian from the 2nd Newton law and then makes additional transformation. So I should obtain the Lagrange equations and then make inverse transformation.

The same calculations can be made in electrodynamics. The problem is that the magnitude of non-compensated force is too weak, less 10^(-12) Newtons for the lab installation. I will present calculations in some time.

The easiest way to obtain non-compensated force is to consider Bohr's model of the hydrogen atom. Assuming the proton is being at rest that is good approximation because of the difference in masses of the particles, we find the forces in the system.

The only EM field we should find is the E field. Following Darwin who derived his Lagrangian for this model, we have

$$

\varphi = \frac{e}{R} {\bf A}=\frac{q}{2cR][ {\bf v}+({\bf v}*{\bf n}){\bf n} ]

$$

so the force acting on the proton by the electron is

$$

{\bf F}_1 = -\frac{e^2{\bf n}}{R^2}-\frac{e^2}{2c^2R}\left[{\bf a}+\left({\bf a}\cdot {\bf n}\right){\bf n}+\frac{3\left({\bf v}\cdot {\bf n}\right)^2-v^2}{R}{\bf n} \right]

$$

and the force created by the proton

$$

{\bf F}_2= \frac{e^2{\bf n}}{R^2}

$$

Because for the circular orbit of the electron $a=\dfrac{v^2}{R}$, $\left({\bf a}\cdot {\bf n}\right)=a$, $\left({\bf v}\cdot {\bf n}\right)=0$, and we obtain

$$

\Delta {\bf F}={\bf F}_1 - {\bf F}_2 = \frac{e^2 v^2}{2c^2R^2}{\bf n}

$$

This force arises from the difference in the EM potentials of the charge being at rest and the moving charge, namely from the retardation effect.

The 3rd Newton law is broken in this system but the law of the total momentum conservation isn't broken yet. However, it isn't hard to change a design of the system in order to provide breaking the law of the total momentum conservation too.

Once more, show in detail where these so-called ``forces'' come from. Again, you would be far better off looking at the actual Darwin Lagrangian.

Just as an obvious question: where do you get that the acceleration on the electron is v^2/R? In what reference frame are you working? How do you define the acceleration? Just writing complicated formulas is not enough to solve such a problem: you must argue what the quantiities you look at mean, and you should correctly derive your equations.

It is no difference if I am near or far from the Darwin Lagrangian. The expression for the E field is right.

The acceleration of the electron moving with |v| = const is only centripetal acceleration and it is equal to v^2/R

Because the Darwin potentials contain no retarded quantities, there is no problem to choose the frame. The only requirement is that it should be inertial.

If you don't know how to calculate the derivatives of the Darwin potentials, you should look at the paper of L. Page and N.I. Adams, Jr., Action and Reaction Between Moving Charges, Am. J. Phys. v. 13, 141 (1945). Their Eq. (3) is the same I use.

The Lagrange approach 'ignores' acceleration o f the source - it is only to explain the difference in the 2nd and the Lgrange approaches.