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A Lagrangian Solution for the Precession of Mercury's Perihelion

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A Lagrangian Solution for the Precession of Mercury's Perihelion International Journal of Astronomy 2014, 3(2): 31-34 DOI: 10.5923/j.astronomy.20140302.01 A Lagrangian Solution for the Precession of Mercury’s Perihelion Adrián G. Cornejo Electronics and Communications Engineering from Universidad Iberoamericana, Col. Santa Mónica, Querétaro, Mexico Abstract A Langrangian solution is proposed for the apsidal precession by considering the precession of a body orbiting and rotating in a rotating reference frame, deriving an analogous equation to the relativistic solution. Keywords Rotating Reference Frame, Lagrangian Solutions, General Theory of Relativity, Precession of Mercury’s Perihelion 1. Introduction distance from the spin axis to the final position of a point P that according to the fixed frame of reference is the radius The rotating reference frame is a frame that rotates about vector named rf while some position of the same point a spin axis (fixed for simplicity) with a given angular according to the rotating reference frame is named r, and velocity. Formalism to describe a rotating reference frame r= rR + , (1) is commonly given by the Lagrangian mechanics. In this f kind of reference frame, an angular velocity of rotation is where R denotes the position from the origin of the rotating present in the particles within the rotating reference frame frame according to the fixed frame. Furthermore, the rotating (like the particles in a whirl as described by the fluid reference frame is described by the Lagrangian [2] for the mechanics or the particles on a solid body in rotation as motion of a particle with mass m in a rotating frame (with its described by the classical mechanics). This additional origin coinciding with the fixed-frame origin) in the presence angular velocity in the periodic motion causes a precession of the potential U(r), defined as towards the sense of rotation, as it is the case of the m 2 gyroscope. This effect is also present in some other natural LU(,)rr= r +Ω× r − ( r ), (2) phenomena where a rotating frame is involved, such as the 2 Coriolis effect and the Foucault’s pendulum. However, where Ω is the angular velocity vector of the rotating could this kind of effect also be present in the apsidal reference frame, and precession? 22 A Langrangian solution is proposed for the apsidal r+Ω× r =r +2 Ω( rr ×) + Ω22 r −( Ω r ). 2 (3) precession (like the precession of Mercury’s perihelion) by considering the precession of a body orbiting and rotating in Moreover, the equation for the canonical momentum is a rotating reference frame, deriving an analogous equation given by to the relativistic solution. ∂L p= =m( rr +Ω× ), (4) ∂r 2. Lagrangian Formulation for the and Rotating Reference Frame dL∂ In this section, certain aspects of the rotating reference =m(rrr +Ω× +Ω× ). (5) frame will be reviewed through the general case of a rotating dt ∂r reference frame and a fixed frame where a body is orbiting Then, the partial derivative is given by and rotating about a spin axis [1]. Let us consider the ∂L =−∇Um()rr − Ω× +Ω×( Ω× r) . (6) * Corresponding author: ∂r [email protected] (Adrián G. Cornejo) Published online at http://journal.sapub.org/astronomy By applying Lagrangian from Eq. (2), we derive the Copyright © 2014 Scientific & Academic Publishing. All Rights Reserved general Euler-Lagrange equation for r term [3], defined as 32 Adrián G. Cornejo: A Lagrangian Solution for the Precession of Mercury’s Perihelion mr=−∇ Um() r − Ω× rr + 2 Ω× +Ω×( Ω× r) .(7) GMm 1 22 V()r=− + mm Ωr − Ω× vrf . (16) r 2 Here, we can see that potential energy generates the fixed-frame acceleration, -RotU = maf, and that the This Lagrangian solution for gravitational potential Euler-Lagrange equation (7) takes the form energy can be written in terms of the angular momentum of a body orbiting in circular motion, which in polar coordinates aarf= − A −Ω×( Ω× r) −2, Ω× vr −Ω× r (8) is defined as 2 which represents the sum of the net acceleration (af -A), Lr= m ω, (17) where centrifugal acceleration is given by -Ω × (Ω × r), and being ω the angular velocity of the body in circular orbit. ac = -2Ω × vr is the Coriolis acceleration that only depends Nevertheless, this angular velocity does not consider any on velocity [4]. Being final velocity vf equivalent to the rotation from a rotating reference frame. Then, considering Coriolis velocity vc, it is defined as that the angular velocity of a body with respect to the rotating v f reference frame can be defined as vvcf= =2. Ω× r ∴Ω = (9) 2r 2 LrΩ =m Ω, (18) In terms of velocity, from Eq. (8) yields being Ω the angular velocity of the rotating reference frame. 2 2 22 2 In addition, the second term in Eq. (16) corresponds to the vrf= v −2. Ω× vr f +Ω × r =( vf −Ω× r) (10) angular velocity of the body in circular motion (as orbiting) Simplifying and reordering, we can write this equation as around the centre, while the third term corresponds to the angular velocity of the body with respect to the rotating vv= +Ω× r. (11) fr reference frame. Then, having this correspondence, Eq. (16) becomes Then, final velocity (like the final orbital velocity vo) of a body orbiting and rotating in a rotating reference frame is the GMm 1 22 addition of both, velocity of the body at the distance r from V()r=− + mmω r − Ω× vrf . (19) the centre and the angular velocity of the rotating reference r 2 frame at the same distance from the spin axis. We can write Eq. (19) in terms of angular momentum throughout its transformation to an equivalent equation that includes terms of the angular momentum, for instance, by 3. Total Energy in the Rotating multiplying the second and third terms, respectively, by Reference Frame some complementary terms which are equals to unit, giving 2 In the rotating coordinate system [5], total energy on the GMm 1 mr mω× vrf V()r= − + mmω22r− Ω×vr.(20) orbital motion is defined as 2 f r 2 mmr ω× vrf EE= + V(r ), (12) K Having angular velocity for the Coriolis effect from Eq. where EK is the kinetic energy and V(r) is the effective (9), gravitational potential energy can be written as gravitational potential energy. From Eq. (10), the total 2 energy for a rotating reference frame is given by mr2ω 2 22 GMm ( ) vvffm rω V ()r =−+− . (21) 112 2 2 ω× EmV=vrrf +() = m( v −Ω× r) + V(). r (13) r22mm r r vrf 22 Now, we will apply some equivalences such as the Gravitational potential energy is described between a equivalence between the escape velocity and the final orbital massive body M and another given mass m orbiting at a velocity [5], which is defined as distance r. Then, the total energy can be written as 2GM 1 GMm = = =2 − Ω× +Ω22 × − vvef2, (22) Em(vff2, vr r) (14) r 2 r and the square of velocity of a body in a circular orbit about a where G is the Newtonian constant of gravity. Rearranging central mass [6], given by this equation, we rewrite the new equation as 22GM = = (23) 112 GMm 22 vv0 f . mEvff= − − + m Ωr − m Ω× vr, (15) r 22r Then, considering that the velocity of a body in circular where second term is equivalent to the effective gravitational motion is given by v = ω×r, replacing Eq. (22) and reducing potential energy, given by common terms, Eq. (21) becomes International Journal of Astronomy 2014, 3(2): 31-34 33 222 22 Coriolis force, given by GMm (mmrωω) vrf ( r ) V ()r =−+− . (24) 2 2 2 23 3GML2 3GM( mr ω) r2mm r vre F()r = = , (29) 24 24 According to the equations (17) and (23), we can write the mmvree vr gravitational potential energy equation from Eq. (24) in having that velocity of the body in rotation is given by v = terms of the angular momentum of a body in circular motion, ω×r and reducing common terms, the angular momentum of giving inertia is written as GMmLL22 GM 3GMmvω V ()r =−+− , (25) MF=×=rr() . (30) 2 23 0 2 r2mm r vre ve and by the radius derivative on Eq. (25), while the angular The angular velocity of precession for a body rotating momentum and the escape velocity are kept constants, the about a central point can be calculated by replacing Eq. (30) net (or absolute) force is given by in Eq. (28), given by 22 dV()r GMm LL3 GM dϕ 31 GMmvω =−=F − + . (26) = . (31) dr r2m r 3 m vr 24 22 e dt vre m ω Thus, the magnitude of the net force is given by Replacing the velocity of the body in rotation by ω×r and GMmLL223 GM reducing common terms, yields F = −+ . (27) 2 3 24 33GMω dt GMdθ rmm r vre dϕ = = , (32) 22 The first and second terms are the gravitational force and rvee rv the centrifugal force, respectively. The third term includes where dθ is the differential of the radial angle covered by the the Coriolis effect present in a rotating reference frame, that rotation of the body in the rotating reference frame. is inversely proportional to the radius to the fourth power (r4), which is also included in the relativistic solution [7]. 5. Analogous Equation to the Relativistic Solution 4. The Precession in a Rotating Now, considering a body in elliptical orbit around a Reference Frame central point in a rotating reference frame (Figure 1), Precession results from the angular velocity of rotation distance from the centre changes according to the ellipse, 2 and the angular velocity that is produced by the torque.
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