Global Journal of Researches in Engineering, A: Mechanical & Mechanics, Volume 22 Issue 1

1/2 1 1 2 1/2 4 e l B × = di r o π µ . (1.4.2) but i d q 1 1 1 1 l v = so, for point charges, we can write for the force on charge 2 due to charge 1 is F e v v e 2 1 0 1 2 2 1 2 2 1 1 2 2 1 2 2 1 2 1 1 4 4 / / / / / ( ) = + × × π ε µ π q q r q q r o (see also page 256 of ref. [7]) More detailed analysis of the electrodynamics shows that the right hand side of the above equations should be multiplied by γ , but is usually omitted for small velocities. The speed of light c o o = 1 ε µ so we have               × ×     + = 1/2 1 2 1/2 2 1/2 21 1/2 4 e v v e F c c r qq o πε γ (1.4.3) For mass elements we assume that               × ×     + −= 1/2 1 2 1/2 2 1/2 2 1 1/2 e v v e F c c r mGm γ (1.4.4) Here c , the speed of gravity waves, is assumed to be the same as that of light. The form of this equation will be taken as a guide; the problem is to choose the most appropriate values for the velocities. The assumptions that are made in the following development are: a) That the field generated by body (1) depends on the velocity of body (1) relative to body (2) i.e. ( v 1/2 ) and b) The force on body (2) depends on the velocity of body (2) relative to the field i.e. ( v 2/1 ). Let v 2/1 = v, therefore v 1/2 = - v. It must be emphasised that force and force fields are inventions solely for the purpose of visualisation. Substituting these values into equation (1.4.4) gives 2/1 = − 1 2 2/12 � / − � � × � × / �� = − 1/2 (1.4.5) For the Newtonian case the acceleration of body (2) relative to body (1) is 1 1 ) ( / 2 1 2 1 / 2 1 / 1 1/2 2 1/2 1 2 µ 12 12 12 F F F F F r r a =      + =       + = − − = − = mm mm m m m m   where 2 1 2 1 mm mm + = µ is known as the reduced mass. For the relativistic case we shall define force as ( ) ( ) 1 1 where 2 2 c v c dt d − =       ⋅ + = = γ µ γ γ γµ vav a v F 2 3 Using this definition of force we obtain, using equation (1.4.5), ( )               × ×     − + −= ⋅ + 1/2 1/2 2 1/2 2 1 2 3 ) ( e v v e vav a c c r mmG c γ γ γ or ( )               ⋅ −      + + −= ⋅ + 1/2 2 1/2 1/2 2 1/2 2 1 2 ) ( e vv e e vav a 2 cc c v r mmG c γ (1.4.6) Where m 1 and m 2 , are the respective rest masses. Change the suffix of the unit vector to r , let K = G(m 1 + m 2 ) and let r = | r 2/1 | be the separation. Equation (1.4.6) now becomes ( ) v e vav a 22 2 2 2 2 2 1 cr Kv c v r K c r r +       + −= ⋅ + γ (1.4.7) Note that this equation is symmetric with regard to the two bodies. The scalar product of equation (1.4.7) with v yields 2 2 2 2 1) ( r Kv c v r −=      + ⋅ γ va By definition, ( ) γ 2 2 2 1 1 = − / / v c therefore © 2022 Global Journals Global Journal of Researches in Engineering (A ) Volume XxXII Issue I Version I 4 Year 2022 Gravitomagnetics a Simpler Approach Applied to Dynamics within the Solar System