Coordinate acceleration without a Force

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SUMMARY

The discussion centers on the concept of coordinate acceleration in the context of general relativity (GR), specifically contrasting geodesic and non-geodesic motion. The geodesic motion is defined by the equation involving Christoffel symbols, while non-geodesic motion incorporates four-momentum and external forces. The user presents two equations for coordinate acceleration, highlighting a version that omits force, leading to the conclusion that without force, motion cannot be classified as non-geodesic. The inquiry seeks clarification on this relationship.

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  • Knowledge of four-momentum and Christoffel symbols
  • Basic grasp of differential equations in physics
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Coordinate acceleration without a Force !

Hi
GR had presented two types of motion , the geodesic motion and the non-geodesic motion . We know that the geodesic motion equation is :

\[<br /> \frac{{d^2 x^\alpha }}{{d\tau ^2 }} + \Gamma _{\beta \mu }^\alpha \frac{{dx^\beta }}{{d\tau }}\frac{{dx^\mu }}{{d\tau }} = 0<br /> \]
and ( as I know ) the equation of the non-geodesic motion in the derivative of the four - momentum :
\[<br /> F^\lambda = \frac{{dP^\lambda }}{{d\tau }} + \Gamma _{\mu \nu }^\lambda U^\mu P^\nu <br /> \]
and from this equation we can calculate the equation of the coordinate acceleration :
\[<br /> a^\lambda = \frac{1}{m}\left( {\frac{{d\tau }}{{dt}}} \right)^2 \left[ {F^\lambda - \left( {\frac{{u^\lambda }}{c}} \right)F^0 } \right] + \left( {\frac{{u^\lambda }}{c}} \right)\Gamma _{\mu \nu }^0 u^\mu u^\nu - \Gamma _{\mu \nu }^\lambda u^\mu u^\nu {\rm }...{\rm Eq(1) }<br /> \]
but I saw and worked with a different equation for the coordinate acceleation :
\[<br /> a^\lambda = \left( {\frac{{u^\lambda }}{c}} \right)\Gamma _{\mu \nu }^0 u^\mu u^\nu - \Gamma _{\mu \nu }^\lambda u^\mu u^\nu <br /> \]
and I saw that this equation is Eq1 with zero four - force . without force the motion won't be non-geodesic any more .
How ?
thanks
 
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Hello !
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