In a paper I was reading, I came across the Newtonian geodesic deviation equation $${\ddot{\eta}}^a + K{^a}{_b}{\eta^b}=0$$ Where $\eta{^a}$ are the components of the 3 -vector separating the two particles and $K{^a}{_b} = \phi{^{,a}}{_{,b}}$ and $\phi$ is the Newtonian potential and commas mean derivatives.

Extending the same idea, the author comments that if the reference orbit is circular with radius R, the geodesic deviation equation in the barycentric frame is given by $${\ddot{\eta}}^a + \frac{GM}{R^5}({R^2\delta{^a}{_b} }- 3{X^a}{X_b}){\eta^b}=0$$ where $X^a$ is the vector $(X,Y,Z)$. I did not quite get how did the second term aka $3{X^a}{X_b}){\eta^b} $was derived. If someone could guide me with the derivation, it would be of great help. Thanks in advance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.