Category: General Relativity

This note can be use as an example of tensor analysis

This article shows the derivation of Newtonian tidal effect from Newton’s equation of motion and gravity, with the notation of tensor analysis and Einstein summation convention. This may serve as an example for beginners in tensor analysis to check their comprehension, meanwhile as a baby version or lead-in to geodesic deviation of general relativity.

Newton’s second law (componentwise)

$$
\frac{d^{2} x^{i}}{d t^{2}} \equiv a^{i} \equiv \frac{F^{i}}{m}
$$

Newtonian equation of gravitation

$$
\frac{F^{i}}{m}=-\eta^{i j} \partial_{j} \phi
$$

combine two equations give:

$$
\frac{d^{2} x^{i}}{d t^{2}}=-\eta i j\left[\partial_{j} \phi\right]_{\vec{x}}
$$

where the subscription $\vec{x}$ denotes the derivative is operated at position $\vec{x}$ similarly we have at $\vec{x}+\vec{n}$ :

$$
\frac{d^{2}\left(x^{i}+n^{i}\right)}{d t^{2}}=-\eta^{i j}\left[\partial_{j} \phi\right]_{\vec{x}+\vec{n}}
$$

make subtraction give:

$$
\frac{d^{2} n^{i}}{d t^{2}}=-\eta^{i j}\left(\left[\partial_{j} \phi\right]{\vec{x}+\vec{n}}-\left[\partial{j} \phi\right]_{\vec{x}}\right)
$$

given $\vec{n}$ is infinitesimal, we have RHS:

$$
\text { RHS }=-\eta^{i j} n^{k}\left[\partial_{k}\left(\partial_{j} \phi\right)\right]_{\vec{x}}
$$

set Cartesian coordinates originated from centre of earth orient the $z$-axis so that is consists with $\vec{x}$ (position vector) then $\vec{X}=\left(X^{1}, X^{2}, X^{3}\right)=(0,0, z)$.

gravitational potential is given by:

$$
\phi=-\frac{G M}{r}=-\frac{G M}{\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}
$$

and thus

$$
\begin{aligned}
\partial_{j} \phi & =\frac{\partial \phi}{\partial x^{j}} \\
& =-\frac{d \phi}{d r} \frac{\partial r}{\partial x^{j}} \\
& =\frac{G M}{r^{3}} x^{j}
\end{aligned}
$$

and

$$
\begin{aligned}
\partial_{k} \partial_{j} \phi & =\frac{\partial}{\partial x^{k}}\left(\frac{G M}{r^{3}} x^{j}\right) \\
& =\frac{\partial}{\partial x^{k}}\left(\frac{G M}{r^{3}}\right) x^{j}+\delta_{k}^{j} \frac{G M}{r^{3}} \\
& =-\frac{3 G M}{r^{4}} x^{j}+\delta_{k}^{j} \frac{G M}{r^{3}}
\end{aligned}
$$

thus.
$$
\begin{aligned}
& \frac{d^{2}}{d t^{2}} n^{x}=-\frac{G M}{r^{3}} n^{x} \\
& \frac{d^{2}}{d t^{2}} n^{y}=-\frac{G M}{r^{3}} n^{y} \\
& \frac{d^{2}}{d t^{2}} n^{z}=\frac{2 G M}{r^{3}} n^{z}
\end{aligned}
$$