I was reading this Wikipedia article which attempts to motivates some concepts key to General Relativity in the Newtonian setting first.

However I was not able to understand one of the equations here.

Context Assumed:

(Below is a clickable screenshot for convenience, but the content can be viewed by reading the first 2-3 paragraphs in the link attached, feel free to mention in comments if you would like me to reproduce the text in whole here):

enter image description here


So I was trying to independently verify the "geodesic equation" of the seperation vector $h$ mentioned in the article. And so I tried to reproduce this system mathematically.

I assume there are two particles of negligibly small mass, called $P_1$ and $P_2$ whose locations at a moment in time are given by

$$ P_1 : \left\{\begin{matrix} x=r\cos(vt) \\ y= 0\\ z = r\sin(vt)\end{matrix} \right| $$

$$ P_1 : \left\{\begin{matrix} x= 0 \\ y= r \cos(vt) \\ z = r\sin(vt)\end{matrix} \right| $$

They can be seen as circularly orbiting at speed $v$ around the origin $(0,0,0)$. lets assume the origin is the location of a point mass with mass $M$.

The centripetal acceleration for either particle then has magnitude $\frac{v^2}{r}$ and we can pick the mass $M$ to be equal to $\frac{v^2r}{G}$ so that the gravitational force equals the centripetal force

With this setup we can now define the seperation vector $H$ as the difference of the locations of $P_1, P_2$.

$$H: \left\{ \begin{matrix} x = r\cos(vt) \\y = -r\cos(vt) \\ z = 0 \end{matrix} \right|$$

Component wise we describe $H$ as $H_x,H_y,H_z$ respectively. It's easy to verify that for any component, (we pick $H_x$ to be concrete) that

$$ \frac{d^2 H_x}{dt^2} + v^2 H_x = 0 $$

If we let $\tau = ct$ then we similarly have

$$ \frac{d^2 H_x}{d\tau^2} + \frac{v^2}{c^2} H_x = 0 $$

But the wikipedia article claims that we should find

$$ \frac{d^2 H_x}{d\tau^2} + \frac{v^2}{r^2 c^2} H_x = 0 $$

And that just isn't consistent with our model.

Where am I going wrong here?

  • 1
    $\begingroup$ $x=r\cdot cos\left( \omega t\right) =r\cdot \cos \left( \dfrac{v}{r}\cdot t\right) $ $\endgroup$
    – Eli
    Sep 23, 2019 at 7:57

1 Answer 1


They can be seen as circularly orbiting at speed $v$...

No. In your equations, $v$ isn’t the speed. It’s the angular velocity, and you should have called it $\omega$. It can’t be a speed because it has dimensions of 1/time, not length/time.


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