# Understanding the geodesic equation in a Wikipedia article

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):

## Problem:

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?

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

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.