# Geodesic deviation in flat space

Suppose that $$x^\mu(t,s)$$ represents a family of curves. Let $$v^\mu$$ represents the the tangent vector to a curve $$x^\mu(t,s_0)$$ with $$s_0$$ fixed that is $$v^{\mu}=\partial x^{\mu} / \partial t$$ and deviation vector is given by $$\xi^{\alpha}=\partial x^{\alpha} / \partial s$$.

In the usual derivation of the geodesic deviation equation it is showed that $$\xi$$ is lie transposed through $$v$$ that is $$L_v\xi=0$$ where $$L$$ stands for the lie derivative. To make the discussion more clear let us suppose we are in flat spacetime with usual Cartesian coordinates $$x^0=t,x^1,x^2,x^3$$.

We choose $$x^\mu(t,s)=(t,st,0,0)$$ and so $$v^\mu=(1,s,0,0)$$ and $$\xi^{\alpha}=(0,t,0,0)$$

we have $$L_v\xi=\left[\frac{\partial}{\partial t}+s\frac{\partial}{\partial x^{1}},t\frac{\partial}{\partial x^{1}} \right]=\frac{\partial}{\partial x^1}-\frac{\partial s}{\partial x^1}\frac{\partial}{\partial x^1}$$

So we choose the parameter $$s$$ for example to $$2x^1$$ we would have $$L_v\xi \ne0$$

• What is $s$? What does it stand for? The length of the curve?
– user303670
Commented Jun 8, 2021 at 6:14
• $s$ is a parameter to label different curves Commented Jun 8, 2021 at 7:26
• The patch of geodesics is supposed to be locally parallel I believe? Your geodesics are not parallel, in fact they all intersect at $t = 0$. Perhaps a better patch that illustrates geodesic deviation is $x^{\mu}(t, s) = (t, s, 0, 0)$. Commented Jun 8, 2021 at 9:25

You are missing a $$t$$, your commutator should be
$$L_v\xi=\left[\frac{\partial}{\partial t}+s\frac{\partial}{\partial x^{1}},t\frac{\partial}{\partial x^{1}} \right]=\frac{\partial}{\partial x^1}-t\frac{\partial s}{\partial x^1}\frac{\partial}{\partial x^1}$$
Keeping in mind that $$t=x^0$$ and $$s=\frac{x_1}{x_0}$$, you have that $$t\frac{\partial s}{\partial x^1}=1$$.