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 $L_v\xi \ne0$

Isn't this a contradiction?

  • $\begingroup$ What is $s$? What does it stand for? The length of the curve? $\endgroup$
    – user303670
    Jun 8, 2021 at 6:14
  • $\begingroup$ $s$ is a parameter to label different curves $\endgroup$ Jun 8, 2021 at 7:26
  • 1
    $\begingroup$ 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 $x^{\mu}(t, s) = (t, s, 0, 0)$. $\endgroup$ Jun 8, 2021 at 9:25

1 Answer 1


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 $t\frac{\partial s}{\partial x^1}=1$.


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