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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?

amilton moreira
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2 Answers2

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Here is a complete rewriting of my answer. I insist that the problem is that you are making confusion between the two "$t$" variables, and I suggest anybody reading this to avoid using the same symbol for two completely different coordinates patch.

I will restrict to the two dimensional case as there is nothing to say about the two last variables. You are considering the function \begin{align} f \colon \mathbb{R}^2 &\to \mathbb{R}^2 \\ (t,s) & \mapsto (t,st) \end{align} This function is, away from the line $\{t=0\}$, a diffeomorphism onto its image: it is injective, and its Jacobian is $t\neq 0$. Use the coordinate system ${x^1,x^2}$ to describe $\mathbb{R}^2$ on the RHS. Then: \begin{align} v(t,s)=\frac{\partial f}{\partial t}(t,s) &= \partial_1|_{f(t,s)} + s \partial_2|_{f(t,s)},\\ \xi(t,s) = \frac{\partial f}{\partial s}(t,s) &= t \partial_2|_{f(t,s)}. \end{align}

Here is the trick: in terms of the coordinates $(x^1,x^2)$, we have $t(x^1,x^2) = x^1$ and $s(x^1,x^2) = \frac{x^2}{x^1}$. Hence, you can write, away from $\{x^1=0\}$, $v$ and $\xi$ as vector fields on the RHS with: \begin{align} v &= \partial_1 + \frac{x^2}{x^1}\partial_2,\\ \xi &= x^1 \partial_2. \end{align} Therefore: \begin{align} [v,\xi] &= \left[\partial_1 + \frac{x^2}{x^1}\partial_2, x^1\partial_2\right] \\ &= \left[\partial_1,x^1 \partial_2\right] + \left[\frac{x^2}{x^1}\partial_2,x^1\partial_2\right]\\ &= \partial_2 + \left[\frac{x^2}{x^1}\partial_2,x^1\partial_2\right]. \end{align} Use the formula $\left[fX,gY\right] = fg[X,Y] + f \left(X\cdot g\right)Y - g \left(Y\cdot f\right)X$, it follows that: \begin{align} \left[\frac{x^2}{x^1}\partial_2,x^1 \partial_2 \right] &= \frac{x^2}{x^1}\cdot x^1 \underbrace{[\partial_2,\partial_2]}_{=0} + \frac{x^2}{x^1}\underbrace{\left( \partial_2 x^1\right)}_{=0}\partial_2 - x^1 \cdot \underbrace{\partial_2\left(\frac{x^2}{x^1} \right)}_{= \frac{1}{x^1}} \partial_2 \\ &= - \partial_2, \end{align} and the result follows.

(This is not surprising: $f$ restricted to $\{t\neq 0\}$ being a diffeomorphism onto its image, by naturality of the Lie bracket: $[v,\xi]=[\partial_tf,\partial_sf] = [f_*\partial_t,f_*\partial_s] = f_*[\partial_t,\partial_s] = f_*(0) = 0.$)

Didier
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Away from $t=0$ you can think of $x$ as a regular parametrization of a surface in your 4D space. The vector fields you call $v$ and $\xi$ are just the coordinate vector fields usually called $x_t$ and $x_s$. Their Lie bracket vanishes because $x_{st}=x_{ts}$.

John Sullivan
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