# The acceleration of a particle moving at fixed Rindler coordinate $X=\frac{1}{g}$ [closed]

Let us define the transformation from the Minkowski coordinates $$x^\alpha=(t,x)$$ to the Rindler coordinates $$x^{\bar\alpha}(T,X)$$ by $$t=X\sinh T,\\ x=X\cosh T.$$ If an object is moving at fixed Rindler coordinate $$X=1/g$$ and $$ds^2$$ for Minkowski coordinates is given by $$ds^2=-dt^2+dx^2$$, I'd like to show that the 2-acceleration $$a^{\bar\alpha}$$ is given by $$a^{\bar\alpha}=(a^T,a^X)=(0,g).$$ I tried to figure it out by using covariant derivatives, because I have successfully obtained the Christoffel symbols for the Rindler coordinates: $$\Gamma^X_{TT}=X,\Gamma^T_{TX}=\Gamma^T_{XT}=\frac{1}{X}$$ After observing the complicated formula $$a^{\bar\alpha}=\frac{d^2}{d\tau^2}x^{\bar\alpha}+\Gamma^{\bar\alpha}_{\bar\nu\bar\lambda}\frac{dx^{\bar\nu}}{d\tau}\frac{dx^{\bar\lambda}}{d\tau},$$ though, I didn't see any promising future for this approach. Am I on the right track? Thank you.

In your notation $$T$$ is proper time and your $$\tau$$ is redundant because it equals $$T\,.$$ Some further notation of yours is a bit confusing. Did you mean $$X=1/a$$ (title) or $$X=1/g$$ (text)? In any case the coordinate change from Rindler $$(T,X)$$ to Minkovski $$(t,x)$$ that I am familiar with is \begin{align} t&=X\sinh(\rho\,T)\,,\\ x&=X\cosh(\rho\,T)\,. \end{align} The world line of the object in Rindler coordinates is $$x^{\overline{\alpha}}=(T,X)$$ where $$X$$ is fixed. It is known [1] that $$\rho=1/X$$ is the proper acceleration of the object. To check this observe that the metric tensor is $$g_{\alpha\delta}=\left(\begin{matrix}-\rho^2X^2 & 0\\0&1\end{matrix}\right)\,,\quad g^{\alpha\delta}=\left(\begin{matrix}-\frac{1}{\rho^2X^2} & 0\\0&1\end{matrix}\right)$$ so that $$g_{TT,X}=-2\rho^2X$$ and all other partial derivatives being zero. Therefore, $$\Gamma^\delta_{\beta\gamma}=\frac{1}{2}g^{\alpha\delta}\left(g_{\alpha\beta,\gamma}+g_{\alpha\gamma,\beta}-g_{\beta\gamma,\alpha}\right)$$ becomes \begin{align} \Gamma^T_{TX}&=\frac{1}{X}\,,&\quad\delta=\beta=T,\gamma=X\,,\\ \Gamma^T_{XT}&=\frac{1}{X}\,,&\quad\delta=\gamma=T,\beta=X\,,\\ \Gamma^X_{TX}&=0\,,&\quad\beta=T,\delta=\gamma=X\,,\\ \Gamma^X_{XT}&=0\,,&\quad\gamma=T,\delta=\beta=X\,,\\ \Gamma^T_{TT}&=0\,,&\quad\beta=\delta=\gamma=T\,,\\ \Gamma^X_{TT}&=\rho^2X\,,&\quad\beta=\gamma=T,\delta=X\,,\\ \Gamma^X_{XX}&=0\,,&\quad\beta=\gamma=\delta=X\,,\\ \Gamma^T_{XX}&=0\,,&\quad\beta=\gamma=X,\delta=T\,. \end{align} Note the factor $$\rho^2$$ in $$\Gamma^X_{TT}$$ that you missed. Therefore,

\begin{align} a^T&=\underbrace{\frac{d^2T}{dT^2}}_{0}+\underbrace{\Gamma^T_{TX}\frac{dT}{dT}\frac{dX}{dT}}_{\frac{1}{X}\cdot 1\cdot 0=0}+\underbrace{\Gamma^T_{XT}\frac{dX}{dT}\frac{dT}{dT}}_{\frac{1}{X}\cdot 0\cdot 1=0}=0\,,\\ a^X&=\underbrace{\frac{d^2X}{dT^2}}_{0}+\underbrace{\Gamma^X_{TT}\frac{dT}{dT}\frac{dT}{dT}}_{\rho^2X\cdot 1\cdot 1}=\rho^2X\,. \end{align} Since $$X$$ is constant equal to $$1/\rho$$ we get the 2-acceleration as $$\boldsymbol{a}=\left(\begin{matrix}0\\\rho\end{matrix}\right)$$ as expected.

• I'm very sorry to keep you waiting. I didn't notice your reply. Yes, $X=\frac{1}{g}$ in the title. I have fixed the inconsistency. Thank you.
– Boar
Commented Nov 21, 2021 at 20:46