Difference between these two Rindler metrics

Question

I am uncertain about the difference between these two Rindler metrics:

$$ds^2 = -\left(1 + \frac{\alpha x}{c^2}\right)^2 c^2dt^2 + dx^2+dy^2+dz^2$$

$$ds^2 = -\left(\frac{\alpha x}{c^2}\right)^2 c^2dt^2 + dx^2+dy^2+dz^2$$

For the first metric, according to this answer, geodesic of a free particle reduce to the familiar form $\frac{dx^2}{dt^2}\approx-\alpha$ under Newtonian limit.

On the other hand, when I use the second metric, I expected to obtain the same result $\frac{dx^2}{dt^2}\approx-\alpha$ but I got

$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)^2=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\alpha\frac{c^2}{\alpha x}=-\frac{c^2}{x}$$

I am not sure how to make sense of this Newtonian limit. I wonder if $x$ is interpreted differently in the second metric.

---

It turns out that there's an issue with how I approximate the $\alpha x/c^2$ term.

An observer in a spacetime described by the second Rindler metric can be likened to an observer standing on an infinite plane of uniform gravitational field. The coordinate $x$ measures vertical distance. I think the plane is located at $x=c^2/\alpha$. An object placed at $x>c^2/\alpha$ is located above the plane. An object placed at $x<c^2/\alpha$ is located below the plane.

If I assume that an object is at rest somewhere near $c^2/\alpha$ at $t=0$, then $\alpha x/c^2\approx1$. Then, the equation of motion becomes

$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)^2=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\alpha\frac{1}{1-\frac{v^2}{c^2}}\approx-\alpha$$

On the other hand, if the object is at rest at $x=\varepsilon$ where $\varepsilon$ is infinitesimally small quantity. We have

$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)^2=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\frac{c^2}{\varepsilon}$$

As $\varepsilon\to0$, the magnitude of coordinate acceleration will be very large. I am not sure how it is interpreted physically...

Answer 1

I must admit that there are points to the question you linked that are unclear to me. In particular as reported by FUGSUZ in this answer the Newtonian limit includes a set of approximations: slow velocities, weak gravitational field and stationary field.

In my opinion, the derivation of FUGSUZ is very clear and there is no point in reporting it here, so I just use the result: $$\frac{d^2 x}{dt^2} - \frac{1}{2} \partial_x h_{tt} = 0, $$

which in the generalization given in here, can be written as

$$m \left(\frac{d^2 x}{dt^2} - \frac{1}{2} \partial_x h_{tt} \right)= f_x$$

where $h_{tt}$ can be considered as a small perturbation $g_{tt} \simeq \eta_{tt} - h_{tt}$.

• First Rindler metric $g_{tt} = -\left(1 + \frac{\alpha x}{c^2} \right)^{2} c^{2} = -c^2 +h_{tt} $ where $$h_{tt} = -2 \alpha x -\left(\frac{\alpha x}{c^2}\right)^2 c^2 \simeq -2 \alpha x$$ $$m \left(\frac{d^2 x}{dt^2} - \frac{1}{2} \partial_x h_{tt} \right) =m \left(\frac{d^2 x}{dt^2} - \frac{1}{2} \partial_x(-2 \alpha x) \right) = m\frac{d^2 x}{dt^2} + m\alpha = f_x $$

• Second Rindler metric $$g_{tt} = -\left(\frac{\alpha x}{c^2} \right)^{2} c^{2} = -\left(1-1+\frac{\alpha x}{c^2} \right)^{2} c^{2} \equiv -\left(1 + \beta(x) \right)^{2} c^{2}$$ where $$\beta(x) = -1+\frac{\alpha x}{c^2}. $$ As before then $$h_{tt} = -2c^2\beta(x)-c^2\beta^2(x) \simeq -2c^2\beta(x)$$ $$m \left(\frac{d^2 x}{dt^2} - \frac{1}{2} \partial_x(-2c^2\beta(x)) \right) = m\frac{d^2 x}{dt^2} + m\alpha = f_x $$

Thus it is clear that the two metrics give the same Newtonian limit as it should be.

Answer 2

The answer to your question is in the correct understanding of the Rindler metric. Everywhere, I have set $c=1$

Consider an accelerating observer with constant acceleration of $\alpha$ in the $x$ direction. We now determine different properties seen by an inertial frame with coordinates $(t,x)$ for this accelerated observer. Let us first look at this observer through an inertial frame. Let the accelerated observer have velocity $(u^t,u^x)$, and acceleration $(a^t,a^x)$ in the inertial frame and let their proper time be $\tau$, then the conditions \begin{align} u^2 = -1 \\ a^\mu u_\mu= 0 \\ a^2 = \alpha^2 \end{align} give us \begin{align} u^t =\cosh{\alpha \tau} \\ u^x =\sinh{\alpha \tau} \end{align} and \begin{align} t = \alpha^{-1}\sinh{\alpha \tau} \\ x = \alpha^{-1}\cosh{\alpha \tau} \end{align} where I have set the that the accelerated observer is at position $(t,x)=(0,\alpha^{-1})$ at $\tau =0$.

It is easy to see that the spacetime is still flat and we are still in Special Relativity.

Now, if we ask how the accelerated observers perceives the spacetime around them, we note the transformation to an accelerated reference system, while certainly allowed in special relativity, is not a Lorentz transformation. Let the accelerated observer have a coordinate system $(T,X)$.

Then one finds(using the usual way of finding lines of constant $T$ and constant $X$) \begin{align} t = X \sinh{\alpha T} \\ x = X \cosh{\alpha T} \end{align} which gives \begin{align} ds^2 &= -dt^2 + dx^2 \\ &=- \alpha^2 X^2 dT^2 + dX^2 \end{align} which is the Rindler metric. Now what is left is to show that an observer at $x=X=\alpha^{-1}$ at time $t=T=0$ feels an acceleration of $\alpha$ which is almost where you left your solution. In the Rindler metric as you have calculated the acceleration is $\frac{1}{X} = \alpha$ at time $t=T=0$. But as the acceleration is constant, this is the same at all times.

Hope this helps!