2
$\begingroup$

I understood from the equivalence principle that an accelerated observer in free space is equivalent to a stationary observer in a gravitational field. As far as I understood further, this means to analyze systems which occur at the earth's surface it is possible to use the Rindler Coordinates. However the 00 element of the rindler metric contains a term $\alpha x$ where $\alpha$ is the proper acceleration and $x$ is the position. It is clear what to substitute for $\alpha$, but what number should one plug in for $x$?

$\endgroup$
2
$\begingroup$

$\let\pd=\partial \def\dxi{\dot\xi} \def\ddxi{\ddot\xi} \def\const{{\rm const.}}$ Let me recall that Rindler coordinates describe a reference frame with constant acceleration in a flat spacetime. But in that frame the apparent gravitational field is not uniform.

I think a brief review of the matter is in order. I'll use my favourite notations: $c=1$, metric signature $(+,-,-,-)$. In usual Minkowsky coordinates $(t,x)$ $$d\tau^2 = dt^2 - dx^2.\tag1$$ Rindler coordinates $(\xi,\eta)$ are defined by $$\eqalign{ t &= \eta\,\sinh \xi \cr x &= \eta\,\cosh \xi \cr} \tag2$$ ($\xi$ timelike, $\eta$ spacelike). For further reference, observe that $$x^2 - t^2 = \eta^2.$$ Substituting (2) into (1) $$d\tau^2 = \eta^2 d\xi^2 - d\eta^2.\tag3$$

Let's consider a point P, moving with $\eta=\eta_1=\const$ From (2) we have for the 4-velocity components of P $$\eqalign{ u^t &= {dt \over d\tau} = {\pd t \over \pd \xi}\,\dxi = x\,\dxi \cr u^x &= {dx \over d\tau} = {\pd x \over \pd \xi}\,\dxi = t\,\dxi \cr}$$ (dot means $d/d\tau$) and since $\left({u^t}\right)^2-\left({u^x}\right)^2=1$ we get $$1 = (x^2 - t^2)\,\dxi^2 = \eta_1^2\,\dxi^2$$ $$\dxi = {1 \over \eta_1}.\tag4$$ We can safely assume $$\xi = {\tau \over \eta_1}$$ and inserting in (2) we find the parametric equations of P's worldline: $$\eqalign{ t &= \eta_1\,\sinh {\tau \over \eta_1} \cr x &= \eta_1\,\cosh {\tau \over \eta_1}.\cr} \tag5$$ We also have $$\eqalign{ u^t &= \sinh {\tau \over \eta_1} \cr u^x &= \cosh {\tau \over \eta_1}.\cr}.\tag6$$

Let's now compute 4-acceleration: $$\eqalign{ a^t &= {du^t \over d\tau} = {1 \over \eta_1}\,u^x \cr a^x &= {du^x \over d\tau} = {1 \over \eta_1}\,u^t \cr}$$ $$\left(a^x\right)^2 - \left(a^t\right)^2 = {1 \over \eta_1^2}.$$ We have shown that a point with constant $\eta$ has constant proper acceleration $1/\eta$ in the inertial frame (hyperbolic motion).

If there is a set of points, all having $\eta=\const$, eqs. (6) shows that they all have the same speed for the same $\xi=\tau/\eta$. At the same time, eq. (3) says that for $\xi=\const$ coordinate $\eta$ measures proper distance.

We may then conclude that if we have a rigid rod kept in constant proper acceleration for all its points, it defines an accelerated frame whose natural coordinates are Rindler's. Two facts happen however:

  • Different points of the rod do have different proper accelerations: $a=1/\eta$. This is nothing but the famous Bell's paradox.
  • Coordinate $\xi$, although being a time coordinate, does not measure proper time of the rod's points. Instead we have $\tau=\eta\,\xi$. This is the equivalent of what is very improperly named "gravitational time dilation."

In other words, if we try to use Rindler coordinates - via equivalence principle - to deal with a gravitational field, we must be careful. On one side, because gravitational fields occurring in nature are not uniform. On the other side, not even Rindler's frame represents a uniformly accelerated frame. Of course if we recall that equivalence principle is only local, i.e. holds only for small regions of spacetime, no problem arises.

$\endgroup$
  • $\begingroup$ Thank you. So what metric does one have to use if one wants to describe the gravitational pull near the earth's surface? $\endgroup$ – user139383 Oct 26 '18 at 16:05
  • $\begingroup$ @user139383 Good question. It deserves a longer answer than is possible in a comment. Stay tuned. $\endgroup$ – Elio Fabri Oct 27 '18 at 12:35
3
$\begingroup$

The Rindler metric describes an homogenous gravitational field (as you say in your title) i.e. a field that is the same everywhere. This means it is only an approximation at the Earth's surface since on Earth the gravitational acceleration changes with height. The Rindler metric will describe the geometry only over a region small enough the the change in $g$ with height is negligible.

With this caveat it's perfectly reasonable to use the Rindler metric as a local approximation. If we write the metric in the form:

$$ \mathrm ds^2 = -\left(1 + \frac{a}{c^2}x \right)^2 c^2 ~\mathrm dt^2 + \mathrm dx^2 $$

then the $x$ coordinate is the distance measured by the observer at the origin and likewise the time coordinate is the time measured by the person at the origin. So in this case you would put the origin at the surface of the Earth and set the acceleration to $g$. So $x$ is zero at the surface.

We need to be careful about the sign of $x$ though actually the obvious choice is the correct one. The acceleration $a$ is the proper acceleration of the observer at $x=0$, and if you are standing on the surface of the Earth your proper acceleration points upwards i.e. away from the centre of the Earth. So upwards is the positive direction.

So to summarise, $x$ is zero at the surface, positive above the surface and negative below the surface.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.