# Rindler coordinates

1. Feb 20, 2013

### nigelscott

I am trying to understand Rindler coordinates at a basic level. Here is what I have so far:

A uniformly accelerated observer will follow a hyperbolic path in a stationary frame.

This is equivalent to having a stationary observer in a uniformly accelerated reference frame. This where Rindler coordinates enter the picture.

The part I am having difficulty with is interpreting the Rindler wedge diagram. I see multiple
hyperbolas spaced out on the x-axis with straight lines of origin 0,0 intersecting them at various points. What exactly is this diagram showing/telling me?

FYI.. I am studying this subject out of interest. My background is in engineering so I understand a fair bit of the math. I just have trouble visualizing what is going on here.

Thanks.

2. Feb 20, 2013

### Naty1

3. Feb 20, 2013

### Naty1

4. Feb 20, 2013

### Staff: Mentor

The hyperbolas are curves of constant Rindler X coordinate, as seen in ordinary Minkowski coordinates. The straight lines through the origin that intersect the hyperbolas are curves of constant Rindler T coordinate, as seen in ordinary Minkowski coordinates.

5. Feb 20, 2013

### George Jones

Staff Emeritus
Just as inertial coordinate are radar coordinates for an inertial observer, Rindler coordinates are radar coordinates for an accelerated observer.

First, a review of how an inertial observer establishes an inertial coordinate system using her wristwatch and light signals. Suppose $P$ is any event in spacetime. The observer continually sends out light signals, and suppose the light signal that reaches event $P$ left her worldline at time $t_1$ according to her watch. Upon reception at $P$ of this signal, $P$ immediately sends a light signal back to the observer, which she receives at time $t_2$. If $x$ is the spatial distance of $P$ from the observer's worldline, then, since the light goes out and back, the light travel a distance $2x$ in a time $\left(t_2 - t_1 \right)$. Thus

$$2x = c \left(t_2 - t_1 )\right.$$
The light spends half the time going out, and half the time coming back. Therefore the time coordinate of event P is the same as the the event on the observer's worldline that is halfway (in time) between the observer's emission and reception events. Consequently, the time coordinate of $P$ is

$$t = \left(t_2 + t_1 \right)/2.$$
It is easy to convince oneself that this operational definition establishes a standard inertial coordinate system.

Note that $t_1$ and $t_2$ are proper times for the observer that sets up the coordinate system.

Assume that an accelerated observer uses the same procedure to establish a non-inertial coordinate system. Consider the case where an observer has constant acceleration. Let $\left(t , x \right)$ be standard coordinates for a global inertial frame. Let the worldline for the accelerated observer be parametrized by her wristwatch (proper) time $T$ so events on her worldline have inertial coordinates

$$\left(t , x \right) = \left(\frac{c}{a} \sinh \left( aT \right) , \frac{c^2}{a} \cosh \left( aT \right) \right).$$
This is one branch of the hyperbola $c^2 t^2 - x^2 = -\left( c^2 / a \right)^2$.

Now set up a coordinate system for the accelerated observer using the light signal procedure given above for an inertial observer. Let $P$ be an event in spacetime that: receives a light signal that left the accelerated observer at proper time $T_1$; sends a light signal that the accelerated observer receives at time $T_2$. The accelerated frame coordinates are

$$\left(t' , x' \right) = \left( \frac{1}{2} \left (T_2 + T_1 \right) , \frac{c}{2} \left(T_2 - T_1 \right) \right).$$
To get a handle on this coordinate system, find what curves of constant $x'$ and curves of constant $t'$ look like in the inertial coordinate system. Let $Q$ and $R$ be the emission and reception events of the accelerated observer, respectively. The inertial coordinates of $P$, $Q$, and $R$ are

$$\left(t_P , x_P \right) = \left(t , x \right)\\ \left(t_Q , x_Q \right) = \left(\frac{c}{a} \sinh \left( a T_1 \right) , \frac{c^2}{a} \cosh \left( a T_1 \right) \right)\\ \left(t_R , x_R \right) = \left(\frac{c}{a} \sinh \left( a T_2 \right) , \frac{c^2}{a} \cosh \left( a T_2 \right) \right)$$
$QP$ lightlike gives

\begin{align} c \left( t_P - t_Q \right) &= \left( x_P - x_Q \right)\\ ct - \frac{c^2}{a} \sinh \left( a T_1 \right) &= x - \frac{c^2}{a} \cosh \left( a T_1 \right), \end{align}
which then gives

$$ct - x = \frac{c^2}{a} \left( \sinh \left( aT_1 \right) - \cosh \left( aT_1 \right) \right) = -\frac{c^2}{a} e^{-aT_1} ~~~~~~(1)$$
Similarly, PR lightlike gives

$$ct + x = \frac{c^2}{a} e^{aT_2}~~~~~~(2)$$
Multiplying (1)*(2) gives

$$c^2 t^2 - x^2 = -\left(\frac{c^2}{a} \right)^2 e^{a \left(T_2 - T_1 \right)} = - \left(\frac{c^2}{a} \right)^2 e^{2ax'}.$$
So, curves of constant $x'$ are hyperbolae in the inertial coordinates. The hyperbolae all have asymptotes $t = x$ and $t = -x$. For $x>0$, these hyperbolae successively become less sharply curved as x increases.

Dividing (2)/(1) and rearrangement gives
$$t = \frac{e^{2at'} - 1}{e^{2at'} + 1} x$$
So, curves of constant $t'$ are straight lines that pass through the origin of the inertial coordinates. As $t' \rightarrow \infty$, the lines approach $t = x$; as $t' \rightarrow -\infty$, the lines approach $t = -x$.

The accelerated coordinate system $\left(t' , x' \right)$ only covers the wedge $x > \left|t \right|$ of the global inertial coordinate system, with the halflines $t = x$ and $t = -x$ playing the roles of horizons.

6. Feb 20, 2013

### Naty1

from the prior post....
These are analogous to the horizons of a black hole...so so teases out Unruh radiation.....analogous to HAwking radiation of a BH.. [which is impossible for me to grasp!!].

Also, 'more ''curvature' means more acceleration....So acceleration at the origin diverges and
gradually lessens at x increases to the right....This is the origin of the 'odd' Born rigidity discussed in one of the links I posted....

7. Feb 21, 2013

### nigelscott

Thanks, I think I have a better understanding after this.

8. Feb 27, 2013

### stevendaryl

Staff Emeritus
That's very nice; I haven't seen that before. However, the coordinate system that I'm familiar with for an accelerated observer $X,T$ relates to the inertial coordinates $x,t$ via:

$x = X cosh(\frac{a T}{c})$
$t = \dfrac{X}{c} sinh(\frac{a T}{c})$

The coordinates that you are talking about are $x',t'$ related to inertial coordinates this way (I think):

$x = exp(x' \frac{a}{c^2}) cosh(\frac{a}{c} t')$
$t = exp(x' \frac{a}{c^2}) sinh(\frac{a}{c} t')$

So your $x'$ is basically the logarithm of the $X$ that I'm familiar with.

9. Mar 8, 2015

### LoicM

A nice way to look at things is to take an analogy with polar coordinates of Euclidian plane. In the 2D Euclidian plane, you can describe a point by (x,y) coordinates, but you can also use polar coordinates (r, theta). The polar coordinates have some problem at O, but they are well suited to study points on a circle. Coordinate lines are then represented by concentric circles of center O ( r=cstt lines), and rays going trough O ( theta = cstt).

A point in 1+1 Minkowski space ( 1 dim of time, 1 dim of space), called an event, can be described with its (t,x) coordinates ( the usual global inertial frame, equivalent to (x,y) in the Euclidian case), but also using Rindler coordinates (T,X), which are well suited for events on located on hyperboles, which indeed correspond to objects which are uniformly accelerated relative to the original frame. The lines on your Rindler diagram are of two kinds : pieces of hyperboles correspond to Rindler's coordinate X=cstt, while the rays going through a single point A correspond to T= cstt. For example, the trajectory of the accelerated observer is the coordinate line X=0 ( for all values of T, which happens to be the proper time of the observer). Rindler coordinates also have problems for events situated beyond the point A, on the negative side. But they have the great interest of allowing to describe the physics of uniformly accelerated observers in the scope of Special Relativity, and they represent an excellent "toy-model" to later study the physics of black holes in General Relativity.

Last edited by a moderator: Mar 8, 2015
10. Mar 14, 2015

### PAllen

Right, Rindler coordinates commonly given are Fermi-Normal coordinates for a uniformly accelerating observer, followed by an origin translation. It happens that simultaneity hyper-surfaces for a uniformly accelerating observer are the same between Radar coordinates and Fermi-Normal coordinates. Thus, George's derivation of radar coordinates differs from the commonly given form (of Rindler coordinates) only by a scaling factor.

11. May 6, 2015

### LoicM

Here are some additional data to clarify my previous post. I am actually looking at Rindler coordinates as some kind of "Hyperbolic Polar" coordinates of the (x,t) plane with Minkowski metric instead of E2 metric. Taking the Observer with proper acceleration a=1 ( in units where c=1), as the "reference Observer" (that is the one whose proper time τ is used as the Rindler time coordinate T), gives as worldline a hyperbola with "Radius" R= x2-ct2=1, and "angle", or rapidity, ψ = τ. The Rindler space coordinate, X of any point (event) is then the "radius" of the hyperbola on which the point is located.

Rindler coordinates (X,T) of any event as in :
are then equal to (R,ψ) : "polar coordinates", just as in :

$x = X cosT=Rcosψ$
$t = X sinT=Rsinψ$

Lines of constant X are "concentric" hyperbolas, and lines of constant T are radiuses. This result would also be obtained by taking 'ict' instead of Minkowski time (Wick rotation), leading to Euclidian-like geometry.

Last edited: May 6, 2015