1. Aug 25, 2008

### snoopies622

I am looking for a derivation of the metric

$$ds^2 = (1+gz/c^2)^2(cdt)^2 - dx^2 - dy^2 - dz^2$$

for a uniform gravitational field.

Does anyone know where I can find one?

2. Aug 26, 2008

### Mentz114

Hi,
I don't think Dr Rindler derived it so much as writing it down. I rediscovered it when attempting to write down a metric for 'Newtonian' space-time - that's another story.

The logic is that acceleration cause time dilation, and for that you need a minimum of g00 depending on a space coordinate. So if bodies in this ST are to experience coordinate acceleration - engineer a g00 to suit.

It helps if the ST has a zero Einstein tensor so it's a vacuum solution - which is the case for Rindlers metric.

M

3. Aug 26, 2008

### DrGreg

By the equivalence principle, the Rindler metric describes not only a uniform gravitational field in "stationary" coordinates, but also zero gravitation in uniformly accelerating coordinates.

In this 2nd scenario, see this post of mine where I state how to convert between inertial coords $(t, x)$ and uniformly accelerated coords $(T, X)$. (To convert to your notation you will need to write the accelerated coords as $(t, z)$.)

(Spot the error: there's a factor of "c" missing from the equation for t in that post.) (Note: all the zero-subscripted quantities are arbitrary constants which can be taken as zero for simplicity.)

From those equations it's easy (for anyone skilled in the calculus of hyperbolic functions!) to calculate the metric. (Remember in inertial coords it's the usual Minkowski form of the metric.)

If you want to see why those equations are correct, see my posts #13,#14,#15 (and correction in post #28) in this thread.

4. Aug 26, 2008

### yuiop

Would I be right in thinking the Rindler metric does not apply to the gravitational field in the vacuum surrounding a typical non rotating uncharged spherical gravitational mass?

5. Aug 26, 2008

### Mentz114

Deriving the Rindler metric

I mentioned earlier that having a zero Einstein tensor would be essential to describe empty space, and it turns out this is just what enables the metric to be found.

$$(ds)^2 = -a(x)(dt)^2 + (dx)^2 + (dy)^2 + (dz)^2$$

where a(x) is to be determined, then the scalar curvature is

$$R^n_{ n} = a(x)^{-2}(\partial_{x}a(x))^{2}-2a(x)^{-1}\partial_{x}(\partial_{x}(a(x)))$$

If we set this to zero, the simple differential equation solves to

$$a(x) = (1+gx)^2$$.

Actually, every element of the Riemann tensor Rm qns is a multiple of the equation above, so this solution zeros them all, and so the Einstein tensor. Only the Christoffel symbols are left.

6. Aug 26, 2008

### DrGreg

You would be right. The Rindler metric is flat, the Schwarzschild metric (one possible coordinate system for the situation you describe) is not flat.

As Mentz114 indicates, there are mathematical formulas for determining whether a given metric is flat or not, but in this case the Rindler metric is derived on the assumption of flatness anyway (at least in my derivation it is). Being "flat" means there is a change of coordinates that yields the Minkowski metric (exactly, everywhere where both coordinate systems are defined), which we can explicitly demonstrate in this case.

However if you zoomed into a small region of spacetime where the "gravitational field was approximately uniform", the Rindler metric would be a valid local approximation for a hovering "stationary" observer, and a better approximation than the Minkowski metric (which is more suited to locally approximating freefalling observers). You would need to rescale space and time by the appropriate local dilation factors to get the right approximation -- i.e. based on the observer's local proper time and local ruler distance.

7. Aug 26, 2008

### snoopies622

Thanks, guys. I'll start analyzing these responses as soon as I have a free moment, probably some time early this evening. In the meantime, Mentz's approach reminds me: is there an on-line program that yields Christoffel symbols, the curvature tensor, etc., when given a metric tensor? I know how to do these calculations, but they can be very tedious.

8. Aug 26, 2008

### yuiop

Thanks for the reply DrGreg ;)

As I understand it, Rindler coordinates are the point of view of the accelerating observers normally shown as hyperbolic arc wordlines on a Minkowski diagram as shown here: http://en.wikipedia.org/wiki/Image:RindlerObserversCartesian.png

Although not shown explicitly in the Minkowski diagram the wordlines of the inertial observers in the Minkowski diagrams are straight vertical lines. In the Rindler diagram the worldlines of the accelerated Rindler observers are straight vertical lines (as is the event horizon which is diagonal in the Minkowski diagram) and the worldlines of the free falling inertial observers are "hyperbolic secant curves" as shown here for a single representative Minkowski observer: http://en.wikipedia.org/wiki/Image:MinkowskiObserverRindler.png

Anyway, the point of this post is that after a quick scan of Google images for "Rindler metric" and "Rindler coordinates", I can not find a single example of a Rindler chart showing multiple Minkowski observer wordlines (one for each accelerated Rindler observer in the Minkowski chart), clearly labeled axes .. and the curved null wordlines of representative light rays. Anyone know where I might find one?

Last edited: Aug 26, 2008
9. Aug 27, 2008

### DrGreg

I decided to draw my own diagram (or, rather, get MATLAB to do it for me), so here it is.

With units of c = 1 lightyear/year and a=0.1 lightyear/year2 (the black observer's proper acceleration), the left hand diagram shows

$$x = X \cosh \frac{aT}{c}$$
$$ct = X \sinh \frac{aT}{c}$$​

The right hand diagram shows the inverse transformation

$$X = \sqrt{x^2 - c^2t^2}$$
$$T = \frac{c}{a}\tanh^{-1}\frac{ct}{x}$$​

the red lines being worldlines of free-falling inertial particles. For technical reasons, I haven't drawn the grid lines near the event horizon at X=-10, nor have I drawn the grid lines for large x, but the lines should really fill the right hand graph.

I haven't drawn the photon worldlines either, but you can imagine them joining the red dots across the diagonals of each distorted "square" of the grid.

Note that both photons and massive particles take an infinite Rindler time T to reach the event horizon, but only a finite Minkowski time t to do so (and then pass through to the purple zone). Nothing can pass through the event horizon in the opposite direction out of the purple zone.

The purple zone looks rather like a black hole according to a Rindler observer (and the yellow zone looks like a white hole).

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10. Aug 28, 2008

### DrGreg

To be pedantic, the equations I gave in post #9 don't quite match the graphs. They should really be

$$x + \frac{c^2}{a} = \left(X + \frac{c^2}{a}\right) \cosh \frac{aT}{c}$$
$$ct = \left(X + \frac{c^2}{a}\right) \sinh \frac{aT}{c}$$
$$X + \frac{c^2}{a}= \sqrt{\left(x + \frac{c^2}{a}\right)^2 - c^2t^2}$$
$$T = \frac{c}{a}\tanh^{-1}\frac{ct}{x + c^2/a}$$​

I shifted the spatial axes to put the observer at zero.

11. Aug 29, 2008

### snoopies622

I just noticed that these transformations look rather like going from polar coordinates to Cartesian and back, except that one uses hyperbolic trig functions instead of the regular kind, and the sign of the $$t^2$$ is negative instead of positive. Hmm. (scratches chin..) Does uniform acceleration in Minkowski spacetime somehow correspond to moving in a circle (with constant radius and angular velocity) in a complex plane?

Last edited: Aug 29, 2008
12. Aug 29, 2008

### DrGreg

More or less, yes. There are lots of parallels between trig and circles in Euclidean space and "hyperbolic trig" and hyperbolas in Minkowski space. You can consider the proper acceleration g of a particle to be the curvature of its worldline in Minkowski space, with "hyperbolic radius of curvature" of $c^2/g$. The singularity at $x = -c^2/g$ is the "centre of curvature". Even the Lorentz transform, written out as a matrix in terms of rapidity $\phi$, looks suspiciously like the rotation matrix of Euclidean geometry, but in hyperbolic form. In terms of 4D geometry the rapidity of an object relative to an observer is the "hyperbolic angle" between their two worldlines. And you can use the metric to calculate the "angle" between two timelike vectors (4-velocities):

$$U_{\alpha}V^{\beta} = \sqrt{U_{\alpha}U^{\alpha}}\sqrt{V_{\beta}V^{\beta}} \cosh \phi$$​

13. Aug 30, 2008

### yuiop

Hi DrGreg,

Sorry about the delay in responding and many thanks for the diagram and equations. I found them useful. I also decided to create a diagram for my own education and have added some detail including the null worldlines of photons in the Rindler chart.

I have colour coded all the worldlines so that they are same in both the Minkowski and Rindler charts to make it easier to follow what is going on. The dashed red lines are the worldlines of the accelerating Rindler observers while the the solid black lines are the worldlines of the inertial Minkowski observers. The dashed green lines are the lines of equal proper time for the Rindler observers and the dashed grey lines are the equal proper time for the Minkowski observers. The solid red lines on the left of both the Minkowski and Rindler charts are the effective event horizon for the accelerated observers as they are unable to observe anything to the left of the horizons. (The free falling inertial observers do not see an event horizon). The solid blue lines are representative null worldlines of light rays. I noticed that the curvature varies with location of the Rindler observer so I have made an animated gif showing how it varies with location of the observer shown by a red dot.

I have only shown the accelerated wordlines in the top half of the chart to reduce clutter a bit.

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14. Sep 1, 2008

### DrGreg

Re: error in post #12

No-one spotted my "deliberate mistake" (OK, typo). This should be

$$U_{\alpha}V^{\alpha} = \sqrt{U_{\alpha}U^{\alpha}}\sqrt{V_{\beta}V^{\beta }} \cosh \phi$$

15. Sep 1, 2008

### DrGreg

Re: Post #13

That's a pretty impressive diagram, kev.

(Although the animation isn't really necessary, in effect all it is doing is zooming the scale(s).)

I recently read in Rindler's book(1) how to take the comparison between the Rindler form of the flat metric and the Schwarzschild metric one step further, by defining a distorted spacelike coordinate

$$R = \frac{1}{2} \left[1 + \left(\frac{aX}{c^2}\right)^2 \right]$$ ......(1)​

(using the version centred on the singularity, i.e. post #9's coordinates instead of post #10's coordinates.) Whereas X is ruler distance, R is not.

Putting $a = c = 1$ for simplicity, we have

$$x = \sqrt{2R - 1} \cosh T$$ ......(2E)
$$t = \sqrt{2R - 1} \sinh T$$ ......(3E)​

(for R > ½) and it can be shown that (dealing with 2 spacetime dimensions only), the metric becomes

$$ds^2 = (2R-1) dT^2 - \frac{dR^2}{2R-1}$$ ......(4)​

This bears some resemblance to the Schwarzschild metric. The analogy goes even further, though.

So far our coordinates cover only the "east quadrant" of the Minkowski diagram, to the "east" of the upper and lower event horizons. They can be extended to cover the west quadrant via the equations

$$x = -\sqrt{2R_w - 1} \cosh T_w$$ ......(2W)
$$t = -\sqrt{2R_w - 1} \sinh T_w$$ ......(3W)​

Not only that, in the north quadrant (the "black hole") we can define (for Rn < ½)

$$x = \sqrt{1 - 2R_n} \sinh T_n$$ ......(2N)
$$t = \sqrt{1 - 2R_n} \cosh T_n$$ ......(3N)​

Note that in this quadrant the lines of constant T are worldlines of inertial particles (with rapidity T) through the origin event, $\sqrt{1 - 2R}$ being the proper time of such particles. Thus T is spacelike and R is timelike, reversing the situation of the east and west quadrants. (Similarly we can define negated coordinates in the south quadrant ("white hole").)

However, all four quadrants share the same metric equation (4).

So we have a single metric equation (4) covering the whole of spacetime except for a coordinate singularity at an event horizon at R = ½, a coordinate R that is spacelike "outside" the event horizon but timelike inside, and coordinate T that is timelike outside but spacelike inside. For a freefalling inertial particle falling through the upper event horizon (e.g. given by x=1) T increases to infinity at the horizon, then decreases beyond the horizon, all in finite proper time.

Ring any bells?

And yet the whole thing describes flat (gravitationless) space time, which can equally be described by inertial Minkowski coordinates.

Reference

1. Rindler, W. (2006 2nd ed), Relativity: Special, General and Cosmological, Oxford University Press, Oxford, ISBN 978-0-19-856732-5, Section 12.4, "The uniformly accelerated lattice", pp.267-272.