Inconsistent forms of the metric in a uniform field

[Altabeh]

Hi, Altabeh -- You and I don't seem to be convincing each other re g=2. Maybe it would help if Mentz114 or George Jones could take a look at this and give an opinion.

Here I take another approach to prove that if

$$\dot{t}=1/\sqrt{g_{tt}}=e^{-gy},$$

then in the metric

$$d\tau^2=e^{2gy}dt^2-dy^2,$$

particles following geodesics would give rise to a bizarre result which is only true if $$g=0$$. (*)

But first off, I'd like to prove my velocity formula for the Rindler's metric

$$ds^2=e^{2\Phi}dt^2-dy^2,$$ (R)

in the case dealing with $$\Phi \propto y$$ where the proportionality constant is taken to be $$g$$.

If $$\Phi =gy$$, then obviously $$(\Phi)' =g$$. (Henceforth we denote the derivative wrt y by a prime and the derivative wrt proper time $$\tau$$ by a dot.)

We then write the geodesic equations in the Rindler's spacetime:

$$\ddot{y} + ge^{2gy}\dot{t}^2=0,$$ (1) and
$$\ddot{t} + 2g\dot{t}\dot{y}=0.$$ (2)

Re-write $$\ddot{y}$$ as

$$\ddot{y}=\dot{(\frac{dy}{dt}\frac{dt}{d\tau})}=\ddot{t}\frac{dy}{dt}+\dot{(\frac{dy}{dt})}\dot{t}.$$ (3)

Introducing (3) into (1) and making use of (2) gives

$$\ddot{y}=-2g\dot{t}\dot{y}\frac{dy}{dt}+\dot{(\frac{dy}{dt})}\dot{t}=-ge^{2gy}\dot{t}^2,$$ (V)

which can be written as

$$-2g{\dot{y}}^2+\dot{v}\dot{t}=-ge^{2gy}\dot{t}^2,$$

in which $$v=dy/dt$$ represents the coordinate velocity. Now divide each side of this equation by $$\dot{t}^2:$$

$$-2g(\frac{\dot{y}}{\dot{t}})^2+\frac{\dot{v}}{\dot{t}}=-ge^{2gy},$$

and since $$(\frac{\dot{y}}{\dot{t}})^2 = v^2$$ and $$a=\frac{\dot{v}}{\dot{t}}$$ representing the coordinate acceleration, we finally obtain

$$v^2=\frac{1}{2g}(a+ge^{2gy})$$ which is a generalized form of my velocity formula for $$g=1.$$

It is time to say why a bizarre result would be reached if your assumption, bcrowell, holds using the Euler-Lagrange method: Divide each side of the metric (R) by $$d\tau^2:$$

$$(\frac{ds}{d\tau})^2=e^{2\Phi}\dot{t}^2-\dot{y}^2\equiv g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}.$$

You know that this must satisfy the Euler-Lagrange equations, i.e.

$$\frac{d}{d\tau}[\frac{\partial}{\partial \dot{x}^{\mu}}(g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu})]=\frac{\partial}{\partial {x}^{\mu}}(g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}).$$

So that for Rindler's metric (R) with $$\Phi = gy$$ we will have

$$\frac{d}{d\tau }(-2\dot{y})=2ge^{2gy}\dot{t}^2;$$
$$\frac{d}{d\tau}(2\dot{t}e^{2gy})=0.$$

A straightforward calculation confirms that these two are respectively given in (1) and (2). So assuming $$\dot{t}=1/\sqrt{g_{tt}}=e^{-gy}$$ and taking a dot derivative of it reveals that

$$\ddot{t}=-g\dot{y}e^{-gy}=-g\dot{y}\dot{t},$$ (4)

which is not abnormous iff particle is instantaneously at rest so $$\ddot{t}=0$$. But instantaneously at rest means that proper acceleration would also vanish instantaneously, thus revealing

$$ge^{2gy}\dot{t}^2=0,$$

from (1). This is again not abnormous iff g=0. (**) I think everything is now clear as sun: Demanding $$dy=0$$ costs so much for you because it does not let the proper acceleration be non-vanishing and consequently it does not preserve the non-flatness of spacetime.

AB

--------------------------------------
(*) g is not 2 because I calculated geodesic equations based on a Rindler's metric with $$g=1$$ then worked your $$\dot{t}$$ into it to get $$g=2$$.

(**) This can be obtained from (V) as well. There you can see if $$v=a=0$$ inspired by your assumption, then $$g=0$$ necessarily.

 

[bcrowell]

Hi, Altabeh -- Re your #36, please note that I have always explicitly said that $\dot{t}=e^{-gy}$ only applies to particles instantaneously at rest. I said that in #19, and I pointed out in #26 that you were drawing incorrect conclusions by applying it to particles that were not instantaneously at rest. You can't use $\dot{t}=e^{-gy}$ in a derivation of geodesics, because it only holds when the coordinate velocity is zero, and the coordinate velocity will not always be zero on a geodesic. For independent verification of my statement that the acceleration equals -g for particles instantaneously at rest, see this page: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html (at "More precisely, time-like geodesics..."; they're doing the special case of g=1).

 

[Mentz114]

The problem for me is that this metric $$d\tau^2=e^{2gy}dt^2-dy^2$$ is unphysical so I can't motivate myself to check Altabeh's calculation.

I would like to work out the proper acceleration of the hovering observer, but the only way I've seen to do this is with frames, and bcrowell's post #19.

Using the metric $ds^2=(1+ax)^2-dx^2 +-dy^2-dz^2$ ( which is a vacuum solution of the EFE) I repeat the latter calculation. The geodesic equation for $x$ is

$$\frac{d^2x}{ds^2}=-a(1+ax)\left(\frac{dt}{ds}\right)^2$$
where $s$ is an affine parameter. Now, like post #19 I take $s$ to be the proper time $\tau$. I have doubts about whether this is a generally valid thing to do, but maybe in this case it is (?). For an observer at rest we have $dt/d\tau=1/\sqrt{g_{tt}}$ which gives
$$\frac{d^2x}{d\tau^2}=\frac{-a(1+ax)}{(1+ax)^2}=\frac{-a}{1+ax}$$
What does this mean ? It is not a constant acceleration as advertised. The problem could be that $s\ne \tau$. Don't we need a worldline with starting conditions to get the relationship between $\tau$ and $s$ ?

It is consistent with this statement from M. Weiss in the page referenced by bcrowell

"My guess is that the Einstein empty-space equations forbid a uniform gravitational field in the above sense. I haven't checked this, though."

Really, is it too much to ask someone to do this ?

George Jones, thank you for your post. It didn't help though. I have the book recommended by Fredrik so maybe in a few years I might understand what you've posted.

 

[bcrowell]

Hi, Lut -- Thanks for your #38, which is very interesting.

Re the use of the proper time as an affine parameter, I'm pretty sure this okay. The proper time is really the prototypical example of an affine parameter. When you're computing geodesics, you generally use the proper time as your affine parameter if you possibly can. The most common reason not to use proper time as your affine parameter is if you're doing lightlike geodesics, which have a proper time that vanishes identically.

I'm not sure it's a problem that the Rindler coordinates give a $d^2 x/d \tau^2$ that's not the same everywhere. I think this is just another instance of the Bell spaceship paradox. The factor of $1+ax$ occurring in the metric is interpreted as a gravitational time dilation by the accelerated observers. When that factor crops up all over the place, that's probably what it represents. If your observer sitting at x=0 looks at a test particle at some other x, he's going to see its x varying with an acceleration of a according to his *own* proper time, because that's how the coordinate system was constructed. When you compute $d^2 x/d \tau^2$ for the test particle, you're using the proper time of the test particle. They shouldn't agree.

My current take on the whole situation is that I agree with DrGreg's #4:

Somebody once sent me a PM to say that it is tricky to find a good GR analogy of a Newtonian uniform gravitational field which is valid globally, and the closest is the 1962 Petrov solution

 

[Mentz114]

If your observer sitting at x=0 looks at a test particle at some other x, he's going to see its x varying with an acceleration of a according to his *own* proper time, because that's how the coordinate system was constructed. When you compute $d^2 x/d \tau^2$ for the test particle, you're using the proper time of the test particle. They shouldn't agree.

So we want to calculate the proper acceleration of the hovering oberver from the point of view of another hovering observer ? Or in terms of coordinate time $t$ ?

These are rhetorical questions I need to think about.

 

[George Jones]

$$\frac{d^2x}{d\tau^2}=\frac{-a(1+ax)}{(1+ax)^2}=\frac{-a}{1+ax}$$

What does this mean ? It is not a constant acceleration as advertised.

Yes, it is.

 

[Mentz114]

George,
Thanks. You're right of course. It should be written,
$$\ddot{x}\mid_{x=x_r}=\frac{-a}{1+ax_r}$$
where $x_r$ is the position the observer is hovering at.

I think I was expecting it to be independent of position, which what I would call 'constant', i.e. same everywhere.

What is puzzling is that this decreases with $x_r$ though.

 

[Altabeh]

Hi, Altabeh -- Re your #36, please note that I have always explicitly said that $\dot{t}=e^{-gy}$ only applies to particles instantaneously at rest.

What exactly do you mean by "instantaneously at rest"? My own understanding of it is that if any particle along any path (geodesic or an ordinary curve) is at rest at a moment i.e. its coordinate-dependent velocity vanishes at a moment, then it is called instantaneously or momentarily at rest particle.

A good example is this:

When you throw something upward, eventually it falls back downward. At the peak of the motion, before starting to head back down again, the velocity is 0. The object is instantaneously at rest for that instant.

I said that in #19, and I pointed out in #26 that you were drawing incorrect conclusions by applying it to particles that were not instantaneously at rest.

The motion along a path in any curved spacetime cannot be pinned down by saying that "a property of that motion" does only belong to the path being traveled by particles. This is not logical at all. I mean you say the property of "instantaneously at rest" is gained if in the metric we put $$dy=0$$. You are applying something to the METRIC of spacetime and claim at the same time that you can't use this in the

...derivation of geodesics, because it only holds when the coordinate velocity is zero, and the coordinate velocity will not always be zero on a geodesic.

This by itself destructible in the sense that now you are assigning the property of "being always zero" for the coordinate velocity to the particles which are supposed to be instantaneously at rest. When talking about an instant, then everything is valid at that instant only. At a later time, the assumption $\dot{t}=e^{-gy}$ is not valid because for a particle moving continuously along some curve, $$dy=0$$ will not hold unless assuming that the particle is not moving (and everything being either proper or coordinate-dependent is zero for the particle) or it is just hovering (impossible!) at a given y! One other scenario is that the particle is instantaneously at rest or its coordinate velocity is zero at an instant but it is on the verge of starting to move again which means that its instantaneous acceleration is not zero! From my velocity formula,

$$v^2=\frac{1}{2g}(a+ge^{2gy})$$

putting v=0, gives the non-zero instantaneous acceleration

$$a=-ge^{2gy_0}$$,

where $$y_0$$ is where particle is at rest. You see that I don't use $$\dot{t}=1/\sqrt{g_{tt}}=e^{-gy}$$ in deriving this correct coordinate-dependent value!

But using your assumption in the last scenario given above and applying it to geodesic equations and assuming the initial condition $$dy/d\tau = 0$$ at some point $$y_0$$, then the proper acceleration at $$y_0$$ along a time-like geodesic passing through it, equals $$-g$$. Well, this is the same as Weiss's claim for g=1:

This spacetime possesses a "uniform gravitation field" *. More precisely, time-like geodesics with the initial condition dx/dτ = 0 at some point P satisfy $$d^2x/d\tau^2 = -1$$ at P. So if a lab-frame observer (that is, (t,x) coordinate system) let's go of an object, she'll see it drop with acceleration 1 (provided she uses a clock that keeps local time dτ, e.g. an atomic clock).

In the long run, I just say I don't see anything contradictory in my calculations and those done by Weiss as long as our understanding of "instantaneously at rest" falls upon the scenario admitting a non-vanishing instantaneous acceleration. Period!

--------------------------------------------------
* Weiss (I assume) has forgotten to add to this sentence a 'locally' as correctly he gives a local representation of what he seeks out to show. As already was talked about, the uniform gravitational field does only exist locally. (See one of my early posts here in which I responded to Mentz114.)

AB

 

[Altabeh]

The problem for me is that this metric $$d\tau^2=e^{2gy}dt^2-dy^2$$ is unphysical so I can't motivate myself to check Altabeh's calculation.

Speaking of which, so why don't people quit studying string theory because it's highly unphysical and just is talked about in papers? I did find your motivation a little bit inelegant, though respected!

AB

 

[Altabeh]

I just realized something interesting about Weiss's idea. He thinks that instantaneously at rest particles are those following (time-like) geodesics in any spacetime with a constant instantaneous acceleration at each point of the geodesic. This is not even wrong: One can picture two scenarios for this nonsense as follows:

1- The motion is discontinuous, in the sense that at each point of the geodesic, particle is supposed to be at rest and suddenly is about to start moving at a very near time later;
2- The motion takes place on a straight line to have a constant acceleration, thus the flatness of spacetime is necessary!

The second scenario is the case $$g=0$$ which happens when a co-moving observer is seeing that the particle's proper velocity is zero along the geodesic at any time. This means that both particle and observer have a zero proper acceleration, too!

The moral: I don't know what to say next because I'm flabbergasted!

AB

 

[bcrowell]

What exactly do you mean by "instantaneously at rest"?

As I said in #37, I mean that it has a zero coordinate velocity.

I just realized something interesting about Weiss's idea. He thinks that instantaneously at rest particles are those following (time-like) geodesics in any spacetime with a constant instantaneous acceleration at each point of the geodesic. This is not even wrong:

If Weiss's treatment is so trivially wrong, then it would surprise me that John Baez, who is a tenured, senior professor of physics and mathematics at a highly regarded research university, would have posted it on his FAQ page. The material on those FAQ pages is distilled from usenet discussions, from back in the days when sci.physics was a healthy forum where lots of knowledgeable people were posting; so it also seems odd to me that a mistake as trivial as you're claiming this is would not have been detected in those discussions. Of course smart, competent people can occasionally make dumb mistakes. Maybe Weiss, Baez, and I are all making the same dumb mistake. If you think that is the case, then I suggest you contact Don Koks, the current maintainer of the FAQ, using the contact information given here http://math.ucr.edu/home/baez/physics/ , and point out the error.

 

[bcrowell]

So we want to calculate the proper acceleration of the hovering oberver from the point of view of another hovering observer ? Or in terms of coordinate time $t$ ?

These are rhetorical questions I need to think about.

Hope it's not too annoying if I tell you what I think are the answers to your rhetorical questions :-)

If an observer A hovering at a constant x wants to know his proper acceleration, he can do it by releasing a test particle P from rest and observing its acceleration, relative to him, as measured with his own clocks and rulers. If P is released at rest, then there is negligible difference between the proper times and proper lengths measured in P's frame and those measured in A's frame.

A second hovering observer B at a different height in the gravitational field will disagree with A's measurements due to gravitational time dilation.

An observer C who is not hovering will typically be in motion relative to A and B (except perhaps at one instant in time), and will therefore see additional special-relativistic effects. These effects are the ones that are involved in Bell's spaceship paradox.

[EDIT] Some further thoughts: If the above is correct, then I think I can confirm that your $d^2x/d\tau^2=-a/(1+ax)$ actually does give an acceleration that is independent of x, for the accelerating observer who constructed the Rindler coords relative to himself at x=0. Your calculation had $\tau$ as the affine parameter used for P's geodesic equation, so your $\tau$ is the proper time of P and (approximately) A. Let's say B is lower in the gravitational field, and the Rindler coords are the ones constructed by B. The sign of your equation shows that positive x is up. Therefore B's clocks run more slowly than A's by a factor of 1+ax. A says his own proper acceleration is too small by a factor of 1+ax. But B looks up at A's little experiment and sees the experiment sped up according to his clock, so to him the acceleration seems to be a, matching his own proper acceleration.

 

[Mentz114]

Ben,

I'm still thinking about #47 ( and #43->#46 !).

Wiki shows how in the Rindler chart, the Minkowski line element becomes
$ds^2=-a^2x^2+dx^2+dy^2+dz^2$.
Repeating the calculation I did previously with this I get $\ddot{x}=-a$.

Which is independent of $x$. The line element is a coordinate transformation from Minkowski space, but the original space and time have been mixed, so the $x$ coordinate is not pure space. I suppose one could consider the line-element to be the result of some source.

It's difficult to put any physical meaning to this.

[aside]
Can we can define 'uniform' in a gravitational field as meaning a ball of coffee-grounds in free-fall will not change shape at all, in any direction ? So neighbouring geodesics stay parallel and two particles released simultaneously wrt to an observer half-way between them but from different 'heights' stay the same distance apart ?

I suspect that the metric above has these properties but I'm too sleepy now to work it out.

 

[Mentz114]

Speaking of which, so why don't people quit studying string theory because it's highly unphysical and just is talked about in papers? I did find your motivation a little bit inelegant, though respected!

AB

Good question ! You should ask the string theorists because I always wondered about that myself.

There seems to be something equally unphysical about 'instantaneously at rest wrt ...' as you have said. That's why I'm trying to understand this in terms of realizable situations with observers who hold their positions with rocket engines, or fall freely or anything in between.

 

[Altabeh]

As I said in #37, I mean that it has a zero coordinate velocity.


If Weiss's treatment is so trivially wrong, then it would surprise me that John Baez, who is a tenured, senior professor of physics and mathematics at a highly regarded research university, would have posted it on his FAQ page. The material on those FAQ pages is distilled from usenet discussions, from back in the days when sci.physics was a healthy forum where lots of knowledgeable people were posting; so it also seems odd to me that a mistake as trivial as you're claiming this is would not have been detected in those discussions. Of course smart, competent people can occasionally make dumb mistakes.

"Instantaneously at rest" is very locally true as in the case of a measurement in a gravitational field unless your instantaneous co-moving observer has a completely ideal clock that is unaffected by the gravitational acceleration! I've not seen anything like this though you can find experimentally proven cases in SR such as transverse Doppler effect, where spacetime is flat and gravity does not exist!

Look at Semay's metric and his achievement in a flat spacetime! That one is acceptable from any point of view, but in this case, I doubt one can actually make escape routes such as use of ideal clocks! Since we are talking about a 2D Rindler's metric, so how can one even think about gravity with a vanishing Weyl tensor? The use of ideal clocks makes the whole thing ultra-unphysical! Coley, A.A. seems to have a good paper "http://www.osti.gov/energycitations/product.biblio.jsp?osti_id=5724386"" which I can't get access to! If you found that paper, please put it here so as for me to be able to understand if Weiss's idea would be considered globally along a time-like geodesic!

Maybe Weiss, Baez, and I are all making the same dumb mistake. If you think that is the case, then I suggest you contact Don Koks, the current maintainer of the FAQ, using the contact information given here http://math.ucr.edu/home/baez/physics/ , and point out the error.

You are talking about 'mistake', but later you change it to 'error'! I said we both agree on calculations, but in the physical interpretation, I keep the path of local use of 'instantaneously at rest' particles, but Weiss changes the discussion towards using ideal clocks to extend the stuff to the whole trajectory (time-like geodesic) being traveled by particles! I don't see any mathematical error to point out!

AB

 

[Altabeh]

Ben,

[aside]
Can we can define 'uniform' in a gravitational field as meaning a ball of coffee-grounds in free-fall will not change shape at all, in any direction?

Not in any direction! Let's say along a time-like geodesic, for instance, because acceleration remains constant and the coffee never splashes!

So neighbouring geodesics stay parallel and two particles released simultaneously wrt to an observer half-way between them but from different 'heights' stay the same distance apart?

Correct if we believe in the equivalence principle and the locally flatness!

I suspect that the metric above has these properties but I'm too sleepy now to work it out.

Both can be found in the above metric because it doesn't admit a gravitational field and so it is flat!

 

[yuiop]

Also, Rindler coordinates are not a uniform gravitational field (at least that's what Rindler's text claims).

I think this is right because if I recall correctly, the acceleration in Rindler coordinates is proportional to 1/x.

... These effects are the ones that are involved in Bell's spaceship paradox...

I think Ben has touched on an important aspect here and I think the physical implications of the effects should be made clear. If we take an array of accelerating rockets in flat spacetime, all with equal proper acceleration in the positive x direction, such that there mutual spatial separation remains constant according to an inertial Minkowski observer, then the mutual spatial separation of the rockets is continually increasing in the x direction, according to the accelerating rocket observers. By the EP, this implies that observers at rest in a uniform gravitational field will see distances constantly increasing as measured by radar signals and any physical rulers that attempt to remain at rest with the uniform gravitational field will be torn apart. This makes defining distance or even defining what "at rest" means very difficult because distance is always changing over time. Certainly, we can not create a physical grid of rulers and clocks in such a space-time.

Also, could someone make it clear if by "uniform gravitational field" we mean a uniform non-zero proper acceleration in a one dimensional way, such as along the x axis, or if we mean uniform non-zero proper acceleration in all 3 directions? I guess the latter would be represented by the FLRW metric on the cosmological scale.

In an old thread we discussed the gravitational fields of some hypothetical objects and I think concluded that an infinitely long cylindrical massive object would have an external gravitational field that varied proportional to 1/r and that a thin planar gravitational mass extending infinitely in the y and z directions would have an external field that is independent of x.

As far as I can tell, the term "uniform gravitational field" is used in the context where we consider a small enough volume of space-time that deviations from uniformity are negligible and this is a local approximation.

[aside]
Can we can define 'uniform' in a gravitational field as meaning a ball of coffee-grounds in free-fall will not change shape at all, in any direction ?

From the above considerations, the ball of coffee grounds will not change shape in a uniform field, according to an inertial observer that is co-free-falling with the coffee grounds and will be expanding according to an accelerating observer that is at rest with gravitational field.

So neighbouring geodesics stay parallel and two particles released simultaneously wrt to an observer half-way between them but from different 'heights' stay the same distance apart ?

Yes, (if the observer is released at the same time). Presumably, simultaneity is not an issue here, because clocks at different heights in a uniform gravitational field run at the same rate.

I suspect that the metric above has these properties but I'm too sleepy now to work it out.

All out of coffee-grounds?

 

[bcrowell]

In an old thread we discussed the gravitational fields of some hypothetical objects and I think concluded that an infinitely long cylindrical massive object would have an external gravitational field that varied proportional to 1/r and that a thin planar gravitational mass extending infinitely in the y and z directions would have an external field that is independent of x.

This is only true in the weak-field limit, if the source isn't rotating. Under those circumstances, you can just find the gravitational field with Gauss's law, and you don't need relativity.

The Petrov solution can be interpreted as the exterior field of an infinitely long cylinder of dust that rotates rigidly at $\omega =c/R$, where R is the radius of the cylinder. Because it's a strong field, you can't find the field using Gauss's law. The external region has a constant scalar curvature everywhere, so there definitely isn't a 1/r field.

Can we can define 'uniform' in a gravitational field as meaning a ball of coffee-grounds in free-fall will not change shape at all, in any direction ? So neighbouring geodesics stay parallel and two particles released simultaneously wrt to an observer half-way between them but from different 'heights' stay the same distance apart ?

I think that's equivalent to saying that there is no tidal curvature, i.e., a vanishing Weyl tensor. If we required that, then the Petrov metric would be disqualified, because it has a nonzero Weyl tensor.

If we want it to be a vacuum solution, then we need the Ricci curvature to vanish. If we demand that the Weyl tensor vanish as well, then we're requiring the space to be flat -- only a flat space has both a vanishing Weyl tensor and a vanishing Ricci tensor. If you want a flat space described in a uniformly accelerating frame, then Rindler coordinates do that, but as we've seen, they don't have all the properties you'd want in a uniform field.

If you change the ball of coffee grounds into two spaceships connected by a thread, then you have Bell's spaceship paradox.

The thing that I'm currently trying to understand about the Petrov metric is how you would define a proper acceleration in it. In e.g., Rindler coordinates this is fairly simple. You define a hovering observer as one whose coordinate velocity is zero, and then that observer releases a test particle initially at rest, and observes its acceleration relative to her. A couple of problems come up when you try to use this definition in the Petrov metric. One is that you have two different coordinates that can play the role of time, depending on r. The metric is stationary but not static, so there is a preferred time coordinate, but this preferred time coordinate is not $\phi$ or t. The other problem is that I'm not even sure material particles can have a coordinate velocity of zero.

I want to play around with this stuff this weekend by doing numerical simulations of geodesics.

 

[bcrowell]

Wiki shows how in the Rindler chart, the Minkowski line element becomes
$ds^2=-a^2x^2+dx^2+dy^2+dz^2$.
Repeating the calculation I did previously with this I get $\ddot{x}=-a$.

Which is independent of $x$. The line element is a coordinate transformation from Minkowski space, but the original space and time have been mixed, so the $x$ coordinate is not pure space. I suppose one could consider the line-element to be the result of some source.

For the line element, I assume you meant
$ds^2=-a^2x^2dt^2+dx^2+dy^2+dz^2$
(with a dt2 in the first term).

This confuses me, because $a^2x^2$ is the same as $(1+ax)^2$ except for adding a constant to the x coordinate. So I don't see how the proper acceleration can come out to be independent of x in the first case, but dependent on x in the second case.

 

[bcrowell]

OK, re Lut's #48, I calculated the Christoffel symbols for both metrics, using Maxima in an attempt to avoid computational errors.

load(ctensor);
dim:2;
ct_coords:[t,x];
lg:matrix([(1+a*x)^2,0],
          [0,-1]);
cmetric();
christof (lcs);

This gives $\Gamma\indices{^x_{tt}}=-a(1+ax)$.

load(ctensor);
dim:2;
ct_coords:[t,x];
lg:matrix([(a*x)^2,0],
          [0,-1]);
cmetric();
christof (lcs);

This gives $\Gamma\indices{^x_{tt}}=-a^2x$.

In both cases, the Christoffel symbol can be interpreted as the instantaneous value of $d^2x/d\tau^2$, where x is the position of a particle that has an instantaneous coordinate velocity of zero, and $\tau$ is the proper time of that particle.

Both of these results for the acceleration vary with x, and they're consistent with each other when you do the coordinate transformation $x \rightarrow x \pm 1/a$.

I think this is all consistent with the time-dilation interpretation I gave in #47.

 

[Altabeh]

I think Ben has touched on an important aspect here and I think the physical implications of the effects should be made clear. If we take an array of accelerating rockets in flat spacetime, all with equal proper acceleration in the positive x direction, such that there mutual spatial separation remains constant according to an inertial Minkowski observer, then the mutual spatial separation of the rockets is continually increasing in the x direction, according to the accelerating rocket observers. By the EP, this implies that observers at rest in a uniform gravitational field will see distances constantly increasing as measured by radar signals and any physical rulers that attempt to remain at rest with the uniform gravitational field will be torn apart. This makes defining distance or even defining what "at rest" means very difficult because distance is always changing over time. Certainly, we cannot create a physical grid of rulers and clocks in such a space-time.

Weiss's idea is flawed physically, as already was discussed, since it suffers lack of a globally compatible definition of constant proper acceleration along geodesics due to dependency on spatial coordinates so that making use of the idea of "instantaneously at rest" particles and instantaneous co-moving observers to overcome this problem for such a curved metric is like you are trying to patch a huge rupture with a highly unphysical tool (ideal clock) around a gravitational field which naturally isn't uniform!

Also, could someone make it clear if by "uniform gravitational field" we mean a uniform non-zero proper acceleration in a one dimensional way, such as along the x axis, or if we mean uniform non-zero proper acceleration in all 3 directions? I guess the latter would be represented by the FLRW metric on the cosmological scale.

In all three dimensions! Here we have been talking about Rindler metric in 2 dimensions, so the second axis would be x or y or z along which the uniformity of gravitational field is considered. The FLRW metric is not static, but dependent on time if we take all coordinates to be fixed and pin the changes to time only which enters in the metric by scale factor, so particles moving on geodesics will either accelerate or decelerate according to which epoch the cosmos is getting through: the epoch at which the universe contracts, or at the one it expands as we see it right now! This is simply seen, for example, for radial motion that the popper acceleration depends on t, so it is not constant at all! Furthermore, FLRW metric is curved and it does not admit any uniform gravitational field everywhere but locally, as expected.

In an old thread we discussed the gravitational fields of some hypothetical objects and I think concluded that an infinitely long cylindrical massive object would have an external gravitational field that varied proportional to 1/r and that a thin planar gravitational mass extending infinitely in the y and z directions would have an external field that is independent of x.

In the case of a weak gravitational field, this can be read off the Poisson equation directly. But if the central or self-gravitating body has a very strong gravity, then this won't hold until one says that spacetime is not closed and at sufficiently large distances, the field starts getting closer to zero in proportionality with 1/r as in the asymptotically flat spacetimes.

As far as I can tell, the term "uniform gravitational field" is used in the context where we consider a small enough volume of space-time that deviations from uniformity are negligible and this is a local approximation.

Completely right! This is because a "uniform gravitational field" within a spacetime makes all points undergo the same changes geometrically and proper accelerations of particles moving along paths change point-by-point equivalently so this means that in order to have such a field, the (time-like) geodesics must be independent of spatial coordinates (in the case of static metrics) which cannot be seen in any curved spacetime throughout the whole manifold except in a very small region of it!

AB

 

[bcrowell]

Also, could someone make it clear if by "uniform gravitational field" we mean a uniform non-zero proper acceleration in a one dimensional way, such as along the x axis, or if we mean uniform non-zero proper acceleration in all 3 directions? I guess the latter would be represented by the FLRW metric on the cosmological scale.

What I have in mind personally is one that has uniformity in all three dimensions. I don't think an FLRW metric qualifies, because an object released with a coordinate velocity of zero (in the usual coordinates) maintains a coordinate velocity of zero indefinitely.

As far as I can tell, the term "uniform gravitational field" is used in the context where we consider a small enough volume of space-time that deviations from uniformity are negligible and this is a local approximation.

Any field is uniform in this weaker local sense. IMO the only thing that makes the problem interesting and hard is that it's so hard to find anything that looks *globally* like a uniform field in GR. The Petrov metric is globally uniform (although it has lots of other strange properties that make it not necessarily the global solution that we'd like to say is *the* uniform field in GR).

 

[yuiop]

This is only true in the weak-field limit, if the source isn't rotating. Under those circumstances, you can just find the gravitational field with Gauss's law, and you don't need relativity.

Your right. We did use Gauss's law and it probably does not apply here outside the weak field limit.

If you change the ball of coffee grounds into two spaceships connected by a thread, then you have Bell's spaceship paradox.

So do you agree that physical rulers would be torn apart in such a spacetime, like the thread in Bell's spaceship paradox?

In all three dimensions! Here we have been talking about Rindler metric in 2 dimensions, so the second axis would be x or y or z along which the uniformity of gravitational field is considered. The FLRW metric is not static, but dependent on time if we take all coordinates to be fixed and pin the changes to time only which enters in the metric by scale factor, so particles moving on geodesics will either accelerate or decelerate according to which epoch the cosmos is getting through: ..

What I have in mind personally is one that has uniformity in all three dimensions. I don't think an FLRW metric qualifies, because an object released with a coordinate velocity of zero (in the usual coordinates) maintains a coordinate velocity of zero indefinitely.

I see now that my casual guess that the FLRW metric might qualify was incorrect. Thanks guys.

OK, so we are talking about constant acceleration in all 3 dimensions. Now there is another problem. Since there is equal acceleration in all directions, how does a particle know which way to accelerate or does everyting simply expand? Do we have to assume a coordinate centre to such a spacetime and assume everything either accelerates directly towards (or away?) from that coordinate centre? It seems such a gravitational system could not be isotropic, in the sense that any arbitary location could be chosen as the centre.

 

[yuiop]

This is probably irrelevant, but there are also the x and z coordinates. I've seen a nice elementary argument by Born that because a measuring rod in a rotating frame of reference experiences a Lorentz contraction when oriented in the transverse direction, by the equivalence principle a rod's length varies with height in a gravitational field. This would presumably apply to x and z, not y.

The only trouble is that we know length contraction does not occur in the x and z directions in the Schwarzschild metric. (I assume you mean horizontal by x and z in the gravitational context)

Here are some comparisons of rotational and gravitational length measurements. All measurements in a small region to a first order aproximation. I am also using z as the radial coordinate as you have done in your later post.

Z COORDINATE:

Radial coordinate of a cylinder:
Parallel to the felt acceleration.
NOT length contracted according to radar or ruler measurements.(A rotating and non rotating observer will measure this to be the same.)
Increasing time dilation in the positive y direction.

Vertical coordinate in a gravitational field:
Parallel to the felt acceleration.
Length contracted according to radar or ruler measurements. (A stationary local observer will measure this distance to be longer than the distance calculated by an observer at infinity.)
Decreasing time dilation in the positive y direction.


As you can see there are obvious differences (and more than I mention here) between the rotating disc and the gravitational field.

On the face of it, Born's statement appears to be incorrect, or he he saying the transverse coordinates (orthogonal to the radius) of a disc are equivalent to the vertical radial coordinate of a gravitational field?

I found the passage in the Born book that I assume is the one I'd seen summarized elsewhere. It's at p. 320 in the 1962 Dover edition. It's part of a discussion of spacetime on a rotating disk, and it's actually very brief. He discusses the impossibility of global clock synchronization, talks about the interpretation in terms of the equivalence principle, and then says:

In a gravitational field a rod is longer or shorter or a clock goes more quickly or more slowly according to the position at which the measuring apparatus is situated.

This seems like somewhat of a leap to me, since he's generalizing from the rotating disk to gravitational fields in general. But it does seem to tie in correctly with the fact that generalizing the 1+1 metric to 3+1 by simply adding $-d x^2 - d y^2$ gives unphysical results.

In the 2+1 carousel setup, rulers oriented in the transverse direction are shorter when they're lower in the gravitational field (closer to the rim). This means that the xx part of the metric should decrease with z. Generalizing to 3+1, it's not obvious to me whether the contraction should apply to both x and y or only to x.

I am going to assume x, y and z in the context of a cylinder rotating around its long axis, as that is easier to visualise than a disc or sphere rotating around two axes at the same time, although that is technically possible. For the sake of argument let's say the x coordinate is circular and goes around the circular perimeter of the cylinder (equivalent to $d\theta *r$ in the Schwarzschild metric) and the y coordinate is distances measured parallel to the long rotational axis of the cylinder.

X COORDINATE:

Circumference of the cylinder:
Orthogonal to the felt acceleration.
Length contracted according to radar and ruler measurements.(A rotating observer will measure the distance between two marks on the perimeter to be longer than the distance measured by the non rotating observer.)

Horizontal coordinate in a gravitational field:
Orthogonal to the felt acceleration.
NOT length contracted according to radar and ruler measurements. (A stationary local observer will measure the distance between two marks to be the same as the distance calculated by an observer at infinity.)

Y COORDINATE:

Parallel to the long rotational axis of the cylinder:
Orthogonal to the felt acceleration.
NOT length contracted according to radar or ruler measurements.(A rotating and non rotating observer will measure this to be the same.)

There is no equivalent of the cylindrical Y coordinate in the Schwarzschild metric. In general there is also no equivalence for length contraction between a rotating system and a gravitational system and that is not what the EP claims. It simply implies that measurements conspire to make it very difficult to determine if you are in a rotational or gravitational field if you restricted to measurements in an infinitesimally small volume of spacetime.

In the case of a rotating sphere, it is impossible to arrange a rotation scheme around any number of axes simultaneously, whereby an asymmetry would be undetectable to an observer at rest with the sphere, so it is difficult to imagine that there is an a rotational equivalent of a 3+1 uniform gravitational field.

If nothing else, I wanted to clear up what you mean by the x, y and z coordinates. In your first post you seam to be using y as your radial coordinate and the later post you seem to using z as the radial coordinate. That might to be due to different usages in the texts you are referring to. Maybe we should have our own coordinates for this thread, such as (p,q,r) for (x,y,z) as defined in this post?

Of course, I may be completely off target here and you are using cartesian coordinates with one coordinate axis being the axis of rotation and the other two being radial axes that are orthogonal to each other and the rotation axis, in which case the two radial axes are completely equivalent as far as length contraction is concerned.

 

[yuiop]

This confuses me, because $a^2x^2$ is the same as $(1+ax)^2$ except for adding a constant to the x coordinate. So I don't see how the proper acceleration can come out to be independent of x in the first case, but dependent on x in the second case.

$(1+ax)^2$ expands to $1 + 2ax + a^2x^2$ and the additional (1+2ax) is not a constant.

 

[Altabeh]

OK, so we are talking about constant acceleration in all 3 dimensions. Now there is another problem. Since there is equal acceleration in all directions, how does a particle know which way to accelerate or does everything simply expand? Do we have to assume a coordinate centre to such a spacetime and assume everything either accelerates directly towards (or away?) from that coordinate centre? It seems such a gravitational system could not be isotropic, in the sense that any arbitrary location could be chosen as the centre.

I think the problem is that the definition of constant proper acceleration in a 4D spacetime is not so much well established but it is rather baffling and thus it takes me a few minutes writing some lines to get you to know how particles can find which way to accelerate.

I assume you've taken a look at Semay's metric introduced http://arxiv.org/pdf/physics/0601179v1" (early posts in this thread can also help you to get acquainted with it):

$$ds^2=(1+ax)^2dt^2-dx^2-dy^2-dz^2.$$

You can simply see that the curvature tensor vanishes for this metric, giving rise to a flat spacetime wherein a uniform gravitational potential can be considered locally which chalks up to equivalence principle. To verify this, let's check out the geodesic equations:

$$\ddot{x}=-a(1+ax)\dot{t}^2,$$
$$\ddot{y}=\ddot{z}=0,$$
$$\ddot{t}=\frac{-2a\dot{x}\dot{t}}{1+ax},$$

where the dot being the derivative wrt proper time. The aim is to show that for this metric, freely falling particles which follow timelike geodesics, are all accelerated the same locally and the proper acceleration is $$\ddot{x}=-a.$$ As the second equations demonstrate, along y and z axes the proper accelerations are all zero. This is why the freely falling particles are able to find the path on which they can accelerate and fall in almost every other metric (sometimes as in Schwarzschild metric it gets so much twisted to get such a result and that is because the $$g_{rr}$$ is not considered to be constant in contrast to this case)!

Solving the last equation for $$\dot{t}$$ gives

$$\dot{t}=\frac{1}{(1+ax)^2}.$$

Introducing this into the first equation yields

$$\ddot{x}=\frac{-a}{(1+ax)^3}.$$

Therefore, for $$ax<<c^2$$, one gets $$\ddot{x}=-a$$ that confirms the particles following timelike geodesics have constant proper acceleration locally. Sounds like particles know how to accelerate in Semay's spacetime! I see people use hovering particles at a constant $$x=x_0$$ to get global proper accelerations along the geodesics throughout the spacetime . Roughly speaking, this makes the motion boil down to a 2D spatial one so neglecting the general 3D motions in space! I think now you are convinced that on planes parallel to the yz plane the motion gains a constant proper acceleration, but as long as the planes turn to have a slope, the constancy of proper acceleration is guaranteed if one uses equivalence principle.

Now suppose the geodesic equations in a special Rindler's metric,

$$ds^2=e^{2ay}dt^2-dx^2-dy^2-dz^2.$$

Following calculations similar to those done for Semay, we can get

$$\ddot{x}=-ae^{-2ax}.$$

This is again for $$ax<<c^2$$, the result shown by Semay and thus we have a locally uniform gravitational field. But note that this metric is not flat, so that would not be possible to make it coincide with Semay's metric up to all orders in $$ax$$ and thus you have to be careful not to think both are flat since the Taylor expansion of $$e^{2ax}$$ is
$$1+2ax$$ up to order 1 in $$ax.$$ Rest assured that in this case, freely falling particles will find the way they can accelerate!

AB

 

[bcrowell]

OK, so we are talking about constant acceleration in all 3 dimensions. Now there is another problem. Since there is equal acceleration in all directions, how does a particle know which way to accelerate or does everyting simply expand?

There are two different issues:

(1) Does the field vary with position in space?

(2) What direction does the field point?

The notion of a uniform field refers to #1, not #2.

 

[bcrowell]

Now suppose the geodesic equations in a special Rindler's metric,

$$ds^2=e^{2ay}dt^2-dx^2-dy^2-dz^2.$$

This is not a vacuum solution when you add the $-dx^2-dy^2$ terms. See Lut's #16 and my #29.

 

[bcrowell]

The only trouble is that we know length contraction does not occur in the x and z directions in the Schwarzschild metric.

The Schwarzschild isn't a uniform field, it's a spherically symmetric field. That clearly is going to have all kinds of nontrivial effects on the transverse stuff. If you want to check whether Born's statement holds in a specific GR example, I think it makes more sense to do it in a GR example that has translational symmetry in the transverse directions. The Petrov metric has that kind of symmetry, and Born's statement does hold for the Petrov metric. (The Petrov metric has four killing vectors. Three of them are translations of the time and transverse space coordinates, while the fourth is a kind of rescaling that connects the radial coordinate to the others.)

On the face of it, Born's statement appears to be incorrect, or he he saying the transverse coordinates (orthogonal to the radius) of a disc are equivalent to the vertical radial coordinate of a gravitational field?

I think he's making the analogy where (r,phi) on the disk corresponds to (vertical,horizontal) in a uniform field.

I don't claim that Born's argument is anything more than a heuristic. It just provides a nice motivation for trying out a metric of the Kasner form, whose results end up being very similar to the Petrov metric. One way to see that it definitely can't be any more than a heuristic is that he's basing it on the analogy with a rotating coordinate system, but the rotating coordinate system has a nonuniform spatial metric (the scalar curvature depends on r).

 

[bcrowell]

Yes, (if the observer is released at the same time). Presumably, simultaneity is not an issue here, because clocks at different heights in a uniform gravitational field run at the same rate.

No, there is gravitational time dilation. They run at different rates.

 

[Altabeh]

This is not a vacuum solution when you add the $-dx^2-dy^2$ terms. See Lut's #16 and my #29.

I think talking about vacuum solutions makes no sense when it comes to local inertia! I just was trying to show that particles move along timelike paths that determine the direction of acceleration in two examples so the problem of "which way to accelerate" is dependent on the geodesic equations and thus the metric itself!

There are two different issues:

(1) Does the field vary with position in space?

(2) What direction does the field point?

The notion of a uniform field refers to #1, not #2.

I think he meant something else! In all static curved spacetimes, the field varies with position and in all flat spacetimes there is no gravitational field unless one makes use of EP and says 'it's just a small region and the flatness is guaranteed within it' so there the spacetime can admit a uniform gravitational field! The notion of the direction of the proper acceleration refers to both (1) and (2)!

AB

 

[yuiop]

No, there is gravitational time dilation. They run at different rates.

Ah, of course. I realized that a while after I posted. I made the basic mistake of assuming equal acceleration equals equal time dilation which is of course wrong.

 

[yuiop]

Ok, let us assume we have a 3D grid of rulers and clocks in flat space. They all accelerate simultaneously according to a non accelerated observer S and they all have equal proper acceleration according to S and constant acceleration (a) as measured by the accelerated grid observers. (All grid points follow parallel paths in S). Now, assuming acceleration in multiple directions can resolved to a single direction and assuming the x-axis can always be chosen such that it aligns with the direction of acceleration, then as far as I can tell the line element is described by:

$$d\tau^2 = (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2$$

Where x and $d\tau$ are measured in S and dt, dx, dy and dz are measured in the accelerating frame. (Could someone check that?)
This would seem to be in agreement with the time dilation factor (1+ax) given by the relativistic rocket FAQ http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html . In other words, when dx=dy=dz=0, dS/dt = (1+ax).

The situation is essentially the Rindler metric with a = constant, substituted for the usual a = 1/x. The situation is also essentially the Bell's rocket paradox so the observers on the accelerated grid will see themselves getting further apart in the x direction (which eventually tears the grid apart) while transverse distances (y and z) remain constant according to the accelerating grid observers (ruler or radar) and the y,z measurements are measured to be the same according to both the accelerated and non accelerated observers. (No transverse length contraction). This can be confirmed by looking at the Rindler metric http://en.wikipedia.org/wiki/Rindler_coordinates where the transformation from Minkowski to Rindler is Y=y and Z=z. The transverse measurements are simply the same as in linear motion in Special Relativity which makes sense in the constext of instantaneous co-moving observers or what is sometimes referred to as instantaneously "at rest" observers which is a bit more confusing.

Since the situation I have described does not have spherical or cylindrical symmetry, it does not need to be local and can be extended arbitrarily far. Interestingly, there are still event horizons in this spacetime, but the location of the event horizon depends upon where the accelerating observer is.

A ball of free-falling coffee grounds in this spacetime will obviously remain unchanged from the point of view of a co-free-falling observer, but to the accelerated observers on the grid the ball of coffee grounds will appear to be shrinking in the x direction.

If the physical description I have given does not match what you have in mind in this thread, please could you give a physical description of what you do have in mind. Also, if it could be made clear which frame the measurements are made in it might help clear up why there appears to be inconsistent forms of the metric.

 

[Altabeh]

Ok, let us assume we have a 3D grid of rulers and clocks in flat space. They all accelerate simultaneously according to a non accelerated observer S...

Yeah, but not in every four dimensional flat spacetime, e.g. non-Minkowski, it would be true for a a non accelerated observer S to observe all particles accelerate simultaneously because the proper acceleration probably depends on the position of particles! This of course does not occur for flat spacetimes for which all Christoffel symbols vanish! I assume that your flat space is like these!

Now, assuming acceleration in multiple directions can resolved to a single direction and assuming the x-axis can always be chosen such that it aligns with the direction of acceleration, then as far as I can tell the line element is described by

$$d\tau^2 = (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2$$

Where x and $d\tau$ are measured in S and dt, dx, dy and dz are measured in the accelerating frame. (Could someone check that?) This would seem to be in agreement with the time dilation factor (1+ax) given by the relativistic rocket FAQ http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html . In other words, when dx=dy=dz=0, dS/dt = (1+ax).

Except for $$\tau$$, $$dt$$ and $$dx^i$$ for i=1,2,3 are all measured by an at-rest observer and here $$d\tau$$ is measured by the accelerating observers. This is reminiscence of time dilation (assuming $$dx^i=0$$) where

$$c^2d\tau^2=c^2(1+ax/c^2)^2 dt^2,$$

and thus

$$d\tau=(1+ax/c^2) dt,$$ with $$(1+ax/c^2)^{-1}=\gamma$$. Nevertheless, all observers moving relative to each other in arbitrary ways agree on the quantity $$d\tau^2$$.

The situation is essentially the Rindler metric with a = constant, substituted for the usual a = 1/x.

I don't see this. Please explain it a little bit!

The situation is also essentially the Bell's rocket paradox so the observers on the accelerated grid will see themselves getting further apart in the x direction (which eventually tears the grid apart) while transverse distances (y and z) remain constant according to the accelerating grid observers (ruler or radar) and the y,z measurements are measured to be the same according to both the accelerated and non accelerated observers.

This may be incorrect because we don't know about the proper velocities along y and z axes relative to an observer moving inertially in the spacetime! They can be anything constant equally since from the two geodesic equations we have $$\ddot{z}=\ddot{y}=0$$ so along a geodesic particles can uniformly move parallel to y and z axes relative to the observers. To a non accelerated (inertial) observer moving in the direction of x axis, you are right.

(No transverse length contraction). This can be confirmed by looking at the Rindler metric http://en.wikipedia.org/wiki/Rindler_coordinates where the transformation from Minkowski to Rindler is Y=y and Z=z. The transverse measurements are simply the same as in linear motion in Special Relativity which makes sense in the constext of instantaneous co-moving observers or what is sometimes referred to as instantaneously "at rest" observers which is a bit more confusing.

The transverse measurements of velocity, as I said, are not the same unless they are zero according to all observers or to the observers that are all moving inertially parallel to the x axis! But instantaneous co-moving observers are also able to overcome the problem as they are moving parallel to the geodesics with the same acceleration and velocity.

Since the situation I have described does not have spherical or cylindrical symmetry, it does not need to be local and can be extended arbitrarily far.

In a flat spacetime, for example the Semay's spacetime, that the geodesic equations are dependent on the position at least the measurement of proper acceleration is always local from every observer's perspective.

AB

 

[bcrowell]

If the physical description I have given does not match what you have in mind in this thread, please could you give a physical description of what you do have in mind.

I don't think it totally works as a uniform field, because an observer hovering at a fixed location relative to the lattice you described can release a test particle from rest, and the acceleration of the test particle is $-a(1+ax)$. This isn't constant, so the field isn't uniform.

I don't think there is any metric in GR that can be called a uniform field and that satisfies every criterion you could put on a wish list. They all have certain undesirable properties. But anyway this is the undesirable property of this particular metric.

Also, if it could be made clear which frame the measurements are made in it might help clear up why there appears to be inconsistent forms of the metric.

The whole notion of a gravitational field is frame-dependent. If you're looking for a metric that in some sense embodies a uniform field with a particular value of g, then all you can hope for is a metric that accomplishes that in one particular set of coordinates. Those are the coordinates you use to write down the line element.

That doesn't mean that there are no frame-independent properties that you should hope for. In particular, I think it's desirable for the scalar curvature to be constant, and for the metric to be vacuum solution. Both the Rindler coordinates and the Petrov spacetime satisfy these criteria.