inconsistent forms of the metric in a uniform field


by bcrowell
Tags: field, forms, inconsistent, metric, uniform
bcrowell
bcrowell is offline
#37
Feb14-10, 09:57 AM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,500
Hi, Altabeh -- Re your #36, please note that I have always explicitly said that [itex]\dot{t}=e^{-gy}[/itex] 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 [itex]\dot{t}=e^{-gy}[/itex] 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/physic...ip_puzzle.html (at "More precisely, time-like geodesics..."; they're doing the special case of g=1).
Mentz114
Mentz114 is offline
#38
Feb14-10, 12:26 PM
PF Gold
P: 4,081
The problem for me is that this metric [tex]d\tau^2=e^{2gy}dt^2-dy^2 [/tex] 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 [itex]ds^2=(1+ax)^2-dx^2 +-dy^2-dz^2[/itex] ( which is a vacuum solution of the EFE) I repeat the latter calculation. The geodesic equation for [itex]x[/itex] is

[tex]
\frac{d^2x}{ds^2}=-a(1+ax)\left(\frac{dt}{ds}\right)^2
[/tex]
where [itex]s[/itex] is an affine parameter. Now, like post #19 I take [itex]s[/itex] to be the proper time [itex]\tau[/itex]. 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 [itex]dt/d\tau=1/\sqrt{g_{tt}}[/itex] which gives
[tex]
\frac{d^2x}{d\tau^2}=\frac{-a(1+ax)}{(1+ax)^2}=\frac{-a}{1+ax}
[/tex]
What does this mean ? It is not a constant acceleration as advertised. The problem could be that [itex]s\ne \tau[/itex]. Don't we need a worldline with starting conditions to get the relationship between [itex]\tau[/itex] and [itex]s[/itex] ?

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
bcrowell is offline
#39
Feb14-10, 02:49 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,500
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 [itex]d^2 x/d \tau^2[/itex] that's not the same everywhere. I think this is just another instance of the Bell spaceship paradox. The factor of [itex]1+ax[/itex] 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 [itex]d^2 x/d \tau^2[/itex] 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
Mentz114 is offline
#40
Feb14-10, 03:02 PM
PF Gold
P: 4,081
Quote Quote by bcrowell
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 [itex]d^2 x/d \tau^2[/itex] 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 [itex]t[/itex] ?

These are rhetorical questions I need to think about.
George Jones
George Jones is offline
#41
Feb14-10, 03:37 PM
Mentor
George Jones's Avatar
P: 6,038
Quote Quote by Mentz114 View Post
[tex]
\frac{d^2x}{d\tau^2}=\frac{-a(1+ax)}{(1+ax)^2}=\frac{-a}{1+ax}
[/tex]

What does this mean ? It is not a constant acceleration as advertised.
Yes, it is.
Mentz114
Mentz114 is offline
#42
Feb14-10, 03:58 PM
PF Gold
P: 4,081
George,
Thanks. You're right of course. It should be written,
[tex]
\ddot{x}\mid_{x=x_r}=\frac{-a}{1+ax_r}
[/tex]
where [itex]x_r[/itex] 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 [itex]x_r[/itex] though.
Altabeh
Altabeh is offline
#43
Feb14-10, 05:27 PM
P: 665
Quote Quote by bcrowell View Post
Hi, Altabeh -- Re your #36, please note that I have always explicitly said that [itex]\dot{t}=e^{-gy}[/itex] 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 [tex]dy=0[/tex]. 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 [itex]\dot{t}=e^{-gy}[/itex] is not valid because for a particle moving continuously along some curve, [tex]dy=0[/tex] 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,

[tex]v^2=\frac{1}{2g}(a+ge^{2gy})[/tex]

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

[tex]a=-ge^{2gy_0}[/tex],

where [tex]y_0[/tex] is where particle is at rest. You see that I don't use [tex]\dot{t}=1/\sqrt{g_{tt}}=e^{-gy}[/tex] 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 [tex]dy/d\tau = 0[/tex] at some point [tex]y_0[/tex], then the proper acceleration at [tex]y_0[/tex] along a time-like geodesic passing through it, equals [tex]-g[/tex]. 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 [tex]d^2x/d\tau^2 = -1[/tex] at P. So if a lab-frame observer (that is, (t,x) coordinate system) lets 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
Altabeh is offline
#44
Feb14-10, 06:04 PM
P: 665
Quote Quote by Mentz114 View Post
The problem for me is that this metric [tex]d\tau^2=e^{2gy}dt^2-dy^2 [/tex] 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
Altabeh is offline
#45
Feb14-10, 06:26 PM
P: 665
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 [tex]g=0[/tex] 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
bcrowell is offline
#46
Feb14-10, 07:05 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,500
Quote Quote by Altabeh View Post
What exactly do you mean by "instantaneously at rest"?
As I said in #37, I mean that it has a zero coordinate velocity.

Quote Quote by Altabeh View Post
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
bcrowell is offline
#47
Feb14-10, 07:18 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,500
Quote Quote by Mentz114 View Post
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 [itex]t[/itex] ?

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 [itex]d^2x/d\tau^2=-a/(1+ax)[/itex] 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 [itex]\tau[/itex] as the affine parameter used for P's geodesic equation, so your [itex]\tau[/itex] 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
Mentz114 is offline
#48
Feb14-10, 10:15 PM
PF Gold
P: 4,081
Ben,

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

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

Which is independent of [itex]x[/itex]. The line element is a coordinate transformation from Minkowski space, but the original space and time have been mixed, so the [itex]x[/itex] 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
Mentz114 is offline
#49
Feb14-10, 10:39 PM
PF Gold
P: 4,081
Quote Quote by Altabeh View Post
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
Altabeh is offline
#50
Feb15-10, 03:33 AM
P: 665
Quote Quote by bcrowell View Post
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 "clocks and gravity" 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
Altabeh is offline
#51
Feb15-10, 03:47 AM
P: 665
Quote Quote by Mentz114 View Post
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
yuiop is offline
#52
Feb17-10, 03:56 PM
P: 3,966
Quote Quote by atyy View Post
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.
Quote Quote by bcrowell View Post
... 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.

Quote Quote by Mentz114 View Post
[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.
Quote Quote by Mentz114 View Post
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.
Quote Quote by Mentz114 View Post
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
bcrowell is offline
#53
Feb17-10, 05:04 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,500
Quote Quote by kev View Post
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 [itex]\omega =c/R[/itex], 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.

Quote Quote by Mentz114 View Post
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 [itex]\phi[/itex] 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
bcrowell is offline
#54
Feb17-10, 05:14 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,500
Quote Quote by Mentz114 View Post
Wiki shows how in the Rindler chart, the Minkowski line element becomes
[itex]ds^2=-a^2x^2+dx^2+dy^2+dz^2[/itex].
Repeating the calculation I did previously with this I get [itex]\ddot{x}=-a[/itex].

Which is independent of [itex]x[/itex]. The line element is a coordinate transformation from Minkowski space, but the original space and time have been mixed, so the [itex]x[/itex] 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
[itex]ds^2=-a^2x^2dt^2+dx^2+dy^2+dz^2[/itex]
(with a dt2 in the first term).

This confuses me, because [itex]a^2x^2[/itex] is the same as [itex](1+ax)^2[/itex] 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.


Register to reply

Related Discussions
Weak Field metric Advanced Physics Homework 4
[SOLVED] Linear forms and complete metric space Calculus & Beyond Homework 12
Question about Differential Forms for Non-Abelian Field Strengths General Physics 1
Uniform Metric vs Box Topology General Math 2
Question about the metric tensor in Einstein's field equations. Special & General Relativity 3