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## inconsistent forms of the metric in a uniform field

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.

 Quote by bcrowell 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

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

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(*) 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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/physic...ip_puzzle.html (at "More precisely, time-like geodesics..."; they're doing the special case of g=1).
 Blog Entries: 3 Recognitions: Gold Member 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.

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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

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 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 $d^2 x/d \tau^2$ 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.

Mentor
 Quote by Mentz114 $$\frac{d^2x}{d\tau^2}=\frac{-a(1+ax)}{(1+ax)^2}=\frac{-a}{1+ax}$$ What does this mean ? It is not a constant acceleration as advertised.
Yes, it is.
 Blog Entries: 3 Recognitions: Gold Member 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.

 Quote by 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.
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) 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!

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* 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

 Quote by 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.
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
 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

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 Quote by Altabeh What exactly do you mean by "instantaneously at rest"?
As I said in #37, I mean that it has a zero coordinate velocity.

 Quote by 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:
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.

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 Quote by Mentz114 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.
 Blog Entries: 3 Recognitions: Gold Member 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.

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 Quote by Altabeh 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.

 Quote by bcrowell 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

 Quote by Mentz114 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!