# Inconsistent forms of the metric in a uniform field

1. Feb 10, 2010

### bcrowell

Staff Emeritus
Let's say we want to write down the metric in a uniform field. I see two ways of going about this.

Method 1: Straightforward arguments using the equivalence principle and photons in elevators show that if a photon with initial energy E rises or falls by dy, then its energy shift is given (ignoring signs) by $dE/E=g dy$. Integration shows that the time dilation factor between two different heights is $\exp(\Delta \Phi)$, where $\Phi$ is the gravitational potential. If two clocks have parallel world-lines at two different heights, then the ratio of the proper times is the square root of the ratio of the time-time elements of the metric, so we have
$$ds^2=e^{2\Delta\Phi}dt^2-dy^2 \qquad [1]$$

Method 2: Start in a frame where the metric is Minkowski. Find the motion of an observer who experiences constant proper acceleration. Transform into this observer's frame by using the tensor transformation law on the metric. The result is
$$ds^2=(1+ay)^2dt^2-dy^2 \qquad [2]$$
This is given in Semay, http://arxiv.org/abs/physics/0601179 . By the equivalence principle, it can also be interpreted as the metric experienced by an observer in a gravitational field with g=a.

If we set $\Phi=gy$, then these two forms are equivalent to the first non-constant order:
$$g_{tt} = 1+2gy+\ldots \qquad$$ ,
but they disagree in their higher-order terms.

What is the reason for this discrepancy?

Rindler's Essential Relativity (2nd ed., 120) suggests using the gravitational redshift to define the gravitational potential. I'm not sure if this is meant to suggest that the potential in a uniform field is not necessarily exactly $gy$, or if it's meant to allow the generalization to nonuniform fields (which he carries out a few pages later).

Do the two forms differ because there's an implied choice of coordinates that is different in the two cases? Is this perhaps related to Bell's spaceship paradox, i.e., to issues in defining the notion that two different objects both experience the same proper acceleration, due to the relativity of simultaneity?

Form [1] (with $\Phi=gy$) has a property that I would consider indispensable for a uniform field, which is that I can't determine my y by local measurements. With form [2], I can take a vertical measuring rod with clocks at each end, and depending on what y I'm at, the ratio of the clocks' rates will be different. This seems physically wrong to me, even in the case where you interpret it as an acceleration rather than a gravitational field. An observer inside an accelerating rocket should not be able to determine what point in the motion he's presently at, using local measurements -- should he?

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. (This argument is apparently made in Born, 1920, Einstein's Theory of Relativity, which was an early popularization of GR. i don't have the book yet, so I'm just doing this from a summary of Born's argument.)

2. Feb 10, 2010

### atyy

Since gravity is curvature and acceleration can never make flat space curved, so I wouldn't expect agreement to all orders.

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

Is $\Phi=gy$ a solution of the field equations?

Last edited: Feb 10, 2010
3. Feb 10, 2010

### bcrowell

Staff Emeritus

Hmm...well, I guess that depends on what you mean by "gravity." You can certainly have a gravitational field in flat space; if that were not true, then it would violate the equivalence principle.

On the other hand, you make a good point about inspecting these two metrics for curvature. Metric [2] is a flat-space metric, since it was derived from a Minkowski metric by a change of coordinates. Metric [1] is not a flat-space metric. Its Ricci tensor has $R_{tt}=g^2e^{2\Phi}$ and $R_{yy}=-g^2$. So that's cool, that establishes that the two metrics are not equivalent under a change of coordinates.

Hmm...okay, WP http://en.wikipedia.org/wiki/Rindler_coordinates says Rindler coordinates are flat-space coordinates with a metric $ds^2=y^2dt^2-dy^2$. I see, so this is the same as the coordinates in Semay's metric [2] except for a trivial linear transformation.

If we assume $\Phi=gy$, and if we also take $g_{tt}=e^{2\Phi}$, then the Ricci tensor I calculated above shows that it's definitely not a vacuum solution.

Hmm...so this does make things a little clearer in my mind. However, I'm still not completely clear on the physical situation. Metric [1] gives a relative time dilation between clocks at y and y+dy that is independent of y, and this seems like a necessary property if we're going to think a certain metric as representing the metric experienced by an observer at rest in a uniform field. On the other hand, the Ricci tensor is also a local observable, and it varies with y for this metric, so by that criterion it seems like [2] is more like the right one. It seems to me that there ought to be a metric and coordinates that satisfy the following properties: (a) it's a flat-space metric; (b) test particles have coordinate acceleration g; and (c) it's not possible to determine your height in the field by local measurements of curvature (which follows trivially from property a) *or* by local measurements relative to the coordinate lattice (i.e., with clocks and rulers that have zero coordinate velocities). Am I finding out that you can't have a, b, and c all in one metric? Need to think about this some more.

4. Feb 10, 2010

### DrGreg

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 (discussed in Gibbons & Gielen (2008), "The Petrov and Kaigorodov-Ozsváth Solutions: Spacetime as a Group Manifold", Class.Quant.Grav.25:165009). To be honest, the details of this go over my head, so don't ask me to explain it, but I pass it on for what it's worth.

(I think that means the answer to atyy's last question is "no", but I'm out of my depth.)

5. Feb 10, 2010

### bcrowell

Staff Emeritus
Interesting. The Gibbons and Gielen paper is over my head, too. As far as I can tell, the Petrov solution is something to do with a rotating cylinder, it has cylindrical symmetry, and it has CTCs. Seems odd that the closest thing to a uniform gravitational field in GR is something that has CTCs...? It makes sense that the paper is talking about particularly symmetric vacuum solutions, and the uniform gravitational field would certainly be a highly symmetric vacuum solution.

6. Feb 11, 2010

### torquil

Did you see section 5? It seems that they disprove the possibility of accelerating a dust cylinder up to the angular velocity at which the CTCs appear.

Torquil

7. Feb 11, 2010

### Altabeh

The meteric (1) is not a generalized form of (2), nor is it of the same nature of the latter in the sense that one is just a flat spacetime while the other isn't. So claiming that the local flatness must be the same up to all orders in (gy) is by no means possible and in general this even is not true when two spacetimes are both curved: This is because we cannot make the metrics locally flat at one point in both of the spacetimes using one single metric transformation at the same time if their curvatures are different everywhere.

So the demand that the metric (2) and (1) must coincide up to order 2 is not even imaginable, let alone all orders. This has some other reasoning, too: The local flatness requires a special metric transformation to modify $$e^{2\Delta \Phi}$$ as it corresponds at some point to $$(1+gy)^2$$ and this may be feasible; but I myslef see no room for it and suggest you to take a look at post #59 in https://www.physicsforums.com/showthread.php?t=373353&page=4" to know what kind of transformation would do help you out with that!

And about how those properties can be satisfied all together, I have to say that since you are assuming the spacetime to be flat (pick the metric (1) and go on with to calculate all components of geodesic equations i.e. equations for t and y), then to satisfy the second condition, you must have some free degrees of freedom (which can be obtained by a metric transformation) to set the second portion of geodesic equation for y equal to $$-{g}=\frac{GM}{y^2}$$, if the center of gravitational source is at y=0 and along with the geodesic equation for t, solve for those free parameters. To get the value of $$\Phi$$ which gives the vaccum solutions of the field, try to make the $$\bar{R}_{\mu\nu}$$ vanish by solving for $$\Phi$$, where the bar over the Ricci scalar indicates that we are in the new coordinate system.

Here it must be recalled that if we take k to be the proper time, then introducing $$d^2t/dk^2$$ and $$d^2y/dk^2$$ of geodesic equations of Semay's metric into $$\bar{g}_{\mu \nu}$$ leads to a metric at least being equivalent to Semay's metric up to order 2 in (gy).

I hope this helps!

AB

Last edited by a moderator: Apr 24, 2017
8. Feb 11, 2010

### atyy

Well, I wouldn't call flat spacetime a gravitational field (except maybe in a spherical shell). But it's true, EP terminology would accept accelerated frames in flat spacetime as in a gravitational field. I would be surprised if there is any accelerated frame that gives a uniform field, since the Rindler frame isn't a uniform field, even though that would be my naive guess because of its constant accelration. On the other hand, the EP is only local, so we'd only need it to be uniform at a point. So I would take any non-geodesic timelike worldline, and apply Fermi normal coordinates (section 3.2 of http://relativity.livingreviews.org/Articles/lrr-2004-6/ [Broken]). The metric will be Minkowski at a point, but the first derivatives will not disappear, and I guess one can count that as a gravitational field (Eq. 125 - 127)?

Last edited by a moderator: May 4, 2017
9. Feb 11, 2010

### Altabeh

A flat spacetime never admits a gravitational field within it so talking about a flat spacetime just inspires the fact that the geometrical shape of spacetime is everywhere the same i.e. the Riemann tensor vanishes everywhere!

What is EP?

GR would accept any accelerated frames in flat spacetimes and this can be seen for instance, for the metric introduced https://www.physicsforums.com/showpost.php?p=2560660&postcount=52".

It does not have a constant acceleration, does it? As I put forth in my last post, the idea of Rindler's metric admitting a uniform field globally is not true as when $$\Phi = \Phi (y)$$, then the geodesic equation would depend on y so it won't be uniform everywhere, but locally. The locally uniform field can be gained by transforming Rindler's metric into a new metric having zero first derivatives in the neighbourhood of a given point.

If you mean you have doubt about Rindler's metric being a gravitational field, you have to remove it as its Riemann tensor does not vanish so it admits a gravitational field!

AB

Last edited by a moderator: Apr 24, 2017
10. Feb 11, 2010

### atyy

Equivalence Principle (which is not a principle)

11. Feb 12, 2010

### bcrowell

Staff Emeritus
I could be wrong (since I don't claim to understand the paper!), but I think what they're saying is that (1) the Petrov solution does have CTCs (bottom of p. 3), (2) it can be interpreted as the vacuum outside a rotating cylinder, and (3) such a cylinder cannot be created from realistic initial conditions in our universe (section 5). I think the spacetime outside the more slowly rotating cylinder isn't classified as a Petrov solution...?

12. Feb 12, 2010

### bcrowell

Staff Emeritus
I found a good discussion of this topic by Weiss: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken]

My metric [1] is a coordinate system in which a test particle with an initial velocity of zero has an initial proper acceleration g. This acceleration is the same at all points in spacetime.

The page is actually a discussion of the Bell spaceship paradox, and Weiss looks at it from three different perspectives, of which the metric [1] is one of them. If we imagine tying two rocks to the upper and lower ends of a thread and dropping them into the spacetime [1], we get Bell's paradox.

He makes a physical argument that the spacetime must be curved, since otherwise there would be no way to explain the fact that two spaceships hovering in this spacetime, one above the other, maintain a constant distance between them (which would violate the fact established by the usual analysis of Bell's paradox in flat spacetime). He computes the curvature, interprets it as arising from an unphysical stress-energy tensor, and says, "My guess is that the Einstein empty-space equations forbid a uniform gravitational field in the above sense. I haven't checked this, though."

Last edited by a moderator: May 4, 2017
13. Feb 12, 2010

### Altabeh

Are sure that an initial vanishing velocity can be supposed at any spatial coordinate say for y=y0? The geodesic equation in this spacetime gives

$$v^2=\frac{\dot{v}}{2\dot{t}^2}+e^{2y}$$,

(where dot represents derivative wrt proper time) for which, $$v$$ can only be zero if $$y_0\rightarrow -\infty$$. I don't think your claim is true because the spacetime is curved so it doesn't admit a uniform gravitational field everywhere.

AB

14. Feb 12, 2010

### bcrowell

Staff Emeritus
Hmm...I get $\Gamma^t_{zt}=g$ and $\Gamma^z_{tt}=-g$, and if I apply the geodesic equation with the affine parameter taken to be the proper time, I get $\ddot{z}=g\dot{t}^2$, which is the same as $\ddot{z}=g$ for an object initially at rest.

Pretty sure, since this spacetime is symmetric in the sense that locations that differ in z have identical properties. You can tell it has this symmetry because adding a constant to z is equivalent to rescaling the time coordinate, but rescaling the time coordinate doesn't have observable effects due to general covariance.

I don't think curvature is inconsistent with a uniform gravitational field. My initial idea coming into this was that my idea of a uniform gravitational field would be a flat spacetime, but I think my initial concept demanded too many different and incompatible things.

15. Feb 12, 2010

### Altabeh

I guess somethig is wrong: Assuming that $e^{2\Phi }= e^{2gy}$, we get

$\Gamma^y_{tt}=-\frac{1}{2}g^{yy}g_{tt,y}={g}e^{2gy},$
$\Gamma^t_{ty}=\frac{1}{2}g^{tt}g_{tt,y}={g}e^{-2gy}e^{2gy}=g.$

This will follow my velocity equation, i.e.

$$v^2=\frac{\dot{v}}{2\dot{t}^2}+e^{2y}$$, (1)

from which one can get the acceleration, but it is not going to be constant globally!

This is completely true according to your calculation. But according to mine, I have a term which depends on y so any translation y-->y+y_0 would give a different acceleraion.

I think you have to re-sketch the whole thing again. The Rindler's metric is by no means able to admit a globally uniform gravitational acceleration, as you can see from my equation (1) unless you are in a very small region of it which again inspires the local flatness and EP, thus neglecting uniformity of field everywhere! At this point, Semay's metric is compatible with your ideas and can be taken into account if one is interested in studying a completely uniform gravitational field with a constant acceleration at any point!

AB

Edit: Equation one can be written as

$$v^2=\frac{{a}}{2\dot{t}}+e^{2y}$$, (1')

where a is the acceleration.

16. Feb 12, 2010

### Mentz114

One connection between metric [1] and [2] is that if you start with $\Delta\Phi$ as a unknown function of $y$ and set the Ricci tensor to zero, the result is the solution of this equation ( writing $P$ for $\Delta\Phi$)

$$\frac{{d}^{2}}{d\,{y}^{2}}\,P+{\left( \frac{d}{d\,y}\,P\right) }^{2}=0 \qquad [1]$$

which is

$$P=log\left( y+k_1\right) +k_2$$

If we set $$k_2=0[/itex] then $exp(2\Delta\Phi)$ is $(1+ay)^2$ up to a scale factor. My ( possibly wrong ) interpretation is that [tex]k_2=0[/itex] is the value of $P$ at $y=0$. So a particle released at $y=0$ does not move. [edit: I just realised that this is obvious from the metric, which is flat at y=0] Another (interesting?) thing is that the non-zero components of the Einstein tensor are $G_{yy}$ and $G_{zz}$ which both take the value of the lhs of [1]. Is there any possible distribution of energy/momentum/pressure that could give this ? I think not. Last edited: Feb 12, 2010 17. Feb 12, 2010 ### Altabeh No it is not! If you set [tex]k_2=0[/itex], then the Taylor expansion of [tex] P=\log\left(y+k_1\right)$$ would have a term with different sign than that of $$1+2ay+a^2y^2$$. So again we are back to the first point that these two never coincide up until Rindler's metric [1] goes into another coordinate system which I talked about it earlier.

[/QUOTE]

Did you calculate components of the Einstein tensor based on bcrowell's Christoffel symbols?

AB

Edit: The Taylor expansion of
$$e^{2P}=e^{2\log\left( y+k_1\right)}$$

must have been considered, so this only coincides Symay's metric, if $$k_1=g=1$$. But remember that then we are again back to the principle that a flat spacetime only accepts a uniform gravitational field.

Last edited: Feb 12, 2010
18. Feb 12, 2010

### bcrowell

Staff Emeritus
Ah, thanks for the correction -- I did have $\Gamma^y_{tt}$ wrong.

I think my conclusion was still correct, because it was based on the symmetry of the metric, not a symmetry of the Christoffel symbols. For the same reason, I think my calculation of the proper acceleration is still correct globally. The spacetime and coordinates have the same properties everywhere, so I don't think the proper acceleration can be different in different places.

Are we getting our wires crossed because I'm talking about proper acceleration and you're talking about coordinate acceleration?

I agree with you about the interpretation. I don't interpret the metric [1] as being a globally uniform gravitational field.

19. Feb 12, 2010

### bcrowell

Staff Emeritus
Here is a more detailed derivation of the proper acceleration. The Christoffel symbols are $\Gamma\indices{^t_{zt}}=g$ and $\Gamma\indices{^z_{tt}}=-ge^{2gz}$. The geodesic equation with the affine parameter taken to be the proper time is $\ddot{z}=ge^{2gz}\dot{t}^2$, where dots represent differentiation with respect to proper time. For a particle instantaneously at rest, $\dot{t}=1/\sqrt{g_{tt}}=e^{-gz}$, so $\ddot{z}=g$.

20. Feb 12, 2010

### bcrowell

Staff Emeritus
With $\Phi=gy$, G's timelike part is zero, and its spacelike parts are all negative. The Weiss page http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken] notes this, and interprets it as showing that this metric is unphysical.
I wonder if one can do anything interesting with a metric of the form $e^Pdt^2-e^Q(dx^2+dy^2)-dz^2$. This would seem to be a reasonable form to look at because of the argument by Born I described in #1. I have the Born book now, so I'll see if I can find the relevant passage.