Inconsistent forms of the metric in a uniform field

In summary, there are two methods for writing down the metric in a uniform field, and they disagree in their higher-order terms. This discrepancy is due to an implied choice of coordinates that is different in the two cases. It is also possible that there is no single metric that satisfies all the desired properties for a uniform field, such as being flat, producing a relative time dilation independent of height, and being indeterminate for determining height by local measurements.
  • #1
bcrowell
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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 [itex]dE/E=g dy[/itex]. Integration shows that the time dilation factor between two different heights is [itex]\exp(\Delta \Phi)[/itex], where [itex]\Phi[/itex] 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
[tex]ds^2=e^{2\Delta\Phi}dt^2-dy^2 \qquad [1][/tex]

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
[tex]ds^2=(1+ay)^2dt^2-dy^2 \qquad [2][/tex]
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 [itex]\Phi=gy[/itex], then these two forms are equivalent to the first non-constant order:
[tex] g_{tt} = 1+2gy+\ldots \qquad [/tex] ,
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 [itex]gy[/itex], 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 [itex]\Phi=gy[/itex]) 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.)
 
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  • #2
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 [itex]\Phi=gy[/itex] a solution of the field equations?
 
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  • #3
Thanks, atyy, for your helpful reply!

atyy said:
Since gravity is curvature [...]
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.

atyy said:
[...] and acceleration can never make flat space curved, so I wouldn't expect agreement to all orders.
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 [itex]R_{tt}=g^2e^{2\Phi}[/itex] and [itex]R_{yy}=-g^2[/itex]. So that's cool, that establishes that the two metrics are not equivalent under a change of coordinates.

atyy said:
Also, Rindler coordinates are not a uniform gravitational field (at least that's what Rindler's text claims).
Hmm...okay, WP http://en.wikipedia.org/wiki/Rindler_coordinates says Rindler coordinates are flat-space coordinates with a metric [itex]ds^2=y^2dt^2-dy^2[/itex]. I see, so this is the same as the coordinates in Semay's metric [2] except for a trivial linear transformation.

atyy said:
Is [itex]\Phi=gy[/itex] a solution of the field equations?
If we assume [itex]\Phi=gy[/itex], and if we also take [itex]g_{tt}=e^{2\Phi}[/itex], 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
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
DrGreg said:
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.)

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
bcrowell said:
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.

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
bcrowell said:
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.

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 [tex]e^{2\Delta \Phi}[/tex] as it corresponds at some point to [tex](1+gy)^2[/tex] 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 [tex]-{g}=\frac{GM}{y^2}[/tex], 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 [tex]\Phi[/tex] which gives the vacuum solutions of the field, try to make the [tex]\bar{R}_{\mu\nu}[/tex] vanish by solving for [tex]\Phi[/tex], 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 [tex]d^2t/dk^2[/tex] and [tex]d^2y/dk^2[/tex] of geodesic equations of Semay's metric into [tex]\bar{g}_{\mu \nu}[/tex] leads to a metric at least being equivalent to Semay's metric up to order 2 in (gy).

I hope this helps!

AB
 
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  • #8
bcrowell said:
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.

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)?
 
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  • #9
atyy said:
Well, I wouldn't call flat spacetime a gravitational field (except maybe in a spherical shell).[/tex]

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!

But it's true, EP terminology would accept accelerated frames in flat spacetime as in a gravitational field.

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".

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.

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 [tex]\Phi = \Phi (y)[/tex], 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.

I guess one can count that as a gravitational field (Eq. 125 - 127)?

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
 
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  • #10
Altabeh said:
What is EP?
Equivalence Principle (which is not a principle)
 
  • #11
torquil said:
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.

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
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."
 
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  • #13
bcrowell said:
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.

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

[tex]v^2=\frac{\dot{v}}{2\dot{t}^2}+e^{2y}[/tex],

(where dot represents derivative wrt proper time) for which, [tex]v[/tex] can only be zero if [tex]y_0\rightarrow -\infty[/tex]. 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
Altabeh said:
The geodesic equation in this spacetime gives
[tex]v^2=\frac{\dot{v}}{2\dot{t}^2}+e^{2y}[/tex],
Hmm...I get [itex]\Gamma^t_{zt}=g[/itex] and [itex]\Gamma^z_{tt}=-g[/itex], and if I apply the geodesic equation with the affine parameter taken to be the proper time, I get [itex]\ddot{z}=g\dot{t}^2[/itex], which is the same as [itex]\ddot{z}=g[/itex] for an object initially at rest.

Altabeh said:
Are sure that an initial vanishing velocity can be supposed at any spatial coordinate say for y=y0?

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.

Altabeh said:
I don't think your claim is true because the spacetime is curved so it doesn't admit a uniform gravitational field everywhere.

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

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

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

This will follow my velocity equation, i.e.

[tex]v^2=\frac{\dot{v}}{2\dot{t}^2}+e^{2y}[/tex], (1)

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

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.

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 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.

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

[tex]v^2=\frac{{a}}{2\dot{t}}+e^{2y}[/tex], (1')

where a is the acceleration.
 
  • #16
One connection between metric [1] and [2] is that if you start with [itex]\Delta\Phi[/itex] as a unknown function of [itex]y[/itex] and set the Ricci tensor to zero, the result is the solution of this equation ( writing [itex]P[/itex] for [itex]\Delta\Phi[/itex])

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

which is

[tex]
P=log\left( y+k_1\right) +k_2
[/tex]

If we set [tex]k_2=0[/itex] then [itex]exp(2\Delta\Phi)[/itex] is [itex](1+ay)^2[/itex] up to a scale factor.

My ( possibly wrong ) interpretation is that [tex]k_2=0[/itex] is the value of [itex]P[/itex] at [itex]y=0[/itex]. So a particle released at [itex]y=0[/itex] does not move. [edit: I just realized 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 [itex]G_{yy}[/itex] and [itex]G_{zz}[/itex] 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.
 
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  • #17
Mentz114 said:
One connection between metric [1] and [2] is that if you start with [itex]\Delta\Phi[/itex] as a unknown function of [itex]y[/itex] and set the Ricci tensor to zero, the result is the solution of this equation ( writing [itex]P[/itex] for [itex]\Delta\Phi[/itex])

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

which is

[tex]
P=log\left( y+k_1\right) +k_2
[/tex]

If we set [tex]k_2=0[/itex] then [itex]exp(2\Delta\Phi)[/itex] is [itex](1+ay)^2[/itex] up to a scale factor.

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)[/tex] would have a term with different sign than that of [tex]1+2ay+a^2y^2[/tex]. 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.

Another (interesting?) thing is that the non-zero components of the Einstein tensor are [itex]G_{yy}[/itex] and [itex]G_{zz}[/itex] 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.
[/QUOTE]

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

AB

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

must have been considered, so this only coincides Symay's metric, if [tex]k_1=g=1[/tex]. But remember that then we are again back to the principle that a flat spacetime only accepts a uniform gravitational field.
 
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  • #18
Altabeh said:
I guess somethig is wrong: Assuming that [itex]e^{2\Phi }= e^{2gy}[/itex], we get

[itex]\Gamma^y_{tt}=-\frac{1}{2}g^{yy}g_{tt,y}={g}e^{2gy},[/itex]
[itex]\Gamma^t_{ty}=\frac{1}{2}g^{tt}g_{tt,y}={g}e^{-2gy}e^{2gy}=g.[/itex]
Ah, thanks for the correction -- I did have [itex]\Gamma^y_{tt}[/itex] wrong.

bcrowell said:
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.

Altabeh said:
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 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?

Altabeh said:
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!

I agree with you about the interpretation. I don't interpret the metric [1] as being a globally uniform gravitational field.
 
  • #19
Here is a more detailed derivation of the proper acceleration. The Christoffel symbols are [itex]\Gamma\indices{^t_{zt}}=g[/itex] and [itex]\Gamma\indices{^z_{tt}}=-ge^{2gz}[/itex]. The geodesic equation with the affine parameter taken to be the proper time is [itex]\ddot{z}=ge^{2gz}\dot{t}^2[/itex], where dots represent differentiation with respect to proper time. For a particle instantaneously at rest, [itex]\dot{t}=1/\sqrt{g_{tt}}=e^{-gz}[/itex], so [itex]\ddot{z}=g[/itex].
 
  • #20
Mentz114 said:
Another (interesting?) thing is that the non-zero components of the Einstein tensor are [itex]G_{yy}[/itex] and [itex]G_{zz}[/itex] 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.

Your #16 is very cool!

With [itex]\Phi=gy[/itex], 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 [itex]e^Pdt^2-e^Q(dx^2+dy^2)-dz^2[/itex]. 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.
 
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  • #21
Altabeh said:
Did you calculate components of the Einstein tensor based on bcrowell's Christoffel symbols?
No.

Altabeh said:
No it is not!
I think in any realistic situation it is. Don't let me disturb your equanimity, I'm no expert and I'm hoping to learn something.

Altabeh said:
But remember that then we are again back to the principle that a flat spacetime only accepts a uniform gravitational field.
I don't think I'm contradicting that. I never actually understood what a 'uniform' field is. Is it [itex]d\Phi/dx=const.[/itex] ?
 
  • #22
bcrowell said:
Your #16 is very cool!

Thanks. I liked your conclusion [itex]\ddot{z}=g[/itex] in #19.

It's too late for me now but I'll have a play with [itex] e^Pdt^2-e^Q(dx^2+dy^2)-dz^2 [/itex] tomorrow. I think this is the 'Newtonian' metric I came across recently.[ edit : no it's nothing like it ]
 
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  • #23
bcrowell said:
Are we getting our wires crossed because I'm talking about proper acceleration and you're talking about coordinate acceleration?.

Yes, my equation shows the coordinate acceleration (which I think is important) and yours belongs to proper one. BUT

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

You again did put a wrong sign for [itex]\Gamma\indices{^z_{tt}}=ge^{2gz}[/itex]! This gives

[itex]\ddot{z}=-ge^{2gz}\dot{t}^2[/itex].

So that for a particle instantaneously at rest

[itex]\dot{t}=1/\sqrt{g_{tt}}=e^{-gz}[/itex], (1)

the proper acceleration is

[itex]\ddot{z}=-g[/itex].

From (1) it is easy to see

[itex]\ddot{t}=-g\dot{z}e^{-gz}=-g\dot{z}\dot{t}[/itex]. (2)

Introducing (2) into the geodesic equation for t,

[itex]\ddot{t}+2\dot{t}\dot{z}=0[/itex]

yields

[itex]g=2[/itex].

This result shows that g is not arbitrarily chosen if your (1) holds. So your conclusion is of no interest because it does not even give Semay's metric up to order 1 in [itex]gz[/itex].

AB

Correction: My velocity formula must be replaced by

[tex]v^2=\frac{1}{2}(a+e^{2y})[/tex],

so that for an initial vanishing velocity [tex]v_0=0[/tex], we have

[tex]a_0=-2e^{2y_0}[/tex].

But this says that the coordinate acceleration is not invariant under the translation [tex]y_0\rightarrow y_0+y[/tex], leading to the verification of the assumption that Rindler's metric does not admit a uniformly accelerated frame for [tex]\Phi = 2y[/tex] globally. (Of course for any other [tex]\Phi = \Phi (y)[/tex] this is not possible. [exercise]!)
 
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  • #24
Mentz114 said:
I think in any realistic situation it is. Don't let me disturb your equanimity, I'm no expert and I'm hoping to learn something.

Not at all! Come and share your ideas with us and if you did, it would be our pleasure to discuss them!

I don't think I'm contradicting that. I never actually understood what a 'uniform' field is. Is it [itex]d\Phi/dx=const.[/itex] ?

A uniform gravitational field is that remains the same at any point of the spacetime admitting it. This occurs only locally in a curved spacetime and consequently for a flat spacetime no such thing exists because no flat spacetime having a gravitational field within it exists in general!

AB
 
  • #25
bcrowell said:
I wonder if one can do anything interesting with a metric of the form [itex]e^Pdt^2-e^Q(dx^2+dy^2)-dz^2[/itex].

Would you mind specifying the interesting things? I mean, do you want us to discuss the same thing about this metric here as we did for Rindler or Semay's metric? My pleasure to think about it and I'll provide all data needed soon!

AB
 
  • #26
Hi, Altabeh,

Altabeh said:
Yes, my equation shows the coordinate acceleration (which I think is important) and yours belongs to proper one.
Good, I'm glad we cleared that up.

Altabeh said:
You again did put a wrong sign for [itex]\Gamma\indices{^z_{tt}}=ge^{2gz}[/itex]!
Ah, thanks very much for the correction!

Altabeh said:
This result shows that g is not arbitrarily chosen if your (1) holds.
You start with two premises:
(a) my equation [itex]\dot{t}=1/\sqrt{g_{tt}}=e^{-gz}[/itex]
(b) your velocity equation
Based on these two premises, you reach a conclusion that neither of us believes can hold, on physical grounds. That implies that either a is false, b is false, or your reasoning based on a and b has a flaw in it. I think a is clearly true. But note that I only claim that a is true for a particle instantaneously at rest. I suspect you're reaching an incorrect conclusion by applying a in conditions where the particle is not instantaneously at rest.

Here's a little more detail on why I think a is clearly true:
[tex]d\tau^2 = e^{2gz}dt^2-dz^2[/tex]
For particle instantaneously at rest, dz=0, so
[tex]d\tau^2 = e^{2gz}dt^2[/tex]
Therefore
[tex]dt/d\tau = e^{-gz}[/tex]
 
  • #27
I'm trying to use the method of frames to work out the acceleration of the hovering observer in space-time [2]. See http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity.

The coframe is

[tex]
\sigma_0 = -(1+ax),\ \sigma_1=1,\ \sigma_2=1,\ \sigma_3=1
[/tex]

and the frame dual to this

[tex]
\mathfb{e}_0=\frac{1}{(1+ax)}\partial_0,\ \mathfb{e}_1=\partial_1,\ \mathfb{e}_2=\partial_2,\ \mathfb{e}_3=\partial_3
[/tex]

The acceleration I want is given by

[tex]
\nabla_{e_0}\left(\mathfb{e}_0 \right)
[/tex]

What does that derivative mean ? Any help appreciated.
 
  • #28
[itex]\nabla[/itex] is a (linear) connection. The best place to read about those is chapter 4 of Lee's "Riemannian manifolds". I'm studing that book myself (and I made it through that chapter), so if you have a specific question, I might be able to answer it.

You left out some dx's in your coframe. It should look something like this:

[tex]\sigma_0 = -(1+ax)dx^0,\ \sigma_1=dx^1,\ \sigma_2=dx^2,\ \sigma_3=dx^3[/tex]

For your immediate needs, it may be sufficient to know that the Christoffel symbol is defined by

[tex]\nabla_{\partial_i}\partial_j=\Gamma_{ij}^k\partial_k[/tex]
 
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  • #29
Hi, Lut -- Sorry, I'm not familiar with the frame technique. Maybe Altabeh could help with this?

Re the possible physical interest of adding the other two space dimensions, and Born's argument, here's something that I hadn't understood correctly until today. If you compute the Einstein tensor in the (t,z) space, using metric [1], it's zero, i.e., it's a vacuum solution. However, if you do the same thing for the metric [itex]d s^2 = e^{2gz}d t^2-d x^2 - d y^2-dz^2[/itex], you get an unphysical result (nonvanishing xx and yy components of the stress-energy tensor, even though the mass-energy density is zero). This does seem to be consistent with Born's argument that something has to happen in the transverse direction.

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 [itex]-d x^2 - d y^2[/itex] 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. Since the Petrov solution, which has cylindrical symmetry, is claimed to be a good GR analog of a uniform field, it seems like it's reasonable to consider applying it only to x. I messed around in maxima a little, and any metric of the form
[tex]d s^2 = e^{2z}d t^2-e^{-2jz}d x^2 - e^{-2kz}d y^2-dz^2 \qquad [1*][/tex]
has constant scalar curvature everywhere, regardless of the choice of j and k. I think this is just a result of the same symmetry of the metric, which is that under [itex]z\rightarrow z+c[/itex], all you get is an unobservable rescaling of the cartesian coordinates.

Now let's say we want to find a vacuum solution of the form [1*]. The xx and yy components of the Einstein tensor are zero if j and k equal [itex](1 \pm \sqrt{3} i)/2[/itex]. If I choose the sign to be opposite for j and k, I get a vacuum solution, but it's a vacuum solution with a complex oscillatory metric, which has no physical significance. However, if you now compare with the Petrov metric,
[tex] ds^2 = -dr^2 - e^{-2r}dz^2+e^r[2\sin\sqrt{3}r d\phi dt-\cos\sqrt{3}r(d\phi^2-dt^2)] [/itex] ,
there are some obvious similarities. You get transverse components that have the same behavior: they decay exponentially with height and also oscillate. The decay constant and period of oscillation are identical in both cases. The big differences are (1) cylindrical symmetry and (2) not being static.
 
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  • #30
Fredrik, thanks for the reply. The coframe is 1-forms, so I guess that makes sense. I mistranscribed them. The lysdexia playing up.

Ben, I look forward to studying your post later.

I'm out until tomorrow when I'll have time to look at both posts again.
 
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  • #31
Here are a few more thoughts about the same topic as my #29. The complex exponential solution can't be converted to a real one by the usual trick of forming linear combinations, since the field equations are nonlinear. One can simply try forms in which the complex exponentials are real decaying exponentials like [itex]{e^\ldots} \cos\ldots[/itex], but then your components are going to go to zero in certain places, leading to an uninvertible metric. The Petrov metric also has [itex]d\phi^2[/itex] and [itex]dt^2[/itex] terms that vanish at certain values of r, but because of the [itex]d\phi dt[/itex] term, the metric is still invertible. At the values of r where this happens, you can easily construct CTCs of the form [itex](t,\phi=k t,r=const,z=const)[/itex].

It's intriguing that we come full circle historically. Born gives his argument in 1920 that there should be transverse length contractions in a gravitational field, based on the rotating carousel argument. This seems like kind of a leap, since the carousel has non-static properties that aren't generic to all gravitational fields. But then when we try to construct the GR equivalent of a uniform, static gravitational field, we end up being led back to a spacetime that is non-static and rotating!

[EDIT] After some digging around on the web, I found the following paper: McIntosh, 'Real Kasner and related complex “windmill” vacuum spacetime metrics,' GRG 24 (1992) 757. On p. 759, they do the same calculation I did in #29. I would like to be able to gain deeper insight into the physical meaning of all this, but I'm not technically sophisticated enough to understand all the content of the McIntosh paper.

[EDIT] Oops, my definition of the CTC wasn't quite right. I should have kept t constant, and it's really a closed *lightlike* curve.
 
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  • #32
Mentz114 said:
I'm trying to use the method of frames to work out the acceleration of the hovering observer in space-time [2]. See http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity.

The coframe is

[tex]
\sigma_0 = -(1+ax),\ \sigma_1=1,\ \sigma_2=1,\ \sigma_3=1
[/tex]

and the frame dual to this

[tex]
\mathfb{e}_0=\frac{1}{(1+ax)}\partial_0,\ \mathbf{e}_1=\partial_1,\ \mathfb{e}_2=\partial_2,\ \mathbf{e}_3=\partial_3
[/tex]

The acceleration I want is given by

[tex]
\nabla_{e_0}\left(\mathfb{e}_0 \right)
[/tex]

What does that derivative mean ? Any help appreciated.

Do you mean mathematically or physically? Here is some of the calculation.

[tex]
\begin{equation*}
\begin{split}
\nabla_{\mathbf{e}_0} \mathbf{e}_0 &= \frac{1}{1+ax} \nabla_{\partial_0} \left( \frac{1}{1+ax} \partial_0 \right) \\
&= \frac{1}{1+ax} \left[ \left( \nabla_{\partial_0} \left( \frac{1}{1+ax} \right) \right) \partial_0 + \frac{1}{1+ax} \nabla_{\partial_0} \left( \partial_0 \right) \right] \\
&= \frac{1}{1+ax} \left[ \left( \partial_0 \left( \frac{1}{1+ax} \right) \right) \partial_0 + \frac{1}{1+ax} \Gamma^\mu {}_{00} \partial_{\mu }\right]
\end{split}
\end{equation*}
[/tex]

In the last line, the components of the connection are with respect the coordinate basis, not with respect to the frame.
 
  • #33
bcrowell said:
Hi, Altabeh,

You start with two premises:

(a) my equation [itex]\dot{t}=1/\sqrt{g_{tt}}=e^{-gz}[/itex]
(b) your velocity equation

Based on these two premises, you reach a conclusion that neither of us believes can hold, on physical grounds.

Based on these two, I reach a conclusion that

1- g=2 in Rindler's metric,

2- The spacetime is NOT invariant under spatial translations which is reasonable, because this metric is curved,

3- Your equation shows that for a particle moving along a geodesic in the Rindler's spacetime with [tex]\Phi = 2y [/tex] or [tex]g=2[/tex], the proper acceleration equals [tex] -2 [/tex] everywhere; but mine demonstrates that the coordinate acceleration is clearly dependent of [tex]y[/tex] and, as expected, is a function of the velocity [tex]v[/tex] (I took [tex]g=1[/tex]) in the form of equation

[tex]2v^2-e^{2y}=a(y,v).[/tex]

(I don't have that much time to prove it now, but later I'll surely do.)

4- Nothing unphysical befalls in either case and my velocity formula follows your equation plus making use of the second component of the geodesic equation, i.e. [tex]\ddot{t}+2\dot{t}\dot{y}=0,[/tex] (*) so that you can observe the correctness of my equation by taking a differential of it and claming an instantaneously at rest particle, thus leading to [tex]dv=da=dy=0[/tex] through

[tex]4vdv-2dye^{2y}=da.[/tex]

Note here all parameters are coordinate-dependent and consequently having [tex]v=dy/dt=0[/tex] gives [tex]da=0.[/tex]

Here's a little more detail on why I think a is clearly true:
[tex]d\tau^2 = e^{2gz}dt^2-dz^2[/tex]
For particle instantaneously at rest, dz=0, so
[tex]d\tau^2 = e^{2gz}dt^2[/tex]
Therefore
[tex]dt/d\tau = e^{-gz}[/tex]

Your problem is that you ignore the equation (*) so you think everything is done as long as [tex]dt/d\tau = e^{-gz}[/tex] appears. Your g is 2 and this makes re-scaling it impossible so as to get Semay's metric.
 
  • #34
If we are all clear about the (t,y) or (t,z) case, I'm going to go into details of your #29 post tomorow which sounds interesting, yet sophisticated, to me!

Btw, happy Valentine's day to all lovers of physics specially GR.

AB
 
  • #35
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.
 

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