Inconsistent forms of the metric in a uniform field

[bcrowell]

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.)

 

[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?

 

[bcrowell]

Thanks, atyy, for your helpful reply!

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.

[...] 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 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.

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 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.

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

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.

 

[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.)

 

[bcrowell]

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.

 

[torquil]

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

 

[Altabeh]

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 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 vacuum 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

 

[atyy]

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/ ). 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)?

 

[Altabeh]

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 \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.

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

 

[atyy]

What is EP?

Equivalence Principle (which is not a principle)

 

[bcrowell]

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...?

 

[bcrowell]

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

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."

 

[Altabeh]

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

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

 

[bcrowell]

The geodesic equation in this spacetime gives
v^2=\frac{\dot{v}}{2\dot{t}^2}+e^{2y},

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.

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.

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.

 

[Altabeh]

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.

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!

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

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

where a is the acceleration.

 

[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)

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

which is

<br /> P=log\left( y+k_1\right) +k_2<br />

If we set k_2=0[/itex] then exp(2\Delta\Phi) is (1+ay)^2 up to a scale factor.<br /> <br /> My ( possibly wrong ) interpretation is that 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 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 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.

 

[Altabeh]

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)

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

which is

<br /> P=log\left( y+k_1\right) +k_2<br />

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

<br /> <br /> No it is not! If you set k_2=0[/itex], then the Taylor expansion of 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&amp;#039;s metric [1] goes into another coordinate system which I talked about it earlier.&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; 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. &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;[/QUOTE]&lt;br /&gt; &lt;br /&gt; Did you calculate components of the Einstein tensor based on bcrowell&amp;#039;s Christoffel symbols?&lt;br /&gt; &lt;br /&gt; AB&lt;br /&gt; &lt;br /&gt; Edit: The Taylor expansion of &lt;br /&gt; &amp;lt;br /&amp;gt; e^{2P}=e^{2\log\left( y+k_1\right)}&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; must have been considered, so this only coincides Symay&amp;#039;s metric, if k_1=g=1. But remember that then we are again back to the &lt;i&gt;principle&lt;/i&gt; that a flat spacetime only accepts a uniform gravitational field.

 

[bcrowell]

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.

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

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 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 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.

 

[bcrowell]

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.

 

[bcrowell]

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.

Your #16 is very cool!

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 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.

 

[Mentz114]

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

No.

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.

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 d\Phi/dx=const. ?

 

[Mentz114]

Your #16 is very cool!

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

It's too late for me now but I'll have a play with e^Pdt^2-e^Q(dx^2+dy^2)-dz^2 tomorrow. I think this is the 'Newtonian' metric I came across recently.[ edit : no it's nothing like it ]

 

[Altabeh]

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

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.

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

\ddot{z}=-ge^{2gz}\dot{t}^2.

So that for a particle instantaneously at rest

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

the proper acceleration is

\ddot{z}=-g.

From (1) it is easy to see

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

Introducing (2) into the geodesic equation for t,

\ddot{t}+2\dot{t}\dot{z}=0

yields

g=2.

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 gz.

AB

Correction: My velocity formula must be replaced by

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

so that for an initial vanishing velocity v_0=0, we have

a_0=-2e^{2y_0}.

But this says that the coordinate acceleration is not invariant under the translation y_0\rightarrow y_0+y, leading to the verification of the assumption that Rindler's metric does not admit a uniformly accelerated frame for \Phi = 2y globally. (Of course for any other \Phi = \Phi (y) this is not possible. [exercise]!)

 

[Altabeh]

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 d\Phi/dx=const. ?

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

 

[Altabeh]

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.

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

 

[bcrowell]

Hi, Altabeh,

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.

You again did put a wrong sign for \Gamma\indices{^z_{tt}}=ge^{2gz}!

Ah, thanks very much for the correction!

This result shows that g is not arbitrarily chosen if your (1) holds.

You start with two premises:
(a) my equation \dot{t}=1/\sqrt{g_{tt}}=e^{-gz}
(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:
d\tau^2 = e^{2gz}dt^2-dz^2
For particle instantaneously at rest, dz=0, so
d\tau^2 = e^{2gz}dt^2
Therefore
dt/d\tau = e^{-gz}

 

[Mentz114]

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

<br /> \sigma_0 = -(1+ax),\ \sigma_1=1,\ \sigma_2=1,\ \sigma_3=1<br />

and the frame dual to this

<br /> \mathfb{e}_0=\frac{1}{(1+ax)}\partial_0,\ \mathfb{e}_1=\partial_1,\ \mathfb{e}_2=\partial_2,\ \mathfb{e}_3=\partial_3<br />

The acceleration I want is given by

<br /> \nabla_{e_0}\left(\mathfb{e}_0 \right)<br />

What does that derivative mean ? Any help appreciated.

 

[Fredrik]

\nabla 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:

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

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

\nabla_{\partial_i}\partial_j=\Gamma_{ij}^k\partial_k

 

[bcrowell]

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 d s^2 = e^{2gz}d t^2-d x^2 - d y^2-dz^2, 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 -d x^2 - d y^2 gives unphysical results.

In the 2+1 carousel setup, rulers oriented in the transverse direction are shorter when they're lower in the gravitational field (closer to the rim). This means that the xx part of the metric should decrease with z. Generalizing to 3+1, it's not obvious to me whether the contraction should apply to both x and y or only to x. 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
d s^2 = e^{2z}d t^2-e^{-2jz}d x^2 - e^{-2kz}d y^2-dz^2 \qquad [1*]
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 z\rightarrow z+c, 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 (1 \pm \sqrt{3} i)/2. 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,
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] ,<br /> 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.

 

[Mentz114]

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.

 

[bcrowell]

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 {e^\ldots} \cos\ldots, but then your components are going to go to zero in certain places, leading to an uninvertible metric. The Petrov metric also has d\phi^2 and dt^2 terms that vanish at certain values of r, but because of the d\phi dt term, the metric is still invertible. At the values of r where this happens, you can easily construct CTCs of the form (t,\phi=k t,r=const,z=const).

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.

 

[George Jones]

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

<br /> \sigma_0 = -(1+ax),\ \sigma_1=1,\ \sigma_2=1,\ \sigma_3=1<br />

and the frame dual to this

<br /> \mathfb{e}_0=\frac{1}{(1+ax)}\partial_0,\ \mathbf{e}_1=\partial_1,\ \mathfb{e}_2=\partial_2,\ \mathbf{e}_3=\partial_3<br />

The acceleration I want is given by

<br /> \nabla_{e_0}\left(\mathfb{e}_0 \right)<br />

What does that derivative mean ? Any help appreciated.

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

<br /> \begin{equation*}<br /> \begin{split}<br /> \nabla_{\mathbf{e}_0} \mathbf{e}_0 &amp;= \frac{1}{1+ax} \nabla_{\partial_0} \left( \frac{1}{1+ax} \partial_0 \right) \\<br /> &amp;= \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] \\<br /> &amp;= \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]<br /> \end{split}<br /> \end{equation*}<br />

In the last line, the components of the connection are with respect the coordinate basis, not with respect to the frame.

 

[Altabeh]

Hi, Altabeh,

You start with two premises:

(a) my equation \dot{t}=1/\sqrt{g_{tt}}=e^{-gz}
(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 \Phi = 2y or g=2, the proper acceleration equals -2 everywhere; but mine demonstrates that the coordinate acceleration is clearly dependent of y and, as expected, is a function of the velocity v (I took g=1) in the form of equation

2v^2-e^{2y}=a(y,v).

(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. \ddot{t}+2\dot{t}\dot{y}=0, (*) 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 dv=da=dy=0 through

4vdv-2dye^{2y}=da.

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

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

Your problem is that you ignore the equation (*) so you think everything is done as long as dt/d\tau = e^{-gz} appears. Your g is 2 and this makes re-scaling it impossible so as to get Semay's metric.

 

[Altabeh]

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

 

[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.

 

[Altabeh]

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

Here I take another approach to prove that if

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

then in the metric

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

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

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

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

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

If \Phi =gy, then obviously (\Phi)&#039; =g. (Henceforth we denote the derivative wrt y by a prime and the derivative wrt proper time \tau by a dot.)

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

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

Re-write \ddot{y} as

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

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

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

which can be written as

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

AB

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

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

 

[bcrowell]

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

 

[Mentz114]

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

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

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

<br /> \frac{d^2x}{ds^2}=-a(1+ax)\left(\frac{dt}{ds}\right)^2<br />
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
<br /> \frac{d^2x}{d\tau^2}=\frac{-a(1+ax)}{(1+ax)^2}=\frac{-a}{1+ax}<br />
What does this mean ? It is not a constant acceleration as advertised. The problem could be that s\ne \tau. Don't we need a worldline with starting conditions to get the relationship between \tau and s ?

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

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

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

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

 

[bcrowell]

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

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

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

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

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

 

[Mentz114]

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

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

These are rhetorical questions I need to think about.

 

[George Jones]

<br /> \frac{d^2x}{d\tau^2}=\frac{-a(1+ax)}{(1+ax)^2}=\frac{-a}{1+ax}<br />

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

Yes, it is.

 

[Mentz114]

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

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

What is puzzling is that this decreases with x_r though.

 

[Altabeh]

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

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

A good example is this:

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

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

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

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

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

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

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

a=-ge^{2gy_0},

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

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

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

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

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

AB

 

[Altabeh]

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

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

AB

 

[Altabeh]

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

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

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

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

AB

 

[bcrowell]

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

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

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

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

 

[bcrowell]

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

These are rhetorical questions I need to think about.

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

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

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

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

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

 

[Mentz114]

Ben,

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

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

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

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

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

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

 

[Mentz114]

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

AB

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

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

 

[Altabeh]

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


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

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

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

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

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

AB