I What is the Effect of Gravity on Einstein's Train in Special Relativity?

[SlowThinker]

So to find the length along, say, the floor of the rocket in pervect's metric, you would have to find the equation describing the floor of the rocket, and use it and the metric to find the correct integral.

Pervect said something like

a metric is the best and ultimately the only thing needed to describe a coordinate system

So I was wondering how that could work. I could imagine some process of constructing spacelike geodesics from the floor up and then measuring the distance between the resulting points, but it seems like a lot of work, even if I knew how to do it.
Plus, the "up direction" is probably under-defined: in your metric perhaps the $-\tau$ direction is the right one, but I'm not so sure about Pervect's.

First of all, lengths along the floor, or ceiling, or any other constant height in the rocket, are only along the $\psi$ direction in my chart, so you can only integrate at a constant $\chi$ (constant $\tau$ is assumed because you're trying to find length in a surface of constant time) in my chart to find lengths along the floor or any other constant height.

So it's more like "you need a metric and a transformation"?

Second, you can't "find" where the ceiling of the rocket is

Oops, sorry for the confusion. I meant the ceiling of the train's room, not the rocket, but didn't say.
I know it's 3 meters above the train's floor, as seen from the outside. Is a metric enough to find/confirm that the passengers also measure 3 meters?

 

[pervect]

So I have a fairly basic question. I'm stuck at page 11 of Carroll's Lectures on GR, that's why I don't know the answer It's incredibly dense with definitions and there are no explanations.

Once you have the metric, how do you, say, measure the length of the floor and ceiling, in the $y\ /\ \psi$ direction (as if using a ruler, not radar)? I think I could integrate at $\chi=0\text{ or }1/g, \tau=const$ to get the length of the floor, but how do you find the ceiling?

The problem comes in several parts. The first part is this - we have a 4-d space-time, but length is a purely spatial concept. So we have some choices to make here about how we eliminate the time part of the problem.

In making these choices, it helps to keep in mind the "simpler" problem of asking how we measure length on a rotating disk. Recall that torque-free gyroscopes mouned on our block do rotate, so we need to worry about this. There is a lot of literature on this topic, under the name Ehrenfest paradox. This simpler problem is not-so-simple as a perusal of the literature will show, there are a lot of different ideas as to how to go about it.

However, in most cases our block won't be rotating very fast. So if we can ignore the relativistic effects associated with the rotation, we can come up with a much simpler approximate method. This simpler approximate method will ignore any relativistic length contraction effects associated with the rotation. We do have to keep in the back of our mind that we are deliberately ignoring these effects, however.

Using my chart, this simpler method is particularly simple. We take a slice of the block at constant coordinate time $\tau$, and we eliminate the time dimension by projecting our 4-d metric onto a 3-d metric. I'm not going to give a very careful treatment and hope that the simple intuitive approacht will be satisfactory. Intuitively, when we eliminate the time dimension from my line element, we get $ds^2 = d\chi^2 + d\psi^2 + dz^2$. This is just the usual metric of Euclidean space. And we know how to measure distances in Euclidean space = distance^2 = $#Delta \chi)^2 + \Delta \psi^2 + \Delta z^2$. The biggest complicating factor is that the block will be curved in these coordinates. So we do need to specify which "distance" we mean - the "straight-line" distance, or the distance along the curved floor.

The more general mathematical technique says that if you have some particular curve in a 3-d space with a 3-d metric, you integrate the line element ds^2 in the 3d metric to get the length of the curve. So measuring the distance consists of specifying what curve you want to find the length of, then integrating the spatial part of the line element ds along that curve. Usually you want the curve to be a straight-line, but perhaps not always, you might want to measure a distance along the floor for instance.

To anticipate some possible objections from Peter, let me say that by "floor of the rocket", we mean the intersection of the worldsheet of the rocket floor (the collection of all worldlines in the floor) intersected by a surface of constant coordinate time. So we are talking about "the" floor of the rocket at some particular time. When you recall that my coordinate time is different from Peter's coordinate time, you can see that the idea of "the floor" is a bit ambiguous because of this issue.

In Peter's chart, the basic ideas are similar, but the concept of "a straight line" turns into the concept of a "geodesic curve", because the space itself is curved. You can read about geeodesics in Caroll, I'm not sure how much sense it will make to you at this point :(. So rather than having a curved floor in a flat space-time, we have a curve where one of the coordinates of the floor is always the same, but the space itself is curved.

In the more general 4-d case, we first need to eliminate the time dimension by one of several techniques. One technique is to use what's called a quotient manifold, which basically says that we consider every point on the quotient manifold to be a worldline in the 4-d manifold. THis process really makes the most sense when we have a metric that is independent of time. Then we need to define what we mean by the "distance between worldlines". While I think this is probably the best technique, I'm not aware of any textbook that goes into it in great detail :(.

 

[pervect]

Pervect said something like

So I was wondering how that could work. I could imagine some process of constructing spacelike geodesics from the floor up and then measuring the distance between the resulting points, but it seems like a lot of work, even if I knew how to do it.
Plus, the "up direction" is probably under-defined: in your metric perhaps the $-\tau$ direction is the right one, but I'm not so sure about Pervect's.So it's more like "you need a metric and a transformation"?Oops, sorry for the confusion. I meant the ceiling of the train's room, not the rocket, but didn't say.
I know it's 3 meters above the train's floor, as seen from the outside. Is a metric enough to find/confirm that the passengers also measure 3 meters?

I don't have a lot of time, but I'll say something briefly that I hope will be helpful. A metric gives you the Lorentz interval between two nearby points, and when the two points are space-like separated, this Lorentz interval can be interpreted as a distance - a proper distance. So the metric does tell you the distance between nearby points.

To get the distance between two points that are not nearby, you need to define a curve that connects them. Then the metric gives you an integral that defines the length of the connecting curve. But some thought still needs to be given to the question of "which curve do you want the length of?"

 

[pervect]

I've got a question for Peter. I recall seeing some corrections in his post on the shear tensor. I wanted to know if the shear tensor for both our metrics, after all corrections are made, are uniformly zero, as my current thinking leads me to believe should be the case.

Why do I think the shear tensor should be zero? http://arxiv.org/abs/1004.1935 suggests that a Born rigid flow should be shear-free.

In this paper we give a new proof, valid for all dimensions, of the classical Herglotz-Noether theorem that all rotational shear-free and expansion-free flows (rotational Born-rigid flows) in Minkowski spacetime are generated by Killing vector fields (isometric flows).

Both of our line elements should be generated by Killing flows, because in both case all of the metric coefficient a function of time. Therefore I'd expect them both to be expansion-free and shear-free. If they are not, there's some problem with my understanding.

On a related note, I believe both of our metrics should represent the same underlying congruence. This congruence is just the integral curve of the time-like Killing vector field. I don't have any formal, rigorous proof that it's unique, but I feel that it should be. My series expansion of Peter's coordinates, which I referred to as $\tau^\prime, \chi^\prime, \psi^\prime$, seems consistent with the idea of them representing the same congruence. Basically the point is that the transformation from $\chi$ to $\chi^\prime$ and from $\psi$ to $\psi^\prime$ is not a function of time, i.e. $\tau$. While $\tau^\prime$ is a function of all of ($\tau, \chi, \psi)$, this only means that our parameteriztion of the congruence is different - the underlying congruence is the same, only the labels (i.e. the coordinates) have changed. Thus I believe our two approaches represent the same abstract object with different labels. The series expansion, being only to a finite order (and a low one, at that) doesn't prove this, but it's consistent with the idea.

What remains to be done is to define an operational notion of "space" without any approximations via the quotient manifold method. There is not and cannot be any closed spatial surface that's perpendicular everywhere to our (presumably shared) time-like congruence of worldlines, due to the fact that the congruences have non-zero vorticity. So there is no "hypersurface-orthagonal" spatial surface. So splitting space-time into space+time via the idea of space-like surfaces of simultaneity is going to be impossible (at least if we demand that our space-like surfaces be closed surfaces). Instead, we are led to the approach that identifies all points on the same worldline as "having the same spatial coordinates". This turns the 4-d space-time metric into a 3-d, time-independent, spatial metric.

Google books finds some exposition of the quotient manfiold idea in Rizzi and Ruggiero's book "Relativity in Rotating Frames", a google search for "Rizzi and Ruggiero quotient manifold" should find it. Wikki has a reference to a 2002 paper by the same authors, but unfortunately I haven't been able to track it down. I presume the book has the same material, though. It's quite specialized and pricey however - as well as being rather advanced. R&R do make the rather interesting claim, on pg 200, that this notion of space, which they refer to as "relative space", is the only sort of space which "having an actual physical meaning form an operational point of view".

On a closing note, if this seems to be "way too much work", I can certainly sympathize. I'm more-or-less happy with ignoring the relativistic effects related to rotation and the subsequent Lorentz contraction due to the rotation, even though it's an approximation. The approximate solution I prefer is given by the line element that I gave earlier.

 

[PeterDonis]

Both of our line elements should be generated by Killing flows

They are. They're generated by the same Killing flow. We are just using different parameterizations of it.

Therefore I'd expect them both to be expansion-free and shear-free.

The congruence (it's the same for both, see below) does indeed have zero expansion and zero shear. It has non-zero vorticity.

I believe both of our metrics should represent the same underlying congruence.

They do. It's been a number of posts now, but I showed this by deriving the same expression in Minkowski coordinates for both 4-velocity fields. From that expression, I computed the kinematic decomposition and verified that the expansion and shear are zero and the vorticity is nonzero. (I had previously thought the shear was nonzero, but that was an error in computation on my part.)

What remains to be done is to define an operational notion of "space" without any approximations via the quotient manifold method.

I agree that this would be a useful thing to do. However, by analogy with the case of the Langevin congruence (the "rotating disk" congruence), I would expect the "space" defined by this method to be non-Euclidean. (Which is not to say it's the same as the "space" in my metric; I don't know whether it is or not.)

 

[DrGreg]

I've been silently following this thread since it started, but only skimming through it without looking too closely at the maths. But recently I've been attempting to derive my own analysis. I kept going wrong; I couldn't rely on my geometrical intuition to find a solution, because my intuition is trained on Euclidean geometry, not Minkowski geometry. I had a breakthrough once I constructed an analogous problem in Euclidean geometry.

First, some terminology and conventions. I shall use (t, x, y) to denote Rindler coordinates. I shall use column vectors to denote components in Minkowski space (the initial comoving inertial frame at time zero). c=1, and the metric signature is +--. (The fourth z coordinate is suppressed throughout).

Next, my Euclidean analogy. Consider a helical ramp: like a spiral staircase, but with the discrete steps replaced by a continuous ramp. The ramp can be described by the equations<br /> \begin{align*}<br /> X &amp;= r \cos \omega \theta \\<br /> Y &amp;= r \sin \omega \theta \\<br /> Z &amp;= z + k \theta<br /> \end{align*}<br />The analogy with the relativity problem is

• radius r is equivalent to Rindler x coordinate (in the direction of the acceleration, "vertical")
• angle \theta is equivalent to Rindler t coordinate (time)
• vertical height above the ramp surface z is equivalent to the sliding block's horizontal coordinate that I will denote \eta. Note this direction, in the Euclidean helix analogy, is in the vertical direction, not perpendicular to the ramp surface.

I worked through this ramp example to find a solution that I could translate into the relativistic sliding block example. The translation involved replacing circular trig functions with hyperbolic functions and changing a few signs.

With that prologue out of the way, here's my relativistic solution. A point in the block is given by<br /> \mathbf{X} = \begin{bmatrix}<br /> x \sinh gt \\<br /> x \cosh gt \\<br /> \eta + vt<br /> \end{bmatrix}<br />So
<br /> <br /> \mathrm{d}\mathbf{X}<br /> =<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix} \mathrm{d}t<br /> +<br /> \begin{bmatrix}<br /> \sinh gt \\<br /> \cosh gt \\<br /> 0<br /> \end{bmatrix} \mathrm{d}x<br /> +<br /> \begin{bmatrix}<br /> 0 \\<br /> 0 \\<br /> 1<br /> \end{bmatrix} \mathrm{d} \eta<br /><br /> \mathrm{d}\mathbf{X}<br /> =<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix} \mathrm{d}t<br /> +<br /> \begin{bmatrix}<br /> \sinh gt \\<br /> \cosh gt \\<br /> 0<br /> \end{bmatrix} \mathrm{d}x<br /> +<br /> \left(<br /> \frac{gx}{g^2x^2-v^2}<br /> \begin{bmatrix}<br /> v \cosh gt \\<br /> v \sinh gt \\<br /> gx<br /> \end{bmatrix}<br /> -<br /> \frac{v}{g^2x^2-v^2}<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix}<br /> \right)<br /> \mathrm{d} \eta<br /><br /> \mathrm{d}\mathbf{X}<br /> =<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix}<br /> \left( \mathrm{d}t - \frac{v \, \mathrm{d} \eta}{g^2x^2-v^2} \right)<br /> +<br /> \begin{bmatrix}<br /> \sinh gt \\<br /> \cosh gt \\<br /> 0<br /> \end{bmatrix} \mathrm{d}x<br /> +<br /> \begin{bmatrix}<br /> v \cosh gt \\<br /> v \sinh gt \\<br /> gx<br /> \end{bmatrix}<br /> \left( \frac{gx \, \mathrm{d} \eta}{g^2x^2-v^2} \right)<br />Edit: there is an error in what follows. Please ignore and read a post corrected lower down which continues from this point.In the above equation note that the three vectors are mutually orthogonal, the first is timelike and the other two are spacelike, therefore we can read off<br /> ds^2 = \left( g^2x^2-v^2 \right)<br /> \left( \mathrm{d}t - \frac{v \, \mathrm{d} \eta}{g^2x^2-v^2} \right)^2<br /> - \mathrm{d}x^2<br /> - \left( \frac{gx}{g^2x^2-v^2} \right)^2 \, \mathrm{d} \eta^2<br />Not only that, but the first vector is parallel to the congruence of "block" worldlines (obtained by putting \mathrm{d}x = \mathrm{d} \eta = 0), and the other two vectors are orthogonal to the congruence and the two vectors' magnitudes are independent of t. Therefore we can say that the metric for the quotient manifold that represents stationary "space" according to a "block" observer is <br /> ds^2 = \mathrm{d}x^2 + \left( \frac{gx}{g^2x^2-v^2} \right)^2 \, \mathrm{d} \eta^2<br />
If you want to take this a step further and find some isotropic coordinates, by my calculation (details omitted here), you can perform the substitution<br /> \begin{align*}<br /> h &amp;= \exp \left( \tfrac{1}{2} g^2x^2 \right) \, x^{-v^2} \, \cos g \eta \\<br /> k &amp;= \exp \left( \tfrac{1}{2} g^2x^2 \right) \, x^{-v^2} \, \sin g \eta \\<br /> \end{align*}<br />and get, I think,<br /> ds^2 = \left( \frac{gx \, \exp \left( -\tfrac{1}{2} g^2x^2 \right) \, x^{v^2}}{g^2x^2-v^2} \right)^2<br /> \left( \mathrm{d}h^2 + \mathrm{d}k^2 \right)<br />
It's possible I've made a mistake in my calculation, but I think the method is correct.

Finally, as a sanity check, along a worldline of the congruence, putting \mathrm{d}x = \mathrm{d} \eta = 0 we get<br /> \frac{\mathrm{d}\mathbf{X}}{\mathrm{d}t}<br /> =<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix}<br />so the 4-velocity is the unit vector<br /> \frac{\mathrm{d}\mathbf{X}}{\mathrm{d}\tau}<br /> =<br /> \frac{1}{\sqrt{g^2x^2-v^2}}<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix}<br />in which the Y-component is time-independent, as it ought to be (for conservation of the Y-component of 4-momentum).

 

[SlowThinker]

<br /> ds^2 = \left( g^2x^2-v^2 \right)<br /> \left( \mathrm{d}t - \frac{v \, \mathrm{d} \eta}{g^2x^2-v^2} \right)^2<br /> - \mathrm{d}x^2<br /> - \left( \frac{gx}{g^2x^2-v^2} \right)^2 \, \mathrm{d} \eta^2<br />

It would make more sense if there was an additional $\left( g^2x^2-v^2 \right)$ factor at the last term. In fact it seems it is used in the subsequent calculation?

Other than that, it's pretty amazing, I think I almost get it

 

[DrGreg]

It would make more sense if there was an additional $\left( g^2x^2-v^2 \right)$ factor at the last term. In fact it seems it is used in the subsequent calculation?

Other than that, it's pretty amazing, I think I almost get it

Oops, I think you are right, which invalidates much of what follows. Stand by for a correction that I will issue shortly...

 

[DrGreg]

There was an error in my previous long post, the final paragraphs should read as follows...

Edit: this is still wrong; see post #164

In the above equation note that the three vectors are mutually orthogonal, the first is timelike and the other two are spacelike, therefore we can read off<br /> ds^2 = \left( g^2x^2-v^2 \right)<br /> \left( \mathrm{d}t - \frac{v \, \mathrm{d} \eta}{g^2x^2-v^2} \right)^2<br /> - \mathrm{d}x^2<br /> - g^2x^2 \, \mathrm{d} \eta^2<br />Not only that, but the first vector is parallel to the congruence of "block" worldlines (obtained by putting \mathrm{d}x = \mathrm{d} \eta = 0), and the other two vectors are orthogonal to the congruence and the two vectors' magnitudes are independent of t. Therefore we can say that the metric for the quotient manifold that represents stationary "space" according to a "block" observer is <br /> ds^2 = \mathrm{d}x^2 + g^2x^2 \, \mathrm{d} \eta^2<br />
If you want to take this a step further and find some isotropic coordinates, by my calculation (details omitted here), you can perform the substitution<br /> \begin{align*}<br /> h &amp;= x \, \cos g \eta \\<br /> k &amp;= x \, \sin g \eta \\<br /> \end{align*}<br />and get<br /> ds^2 = \mathrm{d}h^2 + \mathrm{d}k^2<br />So the quotient space is just flat Euclidean space in which (x, g\eta) are polar coordinates.

Finally, as a sanity check, along a worldline of the congruence, putting \mathrm{d}x = \mathrm{d} \eta = 0 we get<br /> \frac{\mathrm{d}\mathbf{X}}{\mathrm{d}t}<br /> =<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix}<br />so the 4-velocity is the unit vector<br /> \frac{\mathrm{d}\mathbf{X}}{\mathrm{d}\tau}<br /> =<br /> \frac{1}{\sqrt{g^2x^2-v^2}}<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix}<br />in which the Y-component is time-independent, as it ought to be (for conservation of the Y-component of 4-momentum).Thanks to @SlowThinker for rapidly finding the error in my orginal version. This version is much simpler, more credible, and makes more sense. It's also a little surprising as I wasn't expecting the quotient space to turn out flat. (Unless I've made another mistake...)

 

[pervect]

Pervect said something like

So I was wondering how that could work. I could imagine some process of constructing spacelike geodesics from the floor up and then measuring the distance between the resulting points, but it seems like a lot of work, even if I knew how to do it.
Plus, the "up direction" is probably under-defined: in your metric perhaps the $-\tau$ direction is the right one, but I'm not so sure about Pervect's.So it's more like "you need a metric and a transformation"?

I wanted the transformation to "avoid a lot of work". In this particular case, one could, in theory at least, envision using the metric to determine the readings of a virtual "inertial guidance system", mathematical computations which would give you both the linear and rotational accelerations around 3 spatial axes, a total of 6 acceleration components. . This information would be encoded in the Christoffel symbols. If the virtual "inertial guidance system" gave the same readings, the path followed by the observer at the coordinate origin should be the same too. However, while I know what Christoffel symbols correspond to linear and rotational acceleration in my metric (it's derived in MTW), it's not terribly clear to me how to do the same derivation in Peter's metric, it doesn't follow the conventions used in my textbook so I can't apply the textbook results directly. Actually, I could probably take a pretty good guess, but since it would still be a guess, it wouldn't serve the purposes of convincing anyone of anything. So it was much easier and more productive to ask for the already known generating transformation.

The other point you make is a good one. Suppose you have a flat piece of paper. You know the spatial metric is dx^2 + dy^2. But that doesn't change if you rotate the paper, so you don't know whether the x-axis is pointing "up" or "to the right" just given the metric. But you do know that the distance between any two points on the paper is $\sqrt{\Delta x^2 + \Delta y^2}$, regardless of which way the x-axis points.

 

[PeterDonis]

A point in the block is given by
$$
\mathbf{X} = \begin{bmatrix} x \sinh gt \\ x \cosh gt \\ \eta + vt \end{bmatrix}
$$

Comparing this with my post #68, which has $t = \gamma \left( \tau + v \psi \right)$, $x = \chi$, and $y = \gamma \left( \psi + v \tau \right)$, it looks like your $\eta$ is just $\psi / \gamma$, i.e., you've just labeled the points in the block along the $y$ direction with a parameter that is scaled by a factor of $\gamma$ compared to mine. However, I think this may only be true on the bottom of the block, where $gx = 1$. See below.

$$
\mathrm{d}\mathbf{X} = \begin{bmatrix} gx \cosh gt \\ gx \sinh gt \\ v \end{bmatrix} \left( \mathrm{d}t - \frac{v \, \mathrm{d} \eta}{g^2x^2-v^2} \right) + \begin{bmatrix} \sinh gt \\ \cosh gt \\ 0 \end{bmatrix} \mathrm{d}x + \begin{bmatrix} v \cosh gt \\ v \sinh gt \\ gx \end{bmatrix} \left( \frac{gx \, \mathrm{d} \eta}{g^2x^2-v^2} \right)
$$

When these three basis vectors are normalized to be unit vectors, the first and second appear to be the same as $\hat{e}_0$ and $\hat{e}_1$ from my post #68. (My $\hat{e}_0$ is also the 4-velocity that pervect found.) The third vector above, when normalized, is not, however, the same as my $\hat{e}_2$, which in your notation would be

$$
\hat{e}_2 = \frac{1}{\sqrt{g^2 x^2 - v^2}} \begin{bmatrix} gxv \cosh gt \\ gxv \sinh gt \\ 1 \end{bmatrix}
$$

On the bottom of the block, where $gx = 1$, the two basis vectors match; but for other values of $x$ (my $\chi$), they don't. Your basis vector is clearly "nicer" in that, as you say, it remains orthogonal to the other two everywhere, not just where $gx = 1$. However, that raises a question: why didn't I get the same vector for $\hat{e}_2$ in post #68 as you got above? My vector $\mathbf{X}$ there was

$$
\mathbf{X} = \begin{bmatrix} \chi \sinh g \gamma \left( \tau + v \psi \right) \\ \chi \cosh g \gamma \left( \tau + v \psi \right) \\ \gamma v \tau + \gamma \psi \end{bmatrix}
$$

which, after making the substitutions I gave above, gives exactly the same $\mathbf{X}$ as you got, not just on the bottom of the block where $gx = 1$, but everywhere. But taking $\partial_{\psi}$ of the above and normalizing gives my $\hat{e}_2$, not your third basis vector.

I think the answer is that, when taking derivatives to obtain your basis vectors, you assumed that neither of the first two components of your $\mathbf{X}$ were functions of $\eta$, whereas it's obvious that all three components of my $\mathbf{X}$ are functions of $\psi$. I'm not sure how to resolve the apparent discrepancy implied by all of this.

 

[PeterDonis]

therefore we can read off
$$
ds^2 = \left( g^2x^2-v^2 \right) \left( \mathrm{d}t - \frac{v \, \mathrm{d} \eta}{g^2x^2-v^2} \right)^2 - \mathrm{d}x^2 - g^2x^2 \, \mathrm{d} \eta^2
$$

If we expand this out, we get

$$
ds^2 = \left( g^2x^2-v^2 \right) \mathrm{d}t^2 - 2 v \mathrm{d}t \mathrm{d}\eta - \left( g^2 x^2 - \frac{v^2}{g^2x^2 - v^2} \right) \mathrm{d} \eta^2 - \mathrm{d}x^2
$$

If we use the definitions for my $\tau$, $\chi$, and $\psi$ to convert this into an expression in my coordinate chart, I think it comes out the same as what I gave in post #79.

 

[PeterDonis]

the quotient space is just flat Euclidean space in which $(x, g\eta)$ are polar coordinates.

I see the formal analogy here, but I'm not sure if I agree with the physical interpretation. The coordinate $\eta$ is not periodic, as $\theta$ is in polar coordinates; we don't go back to the same point on the block if we go from $\eta = 0$ to $\eta = 2 \pi$ or some other positive constant. But that periodic property is what makes the polar coordinates $(r, \theta)$ a valid alternate representation of a Euclidean plane.

 

[DrGreg]

Oops again. Not my lucky day.

In my haste to correct my error in post #156, I overcorrected in post #159 and still got it wrong. I now think the correct answer for the quotient space metric should be <br /> \mathrm{d}s^2 = <br /> \mathrm{d}x^2<br /> + \frac{g^2x^2}{g^2x^2-v^2} \, \mathrm{d} \eta^2<br />
That can be made isometric but it's a really ugly and complicated expression, so I won't bother here.

 

[PeterDonis]

I now think the correct answer for the quotient space metric should be
$$
\mathrm{d}s^2 = \mathrm{d}x^2 + \frac{g^2x^2}{g^2x^2-v^2} \, \mathrm{d} \eta^2
$$

Expanding the entire metric out would then change what I posted in post #163 to

$$
\mathrm{d}s^2 = \left( g^2 x^2 - v^2 \right) \mathrm{d}t^2 - 2 v \mathrm{d}t \mathrm{d}\eta - \mathrm{d}\eta^2 - \mathrm{d}x^2
$$

I'll have to go back and recheck my computation of how that transforms into my $\tau$, $\chi$, $\psi$ coordinates.

 

[pervect]

The "simpler" approach is confusing me greatly. Do we have a final, corrected, proposed line element for the 4-d metric in "observer" coordinates to check? Even better would be both the 4-d line-element and the transform that give the value of Minkowskii coordinates as a function of "observer" coordinates. Additional useful information would be [change] what values of the observer coordinates make the line element diagonal and non-singular, assuming there are some. This would be what I'd call the "origin" of the coordinate system, while I put it at (0,0), that seems to be the exception rather than the rule in this thread.

 

[PeterDonis]

Do we have a final, corrected, proposed line element for the 4-d metric in "observer" coordinates to check?

What we have, as far as I can tell, are three proposed coordinate charts and associated line elements: (1) mine, (2) yours, and (3) DrGreg's. Here are brief descriptions of how it looks to me like those three charts and line elements were derived, and their key properties:

(1) My approach was to take the transform from Minkowski $T, X, Y$ to Rindler $t, x, y$ coordinates, $T = x \sinh gt$, $X = x \cosh gt$, and "boost" it in the $y$ direction with velocity $v$. By "boost" here I mean defining new coordinates $\bar{\tau}, \bar{\psi}$ using the ansatz $t = \gamma \left( \bar{\tau} + v \bar{\psi} \right)$ and $y = \gamma \left( v \bar{\tau} + \bar{\psi} \right)$. I defined $\bar{\chi} = x$ as a third coordinate just for clarity. I then derived a line element and a set of normalized coordinate basis vectors $\hat{e}_0$, $\hat{e}_1$, $\hat{e}_2$; those three vectors are each in "coordinate" directions $\partial_{\bar{\tau}}$, $\partial_{\bar{\chi}}$, $\partial_{\bar{\psi}}$, but $\hat{e}_0$ and $\hat{e}_2$ are not, in general, orthogonal (though $\hat{e}_1$ is orthogonal to both).

The key property that this chart shares with Rindler coordinates is that the $\bar{\chi}$ direction is the direction of proper acceleration, and surfaces of constant $\chi$ are surfaces of constant "altitude" in the rocket/block. However, the spatial part of this metric is not Euclidean.

(2) Your approach was to derive Fermi normal coordinates centered on the worldline of the bottom center of the block (i.e., the center of the bottom surface of the block, the surface that is in contact with the floor of the rocket), with just one change, that you allowed the spatial basis vectors to rotate instead of being Fermi-Walker transported. You obtained coordinates $\tau, \chi, \psi$ with the key property that the spatial part of the metric in this chart is Euclidean, as it must be for a Fermi normal chart (since all of the effects of acceleration and rotation are put into the terms involving $d\tau$).

The 4-velocity of observers at rest in this chart is the same as that of observers at rest in mine, so both charts describe the same congruence of worldlines "at rest" in the chart, but with different parameterizations. The direction of proper acceleration in this chart is a mixture of the $\chi$ and $\psi$ directions, and surfaces of constant "altitude" in this chart are explicitly curved.

(3) DrGreg's approach was to start with Rindler $t, x, y$ coordinates, and simply define a new coordinate $\eta$ such that $y = \eta + v t$, leaving $t$ and $x$ unchanged. The $\eta$ coordinate is then obviously just a parameter labeling points along the block according to their $y$ coordinate at time $t =0$ in the Rindler chart. DrGreg then found three basis vectors, the first two of which are the same as my $\hat{e}_0$ and $\hat{e}_1$ (and therefore the 4-velocity of observers at rest in his chart is the same as for ours, so all three charts give different parameterizations of the same congruence of worldlines). His third basis vector is different from my $\hat{e}_2$, and I think it is also different from your $\partial_{\phi}$ (which is a unit vector in your chart). But his third basis vector has the key property that it is always orthogonal to the other two.

DrGreg's $t$ coordinate is not equal to proper time even for the bottom center of the block (whereas our $\bar{\tau}$ and $\tau$ coordinates are). However, the spatial part of his metric is Euclidean (see post #165). Also, using his three basis vectors and the form of the line element he gave in terms of three cobasis 1-forms each multiplied by one of the basis vectors, the quotient space metric can be "read off", and is evidently not Euclidean (see post #164).

What does all this mean? First of all, since DrGreg's formulation is the only one that gives three orthonormal basis vectors everywhere, it is the only one that defines a true frame field, and therefore it seems to me to be the best candidate for a "block observer's frame". This is bolstered by the fact that it has both of two key properties that your and my formulations each only have one of: the direction of proper acceleration is always in his $x$ direction (and surfaces of constant $x$ are surfaces of constant "altitude"), and the spatial part of his metric is Euclidean. (I had speculated earlier in this thread that it was not possible to find a single chart with both of those properties; evidently I was wrong.)

However, the quotient space metric derived from his basis vectors is not Euclidean, which means that the "space seen by block observers" is not Euclidean; the Euclidean "space" that appears in his metric is really just a reparameterization of the Euclidean space seen by Rindler observers (who are not moving in the $y$ direction, perpendicular to their proper acceleration). This is analogous to the fact that in Born coordinates on a "rotating disk", the Euclidean spatial metric is the one seen by the non-rotating observer at the center of the disk, while the quotient space metric seen by the observers rotating with the disk is non-Euclidean.

 

[PeterDonis]

Even better would be both the 4-d line-element and the transform that give the value of Minkowskii coordinates as a function of "observer" coordinates.

DrGreg's formulation doesn't completely define "observer coordinates", because, while his $x, \eta$ are constant along observer worldlines, his $t$ is not the same as any observer's proper time, even on the reference worldline at the bottom center of the block. The obvious simple way to remedy that would be to rescale his $t$ coordinate so that it was the same as observer proper time on the reference worldline (and by extension everywhere on the floor of the rocket/bottom of the block, since his $g_{tt}$ only depends on $x$). I'll try that out and see what happens.

However, DrGreg's formulation does define a frame field for "block observers"--his three basis vectors (plus the fourth $\partial_z$ vector). As I noted in my previous post, that is something that neither my nor your formulation did. And the frame field, physically, is really what we need in order to understand the physical experience of a "block observer".

Additional useful information would be [change] what values of the observer coordinates make the line element diagonal and non-singular, assuming there are some.

In DrGreg's formulation, there aren't any, because, as you can see from post #165, the coefficient of the $dt d\eta$ term is constant; it doesn't vanish anywhere. This actually makes sense, because the congruence of worldlines describing the block has nonzero vorticity everywhere; it is different in this regard from the "rotating disk" Langevin congruence, which has zero vorticity at the center of the disk. There is no point corresponding to an "axis of rotation" for the block.

This would be what I'd call the "origin" of the coordinate system, while I put it at (0,0), that seems to be the exception rather than the rule in this thread.

The simple reason for that is that both DrGreg and I started from Rindler coordinates, where the origin is on the Rindler horizon, not on the "reference" worldline--that worldline is characterized by $gx = 1$, not $x = 0$. You started from Fermi normal coordinates, where the spatial origin is on the reference worldline.

However, there is an underlying point here, which I just mentioned above. There is no natural "origin" for the block if we just look at the points in the block, because there is no "axis of rotation" for the block within the block--the congruence describing the block has nonzero vorticity everywhere. The "zero vorticity" altitude in the block would be at the Rindler horizon, since that is what corresponds to the "axis of rotation". That is why Rindler coordinates place the origin at the Rindler horizon--because that makes things look simplest from the "rotation" point of view.

 

[DrGreg]

Expanding the entire metric out would then change what I posted in post #163 to

$$
\mathrm{d}s^2 = \left( g^2 x^2 - v^2 \right) \mathrm{d}t^2 - 2 v \mathrm{d}t \mathrm{d}\eta - \mathrm{d}\eta^2 - \mathrm{d}x^2
$$

I'll have to go back and recheck my computation of how that transforms into my $\tau$, $\chi$, $\psi$ coordinates.

As a check that I haven't made any other errors, this can also be seen directly from my equation <br /> \mathrm{d}\mathbf{X}<br /> =<br /> \begin{bmatrix}<br /> gx \cosh gt \\<br /> gx \sinh gt \\<br /> v<br /> \end{bmatrix} \mathrm{d}t<br /> +<br /> \begin{bmatrix}<br /> \sinh gt \\<br /> \cosh gt \\<br /> 0<br /> \end{bmatrix} \mathrm{d}x<br /> +<br /> \begin{bmatrix}<br /> 0 \\<br /> 0 \\<br /> 1<br /> \end{bmatrix} \mathrm{d} \eta<br />\implies<br /> \mathrm{d}\mathbf{X}<br /> = <br /> \begin{bmatrix}<br /> \cosh gt \\<br /> \sinh gt \\<br /> 0<br /> \end{bmatrix} gx \, \mathrm{d}t<br /> +<br /> \begin{bmatrix}<br /> \sinh gt \\<br /> \cosh gt \\<br /> 0<br /> \end{bmatrix} \mathrm{d}x<br /> +<br /> \begin{bmatrix}<br /> 0 \\<br /> 0 \\<br /> 1<br /> \end{bmatrix} \left( \mathrm{d} \eta + v \mathrm{d} t \right)<br />which is another orthogonal decomposition, using the "Rindler frame" instead of my "block frame".

I agree with your commentary in posts #167 and #168. It was my objective to find the metric of the quotient space, which, in some senses, best represents the geometry of space (not spacetime). I've come across the quotient space for the Langevin congruence, and was aware it could be used whenever you can find a "stationary-but-not-static coordinate system", but I'd never seen any other practical example, and so was curious to make it work.

As you say, my $t$ coordinate doesn't really tie in very well with a "block observer's" proper time. Maybe it's possible to find a better time cooordinate without upsetting the quotient construction.

 

[PeterDonis]

Maybe it's possible to find a better time cooordinate without upsetting the quotient construction.

Well, the obvious ansatz is $t = \gamma \tau$, where $\gamma = 1 / \sqrt{1 - v^2}$, which gives for the line element

$$
ds^2 = \gamma^2 \left( g^2 x^2 - v^2 \right) \mathrm{d}\tau^2 - 2 \gamma v \mathrm{d}\tau \mathrm{d}\eta - \mathrm{d}\eta^2 - \mathrm{d}x^2
$$

This makes $\tau$ the same as proper time along the worldline of any observer on the bottom of the block/floor of the rocket, where $gx = 1$. It still isn't the same elsewhere, but that is similar to the situation in Rindler coordinates, where $t$ only matches proper time on worldlines with $gx = 1$.

I'm not sure how this affects the quotient space construction, though.

 

[DrGreg]

Well, the obvious ansatz is $t = \gamma \tau$, where $\gamma = 1 / \sqrt{1 - v^2}$, which gives for the line element

$$
ds^2 = \gamma^2 \left( g^2 x^2 - v^2 \right) \mathrm{d}\tau^2 - 2 \gamma v \mathrm{d}\tau \mathrm{d}\eta - \mathrm{d}\eta^2 - \mathrm{d}x^2
$$

This makes $\tau$ the same as proper time along the worldline of any observer on the bottom of the block/floor of the rocket, where $gx = 1$. It still isn't the same elsewhere, but that is similar to the situation in Rindler coordinates, where $t$ only matches proper time on worldlines with $gx = 1$.

I'm not sure how this affects the quotient space construction, though.

I think rescaling any of my coordinates by a constant shouldn't affect the method, so rescaling $t$ (and maybe also $\eta$) by $\gamma$ (regarding $v$ as constant) should be OK and makes it more physically relevant at $x = 1/g$.

 

[PeterDonis]

rescaling $t$ (and maybe also $\eta$) by $\gamma$ (regarding $v$ as constant) should be OK and makes it more physically relevant at $x = 1/g$.

I'm not sure $\eta$ should be rescaled. If we rearrange the line element with $\tau$ that I wrote, we get

$$
\mathrm{d} s^2 = \gamma^2 g^2 x^2 \mathrm{d} \tau^2 - \left( \gamma v \mathrm{d} \tau + d \eta \right)^2 - dx^2
$$

which puts all the $\gamma$ factors with the $\mathrm{d} \tau$ factors.

 

[DrGreg]

I'm not sure $\eta$ should be rescaled. If we rearrange the line element with $\tau$ that I wrote, we get

$$
\mathrm{d} s^2 = \gamma^2 g^2 x^2 \mathrm{d} \tau^2 - \left( \gamma v \mathrm{d} \tau + d \eta \right)^2 - dx^2
$$

which puts all the $\gamma$ factors with the $\mathrm{d} \tau$ factors.

A possible reason for rescaling $\eta$ would be that my quotient metric$$
\mathrm{d}s^2 =
\mathrm{d}x^2
+ \frac{g^2x^2}{g^2x^2-v^2} \, \mathrm{d} \eta^2
$$is$$
\mathrm{d}s^2 =
\mathrm{d}x^2
+ \gamma^2 \mathrm{d} \eta^2
$$ locally wherever $gx=1$.

 

[PeterDonis]

A possible reason for rescaling $\eta$ would be that my quotient metric
$$
\mathrm{d}s^2 = \mathrm{d}x^2 + \frac{g^2x^2}{g^2x^2-v^2} \, \mathrm{d} \eta^2
$$
is
$$
\mathrm{d}s^2 = \mathrm{d}x^2 + \gamma^2 \mathrm{d} \eta^2
$$
locally wherever $gx=1$.

So if we define $\psi = \gamma \eta$, then the quotient space metric becomes

$$
\mathrm{d}s^2 = \mathrm{d}x^2 + \frac{g^2x^2 \left( 1 - v^2 \right)}{g^2x^2 - v^2} \, \mathrm{d} \psi^2
$$

and the full metric becomes

$$
\mathrm{d}s^2 = \frac{g^2 x^2 - v^2}{1 - v^2} \mathrm{d}\tau^2 - 2 v \mathrm{d}\tau \mathrm{d}\psi - \left( 1 - v^2 \right) \mathrm{d}\psi^2 - \mathrm{d}x^2
$$

This still factors into something fairly similar to what I had before:

$$
\mathrm{d}s^2 = \gamma^2 g^2 x^2 \mathrm{d}\tau^2 - \left( \gamma v \mathrm{d}\tau + \frac{1}{\gamma} \mathrm{d}\psi \right)^2 + \mathrm{d}x^2
$$

 

[pervect]

[QUOTE="PeterDonis, post: 5274110, member: 197831"

$$
\mathrm{d}s^2 = \frac{g^2 x^2 - v^2}{1 - v^2} \mathrm{d}\tau^2 - 2 v \mathrm{d}\tau \mathrm{d}\psi - \left( 1 - v^2 \right) \mathrm{d}\psi^2 - \mathrm{d}x^2
$$

What I see here is a metric where $g_{\tau\psi}=v$ is never zero at any point, so that $\partial_\tau$ is never orthogonal to $\partial_\psi$ because the dot product of the vectors $\partial_\tau$ and $\partial_\psi$ is not zero as it would need to be for them to be orthogonal

[add]
I.e. if we denote the dual vector to $\partial_t$ as dt, and the dual vector to $\partial_\psi$ as $d\psi$, the only term in the metric that contributes to the dot product is the $dt\,d\psi$ term, since the product of the vector $\partial_\mu$ and the one form $d \nu$ is the kronecker delta $\delta^\mu_\nu$

 

[PeterDonis]

What I see here is a metric where $g_{\tau\psi}=v$ is never zero at any point, so that $\partial_\tau$ is never orthogonal to $\partial_\psi$

Yes. But the third basis vector in DrGreg's frame field is not $\partial_\psi$ (or $\partial_\eta$ in his original coordinates). It's a linear combination of $\partial_\psi$ and $\partial_\tau$ that is orthogonal to $\partial_\tau$. (I am actually speaking loosely, the frame field vectors are unit vectors and the coordinate basis vectors aren't, at least not all of them, but hopefully you understand what I mean.)

 

[DrGreg]

What I see here is a metric where $g_{\tau\psi}=v$ is never zero at any point, so that $\partial_\tau$ is never orthogonal to $\partial_\psi$ because the dot product of the vectors $\partial_\tau$ and $\partial_\psi$ is not zero as it would need to be for them to be orthogonal

It's inevitable, for any spacetime coordinate system (with one timelike coordinate and 3 spacelike coordinates) in which all points within the block are at rest, that the surfaces of constant time cannot be everywhere orthogonal to the worldlines of those points. The torsion makes it impossible, and you'll always have $g_{0i} \neq 0$ for at least one $i \neq 0$.

That's why we have to resort to the quotient space method to find a metric for a global space (not spacetime) in which the distance is everywhere locally compatible (to first order, ignoring curvature) with spatial distance in the local comoving inertial frame.

 

[pervect]

Yes. But the third basis vector in DrGreg's frame field is not $\partial_\psi$ (or $\partial_\eta$ in his original coordinates). It's a linear combination of $\partial_\psi$ and $\partial_\tau$ that is orthogonal to $\partial_\tau$. (I am actually speaking loosely, the frame field vectors are unit vectors and the coordinate basis vectors aren't, at least not all of them, but hopefully you understand what I mean.)

If you mean

So if we define $\psi = \gamma \eta$, then the quotient space metric becomes

$$
\mathrm{d}s^2 = \gamma^2 g^2 x^2 \mathrm{d}\tau^2 - \left( \gamma v \mathrm{d}\tau + \frac{1}{\gamma} \mathrm{d}\psi \right)^2 + \mathrm{d}x^2
$$

and we take $\psi^\prime = \left( \gamma v \mathrm{d}\tau + \frac{1}{\gamma} \mathrm{d}\psi \right)$ and $\tau^\prime = \gamma g x$
the thing that disturbs me about it is that $(d\tau^\prime)^2 - (d\psi^\prime)^2 - dx^2$ IS diagonal everywhere. Which suggests to me that it isn't rotating, because a rotating metric shouldn't be diagonal everywhere - one can make it diagonal at a point, but there isn't any closed orthogonal spatial hypersurface, so the line element can't be diagonal everywhere.

I'd suggest first writing down exactly what congruence we're talking about as the new proposal for the sliding block congruence - is it the congruence given by the set of worldlines $\psi^\prime, \chi$ = constant, $\tau^\prime = -\infty ... \infty$ above, or did I misunderstand? After we're clear on what the congruence is, then we can comparing the vorticity (at a minimum) of this new congruence and compare the results to your congruence and mine (which I think we agreed was the same congruence). If the vorticity doesn't match, the new congruence can't be the same congruence as yours and mine. I think is fairly trivial to say, though that the congruence generated by $(d\tau^\prime)^2 - (d\psi^\prime)^2 - dx^2$ isn't rotating, i.e. it leads to a zero vorticity.

 

[PeterDonis]

If you mean

I think you mis-copied my previous post (post #174) in the quote just below this; the quotient space metric only has spatial components, there is no $\mathrm{d} \tau$ in it. In post #174, the metric that appears in your quote is not the quotient space metric; it is at the end of the post, after the words "this still factors into something fairly similar to what I had before".

The frame field I am referring to is the following tetrad (expressed in the Minkowsi coordinate basis but using DrGreg's coordinates in the coefficients) that DrGreg used to derive his quotient space metric:

$$
\hat{e}_0 = \frac{1}{\sqrt{g^2 x^2 - v^2}} \left( g x \cosh g t \, \partial_T + g x \sinh g t \, \partial_X + v \, \partial_Y \right)
$$

$$
\hat{e}_1 = \sinh g t \, \partial_T + \cosh g t \, \partial_X
$$

$$
\hat{e}_2 = \frac{1}{\sqrt{g^2 x^2 - v^2}} \left( v \cosh g t \, \partial_T + v \sinh g t \, \partial_X + g x \, \partial_Y \right)
$$

$$
\hat{e}_3 = \partial_Z
$$

This tetrad is obviously orthonormal, and $\hat{e}_0$ is the same 4-velocity field that you and I derived; $\hat{e}_2$ is the linear combination of $\partial_\tau$ and $\partial_\psi$ that I referred to.

 

[sweet springs]

Hi. Let me draw two figures to confirm my understanding of the problem.

 

[sweet springs]

Best.

 

[SlowThinker]

Hi. Let me draw two figures to confirm my understanding of the problem.

The Frame of the Earth looks correct, but It seems your train is moving to the left, which is somehow not the way I imagined it.
And the picture in train's frame is obviously not consistent with the Earth frame, or as we call it, the Rocket frame.

Also now we (or rather, Peter, Pervect and DrGreg) are at the point of trying to figure out how clocks and rulers deform at various places of the train. The original question has been answered long ago.

 

[sweet springs]

Thanks. I will read the thread carefully to find the answer you found..

And the picture in train's frame is obviously not consistent with the Earth frame, or as we call it, the Rocket frame.
.

I draw the Rocket frame firure.

 

[pervect]

$$
\hat{e}_0 = \frac{1}{\sqrt{g^2 x^2 - v^2}} \left( g x \cosh g t \, \partial_T + g x \sinh g t \, \partial_X + v \, \partial_Y \right)
$$

$$
\hat{e}_1 = \sinh g t \, \partial_T + \cosh g t \, \partial_X
$$

$$
\hat{e}_2 = \frac{1}{\sqrt{g^2 x^2 - v^2}} \left( v \cosh g t \, \partial_T + v \sinh g t \, \partial_X + g x \, \partial_Y \right)
$$

$$
\hat{e}_3 = \partial_Z
$$

This tetrad is obviously orthonormal, and $\hat{e}_0$ is the same 4-velocity field that you and I derived; $\hat{e}_2$ is the linear combination of $\partial_\tau$ and $\partial_\psi$ that I referred to.

I've been looking at this and feel like I'm still missing something. Do you happen to have an expression for the basis vectors $\hat{e}_1, \hat{e}_2, \hat{e}_3$ in terms of Minkowskii coordinates $T, X, Y, \partial_T, \partial_X, \partial_Y$ and or in terms of $t,x,y,\partial_t, \partial_x, \partial_y$? At the moment, you have the expressions written in terms of $t,x,y$ multiplying partial derivative written in terms of $T,X,Y$, so there are two sets of variables here. Eliminating one of the two sets of variables would allow me to proceed with things like finding (and comparing) the Christoffel symbols with this approach to the other approaches. I'm just not quite seeing how $t,x,y$ are defined in terms of $T,X,Y$ given the above information. I presumet the needed information may be buried somewhere in the thread, but it's not quite coming together for me.

 

[PeterDonis]

Do you happen to have an expression for the basis vectors $\hat{e}_1, \hat{e}_2, \hat{e}_3$ in terms of Minkowskii coordinates $T, X, Y, \partial_T, \partial_X, \partial_Y$

That's easy enough to derive from what I posted already, since $t, x$ are the same as for Rindler coordinates and we know how those transform to Minkowski coordinates. Here's what we end up with (with $\hat{e}_0$ included for completeness):

$$
\hat{e}_0 = \frac{1}{\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}} \left( g X \, \partial_T + g T \, \partial_X + v \, \partial_Y \right)
$$
$$
\hat{e}_1 = \frac{T}{\sqrt{X^2 - T^2}} \, \partial_T + \frac{X}{\sqrt{X^2 - T^2}} \, \partial_X
$$
$$
\hat{e}_2 = \frac{1}{\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}} \left( \frac{v X}{\sqrt{X^2 - T^2}} \, \partial_T + \frac{v T}{\sqrt{X^2 - T^2}} \, \partial_X + g \sqrt{X^2 - T^2} \, \partial_Y \right)
$$
$$
\hat{e}_3 = \partial_Z
$$

 

[PeterDonis]

or in terms of $t,x,y,\partial_t, \partial_x, \partial_y$

This is pretty straightforward too, and turns out to be, using coordinates $t, x, y$ (note that $\eta$ is the coordinate that DrGreg originally used, but we can write $\partial_y$ instead of $\partial_\eta$, with $y$ being the Rindler $y$ coordinate, since the two partial derivatives are the same and $y$ never appears explicitly so we don't have to worry that it isn't the same as $\eta$):

$$
\hat{e}_0 = \frac{1}{\sqrt{g^2 x^2 - v^2}} \partial_t + \frac{v}{\sqrt{g^2 x^2 - v^2}} \partial_y
$$
$$
\hat{e}_1 = \partial_x
$$
$$
\hat{e}_2 = \frac{v}{g x \sqrt{g^2 x^2 - v^2}} \partial_t + \frac{gx}{\sqrt{g^2 x^2 - v^2}} \partial_y
$$
$$
\hat{e}_3 = \partial_z
$$

Note that we use the Rindler metric to compute norms and inner products for the above, so, for example, $\hat{e}_0 \cdot \hat{e}_2 = g^2 x^2 \left( \hat{e}_0 \right)^t \left( \hat{e}_2 \right)^t - \left( \hat{e}_0 \right)^y \left( \hat{e}_2 \right)^y = \left( v g x - v g x \right) / \left( g^2 x^2 - v^2 \right) = 0$ as desired.

 

[PeterDonis]

the 4-velocity of observers at rest in his chart is the same as for ours

On looking back over things, I realized that this statement is not correct as it stands. The correct statement is that the 4-velocity field described by the first of DrGreg's basis vectors is the same as the 4-velocity field pervect and I both found previously. But the observers whose worldlines are integral curves of this 4-velocity field are not at rest in DrGreg's chart, since that chart is just the Rindler chart (with some possible rescalings of coordinates), and the block is not at rest in the Rindler chart (the rocket is). These observers are at rest in my and pervect's charts.

That is why, for example, the 4-velocity field $\hat{e}_0$ has only one component in my and pervect's charts (it's just $\partial_\tau$ normalized), whereas it has two components in DrGreg's chart, as is evident from post #186.

 

[PeterDonis]

And just to expand a bit more on post #187, the transform to the $\bar{\tau}, \bar{\chi}, \bar{\psi}$ coordinates I used in my chart is $t = \gamma \left( \bar{\tau} + v \bar{\psi} \right)$, $y = \gamma \left( \bar{\psi} + v \bar{\tau} \right)$, which leads to the following expressions for DrGreg's basis vectors:

$$
\hat{e}_0 = \frac{1}{\gamma \sqrt{g^2 \bar{\chi}^2 - v^2}} \partial_{\bar{\tau}}
$$
$$
\hat{e}_1 = \partial_{\bar{\chi}}
$$
$$
\hat{e}_2 = \frac{\gamma v \left( 1 - g^2 \bar{\chi}^2 \right)}{g \bar{\chi} \sqrt{g^2 \bar{\chi}^2 - v^2}} \partial_{\bar{\tau}} + \frac{\gamma \sqrt{g^2 \bar{\chi}^2 - v^2}}{g \bar{\chi}} \partial_{\bar{\psi}}
$$
$$
\hat{e}_3 = \partial_z
$$

The metric in post #79 would be used to compute norms and inner products for the above. Note that this $\hat{e}_2$, as I've noted before, has both $\partial_{\bar{\tau}}$ and $\partial_{\bar{\psi}}$ components. Note also that, on the floor of the rocket/ bottom of the block, where $g \bar{\chi} = 1$, the $\partial_{\bar{\tau}}$ component of $\hat{e}_2$ vanishes; this is consistent with everything we've said before.

 

[pervect]

This is pretty straightforward too, and turns out to be, using coordinates $t, x, y$ (note that $\eta$ is the coordinate that DrGreg originally used, but we can write $\partial_y$ instead of $\partial_\eta$, with $y$ being the Rindler $y$ coordinate, since the two partial derivatives are the same and $y$ never appears explicitly so we don't have to worry that it isn't the same as $\eta$):

$$
\hat{e}_0 = \frac{1}{\sqrt{g^2 x^2 - v^2}} \partial_t + \frac{v}{\sqrt{g^2 x^2 - v^2}} \partial_y
$$
$$
\hat{e}_1 = \partial_x
$$
$$
\hat{e}_2 = \frac{v}{g x \sqrt{g^2 x^2 - v^2}} \partial_t + \frac{gx}{\sqrt{g^2 x^2 - v^2}} \partial_y
$$
$$
\hat{e}_3 = \partial_z
$$

Note that we use the Rindler metric to compute norms and inner products for the above, so, for example, $\hat{e}_0 \cdot \hat{e}_2 = g^2 x^2 \left( \hat{e}_0 \right)^t \left( \hat{e}_2 \right)^t - \left( \hat{e}_0 \right)^y \left( \hat{e}_2 \right)^y = \left( v g x - v g x \right) / \left( g^2 x^2 - v^2 \right) = 0$ as desired.

I concur with this result, which I eventually got from first principles, and the important Christoffel symbols in this basis have the expected value at gx=1, i.e.
$\Gamma^{\hat{2}}{}_{\hat{1}\hat{1}} = \frac{g^2 x}{g^2 x^2 - v^2}$ = "linear acceleration" = $\frac{g}{1-v^2}$ at gx=1

$\Gamma^{\hat{2}}{}_{\hat{1}\hat{3}} = -\Gamma^{\hat{3}}{}_{\hat{1} \hat{2}}$ = "rotation (omega)" = $\frac{gv}{g^2 x^2-v^2}$ which is $\frac{gv}{1-v^2}$ at gx=1

Some spatial Christoffel symbols have non-zero values, as well, this isn't an exhaustive list of all the non-zero Christoffel symbols, but a basic sanity check of the proper acceleration and proper rotation.

t,x,y are Rindler coordinates chosen with the metric having the form $g^2 x^2 dt^2 - dx^2 - dy^2$

 

[PeterDonis]

Christoffel symbols in this basis

I assume this means these are not the Christoffel symbols in the Rindler coordinate basis, since only two of those are nonzero, and the one that might correspond to one you listed, $\Gamma^x{}_{tt}$, doesn't have the value you give for $\Gamma^{\hat{2}}{}_{\hat{1} \hat{1}}$. I assume these are the symbols in the non-coordinate basis given by the vectors I listed?

 

[sweet springs]

Hi. Let me confirm my understanding the situation. In Rindler coordinates
$$ds^2 = (1+\frac{gz}{c^2})^2 c^2dt^2 - dx^2 - dy^2 - dz^2, \; \forall z>-\frac{c^2}{g}, \forall t, x, y \,.$$

Lorentz transformation,
$$
t' = \gamma \left(t - \frac{v x}{c^{2}} \right)
x' = \gamma (x - v t)\,
y' = y\,
z' = z\,$$
where
$$\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}$$
or
$$
t = \gamma \left(t' + \frac{v x'}{c^{2}} \right)
x = \gamma (x' + v t')\,
y = y'\,
z = z'\,$$

Putting them toghether we find the term dxdt in ds^2 does not vanish. This term causes the different manner of forward and backward lights.

 

[PeterDonis]

sweet springs, please note that the delimiter for LaTeX equations is two dollar signs $$, not one. I used magic moderator powers to edit your post #191 to correct that.

 

[PeterDonis]

Lorentz transformation

This is basically the transformation I used to go from Rindler coordinates to my coordinates, except that I used $\tau, \chi, \psi$ instead of $t', x', y'$. But note that this is not, strictly speaking, a Lorentz transformation, because it's not a transformation between inertial frames. It just happens to look the same formally as a Lorentz transformation (and of course I chose that form of transformation for that reason).

the term dxdt in ds^2 does not vanish

In my chart, it's actually the term $dy' dt'$ (or $d\psi d\tau$ in my nomenclature) that doesn't vanish.

 

[pervect]

Hi. Let me confirm my understanding the situation. In Rindler coordinates
$$ds^2 = (1+\frac{gz}{c^2})^2 c^2dt^2 - dx^2 - dy^2 - dz^2, \; \forall z>-\frac{c^2}{g}, \forall t, x, y \,.$$

Lorentz transformation,
$$
t' = \gamma \left(t - \frac{v x}{c^{2}} \right)
x' = \gamma (x - v t)\,
y' = y\,
z' = z\,$$
where
$$\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}$$
or
$$
t = \gamma \left(t' + \frac{v x'}{c^{2}} \right)
x = \gamma (x' + v t')\,
y = y'\,
z = z'\,$$

Putting them toghether we find the term dxdt in ds^2 does not vanish. This term causes the different manner of forward and backward lights.

This is close, but you can't really use the Lorentz transform with the Rindler metric. You can use it in a momentarily co-moving inertial frame, though. While we have mostly been considering the rocket to be accelerating in the "x" direction in this thread, your metric has it accelerating in the "z" direction. I will use your conventions rather than the ones in this thread.

Using your coordinate conventions, then, at t=0 in the momentarily co-moving inertial reference frame (MCIRF) on the rocket floor, a frame we will call S. We will assume z in this frame is constant everywhere at t=0. We will take this constant to be z=0. Because we are assuming the rocket is accelerating in the z direction, we can write an approximate expression for the position of the rocket floor at time t in this MCIRF, which is $z \approx .5 g t^2$. The expression is approximate because we've ignored relativistic effects, but it is sufficient for our purposes here.

If we consider a different inertial frame S' moving in the x direction at some velocity v relative to frame S (again using your coordinate conventions), then because simultaneity is relative, the time t' in frame S' is not the same as the time t in frame S. Given the previous approximate expression, $z \approx .5 g t^2$ in S, we can use the Lorentz transform you wrote to find z' in terms of t'. I won't go through the math in detail, but doing so we find that z' is a function of x', that z'=0 when x'=0, and that z'>0 for x' not equal to zero.

While we have been using much more advanced techniques than MCIRF's in this thread, I think that they are useful in bringing to light some of the issues at a basic level.

 

[pervect]

I found an arxiv paper on the quotient space method in rotating frames that might be useful. It's by one of the same author that I quoted a google books reference for earlier, Ruggiero, and is at http://arxiv.org/abs/gr-qc/0309020. I haven't worked through it in detail, but it has a nice, careful, operational description of how to measure distances on a rotating platform which could be useful in the context of how to measure distances on our sliding block (which, as we have seen, also rotates, at least relative to gyroscopes).

Ruggiero's paper sets up the definition of the quotient space in more detail, which the author refers to as "Relative space". The author goes through the actual exchange of light signals (in accordace with the SI defintion of the meter as "the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.")

 

[sweet springs]

Thanks for your interest and comments to my post #191.
I correct dxdt to dx'dt' there.
I complete the calculation of ds~2.

<br /> ds^2=\gamma^2[(1+\frac{gz}{c^2})^2-\frac{v^2}{c^2}]c^2dt&#039;^2\,<br /> +\gamma^2[(1+\frac{gz}{c^2})^2\frac{v^2}{c^2}-1]dx&#039;^2\,<br /> +2v\gamma^2[(1+\frac{gz}{c^2})^2-1]dx&#039;dt&#039;\,<br /> -dy&#039;^2-dz&#039;^2<br />
or
<br /> ds^2=\gamma^2[(1+\frac{gz}{c^2})^2-\frac{v^2}{c^2}]c^2dt&#039;^2\,<br /> +\gamma^2[(1+\frac{gz}{c^2})^2\frac{v^2}{c^2}-1]dx&#039;^2\,<br /> +2v\gamma^2\frac{gz}{c^2}(2+\frac{gz}{c^2})dx&#039;dt&#039;\,<br /> -dy&#039;^2-dz&#039;^2<br />

Doesn't it make sense at all?

 

[sweet springs]

I correct z to z'.

<br /> ds^2=\gamma^2[(1+\frac{gz&#039;}{c^2})^2-\frac{v^2}{c^2}]c^2dt&#039;^2\,<br /> +\gamma^2[(1+\frac{gz&#039;}{c^2})^2\frac{v^2}{c^2}-1]dx&#039;^2\,<br /> +2v\gamma^2[(1+\frac{gz&#039;}{c^2})^2-1]dx&#039;dt&#039;\,<br /> -dy&#039;^2-dz&#039;^2<br />

or

<br /> ds^2=[1+\gamma^2\frac{gz&#039;}{c^2}(2+\frac{gz&#039;}{c^2})]\, c^2dt&#039;^2\,<br /> +[-1+\frac{v^2}{c^2}\gamma^2\frac{gz&#039;}{c^2}(2+\frac{gz&#039;}{c^2})]dx&#039;^2\,<br /> +2v\gamma^2\frac{gz&#039;}{c^2}(2+\frac{gz&#039;}{c^2})dx&#039;dt&#039;\,<br /> -dy&#039;^2-dz&#039;^2<br />

or
<br /> ds^2=(1+\gamma^2\zeta)\, c^2dt&#039;^2\,<br /> +(-1+\frac{v^2}{c^2}\zeta)dx&#039;^2\,<br /> +2v\zeta\, dx&#039;dt&#039;\,<br /> -dy&#039;^2-dz&#039;^2<br />
where
$$ \zeta=\gamma^2\frac{gz'}{c^2}(2+\frac{gz'}{c^2})$$

I find a poor point. The speed of train changes accorging to z due to time dilation effect of gravity.
Let v be speed of the train at floor level z'=0.
I appreciate if you point out other difficulties.

 

[sweet springs]

correction:
<br /> ds^2=(1+\zeta)\, c^2dt&#039;^2\,<br /> +(-1+\frac{v^2}{c^2}\zeta)dx&#039;^2\,<br /> +2v\zeta\, dx&#039;dt&#039;\,<br /> -dy&#039;^2-dz&#039;^2<br />
where
$$ \zeta=\gamma^2\frac{gz'}{c^2}(2+\frac{gz'}{c^2})$$

 

[pervect]

Oops again. Not my lucky day.

In my haste to correct my error in post #156, I overcorrected in post #159 and still got it wrong. I now think the correct answer for the quotient space metric should be <br /> \mathrm{d}s^2 =<br /> \mathrm{d}x^2<br /> + \frac{g^2x^2}{g^2x^2-v^2} \, \mathrm{d} \eta^2<br />
That can be made isometric but it's a really ugly and complicated expression, so I won't bother here.

I get what I believe is the same answer as Dr. Greg now, let me sketch the process.

Start with the following variant of the RIndler metric:

$ds^2 = -g^2 x^2 dt^2 + dx^2 + dy^2$

Make the substitution t'=t, x'=x, y'=y-vt, z' to arrive at a new metric in which the sliding block has constant coordinates. This new metric doesn't give much insight into the physics, because it's not orthonormal, but it is convenient for understanding the spatial geometry.

$ds'^2 = -(g^2 x'^2 - v^2) dt'^2 + dx'^2 + dy'^2 + 2 v dy' dt'$

Apply the methods from Ruggiero's paper http://arxiv.org/abs/gr-qc/0309020. In section 3.2, rather than using Ruggiero's line element, we use

$ds'^2 = {\it gtt}\,{{\it dt'}}^{2}+{\it gxx}\,{{\it dx'}}^{2}+{\it gyy}\,{{\it dy'}}^{2}+2\,{\it gty}\,{\it dy'}\,{\it dt'}$

We follow Ruggerio's procedure as in section 3.2.

3.2 The local spatial geometry of the rotating frame

We can introduce the local spatial geometry of the disk, which defines the
proper spatial line element, on the basis of the local optical geometry. To this
end we can use the radar method...
Let Π be a point in the rotating frame, where a light source, a light absorber
and a clock are lodged; let Π' be a near point where a reflector is lodged. The
world-lines of these points are the time like helices
...
(see figure 1). A light signal is emitted by the source in Π and propagates along the null world-
line toward Π' here it is reflected back to Π (along the null world-line where it is finally absorbed. Let
dτ be the proper time, read by a clock in Π, between the emission and absorption events: then,
according to the radar method, the proper distance between Π and Π'is defined by

$d\sigma = \frac{1}{2} c d\tau$

Now, we are going to parameterize these events, using the coordinates adapted
to the rotating frame, in order to obtain the explicit expression of the proper
spatial line element ...

The space-time intervals between the events of emissionE and reflection R, and between the events of reflectionR and absorption A, are null. Hence, by setting ds'^2 = 0 we can solve for dt , and obtain the two coordinate time ... [[for emission and absorption]]

Solving for the coordinate time of absorption and emission, and converting the time difference from coordinate time to proper time, and dividing by 2 to convert the round-trip time to the one-way time, we get:

$$ A = 1/2\,{\frac {-2\,{\it gty}\,{\it dy}+2\,\sqrt {{{\it gty}}^{2}{{\it dy
}}^{2}-{\it gtt}\,{\it gxx}\,{{\it dx}}^{2}-{\it gtt}\,{\it gyy}\,{{
\it dy}}^{2}}}{{\it gtt}}}$$
$$ E =
1/2\,{\frac {-2\,{\it gty}\,{\it dy}-2\,\sqrt {{{\it gty}}^{2}{{\it dy
}}^{2}-{\it gtt}\,{\it gxx}\,{{\it dx}}^{2}-{\it gtt}\,{\it gyy}\,{{
\it dy}}^{2}}}{{\it gtt}}}$$

Note some minor typo's in the original paper when we compare our solution to theirs, the typos are in terms which cancel out so are not important to the end result.

Then the coordinate time $\delta t$ it takes to travel the round trip is A-E, the proper time is $\sqrt{|gtt|} \delta t$ and the
proper distance is $d\tau / 2$ (since we are assuming that c=1)

This gives us the expression for $\tau^2$:
$$-{\frac {{{\it gty}}^{2}{{\it dy}}^{2}-{\it gtt}\,{\it gxx}\,{{\it dx}
}^{2}-{\it gtt}\,{\it gyy}\,{{\it dy}}^{2}}{{\it gtt}}}$$

Substituting in the values for g** gives

$$length^2 = dx^2 + \frac{g^2 x^2}{g^2 x^2 - v^2} dy^2 = dx^2 + \frac{dy^2}{1- \left( \frac{v}{gx} \right)^2} $$

Note that the factor multiplying dy^2 is rather similar $\gamma^2$, except that we replace v by the (v / gx), which is the ratio of the coordinate speed of the block v to the coordinate speed of light, gx.

Note that because of the form of this line element, dy represents a longer distance for x=1/g than it does when x>1/g. This gives the coordinate independent fact that the shortest distance between two points on the block at x=0 does not lie along the floor at x=1/g, but rather curves upwards, so that it goes up (x>1/g), then back down.

 

[pervect]

I correct z to z'.I find a poor point. The speed of train changes accorging to z due to time dilation effect of gravity.
Let v be speed of the train at floor level z'=0.
I appreciate if you point out other difficulties.

You should find that the coordinate speed of the train dx/dt, x and t being rindler coordinates, does not vary. The proper speed of the train, dx/d$\tau$, does vary with z - it slows down with increasing z.