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

[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.

$$ds^2=\gamma^2[(1+\frac{gz}{c^2})^2-\frac{v^2}{c^2}]c^2dt'^2\, +\gamma^2[(1+\frac{gz}{c^2})^2\frac{v^2}{c^2}-1]dx'^2\, +2v\gamma^2[(1+\frac{gz}{c^2})^2-1]dx'dt'\, -dy'^2-dz'^2$$
or
$$ds^2=\gamma^2[(1+\frac{gz}{c^2})^2-\frac{v^2}{c^2}]c^2dt'^2\, +\gamma^2[(1+\frac{gz}{c^2})^2\frac{v^2}{c^2}-1]dx'^2\, +2v\gamma^2\frac{gz}{c^2}(2+\frac{gz}{c^2})dx'dt'\, -dy'^2-dz'^2$$

Doesn't it make sense at all?

 

[sweet springs]

I correct z to z'.

$$ds^2=\gamma^2[(1+\frac{gz'}{c^2})^2-\frac{v^2}{c^2}]c^2dt'^2\, +\gamma^2[(1+\frac{gz'}{c^2})^2\frac{v^2}{c^2}-1]dx'^2\, +2v\gamma^2[(1+\frac{gz'}{c^2})^2-1]dx'dt'\, -dy'^2-dz'^2$$

or

$$ds^2=[1+\gamma^2\frac{gz'}{c^2}(2+\frac{gz'}{c^2})]\, c^2dt'^2\, +[-1+\frac{v^2}{c^2}\gamma^2\frac{gz'}{c^2}(2+\frac{gz'}{c^2})]dx'^2\, +2v\gamma^2\frac{gz'}{c^2}(2+\frac{gz'}{c^2})dx'dt'\, -dy'^2-dz'^2$$

or
$$ds^2=(1+\gamma^2\zeta)\, c^2dt'^2\, +(-1+\frac{v^2}{c^2}\zeta)dx'^2\, +2v\zeta\, dx'dt'\, -dy'^2-dz'^2$$
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:
$$ds^2=(1+\zeta)\, c^2dt'^2\, +(-1+\frac{v^2}{c^2}\zeta)dx'^2\, +2v\zeta\, dx'dt'\, -dy'^2-dz'^2$$
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 $$\mathrm{d}s^2 = \mathrm{d}x^2 + \frac{g^2x^2}{g^2x^2-v^2} \, \mathrm{d} \eta^2$$
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.

 

[sweet springs]

Start with the following variant of the RIndler metric:
$ds^2 = -g^2 x^2 dt^2 + dx^2 + dy^2$

Putting g=0, this formula does not go to Minkowsky. No problem?

 

[pervect]

Putting g=0, this formula does not go to Minkowsky. No problem?

Not really a problem, just a matter of differing convention. The above metric is Minkowskii near x=1/g rather than x=0. Take a look at the (current as of this post) Wikki entry on Rindler coordinates, i.e. https://en.wikipedia.org/w/index.php?title=Rindler_coordinates&oldid=676502776, which has the same line element I used above. As I recall this form of the line element is the form originally used by Rindler in his textbook. One can transform the line element from the Wikki form into the line element you use by the transform x' = (x+1/g). We have dx = dx' but the choice of origin is different. The form you quote is quite common as well, and is used in textbooks such as MTW - I used it myself in some of my earlier posts in this thread. It is correct useful to note that MTW's form of the line element works correctly when g=0 and the alternate form used in the wikki does not work well if g=0, as the transform x'=x+1/g becomes undefined when g=0.

 

[sweet springs]

Hi. 　　　Lorentz transgfotmation is applicable to Rindler coordinate.

1. Rindler coordinate has a special height z'=0 where proper time of clock at rest coincide with system coordinate time t.
With constraint that movement is on the plane z'=0, space-time is Minkowsky. Lorentz transformation gives coordinates of the train.

2. With constraint that movement stays on not only z'=0 but z'= const provide the same result i.e. validity of Lorentz transformation, just by scaling time.

3. We can combine these planes thought at 1. and 2. to make 3D space body. Thus Lorentz transformation is applicable to Rindler coordinate with no constraints.

Thus

correction:
$$ds^2=(1+\zeta)\, c^2dt'^2\, +(-1+\frac{v^2}{c^2}\zeta)dx'^2\, +2v\zeta\, dx'dt'\, -dy'^2-dz'^2$$
where
$$ \zeta=\gamma^2\frac{gz'}{c^2}(2+\frac{gz'}{c^2})$$

stands. How is this idea?

 

[pervect]

Hi. 　　　Lorentz transgfotmation is applicable to Rindler coordinates

I'm not sure exactly what you're trying to say, but I was pleasantly surprised as to how simple the Rindler metric became when we applied a transform formally similar to the Lorentz transform.

i.e. if we start with

$ds^2 = -gx^2 + dx^2 + dy^2$

and apply the transform $\quad \gamma = 1/\sqrt{1-v^2} \quad t' = \gamma (t-v\,y) \quad x'=x \quad y' = \gamma (y-v\,t) )$ we get a rather nice-looking metric
$$ \left[ \begin {array}{ccc} -{\frac {-{g}^{2}{x}^{2}+{v}^{2}}{-1+{v}^{2}}}&0&{\frac {v \left( {g}^{2}{x}^{2}-1 \right) }{-1+{v}^{2}}}
\\0&1&0\\{\frac {v \left( {g}^{2}{x}^{2}-1 \right) }{-1+{v}^{2}}}&0&{\frac {{g}^{2}{x}^{2}{v}^{2}-1}{-1+{v}^{2}}}\end {array} \right]$$

which is orthonormal at gx = 1, so we can interpret the physics there fairly easily.

If we use the other form of the Rindler metric, and apply the same transform, we get a similarly nice-looking metric that's orthonormal at x=0

 

[sweet springs]

I am happy to share beauty of the metric with you.

I'm not sure exactly what you're trying to say,

I will add some words here.

Mechanics of Billiard on the horizontal flat tables satisfies SR or Lorentz transformation even under the presence of gravity of the Earth.
Say there are stories of Billiard tables where ball movements satisfies SR on their proper layers.
Say balls on all the layers move in a perpendicular line. The perpendicular line represents wall or column of the train.
Their velocities measured at their proper tables are different due to clock pace change by height.
Though different Lorentz transformation of different speed for layers are applied, they all are expressed by one parameter v, the speed at z'=0 or any other height prefixed.

 

[pervect]

I am happy to share beauty of the metric with you.
I will add some words here.

Mechanics of Billiard on the horizontal flat tables satisfies SR or Lorentz transformation even under the presence of gravity of the Earth.
Say there are stories of Billiard tables where ball movements satisfies SR on their proper layers.
Say balls on all the layers move in a perpendicular line. The perpendicular line represents wall or column of the train.
Their velocities measured at their proper tables are different due to clock pace change by height.
Though different Lorentz transformation of different speed for layers are applied, they all are expressed by one parameter v, the speed at z'=0 or any other height prefixed.

I agree. We imagine the train as a rigid object - the mathematical treatment of this notion is "Born rigidity". The coordinate velocity v as measured by an "rindler observer" is constant, but the value of this constant v depends on the location (height) of the observer, the origin of the coordinate system. The velocity of the train relative to a point on the floor it is passing over is equivalent to the notion of the proper velocity of the train, which, as you say, does vary with height.

 

[sweet springs]

Hi.
In Rindler system let us prepare the coordinates of various xy- speeds that share the same z-axis in an instant.
Let us build a pile along the z-axis reaching down to the event horizon plane of black hole. There time is frozen, so the pile does not move.
As time goes, the coordinates depisperse. There is unike coordinate that keeps the pile as z-axis. Z-axes of other coordinates move to other piles that were built in similar way in other places. Thus we can identify unike coordinate in Rindler system in spite of presumed equality of the coordinates.

 

[PeterDonis]

sweet springs, I have no idea what you're trying to say in post #207.

 

[sweet springs]

I draw a picture to explain my idea.

 

[SlowThinker]

Thus we can identify unike coordinate in Rindler system in spite of presumed equality of the coordinates.

Yes, in gravitational field, or in an accelerating rocket, one speed of train is preferred among others.
I think that it means that this:

Mechanics of Billiard on the horizontal flat tables satisfies SR or Lorentz transformation even under the presence of gravity of the Earth.

is not true.