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

[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:
<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
$$ \zeta=\gamma^2\frac{gz'}{c^2}(2+\frac{gz'}{c^2})$$

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.

 

[PeterDonis]

I draw a picture to explain my idea.

Your picture doesn't make it any clearer. Can you describe what you are talking about using math?

 

[sweet springs]

is not true.

In Ribdler metric $ds^2=g_{00}dt^2-dx^2-dy^2-dz^2$ when dz=0 or in other words for motion in xy plane, g_{00} is constant.
Rescaling time $dT=\sqrt{g_{00}}t$, $ds^2=dT^2-dx^2-dy^2,dz=0$. It looks like SR case. Lorentz transformation for xy plane motion is applicable, I assume.

 

[pervect]

Some things I've just noticed:

The line element we get by applying the Lorentz transform to the Rindler metric is the same as the one peter derives in #79, https://www.physicsforums.com/threads/gravity-on-einsteins-train.835994/page-4#post-5256426, with different names for the variabiles, $\chi -> x$ and $\psi ->y$.

For this line element (using the t,x,y form), $g_{yy}$ = 0 when vgx = 1, i.e.when x = 1/gv, the vector (t,x,y) = (0,0,1) = $\partial_y$ is a null vector. Recall that the length of a vector $u^i$ is $g_{ij} u^i u^j$. So $g_{yy} = 0$ at x = 1/vg makes $\partial_y$ null. For highly relativistic values of v, this happens only slightly "above" the origin at x=1/g, i.e if v =.99c and g=1, the origin is at x=1 and $\partial_y$ becomes timelike at x $\approx$ 1.0101.

Something I had noticed before, which may be related to Sweet springs point (?).

$g_{tt}$ goes to 0 at gx=v, making (1,0,0) -> $\partial_t$ a null vector. The way I would describe this in words may or may not be helpful. If we look at the coordinate velocity in rindler coordinates dy/dt, it must be constant to have a rigid congruence, i.e to keep the distance between two points on the block a constant so they don't change distance as time evolves. But at x = v/g, this required coordinate velocity is the same as the coordinate velocity of light at this value of x. So a point moving rigidly with respect to the other points on the block / congruence would need to move at the speed of light, which is not possible for a material object.

So to summarize, for a block slicing at v = .99c, with g=1, the coordinate chart we (either Peter's or the identical one obtained via just using the Lorentz transform) is well-behaved only for .99 < x < 1.0101. For my coordinate chart in #90, we still need x > .99, but because of the different construction there isn't any upper bound on x or y.

 

[pervect]

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.

This statement is rather suspect. Perhaps I'll get more time to write a detailed account of the Christoffel symbols and their significance. An informal summary would be that if you ignore the vertical direction normal to table, there aren't any noticeable effects, but if you do include the vertical direction and measure the weight of the balls, or the behavior of gyroscopes mounted on the table, you DO notice measurable effects, the weight of the balls varies depending on which direction they are moving, and the gyroscopes attached to the table precess. There's an article describing the precession effects, called Thomas precession, much earlier in the thread which I won't requote unless someone is interested enough to ask.

 

[PeterDonis]

The line element we get by applying the Lorentz transform to the Rindler metric is the same as the one peter derives in #79

Yes. That was how I originally derived my line element.

For this line element (using the t,x,y form), $g_{yy} = 0$ when $vgx = 1$, i.e.when $x = 1/gv$, the vector $(t,x,y) = (0,0,1) = \partial_y$ is a null vector. Recall that the length of a vector $u^i$ is $g_{ij} u^i u^j$. So $g_{yy} = 0$ at $x = 1/vg$ makes $\partial_y$ null.

Yes. I commented on this in post #80. Note that there are also two other "threshold" values of $x$ (or $\chi$ in my notation). At $x = 1/g$, the "cross" term $g_{\tau \psi}$ vanishes, so the metric is orthogonal. (This $x$ coordinate corresponds to the floor of the rocket.) And at $x = v / g$, $g_{\tau \tau}$ vanishes; i.e., this is the "Rindler horizon" for the block. (The Rindler horizon is at $x = 0$ for the rocket.)

 

[sweet springs]

At $x = 1/g$, the "cross" term $g_{\tau \psi}$ vanishes, so the metric is orthogonal. (This $x$ coordinate corresponds to the floor of the rocket.) And at $x = v / g$, $g_{\tau \tau}$ vanishes; i.e., this is the "Rindler horizon" for the block. (The Rindler horizon is at $x = 0$ for the rocket.)

x=1/g is a specific value. What kind of observation would tell us the x value of where we are in Rindler system ?

 

[sweet springs]

Your picture doesn't make it any clearer. Can you describe what you are talking about using math?

Let a train in Rindler coordinate be on the event horizon. The train keeps being at rest due to freezed time.
We can identify the train on the rocket floor that share the same x and y coorinate with the train on the event horizon, in which the forward light and the backward light goes down same amount.
The train on the event horizon is used as a corner stone on which we could build up a pile to be z-axis (x,y)=(0,0) or parallel collums (x,y)=(a,b), constants.

No formula but more explanation about my idea.

 

[PeterDonis]

What kind of observation would tell us the x value of where we are in Rindler system ?

If you measure the proper acceleration $a$ of an object at rest in Rindler coordinates, then the $x$ coordinate of that object is $x = 1 / a$ (or $c^2 / a$ in conventional units).

 

[PeterDonis]

Let a train in Rindler coordinate be on the event horizon. The train keeps being at rest due to freezed time.

No, this is not correct. The Rindler horizon is a null surface; it is impossible for anything to be at rest there. The horizon is not actually covered by Rindler coordinates; the transformation from Minkowski coordinates (where the horizon is just the line $T = X$) to Rindler coordinates is singular on the horizon.

 

[Dale]

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