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

[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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