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Passionflower

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Assume we have a non rigid ball.

When this ball is radially free falling in a Schwarzschild metric the height increases while the width decreases due to tidal effects. How do we calculate the ruler width and height in terms of R and m?

When we have two of those balls radially free falling but one is falling at escape velocity (free fall from infinity) and the other is not, are the widths and heights identical or if not how do they relate to each other in terms of the difference in coordinate speeds?

If one of those balls is in a circular orbit with the same side always facing the 'center' of gravitation, does the width and height stay the same and thus can we say this ball is in an inertial frame, or is there precession? And what if this ball is instead in an elliptical orbit?

How in all those cases how does the radar distance relate to the ruler distance?

Let's first assume:

[tex]

d\theta = 0, \, d\phi = 0

[/tex]

[tex]

r_s = 2m

[/tex]

I know the radar distance of the length in coordinate time is:

[tex]

\Delta t = 2 \left(r_2 - r_1 + r_s \ln \left[ \frac{r_s + r_2}{r_1 - r_s}\right]\right)

[/tex]

The conversion factor from coordinate time to proper time is:

[tex]

\sqrt{1-r_s/r}

[/tex]

So clearly the radar distance depends on the location of the observer on the ball.

Now is the ruler distance this?

[tex]

\Delta h = \sqrt{r_1 (r_1 - r_0)} - \sqrt{r_2(r_2-r_0)} + r_0 \ln { {\sqrt{r_1} + \sqrt{\sqrt{r_1} - r_0} \over {\sqrt{r_2} + \sqrt{\sqrt{r_2} - {r_0}}}

[/tex]

Is that right?

So now how does this relate, if at all to the tidal effect? Suppose we have a ball of height 1 with r

But that is just the height, I do not know how to get the widths.

Now do those formulas work regardless of the coordinate speed of the ball or do they only apply when the ball travels at escape velocity?

When this ball is radially free falling in a Schwarzschild metric the height increases while the width decreases due to tidal effects. How do we calculate the ruler width and height in terms of R and m?

When we have two of those balls radially free falling but one is falling at escape velocity (free fall from infinity) and the other is not, are the widths and heights identical or if not how do they relate to each other in terms of the difference in coordinate speeds?

If one of those balls is in a circular orbit with the same side always facing the 'center' of gravitation, does the width and height stay the same and thus can we say this ball is in an inertial frame, or is there precession? And what if this ball is instead in an elliptical orbit?

How in all those cases how does the radar distance relate to the ruler distance?

Let's first assume:

[tex]

d\theta = 0, \, d\phi = 0

[/tex]

[tex]

r_s = 2m

[/tex]

I know the radar distance of the length in coordinate time is:

[tex]

\Delta t = 2 \left(r_2 - r_1 + r_s \ln \left[ \frac{r_s + r_2}{r_1 - r_s}\right]\right)

[/tex]

The conversion factor from coordinate time to proper time is:

[tex]

\sqrt{1-r_s/r}

[/tex]

So clearly the radar distance depends on the location of the observer on the ball.

Now is the ruler distance this?

[tex]

\Delta h = \sqrt{r_1 (r_1 - r_0)} - \sqrt{r_2(r_2-r_0)} + r_0 \ln { {\sqrt{r_1} + \sqrt{\sqrt{r_1} - r_0} \over {\sqrt{r_2} + \sqrt{\sqrt{r_2} - {r_0}}}

[/tex]

Is that right?

So now how does this relate, if at all to the tidal effect? Suppose we have a ball of height 1 with r

_{1}and r_{2}near infinity, then at a given R how much is h of the ball?But that is just the height, I do not know how to get the widths.

Now do those formulas work regardless of the coordinate speed of the ball or do they only apply when the ball travels at escape velocity?

*Please no approximations I like to get the exact numbers*
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