Long planet and Galileo thought experiment

[DukeL]

One has to have a reasonably long horizontal separation in order to have any measurable drop and especially a measurable difference in drop time. Then spatial curvature bedevils the test - i.e., it is no longer a pure SR problem and the equivalence principle only applies approximately. Only in the OP's fictional "very long, flat Earth" (essentially infinitely large) can the EP be applied directly in this type of test.

In your calculating post above, you said that when you increased the size of the black hole without limit, the acceleration with a 0.8c horizontal velocity remained about 2.3 times the 'straight drop' acceleration. This does not correlate with your "essentially infinitely large" statement above, not so?

I think part of the problem is that your calculation in a Cartesian frame does not work in the Schwarzschild spacetime around a practical black hole of any size.

 

[DukeL]

During the course of this thread I have come to the conclusion that the equivalence principle is a "one way relationship". While an observer in a closed artificially accelerated lab could not be certain that he is not experiencing acceleration due to the proximity of a massive gravitational body, the reverse is not always true.

Hi kev, I guess if a lab is really small and the massive body is really big, neither observer would be able to say for sure which is which. If the hole is arbitrarily big, doesn't the tidal forces at the horizon become arbitrarily small in a small lab?

 

[Jorrie]

In your calculating post above, you said that when you increased the size of the black hole without limit, the acceleration with a 0.8c horizontal velocity remained about 2.3 times the 'straight drop' acceleration. This does not correlate with your "essentially infinitely large" statement above, not so?

Hi DukeL.

Yep, but I think the problem is that one cannot practically "increase the size of the black hole without limit" in physical calculations (they blow up). I suppose one can do it in the formulas, which will then give you a uniform gravitational field and the equivalence principle applies.

I think part of the problem is that your calculation in a Cartesian frame does not work in the Schwarzschild spacetime around a practical black hole of any size.

Agreed, but I have checked this roughly and the error introduced is about 4% in the time difference, relatively small compared to the 33% time difference between the two different 'drops'.

I'm pretty sure (but haven't checked) that if you do a full relativistic orbit calculation at r = 1.0001Rs, with and without d\phi/d\tau = 0.8c/r angular starting velocity, you will roughly get the same drop time difference (say for a decrease in r of one meter).

-J

 

[DukeL]

I'm pretty sure (but haven't checked) that if you do a full relativistic orbit calculation at r = 1.0001Rs, with and without d\phi/d\tau = 0.8c/r angular starting velocity, you will roughly get the same drop time difference (say for a decrease in r of one meter).

Jorrie, I've tried to get my head around this by taking pervect's original derivation, simplified for zero radial velocity, i.e.

<br /> <br /> \frac{d^2 r}{dt^2} = \left( r-2\,m \right) \left(v_{\phi}^{2}-{\frac {m}{{r}^{3}}} \right) <br /> <br />

Looking at it, the angular velocity v_{\phi} of the object can only have any significant influence if v_{\phi}^2 it is in the same order of mag. as m/r^3, because they are subtracted.

Using your 15 trillion suns black hole at radial range r=1.0001Rs and v_{\phi}= 0.8/(r-Rs), I get the following values:

v_{\phi}^2\approx 10^{-38} and m/r^3 \approx 10^{-34} \gg v_{\phi}^2 (in geometric units meter^{-2}).

If I haven't made any calculation errors, this seems to rule out any significant influence of the angular velocity on the radial acceleration in your example, casting doubt on your calculations.

 

[yuiop]

In the example of the accelerating rocket experiencing constant proper born rigid acceleration as plotted on a Minkowski diagram, the acceleration experienced by the observers is proportional to 1/R. Normally gravitaitional acceleration is proportional to 1/R^2. I have recently discovered that the gravity perpendicular to a an infinitely long cylnder attenuates proportional to 1/R. This is an interesting coincidence and adds to the possibility of the spacetime of the cylindrical gravitational body being an exact equivalent of the accelerating rocket. It would be interesting if someone could figure out the gravitational potential of the cylinder and what implications that has to adapting the Schwarzschild metric from the spherical to the cylindrical case.

 

[Jorrie]

I have recently discovered that the gravity perpendicular to a an infinitely long cylinder attenuates proportional to 1/R. This is an interesting coincidence and adds to the possibility of the spacetime of the cylindrical gravitational body being an exact equivalent of the accelerating rocket.

I guess this is correct for the spatial dimension orthogonal to the acceleration vector and parallel to the cylinder's long axis, but not in the other orthogonal dimension (orthogonal also to the cylinder's axis), where the spatial curvature remains. In the rocket's case, there is no curvature in either of the two orthogonal directions.

-J

 

[yuiop]

I guess this is correct for the spatial dimension orthogonal to the acceleration vector and parallel to the cylinder's long axis, but not in the other orthogonal dimension (orthogonal also to the cylinder's axis), where the spatial curvature remains. In the rocket's case, there is no curvature in either of the two orthogonal directions.

-J

I see your point and this presents the equivalence principle with a problem because there appears to be no (even hypothetical) gravitational body that can duplicate all the measurements made inside the cabin of the accelerating spaceship.

The gravity orthogonal to an infinite flat slab does not diminish at any distance from the slab so we can not use that hypothetical gravitational body either.

How do we save the equivalence principle?

 

[Jorrie]

The gravity orthogonal to an infinite flat slab does not diminish at any distance from the slab so we can not use that hypothetical gravitational body either.

How do we save the equivalence principle?

I guess a very large spherical body does approximate the uniform gravitational field that corresponds to the equivalence principle as far as tidal effects are concerned. I suppose even the inverse square law would be very difficult to observe in a small lab.

I do not think there is a body representing a pure uniform gravitational field - it does not even help to make it spherically 'infinite', because then the gravitational field will be the same everywhere, like for the 'infinite' square slab - actually, you can't be outside an infinite 3D thing, so there might be no gravitational effects, not so?

-J

 

[Jorrie]

If I haven't made any calculation errors, this seems to rule out any significant influence of the angular velocity on the radial acceleration in your example, casting doubt on your calculations.

Hi DukeL.

It does not look like you made any computational errors, but I'm still battling to convert what you wrote (for the Schw. coordinate frame) to the local frame. Will come back to you later...

-J

 

[Jorrie]

Hi again DukeL.

It does not look like you made any computational errors, but I'm still battling to convert what you wrote (for the Schw. coordinate frame) to the local frame. Will come back to you later...

It appears that is not necessary to worry about the conversion between Schwarzschild and local coordinates in the solution, after all. The issue lies in the centripetal acceleration inherent in polar coordinates, e.g., one cannot say that an object moving with d\phi/dt &gt; 0, but with dr/dt=0 has zero acceleration. One can quite legally replace the polar coordinates with rectangular x,y,z coordinates and see that, as I indicated here:

To convert this to a Cartesian coordinate acceleration, we firstly have to add the negative centripetal acceleration rv_\phi^2 (because it is inherent in spherical coordinates): (Note should have read -rv_\phi^2)

Eq. 3
<br /> \frac{d^2 r}{dt^2} = \left( r-2\,m \right) \left(v_{\phi}^{2}-{\frac {m}{{r}^{3}}} \right) - r^2v_\phi^2 = -\frac {m}{r^2}\,-\, 2 mv_\phi^2\,+\, \frac {2m^2}{r^3} <br />(Note, the r^2v_\phi^2 should have read rv_\phi^2)

with a final, most useful form:

Eq. 4
<br /> \frac{d^2 r}{dt^2} = -\frac {m}{r^2}\left(1 - \frac {2m}{r} + 2r^2v_\phi^2\right) <br />

Strictly, one should also replace r with y and rv_{\phi} with v_x=dx/dt, the horizontal velocity, but that this is not necessary to see the effect that we are after.

Eq. 4 should hence be written as:

<br /> \frac{d^2 y}{dt^2} = -\frac {m}{y^2}\left(1 - \frac {2m}{y} + 2v_x^2\right) <br />

I found the following acceleration values for the case that you calculated for the distant observer's frame:

\frac{d^2 r}{dt^2}= -1.13\times 10^{-21}

-rv_\phi^2 = -1.45\times 10^{-21}

\frac{d^2 y}{dt^2} = -2.58\times 10^{-21}

All in geometric units of meter^{-1}, indicating the source and magnitude of the increase in acceleration in the presence of angular velocity. Does this make any sense?

-J

 

[pervect]

Rather than attempt to go through this long thread, I'd like to suggest a better model of a "flat planet" - an accelerating spaceship. In particular, I mean the metric I describe
here.

It should be reasonably clear that the proper time of an object dropped from some height in this metric is constant, but that the coordinate time it takes to fall varies, just from considerations of symmetry.

 

[Jorrie]

Rather than attempt to go through this long thread, I'd like to suggest a better model of a "flat planet" - an accelerating spaceship. In particular, I mean the metric I describe
here.

It should be reasonably clear that the proper time of an object dropped from some height in this metric is constant, but that the coordinate time it takes to fall varies, just from considerations of symmetry.

Thanks for the link. I think the thread has concluded that Galileo was right for a 'flat planet' or uniformly accelerated lab, but there is a nagging question about what happens very near a supermassive black hole's horizon with a 0.8c horizontal velocity ball. Does it drop faster then a ball released from rest?

I've done some https://www.physicsforums.com/showpost.php?p=1689919&postcount=70" based on your earlier derivation of the radial and angular accelerations in a Schwarzschild orbit and came to the (perhaps erroneous) conclusion that the horizontally moving ball 'falls' faster in a local Cartesian frame.

-J