Force experienced on a curved geodesic path

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Main Question or Discussion Point

Can a person inside a spaceship falling freely on a geodesic path, experience the same just like a person inside a car experience a force on a turn on earth i.e when the geodesic path is no more straight near a huge planet.

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Answers and Replies

  • #2
Orodruin
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No. By definition the geodesic has zero proper acceleration.
 
  • #3
Ibix
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One minor caveat to Orodruin's response - if the spaceship is large enough you will experience tidal forces. This would leave you pressed up against the side of the ship furthest or nearest the mass. Or, more uncomfortably, with one end of your body pressed against one side and the other end against the other.

This will only be a significant effect for a planet-sized ship (that's why we get tides on Earth) or very close to a small black hole.
 
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One minor caveat to Orodruin's response - if the spaceship is large enough you will experience tidal forces. This would leave you pressed up against the side of the ship furthest or nearest the mass. Or, more uncomfortably, with one end of your body pressed against one side and the other end against the other.

This will only be a significant effect for a planet-sized ship (that's why we get tides on Earth) or very close to a small black hole.
Little confused, this will only happen near a huge planet if the spaceship is large, because you mention tides on Earth due to moon and close to a black hole.

[edit] If so what is the role of a large ship in which one can experience a force on a turn due to spacetime curvature.
 
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  • #5
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If so what is the role of a large ship in which one can experience a force on a turn due to spacetime curvature.
In a curved spacetime, nearby geodesics are not quite exactly parallel ("geodesic deviation"), so nearby masses in free fall will want to move apart or be forced together. If the two masses are the opposite ends of some object, that object will experience crushing or stretching forces. The larger the volume of space we're considering, the greater the effect so it will be more pronounced inside a very large ship than a very small one.

Except under very extreme conditions tides are most easily analyzed using Newtonian gravity: use Newton's law to compute the magnitude (slightly different if ##r## is slightly different) and the direction (slightly different for any two points not on the same radius) of the force vectors on two nearby masses in the gravitational field of a planet. The GR model of tidal forces as geodesic deviation is a lot easier to follow after you've worked through the Newtonian equivalent.
 
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Ibix
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Little confused, this will only happen near a huge planet if the spaceship is large, because you mention tides on Earth due to moon and close to a black hole.

[edit] If so what is the role of a large ship in which one can experience a force on a turn due to spacetime curvature.
You always get tidal forces. In Newtonian terms, this is because the parts of the ship nearer the gravitating mass feel a stronger force than those further away. In GR terms, the reasoning is a bit different (geodesic deviation) but the effect is the same, at least qualitatively.

However, the effect is utterly negligible unless you are really, really large (the size of a planet) or in a gravitational field that changes a lot over a short distance (near a small black hole). Whether you are turning or not isn't relevant.
 

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