General relativity and tidal forces

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SUMMARY

This discussion centers on the relationship between general relativity (GR) and tidal forces, emphasizing that tidal forces can be understood through the geometry of spacetime rather than traditional forces. It is established that tidal forces are components of the Riemann curvature tensor, and under certain conditions, the Newtonian approximation of tidal forces remains valid. The conversation also highlights the complexities of bodies in tidal lock and the implications of different geodesics on structural integrity. Key references include MTW's "Gravitation" and the Bel Decomposition for a deeper understanding of tidal forces in GR.

PREREQUISITES
  • Understanding of general relativity (GR) principles
  • Familiarity with the Riemann curvature tensor
  • Knowledge of tidal locking and its implications
  • Basic grasp of Newtonian gravity and its limitations
NEXT STEPS
  • Study the Bel Decomposition of the Riemann curvature tensor
  • Explore MTW's "Gravitation" for comprehensive insights on tidal forces
  • Investigate the effects of tidal locking on celestial bodies
  • Learn about modern techniques for measuring the gravity tensor, including superconductors and SQUIDs
USEFUL FOR

Astronomers, physicists, and students of general relativity seeking to understand the interplay between tidal forces and spacetime geometry, as well as those interested in the dynamics of celestial bodies in tidal lock.

  • #61
Nugatory said:
The relativistic effects that matter across a light-year (about 1016 meters) are completely irrelevant at the diameter of the moon (less than 107 meters. There are simply no relativistic phenomena involved in understanding the tidal lock of these systems (yes, you CAN solve them using the methods of the GR gravitational theory... but the first step in that solution is to apply the weak-field approximation and that gives you the same results as Newtonian gravity).

Bandersnatch is correct. If no torque is applied to the non-relativistic rotating rod it will be straight (although if has been elastically deformed by an applied torque it will take a moment for the deformation to relax and the rod to regain that undeformed straight shape). The only reason for the outer edge to lag would be drag, atmospheric or otherwise.

I was not talking about relativistic effects, i was talking about deformation due to torque. I think the point of our misunderstanding is that you consider that the torque is not constant, but if the force from gravity is constant, so is the torque. And this force is always greater on the near side. Think of it this way: the law of inertia states the if there are no forces on a body, it will remain in its rest state, either traveling at a constant linear velocity or standing still (which are the same thing). To maintain a stable curved path, like an orbit, the centripetal acceleration must be constant. This constant force generates a constant torque, like in a mery go round. If you don't hold on to something, you will fly off at the tangent. This must mean a constant centripetal force, and if the body is not falling towards the central mass, but orbiting, there is a constant force holding it in its curved path. This constant force is gravity, and it generates a constant torque. Why do you believe the torque ceases?
 
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  • #62
altergnostic said:
I agree, but you are considering the body after it is in tidal lock. The point is that there is a torque for the body to reach tidal lock, and my argument is that the body will reach tidal lock but will suffer shearing.
You're argument is wrong, and has gone on far too long.

I haven't seen tide lock explained by heat the way you state, could give me a reference?
This is old, old stuff, some going back to George Darwin and A.E.H. Love. More recently, but still old stuff,

Goldreich P. and Soter S. (1966). Q in the Solar System. Icarus 5, 375-389.
MacDonald, G.J.F (1964). Tidal Friction, Rev. Geophsys 2, 467-541.

You don't even understand the basics, and you are trying to argue? Thread locked.
 

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