0
$\begingroup$

For planetary motion I can understand that the planets move along the geodesics e.g. the warped space-time geometry. Imagine that the moon gets suddenly stopped by some external force and comes to rest momentarily, which force will bring it to motion towards earth. And how can we describe "Normal force" as mentioned in the subject.

Please excuse me for my ignorance in advance but an excessively mathematical explanation would be very difficult for me to understand.

$\endgroup$
4
  • $\begingroup$ I'm not sure what you're asking here. It seems you're worried about the normal force, the one that impedes one object to enter another, and in the gravitational case maintains a person standing on the Earth. The normal force is just a by-product of quantum mechanical interactions and has nothing to do with general relativity. If you're concerned with how to describe near-earth gravity (as a constant force) in general relativity then you can do no better than Feynman's lecture 42 of volume 2, freely available here feynmanlectures.caltech.edu/II_42.html $\endgroup$ Jun 6, 2014 at 19:00
  • $\begingroup$ @cesaruliana, You've got me right, I can try to clarify my doubt again - "according to Einstein gravity is not a force but an illusion e.g. as moon moves around earth in a geodesic formed by warping of spacetime by mass of earth but not by the so called illusionary gravitational pull. So exactly how can we describe the normal force as I mentioned?", thanks for the link I will go through it. $\endgroup$
    – Yogesh
    Jun 12, 2014 at 7:56
  • $\begingroup$ I was very much looking for the non Newtonian explanation of normal force which is in conformance with the Einstein's ideas about gravity. $\endgroup$
    – Yogesh
    Jun 12, 2014 at 8:05
  • $\begingroup$ @ Yogesh, the normal force is non-gravitational and therefore does not have a geometrical description in general relativity, in the same way as electromagnetism does not. In fact from the point of view of GR a person standing on the surface of Earth is not in equilibrium with respect to forces, there is only the normal force pointing against the ground. What happens is that in GR free-fall is inertial behavior, therefore in order to stand still one must be in non-inertial motion, namely subject to net forces, in this case the normal. Is that what you're asking? $\endgroup$ Jun 12, 2014 at 15:32

1 Answer 1

1
$\begingroup$

There is a somewhat mathematical description in twistor59's answer to What is the weight equation through general relativity?. I don't think you're going to get any real idea of what is going on without some maths.

An alternative way of looking at the question is the river model - see my answer to If you shoot a light beam behind the event horizon of a black hole, what happens to the light? for a discussion of this. In this model spacetime is flowing inwards towards the planet/black hole/whatever and freely falling observs are carried along with the flow just as you would be carried along by the water if you were floating in the river. Someone on Earth's surface is standing still against the flow, and there therefore feels a force just as (though the analogy is getting very loose here!) you feel the force of the water when standing still in a flowing river.

Be cautious about analogies like this though. Spacetime isn't a material like water and doesn't flow. It's the coordinate system that's flowing.

$\endgroup$
1
  • $\begingroup$ Thanks John, I will try to digest these pointers first and then get back. unfortunately I cannot up vote you as I have joined this forum just now. $\endgroup$
    – Yogesh
    Jun 6, 2014 at 9:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.