Is Gravity a force, or not a force?

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Discussion Overview

The discussion revolves around the nature of gravity, specifically whether it should be classified as a force or as a result of the curvature of spacetime. Participants explore various theoretical frameworks, including General Relativity and Quantum Field Theory, and how these frameworks interpret gravity's role in the universe.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • Some participants suggest that gravity is not a force but rather a consequence of the curvature of spacetime, as described by General Relativity.
  • Others argue that in Newtonian physics, gravity is indeed considered a force, and this perspective is supported by the idea of a graviton in Quantum Field Theory.
  • A participant mentions the equivalence principle and how it complicates the definition of gravity as a force, proposing that "freely falling" defines zero acceleration instead.
  • Concerns are raised about the reliability of defining "acceleration due to gravity" in scenarios with multiple gravitational sources or in complex cosmological models.
  • There is a discussion about how mass generates curvature in spacetime, which in turn affects the motion of objects, leading to the sensation of gravity as a force.
  • One participant expresses frustration over the apparent conflict between General Relativity and theories that incorporate gravitons, indicating a perceived contradiction between the two frameworks.

Areas of Agreement / Disagreement

Participants express differing views on whether gravity is a force or a result of spacetime curvature, indicating that multiple competing perspectives remain unresolved. There is no consensus on the interpretation of gravity across the discussed theories.

Contextual Notes

Limitations include the dependence on theoretical frameworks, such as General Relativity and Quantum Field Theory, and the challenges in defining gravitational effects in complex scenarios. The discussion highlights the ambiguity in definitions and the implications of different models.

  • #31
A.T. said:
Newton's model : Gravity is an interaction force
Einsteins's model : Gravity is an inertial force

For the difference see: http://en.wikipedia.org/wiki/Fictitious_force
I would say that Newtonian gravity is also an inertial force, i can always go to a frame to make it vanish, just like e.g. the cCoriolis force. Why do you make a distinction?

Edit it is a matter of definition, i see.
 
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  • #32
pervect said:
Mentz114 said:
If a test particle is released from rest ( wrt to the source of the field), it experiences an acceleration that can easily be defined.
I don't think I agree with this. Maybe what you are saying is that if you have a static gravitational field, you can define a co-located static observer, and that you can then define the relative acceleration of the test particle relative to the static observer. That I would agree with, unfortunately it's only a special case and not a general definition. You need some notion to replace the static observer to measure your acceleration relative to, and it's not clear what to replace it with.
Can you give an example where it's not possible to define a static observer ? This is not a challenge, but I'd like to know because I can't think of any that don't involve null coordinates.
 
  • #33
Mentz114 said:
Can you give an example where it's not possible to define a static observer ? This is not a challenge, but I'd like to know because I can't think of any that don't involve null coordinates.

Well, the universe as a whole doesn't have a static observer - the metric doesn't have a timelike killing vector.

A binary star would be another example, again, no timelike killing vector. To demonstrate that the metric is a function of time when you don't have a full GR solution, consider the Newtonian approximation where you have two equal mass stars, and ask if the Newtonian potential U and/or the tidal forces are constant. Consider a test object when the binaries and the test object are all inline

() () x

and another when they aren't

()
x
()
 
  • #34
Mentz114 said:
Can you give an example where it's not possible to define a static observer ? This is not a challenge, but I'd like to know because I can't think of any that don't involve null coordinates.
What about inside the event horizon of a Schwarzschild black hole?
 
  • #35
DrGreg said:
What about inside the event horizon of a Schwarzschild black hole?
That is scraping the barrel !
 
  • #36
pervect said:
Well, the universe as a whole doesn't have a static observer - the metric doesn't have a timelike killing vector.

A binary star would be another example, again, no timelike killing vector. To demonstrate that the metric is a function of time when you don't have a full GR solution, consider the Newtonian approximation where you have two equal mass stars, and ask if the Newtonian potential U and/or the tidal forces are constant.
...
()
That's food for thought. In a purely Newtonian terms there would be a COM or frames in which the bodies are at rest. In a multi-body GR solution I can see things are not so straightforward although we (presumably) could still calculate the proper acceleration of a test-particle worldline.
 
  • #37
Mentz114 said:
That's food for thought. In a purely Newtonian terms there would be a COM or frames in which the bodies are at rest. In a multi-body GR solution I can see things are not so straightforward although we (presumably) could still calculate the proper acceleration of a test-particle worldline.

For a binary system, even in COM frame, the potential is time varying. In Newtonian approximation, it would be strictly periodic, but in GR not. Due to GW, the metric would be time varying and aperiodic.
 
  • #38
PAllen said:
For a binary system, even in COM frame, the potential is time varying. In Newtonian approximation, it would be strictly periodic, but in GR not. Due to GW, the metric would be time varying and aperiodic.
I understand that a multi-body solution in GR would be time-varying. Does that in principle mean we can't define some coordinate system to write a curve x^\mu(\tau) ?
 
  • #39
Mentz114 said:
I understand that a multi-body solution in GR would be time-varying. Does that in principle mean we can't define some coordinate system to write a curve x^\mu(\tau) ?

But how do you distinguish which lines are static?
 
  • #40
PAllen said:
But how do you distinguish which lines are static?
I don't know. I was just asking if I could have worldlines. It's been pointed out to me ( three times ) that if stuff is whizzing around then defining 'static' is problematic and I understand that.
 
  • #41
Mentz114 said:
I don't know. I was just asking if I could have worldlines. It's been pointed out to me ( three times ) that if stuff is whizzing around then defining 'static' is problematic and I understand that.

But how could you not have world lines? Even inside an event horizon you have world lines. They may end in finite proper time, but for universe with a big crunch, all world lines end in finite proper time. So, yes, you can always have world lines with varying proper acceleration profiles. You can, in many cases, (IMO) invent various heuristic criteria to consider some pseudo-static (to invent a term)*. But only in very special spacetimes can you pick out a unique family satisfying a geometric criterion for being static.

*This is an idea I've played around with and discussed a few times on these forums. Even for this, I've so far found it necessary to assume asymptotic flatness and 'well behaved' local geometry. And I haven't achieved complete success formalizing these ideas.
 
Last edited:
  • #42
PAllen said:
But how could you not have world lines? Even inside an event horizon you have world lines. They may end in finite proper time, but for universe with a big crunch, all world lines end in finite proper time. So, yes, you can always have world lines with varying proper acceleration profiles. You can, in many cases, (IMO) invent various heuristic criteria to consider some pseudo-static (to invent a term)*. But only in very special spacetimes can you pick out a unique family satisfying a geometric criterion for being static.

*This is an idea I've played around with and discussed a few times on these forums. Even for this, I've so far found it necessary to assume asymptotic flatness and 'well behaved' local geometry. And I haven't achieved complete success formalizing these ideas.

I was being deliberately obtuse. This is a good answer, thanks. We're probably hijacking this thread in any case.
 

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