Is Gravity a force, or not a force?

Dale
Mentor
I am with A.T. on this. The only effects of gravity that cannot be transformed away are tidal effects. Other than that, the force of gravity is an inertial force in GR. Inertial forces are related to the Christoffel symbols in a given coordinate system and can always be set to 0 at any event in spacetime and along any geodesic by judicious choice of coordinate system.

The fact that two objects collide is not sufficient to assert a non-inertial interaction force. After all, two non-interacting inertial objects can collide as well. You can even design coordinate systems where they were initially "at rest" wrt each other and accelerated towards each other due to inertial forces.

Kirk: You might find some value in this discussion....at least first and last section [Where are we]...This offers some interesting perspectives on some fundamental theories in physics...broader than gravity, but nevertheless inmsightful.

Unfinished revolution
Introductive chapter of a book on Quantum Gravity, edited by Daniele Oriti,
to appear with Cambridge University Press
Carlo Rovelli

http://arxiv.org/pdf/gr-qc/0604045v2.pdf
excerpt:
....In fact, our present understanding of the physical world at the fundamental level is in a state of
great confusion. The present knowledge of the elementary dynamical laws of physics is given by the
application of QM to fields, namely quantum field theory (QFT), by the particle–physics Standard
Model (SM), and by GR. This set of fundamental theories has obtained an empirical success nearly
unique in the history of science: so far there isn’t any clear evidence of observed phenomena that
clearly escape or contradict this set of theories —or a minor modification of the same, such as a
neutrino mass or a cosmological constant.1 But, the theories in this set are based on badly selfcontradictory
assumptions. In GR the gravitational field is assumed to be a classical deterministic
dynamical field, identified with the (pseudo) riemannian metric of spacetime: but with QM we have
understood that all dynamical fields have quantum properties. The other way around, conventional
QFT relies heavily on global Poincar´e invariance and on the existence of a non–dynamical background
spacetime metric: but with GR we have understood that there is no such non–dynamical background
spacetime metric in nature.
In spite of their empirical success, GR and QM offer a schizophrenic and confused understanding
of the physical world.
Many here will, I suspect, strongly disagree with the above comments...but that's one of the things that makes science especially valuable.

Dale
Mentor
Many here will, I suspect, strongly disagree with the above comments.
I disagree with the use of the word "selfcontradictory". GR does not contradict itself. QFT does not contradict itself. So neither is self-contradictory.

They each contradict the other, not themselves, so I would have said "mutually-contradictory" rather than "self-contradictory". As such, I think that he is over stating it, but overlooking the hyperbole the content of his comments seems valid.

PeterDonis
Mentor
2019 Award
Many here will, I suspect, strongly disagree with the above comments...
Not disagree, exactly, but the comments do leave something out. The argument that GR and QFT are mutually contradictory, as it stands, only has force if one or both of them are supposed to be a "final theory". If both are just approximate "effective theories" valid in limited domains, then there is no issue other than the obvious one of trying to find what lies underneath the effective theories. Rovelli's comments could just as well be interpreted as trying to figure out something about what the underlying theory might be like. (Which is not to say that that's how he meant them to be interpreted.)

pervect
Staff Emeritus
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.

haushofer
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.

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.

pervect
Staff Emeritus
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
()

DrGreg
Gold Member
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?

What about inside the event horizon of a Schwarzschild black hole?
That is scraping the barrel !

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.

PAllen
2019 Award
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.

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 $x^\mu(\tau)$ ?

PAllen
2019 Award
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?

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.

PAllen
2019 Award
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.

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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.