Do Fictitious Forces Still Matter in General Relativity?

[zoltrix]

hello

it is well known that gravitationl force is actually a fictitious force
generally speaking,are fictitious forces still necessary in general relativity ?
the fictitious forces which we experience on a bus or on a car can also be understood as due to the spacetime distortion ?

 

[PeterDonis]

it is well known that gravitationl force is actually a fictitious force

Sort of. See below.

generally speaking,are fictitious forces still necessary in general relativity ?

Fictitious forces are never necessary; they are artifacts of your choice of coordinate system, and you can always make them vanish by choosing an appropriate coordinate system. That is true in both Newtonian mechanics and GR.

The difference between Newtonian mechanics and GR is that GR recognizes that what is called the "force of gravity" in Newtonian physics, or at least its first order component (what we usually think of as "acceleration due to gravity"), can also be made to vanish by choosing an appropriate coordinate system, so it behaves just like fictitious forces in that respect.

the fictitious forces which we experience on a bus or on a car can also be understood as due to the spacetime distortion ?

No. As above, fictitious forces are artifacts of your choice of coordinates. But spacetime curvature is not; it is an invariant, represented by the Riemann curvature tensor, and physically it corresponds to tidal gravity--the (second order) part of gravity that cannot be made to vanish by choosing an appropriate coordinate system.

 

[Dale]

the fictitious forces which we experience on a bus or on a car can also be understood as due to the spacetime distortion ?

Fictitious forces have nothing to do with spacetime curvature. The fictitious forces arise from the coordinates we choose, so they reflect our arbitrary choice of coordinates, not some measurable feature of the world. The spacetime curvature, in contrast, is a measurable feature of the world. It exists regardless of the coordinates we choose.

Mathematically, the fictitious forces are contained in the Christoffel symbols. These depend on the coordinate system and can always be made to go to zero at any given event.

 

[vanhees71]

Imho gravity is not simply a "fictious force" (or perferrably I'd call them "inertial forces" in a non-inertial reference frame) but it's a true interaction with the special property of all the fundamental interactions that the equivalence principle applies to it. If it were a fictitious force your spacetime were flat Minkowski space, where you always can introduce a global inertial frame without any inertial forces present you'd may misinterpret as gravitational interaction.

If there is a real gravitational field spacetime is not flat anymore and there are only local inertial frames, i.e., you can transform away some inertial part of the gravitational field at any point of spacetime (the equivalence principle in the form leading to standard GR) such that the Christoffel symbols vanish at this one point but you still have "tidal fields/forces", which you cannot "gauge away", because a non-zero curvature tensor has non-zero components wrt. any frame, including also any local inertial frame.

 

[cianfa72]

If there is a real gravitational field spacetime is not flat anymore and there are only local inertial frames, i.e., you can transform away some inertial part of the gravitational field at any point of spacetime (the equivalence principle in the form leading to standard GR) such that the Christoffel symbols vanish at this one point but you still have "tidal fields/forces", which you cannot "gauge away", because a non-zero curvature tensor has non-zero components wrt. any frame, including also any local inertial frame.

In other words, in a real gravitational field (i.e. curved spacetime) for each point (event) you can choose a local coordinate chart such that Christoffel symbols vanish only at that given point. In that coordinate chart Christoffel symbols do not not vanish at nearby points (events).

So tidal forces/geodesic deviation manifests itself as the impossibility to bring null Christoffel symbols in a finite region of spacetime around that point.

 

[Dale]

you can choose a local coordinate chart such that Christoffel symbols vanish only at that given point. In that coordinate chart Christoffel symbols do not not vanish at nearby points (events).

Yes, and not only that but the Riemann curvature tensor does not vanish even at that given event.

 

[Dale]

Imho gravity is not simply a "fictious force" (or perferrably I'd call them "inertial forces" in a non-inertial reference frame) but it's a true interaction with the special property of all the fundamental interactions that the equivalence principle applies to it.

I agree with this. The force of gravity, $mg$, is an inertial force (I also prefer that term), but there is more to the gravitational interaction than just the force. The curvature of spacetime is a physical thing that cannot be transformed away and includes tidal gravity and all of the measurable gravitational phenomena.

 

[vanhees71]

I'd be careful with the word "force" because of its Newtonian conotation of "action at a distance". In relativistic physics you rather have interactions, which are described by local field theories (except alternative formulations like Feynman-Wheeler absorber theory, but that's for sure not mainstream physics, and I don't know, whether anybody still works along that line of thought).

 

[Dale]

I'd be careful with the word "force" because of its Newtonian conotation of "action at a distance".

Good point. Of course, the Christoffel symbols and the associated $mg$ inertial forces do exhibit non-physical action at a distance. So the word “force” fits that part well.

 

[PeterDonis]

Imho gravity is not simply a "fictious force" (or perferrably I'd call them "inertial forces" in a non-inertial reference frame) but it's a true interaction

Discussion of this really belongs in either the quantum physics forum, or more likely the Beyond the Standard Model forum, since any theory of gravity as such an interaction goes beyond our current Standard Model of particle physics. This is the relativity forum, so the GR viewpoint is the one that is on topic here.

I would note, though, that even on the "gravity as an interaction like the other Standard Model interactions" viewpoint, "fictitious forces" are still the same as the "acceleration due to gravity"; you can still make all of them vanish by a suitable choice of coordinates. The "gravity as an interaction" viewpoint still only comes into play when you look at spacetime curvature and view it as an emergent property of the underlying interaction. Once you do that, the "gravity interaction" you are talking about is no longer the Newtonian "force of gravity".

If it were a fictitious force your spacetime were flat Minkowski space

I disagree with this claim. Fictitious forces can appear in a curved spacetime in suitably chosen local coordinates on a local patch of spacetime that can be treated as flat. There is no need for spacetime to be globally flat.

 

[Dale]

Discussion of this really belongs in either the quantum physics forum, or more likely the Beyond the Standard Model forum, since any theory of gravity as such an interaction goes beyond our current Standard Model of particle physics.

I don’t think that we need to restrict the word “interaction” to refer exclusively to quantum mechanics.

 

[PeterDonis]

I don’t think that we need to restrict the word “interaction” to refer exclusively to quantum mechanics.

In this particular case, I think we have to, because in GR, gravity is simply not an "interaction". There is a link between spacetime geometry and stress-energy, but that's not an interaction between different pieces of matter.

 

[vanhees71]

Discussion of this really belongs in either the quantum physics forum, or more likely the Beyond the Standard Model forum, since any theory of gravity as such an interaction goes beyond our current Standard Model of particle physics. This is the relativity forum, so the GR viewpoint is the one that is on topic here.

I'm not talking about QT or QFT here, just classical theory of the gravitational interaction.

I would note, though, that even on the "gravity as an interaction like the other Standard Model interactions" viewpoint, "fictitious forces" are still the same as the "acceleration due to gravity"; you can still make all of them vanish by a suitable choice of coordinates. The "gravity as an interaction" viewpoint still only comes into play when you look at spacetime curvature and view it as an emergent property of the underlying interaction. Once you do that, the "gravity interaction" you are talking about is no longer the Newtonian "force of gravity".

In both the "physical" and the "geometrical" approach you end up with the same theory, GR or, if you include, spinor fields, Einstein-Cartan theory. I think it's good to have both points of view at hand.

From both the geometric as well as the interaction point of view the gravitational interaction is not only just inertial forces in accelerating frames, as detailed above.

I disagree with this claim. Fictitious forces can appear in a curved spacetime in suitably chosen local coordinates on a local patch of spacetime that can be treated as flat. There is no need for spacetime to be globally flat.

This I don't understand, because the curvature tensor is a tensor and as such zero (flat spacetime) or not, independent of the chosen coordinates.

 

[PeterDonis]

I'm not talking about QT or QFT here, just classical theory of the gravitational interaction.

And my point is that the classical theory you refer to, namely GR, is not a theory of "the gravitational interaction". It is a theory of spacetime curvature and its relationship to stress-energy. There is no "gravitational interaction" in GR of the kind you are talking about.

the curvature tensor is a tensor

Yes, but the curvature tensor does not describe any "gravitational interaction". It describes the curvature of spacetime, i.e., geodesic deviation.

 

[PeterDonis]

In both the "physical" and the "geometrical" approach you end up with the same theory, GR

The "physical" approach you refer to here, I assume, is the classical theory of a massless spin-2 field. This theory is fine as an approximation (for example, using the linearized theory to study gravitational waves), but not as a justification for any fundamental claims about a classical "gravitational interaction". (The quantum version of this theory, considered as an "effective field theory", might provide a justification for claims about a quantum gravitational interaction, but we've agreed that's not what we're talking about in this thread.) The massless spin-2 field approach assumes a flat background spacetime; this background spacetime is unobservable, which is one problem, but a more serious problem is that this assumption does not work for solutions, like FRW spacetime, that do not have the same conformal structure as flat Minkowski spacetime.

 

[vanhees71]

This is all semantics, and I don't think we have to argue about it further. The point is whether you consider GR as a theory of a gravitational interaction, which is reinterpreted in terms of spacetime geometry or the other way around doesn't make any difference. You end up with the same classical field theory, i.e., GR.

It's the same with classical electrodynamics. You can start as usual with the physical phenomena, i.e., a description of the interaction between electrically charged particles, coming to the conclusion that it is an Abelian gauge theory (geometry) or you start with geometry, i.e., the description as a gauge model, leading to the same description of the physical phenomena. It's just different approaches leading to Maxwell's classical electrodynamics.

It's always good to have many approaches to the understanding of a theory!

 

[cianfa72]

The force of gravity, $mg$, is an inertial force (I also prefer that term), but there is more to the gravitational interaction than just the force. The curvature of spacetime is a physical thing that cannot be transformed away and includes tidal gravity and all of the measurable gravitational phenomena.

I take it as the definition of inertial force: namely it can be always transformed away even in a local patch of curved spacetime choosing a suitable local coordinate chart.

 

[vanhees71]

There's no necessity to introduce fictitious forces in GR, because it's formulated fully covariantly, i.e., there are no more distinguished frames of reference, i.e., you just write down the manifestly covariant equations of motion for a point particle
$$m (\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma})=F^{\mu},$$
where $F^{\mu}$ are all (non-gravitational) "forces", e.g.,
$$F^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu}$$
for the force on a charge $q$ due to an electromagnetic field $F^{\mu \nu}$. Here $x^{\mu}$ are arbitrary coordinates, ${\Gamma^{\mu}}_{\rho \sigma}$ the Christoffel symbols, and the world line is parametrized with an arbitrary affine parameter. For massive particles you can choose its proper time as this parameter by using the constraint equation
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=c^2.$$
The gravitational interaction due to the presence of an gravitational field or inertial forces due to the fact that the coordinates describe a non-inertial reference frame, is already included in the equation of motion (via the Christoffel symbols, i.e., the connection of the pseudo-Riemannian spacetime manifold).

 

[Dale]

There's no necessity to introduce fictitious forces in GR, because it's formulated fully covariantly, i.e., there are no more distinguished frames of reference, i.e., you just write down the manifestly covariant equations of motion for a point particle
$$m (\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma})=F^{\mu},$$
where $F^{\mu}$ are all (non-gravitational) "forces", e.g.,
$$F^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu}$$
for the force on a charge $q$ due to an electromagnetic field $F^{\mu \nu}$. Here $x^{\mu}$ are arbitrary coordinates, ${\Gamma^{\mu}}_{\rho \sigma}$ the Christoffel symbols, and the world line is parametrized with an arbitrary affine parameter.

The Christoffel symbols include the fictitious forces.

 

[vanhees71]

That's what I said in the last sentence of the quoted posting ;-).

 

[cianfa72]

you just write down the manifestly covariant equations of motion for a point particle
$$m (\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma})=F^{\mu},$$
where $F^{\mu}$ are all (non-gravitational) "forces", e.g.,
$$F^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu}$$ for the force on a charge $q$ due to an electromagnetic field $F^{\mu \nu}$. Here $x^{\mu}$ are arbitrary coordinates, ${\Gamma^{\mu}}_{\rho \sigma}$ the Christoffel symbols, and the world line is parametrized with an arbitrary affine parameter. For massive particles you can choose its proper time as this parameter by using the constraint equation
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=c^2.$$

As far as I can tell the parameter employed enters in the equation (e.g. proper time for massive particles) as the parameter w.r.t. the derivatives are taken, right ?

 

[vanhees71]

Yes, the dot means the derivative with respect to this affine parameter.