# Forces in Gr

1. Jul 26, 2012

### pervect

Staff Emeritus
To avoid hijacking an existing thread, I wanted to start a new one on how "gravitational forces" are represented in GR.

There doesn't seem to be a lot on this in the intro textbooks, alas, which mostly deal with the issue by avoiding it. Which suggests there could be some non-obvious problems, or at least grounds for long arguments.

But, forging ahead, nonetheless, I would like to propose an idea, which seems reasonable to me, to see if it will stay afloat, and perhaps the discussion (if we get one) will clarify things a bit.

The basic idea is that gravitational forces in GR are represented by Chrsitoffel symbols - or rather a subset of them.

This does represent the forces one experiences in an accelerating elevator as 'real' forces - but this is exactly the goal we want to achieve, I think, from the principle of equivalence.

To sketch the mathematical, if we consider the reference frame of an accelerating observer in flat space time, ala MTW we find that the metric is

ds^2 = (1 + 2g x) dt^2 - dx^2 - dy^2 - dz^2

see for instance MTW, pg 331, or, with some sign differences, http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3A0901.4465 [Broken], above (20), where we've simplified things by assuming no rotation.

and we can identify $\Gamma^{x}{}_{tt} = -g$ from the usual expression

$$\Gamma^{c}{}_{ab} = \frac{1}{2} g^{cd} \left( \partial_{a} g_{bd} + \partial_{b} g_{ad} - \partial_{d} g_{ab} \right)$$

Curved space-time introduces second order corrections to the metric, which won't affect the values of the Christoffel symbols.

So the basic idea is that "gravitational forces" are represented by $\Gamma^{x}{}_{tt}, \Gamma^{y}{}_{tt}. \Gamma^{z}{}_{tt}$, three of the Christoffel symbols.

We can also note, ala MTW pg 330, that the Fermi-Walker transport law _requires_ that $\Gamma^{\hat{x}}{}_{\hat{t}\hat{t}} = \Gamma^{\hat{t}}{}_{\hat{x}\hat{t}} = a^j$ along the worldline of the accelerating observer, where $a^j$ is the 4-acceleartion of the observer.

So, basically, Christoffel symbols carry information about the acceleration and the rotation of the worldline, and no information about the curvature of the space-time (that information is in second order terms in the metric, and comes from the Riemann tensor).

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2. Jul 26, 2012

### PAllen

- Even for a given body, undergoing some general motion, the Christoffel symbols have an arbitrary element due to coordinate choice. To minimize this, perhaps you want to specify a particular type coordinates for this goal of force definition, that are built by a standard recipe from the body's motion (potentially including rotation).

- Despite the equivalence principle, and the corresponding fact that all non-tidal 'gravitational forces' in GR are inertial forces, many people still want to distinguish the inertial force felt in a turning car from the inertial force felt standing on the ground. Your definition won't make them happy. Addressing this goes in two related directions: trying to define a class of 'effectively stationary observers' or trying to define a special class of coordinate systems for making this distinction (Christoffel symbols in the 'right coordinates' are gravitational force).

3. Jul 27, 2012

### Staff: Mentor

I'm not sure this will work as you state it, because the "right" choice of coordinate system will depend on the spacetime, not the body's motion. That's certainly true in the "canonical" case of falling bodies near the Earth; the "right" coordinates in which the Christoffel symbols give the gravitational "force" are static coordinates (i.e., Schwarzschild coordinates) centered on the Earth, and those are determined by the timelike Killing vector of the spacetime, not the motion of the body.

4. Jul 27, 2012

### Mentz114

This seems sensible conclusion from that metric ( which is not a solution of the EFE).

Last edited: Jul 27, 2012
5. Jul 27, 2012

### PAllen

That's a point if you want to define a global sense of 'force of gravity'. That is what I was describing in my second bullet. In the first, I was taking a local point of view that would apply to even the 'frame' of a spinning top in rocket far from any mass. The idea was you build a possibly rotating Fermi-Normal frame from the object's motion. Then, SR or GR, Christoffel symbols in this frame describe local inertial forces. My point in the first bullet was that if you don't specify some recipe like 'a rotating Fermi frame', Pervect's concept seems ill defined (to me).

6. Jul 27, 2012

### Staff: Mentor

I think "Fermi normal coordinates" only apply along a geodesic (and in such coordinates along a geodesic, all the Christoffel symbols are zero). I think the more general method would be to compute the frame field along the object's worldline, and then equate the proper acceleration vector derived from the frame field to the "acceleration due to gravity" (more precisely to minus the "acceleration due to gravity"). In local coordinates matched to the frame field, this would correspond to evaluating the Christoffel symbols that pervect is using ($\Gamma^{i}_{00}$ for each spatial index $i$).

However, I'm still not sure this would end up matching people's intuitive sense of what "gravitational force" is except in special cases where the object was moving along a particular type of worldline (such as an orbit of a timelike Killing vector field).

7. Jul 27, 2012

### Staff: Mentor

Yes, it is. It's just the Minkowski metric in a different coordinate chart (the Rindler chart), at least if I'm remembering that portion of MTW correctly (I don't have it handy right now to check).

8. Jul 27, 2012

### Staff: Mentor

Thinking it over some more, I'm not sure this makes sense globally. The Christoffel symbol is indeed as given here, but it only corresponds to the observed "acceleration due to gravity" along one particular observer's worldline--the observer whose proper time is equal to the coordinate time $t$ in this chart. Other observers at rest in this chart (the other Rindler observers) have different proper accelerations and so will observe a different "acceleration due to gravity", but the Christoffel symbols in the global chart will be the same.

PAllen's suggestion of using local coordinates (or the local frame field, as I posted) would work because different Rindler observers will have different frame fields, and hence different local coordinates centered on their worldlines. (I see that you refer to this at the end of the OP, but there you use the "hatted" Christoffel symbols, which are *not* the same as the global ones you use earlier in the OP.)

9. Jul 27, 2012

### PAllen

This is not what I learned from MTW. The following online source agrees with me:

http://relativity.livingreviews.org/open?pubNo=lrr-2004-6&amp;page=articlesu18.html [Broken]

(see section 3.2). The whole point of Fermi-Normal coordinates is to describe the local experience of a non-inertial world line.

Again, the point of my first bullet was to go along with Pervect's idea of treating motion in a car as gravity just as much as standing on the ground. My only point was if a coordinate dependent quantity was to be used as a proposed definition of inertial forces in a car, something needed to be said about construction of canonic coordinates to express it in.

It was only in my second, independent, bullet that I raised the question of distinguishing 'inertial forces you would like to think of as gravity' from other inertial forces.

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10. Jul 27, 2012

### bcrowell

Staff Emeritus
I'm sitting in a chair. We can interpret the Christoffel symbols as measures of the gravitational force holding me against the chair, but we can equally well say that there is no gravitational force, and the Christoffel symbols measure the chair's force on me. Really all they're telling me is that if I fix a coordinate system relative to the chair, then an object with zero coordinate velocity in those coordinates has a certain proper acceleration. They're not telling us about physics or forces or gravity, they're telling us about an arbitrarily chosen coordinate system.

Well, those people are being unreasonable and/or don't understand GR very well, so I don't think making them happy is a good criterion. Clearly the two examples (turning car, standing on the ground) have to be described the same way if we take the equivalence principle seriously. I would describe the passenger in the turning car as experiencing an outward gravitational field -- in coordinates fixed to the car. This kind of thing just shows that the gravitational field is not a physically meaningful thing to worry about.

11. Jul 27, 2012

### Staff: Mentor

You're right, I was misunderstanding the terminology. What I was calling "local coordinates matched to the frame field" are actually Fermi Normal coordinates. Sorry for the mixup on my part.

Last edited by a moderator: May 6, 2017
12. Jul 27, 2012

### Staff: Mentor

I tend to lean towards this point of view as well. The OP says it's a goal to "represent the forces one experiences in an accelerating elevator as 'real' forces", but to me the "real" force felt by the observer at rest in an accelerating elevator is not "gravity", it's the elevator pushing on the observer. A freely falling object inside the elevator does not have any "force" on it. In other words, on this view force is always accompanied by proper acceleration, and it's always due to something that can cause proper acceleration.

I understand that many people don't want to take this view of "force". (I've lost count of the number of threads where this issue has arisen in one form or another.) The OP's position appears to be to try to come up with a different definition of "force" that allows gravity to be a force, so we can say that the freely falling rock here on Earth is pulled down by gravity, as is a freely falling object inside an accelerating spaceship. I see the intuitive motivation for this, but on balance I would prefer to try to convince people that the "force means proper acceleration" view works better.

13. Jul 27, 2012

### pervect

Staff Emeritus
I was re-reading MTW, and the online reference and perhaps I've conflated Fermi coordinates with Fermi Normal coordinates.

http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3A0901.4465 [Broken]

If I've gotten it right this time, what I want are the Fermi coordinates - ones defined "by the worldline of an apparatus."

This was more or less the procedure that MTW was adopting - and the simply expressed result of their procedure was that the acceleration could be read directly off the Christoffel symbols. Basically, when you require that your time coordinate be the proper time of an apparatus, you can't parallel transport your basis vectors, but you need to adopt a transport law.

You want the transport law to generate local Lorentz transformations. The infinitesimal generator of these local Lorentz transformations is just conceptually a rate of boost, i.e. an accleleration, and a rate of rotation, because a Lorentz transformation can be generalized as a boost and a rotation.

And the transport law determines the Christoffel symbols, which in turn give us the acceleration.

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14. Jul 27, 2012

### pervect

Staff Emeritus
I may have to retract my retraction. Thanks for the reference.

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15. Jul 27, 2012

### pervect

Staff Emeritus
I would say the traditional frame has been that of a static observer. So traditional gravity is the value of the Christoffel symbols in the frame of a static observer. (Possibly, a stationary observer is sufficient.)

This leaves you in a pickle when you don't have a static (or perhaps stationary) metric, however.

16. Jul 27, 2012

### pervect

Staff Emeritus
Next up on the sanity check list. If forces really are Christoffel symbols, and transform as such, why do 4-accelerations transform as tensors, when Christoffel symbols do not transform like tensors?

I have yet to demonstrate explicitly that Christoffel symbols do, in fact, transform as tensors under Lorentz Boosts, but it seems like the right answer.

This leaves me, at least, with the idea that 4-accelerations transform as tensors under a properly restricted set of transformation laws (i.e. Lorentz boosts), but not under general non-linear transformations, like x' = x + at^2, for instance - or the relativistic equivalent x' = cosh(a tau), t' = sinh(a tau).

17. Jul 27, 2012

### Staff: Mentor

Well, by this definition of "forces", forces don't correspond to 4-accelerations, so you can't expect their transformation properties to be the same. They are simply different things on this view.

18. Jul 27, 2012

### bcrowell

Staff Emeritus
Hmm...let's explore a few cases.

Inside a black hole's event horizon, there are no static or stationary observers, so we can't define gravitational force. I suppose this might be OK.

An FRW cosmological model isn't static or stationary, so we can't define the gravitational force. This seems wrong to me. Any sensible definition of gravitational force should give zero by isotropy.

In the field of a rotating body, we have effects like frame-dragging, so the spacetime is stationary but not static. It seems like you'd want to be able to define the gravitational force on a person standing on the rotating earth's surface, so we can't demand staticity.

But if we can't demand staticity, then it seems like we have too much ambiguity. We can have observer A who is orbiting the moon in a ground-skimming orbit, and observer B who is sitting on the surface of the moon. A says g=0, B says 1.6 m/s2. Who is right?

The only option I see is to embrace the equivalence principle and stop trying to define gravitational forces except as properties of arbitrarily chosen coordinate systems or frames of reference, which makes gravitational forces uninteresting to talk about.

19. Jul 27, 2012

### Staff: Mentor

I think you could still make a "gravitational force" scheme work for a stationary but not static spacetime; the stationary spacetime still has a timelike Killing vector field, so there is still a notion of "static observers". It's just that the "gravitational force" those observers see now has an extra component; it's not directly radial any more. I would have to do the computation to see how that works out in terms of Christoffel symbols, though.

I'm not sure isotropy implies that the "force" has to be zero; I think it only implies that the "force" has to be the same in all spatial directions. The problem I see is that the observed "gravitational force" is obviously *not* zero, although it is the same in all directions. But all of the $\Gamma^{i}_{00}$ Christoffel symbols *are* zero, identically, because the metric is independent of all the spatial coordinates.

So whatever scheme we come up with to represent "gravitational force" mathematically in this spacetime, it will have to be different than the one that works for static (and possibly stationary, if I'm right in what I said above) spacetimes. And in other spacetimes with different properties, we may need another scheme still, if there's any that works at all.

20. Jul 27, 2012

### PAllen

Isn't the proper acceleration of comoving observers in the lambda-CDM model also zero? If so, why not consider the 'gravitational force' cosmically zero, varying only by matter lumpiness? In effect, the comoving coordinates represent the stationary background, and places where lumpiness causes Christoffel symbols to deviate from zero represent the inertial force we choose to call gravity?

21. Jul 27, 2012

### Staff: Mentor

Yes.

Because the comoving observers see each other as accelerating (in the coordinate sense)--accelerating away from each other if dark energy dominates (as it does now in our universe), or accelerating towards each other if ordinary matter dominates (as it did until a few billion years ago in our universe). So by the notion of "gravitational force" that I understand the OP to be proposing (which is basically the Newtonian notion of "force" corresponding to coordinate acceleration), there must be such a force present even in the case of perfect isotropy and homogeneity.

22. Jul 28, 2012

### PAllen

I see. Then this is a really strong case for how only in limited cases can you make Newtonian analogies in GR. You have acceleration by so basic an observation as increasing red shift with time, yet both the relevant Christoffel symbols (the ones associated with acceleration) and proper acceleration are zero. You have observable acceleration without force (on either observer or observed). How non-Newtonian can you get! [edit: But wait: can we not say this effect is in the same family as the tidal forces we ignore in any argument like the OP?].

Despite this, there is a close relation between Christoffel symbols expressed in Fermi-normal coordinates built on a world line, and proper acceleration of the world line. This equivalence extends out for some distance, describing e.g. the inertial forces inside an arbitrarily moving lab of 'some size'. In that limited sense, I can still see validity of the OP.

23. Jul 28, 2012

### Staff: Mentor

Hmm. It's true that this acceleration is due to spacetime curvature, though it's a different sort of manifestation of curvature than ordinary tidal forces (the latter are due to Weyl curvature, but the former is due to Ricci curvature--the Weyl curvature is zero everywhere, at least in the idealized case of exact isotropy and homogeneity). Whereas the kind of acceleration directly mentioned in the OP can be present even in flat spacetime (as it is in the example given in the OP).

So if we adopt the view that we're ignoring curvature by restricting attention to a sufficiently small patch of spacetime, we could say, as you propose, that the "gravitational force" seen by a comoving observer is just zero because the relevant Christoffel symbols are zero. But that still doesn't seem very intuitively appealing to me. I'll have to think about this some more.

24. Jul 28, 2012

### pervect

Staff Emeritus
It's a non-vacuum solution as written, with g_00 = 1+2gz. I threw it into GRTensor - it's got rho=0, but some funky negative pressures.

The familiar g_00 = (1+gz)^2 = 1 + 2gz + g^2z^2 is the vacuum solution for an accelerated observer.

The Christoffel symbols for the two have the same values *at the origin*, i.e.

$\Gamma_{ztt} = g$ for the first line element, while $\Gamma_{ztt}=g*(1+gz)$ for the second. So they're only the same at z=0.

So the extra nonlinear term g^2*z^2 is important for the curvature (making the second one a flat vacuum solution of an accelerating observer), and it has some effect on the Christoffel symbols not at the origin, but I don't see that it affects the argument that the connection coefficients where you are completely determine the gravity you feel.

It does suggest that I emphasize the point that you compute the Christoffel symbols at the origin more carefully, though.

25. Jul 28, 2012

### Mentz114

I was agreeing with you about the Christoffels, but having zero energy density and non-zero pressures breaks the energy condition.

It's possible to generalize this somewhat. We have
$${\Gamma^\mu}_{00}=(1/2)g^{\mu k}\left( g_{0k,0} + g_{k0,0} - g_{00,k} \right)$$
if the metric has no explicit time dependence this is
$${\Gamma^\mu}_{00}=-(1/2)g^{\mu k}g_{00,k}$$
and if the metric is diagonal
$${\Gamma^\mu}_{00}=-(1/2)g^{\mu \mu}g_{00,\mu}$$
with no summation, so ${\Gamma^x}_{00}=-(1/2)g^{x x}g_{00,x}$ etc.

This has the form of a force if g00 is a potential.

Last edited: Jul 28, 2012