# Lagrangian Point in General Relativity

• B
Mentor
That's one possible point of view.

It's the only possible point of view as far as classical GR is concerned. Remember, this is the relativity forum.

For me the gravitational interaction is an interaction as any other.

From the standpoint of quantum field theory, this is a valid point of view. But that would apply to discussions in the quantum forum, not this one.

Gold Member
2022 Award
Also as a classical field theory the gravitational interaction is an interaction. I hope you can agree to the agreement @Dale and I seem to have found now.

From a QFT point of view I'd rather say gravitation is not completely described by GR but by Einstein-Cartan theory, because there are Dirac fields needed to describe the phenomena.

Mentor
From your point of view all phenomena we call "gravitation" were just inertial forces in non-inertial reference frames in Minkowski space.

I have said no such thing. Nor does GR.

In the geometrical interpretation of the gravitational field

Which is what I was using.

As I said before, all this is just semantics and doesn't lead to much deeper understanding of GR.

I disagree. So does, for example, Misner, Thorne, and Wheeler; in one of their early chapters they say that understanding how the various local inertial frames "fit" into the curved global geometry is fundamental to understanding GR.

Gold Member
2022 Award
Then we agree at least so far, and I've misunderstood your previous statement.

Mentor
Also as a classical field theory the gravitational interaction is an interaction.

In the sense that the field equation is nonlinear, yes. But this sense of "interaction" does not make gravity in GR a "force" in the Newtonian sense. Nor does it make it a "force" that is detectable by local measurements. So one has to be very careful using such ordinary language terms with reference to gravity in GR.

• Dale
Mentor
But then you should admit that it's wrong to say gravitational forces (I prefer the term gravitational interaction though since forces are something I want to restrict to the use in Newtonian physics only) are purely fictitious.

I disagree. The part of "gravity" that cannot be transformed away is spacetime curvature, which appears physically as tidal gravity--geodesic deviation. That is not a "force" or an "interaction". Unless you want to say that the nonlinearity of the Einstein Field Equation producing spacetime curvature counts as an "interaction"; but, as I noted in my previous post, you have to be very careful using these ordinary language terms in this domain.

Gold Member
2022 Award
I disagree. The part of "gravity" that cannot be transformed away is spacetime curvature, which appears physically as tidal gravity--geodesic deviation. That is not a "force" or an "interaction".
Well, then we have a pretty different understanding of the word "force" (which is a Newtonian notion) or "interaction". The very point is that there are "tidal forces" which cannot be understood as "inertial forces".

Instead of saying "tidal forces" (a Newtonian notion) I'd indeed rather prefer to say it's curvature of spacetime, describing the gravitational interaction (as the electromagnetic field describes the electromagnetic interaction).

Mentor

I don't understand what you mean by this.

The very point is that there are "tidal forces"

No, there is tidal gravity--geodesic deviation. That is not a force.

Objects which are held together by internal (non-gravitational) forces will be subjected to stresses due to tidal gravity, because the internal forces prevent their individual parts from following geodesic worldlines. But although these internal forces are sometimes called "tidal forces", that is a misnomer, since they are not due to tidal gravity (spacetime curvature) itself. They are due to the non-gravitational forces between the parts of the object.

curvature of spacetime, describing the gravitational interaction (as the electromagnetic field describes the electromagnetic interaction)

But the "interaction" due to the EM field causes a force (the Lorentz force) that can be detected and felt locally by accelerometers. The "interaction" due to spacetime curvature does not. So in classical terms, describing spacetime curvature as an "interaction" at all makes no sense. The actual analogy between spacetime curvature and the EM field is geometric: the EM field is the curvature associated with the connection ##A_\mu## (the EM 4-potential).

Again, please bear in mind that this is the relativity forum, not the quantum physics forum.

Staff Emeritus
Maybe we can agree that classical gravitation can be regarded as a Lagrangian field theory, with an associated action? Note that I'm not saying this is necessarily the only approach, just one possible approach.

Another thing that I would hope we could agree on is that we don't have a well-accepted theory of quantized gravitation at this time.

I don't want to go too much further unless we have more agreement, but it seems appropriate to add one further thing, that the Lagrangian methods are also applicable to the Newtonian three body problem, as one might expect from the name "Lagrange points".

• vanhees71
Mentor
Maybe we can agree that classical gravitation can be regarded as a Lagrangian field theory, with an associated action?

No problem here. But note that such an interpretation does not, of itself, say anything about "forces", and the only thing it says about "interactions" is whatever "interaction" terms are in the Lagrangian. For gravity, the only term is the Ricci scalar term, and the only "interaction" in that term is the nonlinearity in the Ricci scalar.

• vanhees71
Gold Member
2022 Award
I don't understand what you mean by this.

No, there is tidal gravity--geodesic deviation. That is not a force.

Objects which are held together by internal (non-gravitational) forces will be subjected to stresses due to tidal gravity, because the internal forces prevent their individual parts from following geodesic worldlines. But although these internal forces are sometimes called "tidal forces", that is a misnomer, since they are not due to tidal gravity (spacetime curvature) itself. They are due to the non-gravitational forces between the parts of the object.

But the "interaction" due to the EM field causes a force (the Lorentz force) that can be detected and felt locally by accelerometers. The "interaction" due to spacetime curvature does not. So in classical terms, describing spacetime curvature as an "interaction" at all makes no sense. The actual analogy between spacetime curvature and the EM field is geometric: the EM field is the curvature associated with the connection ##A_\mu## (the EM 4-potential).

Again, please bear in mind that this is the relativity forum, not the quantum physics forum.
An interaction is described by fields. Eg. in electromagnetism a charged body has an em. field arond it and the "force" acting on another (test) charge due to the presence of the field at the point of this charge. It's a local concept and in accordance with the relativistic causality structure which forbids the concept of Newtonian action-at-a-distance forces.

The gravitational interaction is not different: The sources are energy-momentum-stress distributions causing the gravitational field and the "force" acting on a test particle is caused by this field.

I agree with your interpretation of the connection, which is mathematically fully analogous for both em. and grav. interactions, because they are both gauge theories.

All this is not related to field quantization. There's no successful QFT of the grav. interaction yet. So I can't say anything about it.

Staff Emeritus
No problem here. But note that such an interpretation does not, of itself, say anything about "forces", and the only thing it says about "interactions" is whatever "interaction" terms are in the Lagrangian. For gravity, the only term is the Ricci scalar term, and the only "interaction" in that term is the nonlinearity in the Ricci scalar.

The question that started out the thread was:

If gravity is a pseudo force, what is going on at the Lagrangian points in GR terms?

And the answer I'm proposing to this question is "one doesn't need forces, all one needs is a Lagrangian, and the Euler Lagrange equations". A side observation is that one probably already used Lagrangian methods in the Newtonian 3 body problem already.

Some important conceptual issues that I haven't covered are how to reduce the continuous field problem, involving partial differential equation, to linear differential equations involving a finite number of variables- the positions of the planets in an n-body problem, for instance.

This necessarily involves some approximation schemes. The electromagnetic analogy would be having an electromagnetic problem with some collection of charges, regarded as pointlike particles. How do we evolve the state of the system? If we ignore radiation we can imagine writing ordinary differential equations where we approximate the state of the system and it's Lagrangian by the positions of the particles. But this will be an approximation to the full system using the full field theory and Maxwell's equations.

This is necessarily an approximation. The techniques to do this for GR is based on approximation schemes such as the various orders of PPN theory. The full field theory approach is the most accurate, but too computationally intense to be applied to many-body problems like the Solar system.

It may be the least part of the problem, but I also wanted to add that for a single point particle, one can make the Lagrangian a bit less abstract by relating it to proper time. The Lagrangian for a test particle expressed in some agreed-on time and space coordinates is just the proper time as expressed in those coordinates. The standard approach to writing the Euler-Lagrange equations still needs us to single out a time coordinate and space coordinates.

As a consequence, the equations of motion that extremize the Lagrangian of the test particle are just the equations that extremize it's proper time.

• vanhees71
Gold Member
2022 Award
Indeed, there's not even a fully selfconsistent description of two point charges in classical em. For the case of gravitation it's even more complicated to find a fully self-consistent solution involving two compacr objects or black holes. Everything relies on the approximations you mention, and these can be pretty accurate descriptions, but they are no self-consistent formulations, not to mention even solutions.

Gold Member
2022 Award
No problem here. But note that such an interpretation does not, of itself, say anything about "forces", and the only thing it says about "interactions" is whatever "interaction" terms are in the Lagrangian. For gravity, the only term is the Ricci scalar term, and the only "interaction" in that term is the nonlinearity in the Ricci scalar.
There are of course the interaction terms between the ##g_{\mu \nu}## and the fields describing matter and radiation.

Mentor
An interaction is described by fields.

In classical physics, for interactions other than gravity, yes.

The gravitational interaction is not different: The sources are energy-momentum-stress distributions causing the gravitational field and the "force" acting on a test particle is caused by this field.

No, this is wrong. There is no gravitational analogue of the Lorentz force equation in EM. The Einstein Field Equation links spacetime curvature to stress-energy and is the gravitational analogue of Maxwell's Equations, but Maxwell's Equations are field equations, not force equations.

Mentor
There are of course the interaction terms between the ##g_{\mu \nu}## and the fields describing matter and radiation.

Calling these "interaction" terms is, IMO, a misnomer, classically speaking. Classically speaking, the only reason the metric appears in these terms is to lower or raise indexes on the fields in order to form scalars.

Gold Member
2022 Award
Sure, but there are also equations of motion for the other fields (or for the point-particle coordinates, if you use the point-particle model), and of course they contain the interaction with the gravitational field (the pseudometric).

Of course you are right in saying that these interactions are usually written on the left-hand side of the equation of motion. There the coupling to the gravitational field occurs in the terms with the Christoffel symbols. For point particles you get (using natural units with ##c=1##)
$$m \mathrm{D}_{\tau} u^{\mu}=K^{\mu},$$
where
$$\mathrm{D}_{\tau} u^{\mu}=\mathrm{d}_{\tau} u^{\mu} + {\Gamma^{\mu}}_{\rho \sigma} u^{\rho} u^{\sigma}, \quad u^{\mu} = \mathrm{d}_{\tau} x^{\mu}.$$
The ##K^{\mu}## is the Minkowski force of all other interactions.

Now, if the interaction with the gravitational field where just as inertial forces in Newtonian mechanics, you could find coordinates, where the Christoffel symbols vanish identically, but that's not the case when "true gravitational fields" are present. Then you can make the Christoffel symbols vanish only at one arbitrary point in spacetime (weak equivalence principle).

That the Christoffel symbols occur in the description of the interaction of matter with the gravitational field is indeed natural from the point of view of gauge theories since it's the connection. That's analogous to the em. field, where the gauge connection is given by the Faraday tensor and the corresponding Minkowski force is
$$K_{\text{em}}^{\mu}=q F^{\mu \nu} u_{\nu}.$$

In Newtonian physics it's the same. What is usually called "inertial forces" belongs to the "covariant time derivative" for vectors wrt. a non-inertial reference frame. E.g., for a rotating frame
$$m \mathrm{D}_t \vec{v} = m (\mathrm{d}_t \vec{v}+\vec{\omega} \times \vec{v})=\vec{F}$$
with
$$\vec{v}=\mathrm{D}_t \vec{x} = \mathrm{d}_t \vec{x} + \omega \times \vec{x}$$
and ##\vec{F}## the "true forces".

It's only convention to bring the terms different from ##m \mathrm{d}_t^2 x## from the left to the right-hand side and call them "inertial forces". Of course, in Newtonian physics you can make all these inertial forces vanish, because there's by assumption always global inertial frames.

$$G_{\mu \nu}=\kappa T_{\mu \nu},$$