# Can the Christoffel connection be observed?

## Main Question or Discussion Point

[Moderator's note: Spun off from a previous thread about Maxwell's Equations and QFT.]

What would be more appropriate to compare is the vector potential A in QED and the Christoffel connection in GR.
Perhaps, but Christoffel connection is also an observable. You feel it in your whole body when you accelerate. One usually calls it the inertial force.

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A. Neumaier
2019 Award
Christoffel connection is also an observable. You feel it in your whole body when you accelerate.
I think you feel the curvature tensor, not the Christoffel connection.

I think you feel the curvature tensor, not the Christoffel connection.
No. You feel the inertial force even if you accelerate in flat spacetime.

• Boing3000
A. Neumaier
2019 Award
No. You feel the inertial force even if you accelerate in flat spacetime.
But acceleration is caused by gravitation = curvature, which is absent in flat spacetime. Or it is caused by other forces, and then your claim that it means ''feeling the Christoffel connection'' is not valid, either.

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But acceleration is caused by gravitation = curvature, which is absent in flat spacetime.
In this case I would not talk about acceleration, but about motion along geodesic. The equation for motion along geodesic is
$$\frac{d^2X^{\mu}}{d\tau^2}+\Gamma^{\mu}_{\alpha\beta} \frac{dX^{\alpha}}{d\tau} \frac{dX^{\beta}}{d\tau}=0$$
which, as you can see, depends on the Christoffel connection and not on the curvature.

Or it is caused by other forces, and then your claim that it means feeling the Christoffel connection is not valid, either.
Even if it is true that it is caused by other forces (provided e.g. by the rocket engine), quantitatively it is equal to (certain components of) the Christoffel connection. So by measuring the force by dynamometer you can measure (certain components of) the Christoffel connection.

Let me make it more precise. If a non-gravitational force $F^{\mu}$ acts on the particle, the equation above generalizes to the relativistic Newton equation
$$m\left( \frac{d^2X^{\mu}}{d\tau^2}+\Gamma^{\mu}_{\alpha\beta} \frac{dX^{\alpha}}{d\tau} \frac{dX^{\beta}}{d\tau} \right)=F^{\mu}$$
We want to consider this equation from the point of view of observer comoving with the particle, so we take coordinates in which the trajectory obeys $X^0=\tau$, $X^i=const$. Hence the equation above reduces to
$$m\Gamma^{\mu}_{00}=F^{\mu}$$
The dynamometer measures the spacial components of this, $F^i=m\Gamma^i_{00}$. This is the weight felt by an observer staying in a rocket or on Earth.

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TeethWhitener
Gold Member
acceleration is caused by gravitation = curvature
Motion along a geodesic produces no measurable acceleration (edit: in the free-falling frame), even if the curvature is nonzero.

George Jones
Staff Emeritus
Gold Member
We are getting far off-topic, but, by

But acceleration is caused by gravitation = curvature
do you mean geodesic deviation (and related "tidal force")?

A. Neumaier
2019 Award
Let me make it more precise. If a non-gravitational force $F^{\mu}$ acts on the particle, the equation above generalizes to the relativistic Newton equation
$$m\left( \frac{d^2X^{\mu}}{d\tau^2}+\Gamma^{\mu}_{\alpha\beta} \frac{dX^{\alpha}}{d\tau} \frac{dX^{\beta}}{d\tau} \right)=F^{\mu}$$
We want to consider this equation from the point of view of observer comoving with the particle, so we take coordinates in which the trajectory obeys $X^0=\tau$, $X^i=const$.
Here you fix a particular curved coordinate system, which is the gravity (massless spin 2) analogue of fixing a gauge in a gauge (massless spin 1) theory.
This shows that you are not measuring the Christoffel connection itself but a member of its equivalence class under the diffeomorphism group, which is the gravity analogue of the gauge group.
Hence the equation above reduces to
$$m\Gamma^{\mu}_{00}=F^{\mu}$$
The dynamometer measures the spacial components of this, $F^i=m\Gamma^i_{00}$. This is the weight felt by an observer staying in a rocket or on Earth.
You showed that an observer can measure 4 particular components of the Christoffel connection in a very specific coordinate system. This yields no information at all about the Christoffel connection in any coordinate system where $X_0\ne \tau$, since for the latter you'd need to know also the derivatives of the Christoffel symbol, which are not available. It is analogous to the fact that from having measured only the vector potential in a special gauge at a given point you cannot tell anything about the vector potential in any other gauge.

This what you did is far from having measured the connection. The latter is impossible.

• dextercioby
A. Neumaier
2019 Award
Motion along a geodesic produces no measurable acceleration (edit: in the free-falling frame), even if the curvature is nonzero.
But in the frame of the ''feeling'' observer, sitting on a nongeodesic curve.

TeethWhitener
Gold Member
But in the frame of the ''feeling'' observer, sitting on a nongeodesic curve.
I don’t understand what this is supposed to mean. If you’re not following a geodesic, by definition an external force is acting on you. This is true in flat space as well as curved space.

haushofer
My 2 cents: you can regard the connection as a gauge field. Can we observe gauge fields classically? I'd say we can only observe gauge invariant quantities.

Edit: what A.Neumaier says, I guess.

Some time ago I posted this post to the forum. (the account has no avatar because it was one that was lost).

https://www.physicsforums.com/threa...-aceleration-from-christoffel-symbols.923208/

My idea was: if the riemman tensor is equal to zero in the flat space, why the christoffel symbols do not cancel out in spherical coordinates. Immediately I realized that maybe they could be related to the forces of inertia. Then in a rotating sphere like the earth we can perhaps interpret these symbols with those accelerations. I do not know if this is what you were looking for.

Edit: In the two examples of motion descibed above, I suposse the trajectory of the object, then of course involves the existence of reactions forces that make these trajectories possibles (for vanishing these acelerations). I don't have response for these two examples and I don't know if in these examples i'ts all alright.

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martinbn
This discussion is very strange. How can a coordinate dependent quantity be an observable! Why would it matter if I use Cartesian or polar coordinates, or some other!

• pervect and alejandromeira
This discussion is very strange. How can a coordinate dependent quantity be an observable! Why would it matter if I use Cartesian or polar coordinates, or some other!
Yes, of course!!!
My English is not very well, then in the above post perhaps I should have said, "interpret these symbols as a relationship with those accelerations". But in any case, what is certain is that from the symbols of christoffel you can obtain these accelerations, without the need to invoke cross vector products!!!

Nontensor such as Christoffel symbol or energy-momentum pseudo tensor is NOT tensor. They do not have invariant values under change of coordinate system. However, in some coordinate they seem to exist e.g. centrifuge force, Coriolis force, downward gravity in accelerating rocket or gravity potential energy mgh.

• Demystifier
This discussion is very strange. How can a coordinate dependent quantity be an observable! Why would it matter if I use Cartesian or polar coordinates, or some other!
An observable is something observed by an observer, right? In the theory of relativity (both special and general), the choice of coordinates is not only a matter of computational convenience. Different observers have different natural coordinates associated with them. Different observers perceive the physical world differently, and this fact can well be described by different coordinates. A standard example is Lorentz contraction, which (as far as I am aware) cannot be described in a Lorentz covariant way. Another example is Unruh effect, which (as far as I am aware) cannot be described in a general covariant way. Or if someone knows covariant descriptions of those effects, I would be very happy to see it.

It doesn't mean that it matters whether you use Cartesian or polar coordinates for the spacial part of spacetime. But it matters whether you split spacetime into space and time one way or another. Different splits may correspond to different observers and hence different observables.

martinbn
Of course observed things depend on the reference frame and on the 3+1 split, but they are invariant. If an observer measures the energy of a particle. What he measures is the inner product of his 4-velocity with that of the particle. If we choose a frame where the timelike vector is the 4-velocity of the observer and three spacelike orthogonal to it, that inner product will be the first component of the 4-velosity of the particle. If we choose a different one it will be a different component or combination of components, but that is just a different representation of the same invariant thing. For a different observer the energy of the particle will be in general different. Not because of choice of coordinates but because it is a different inner product, a different invariant. It is very different when it comes to the connection coefficients.

stevendaryl
Staff Emeritus
Different observers have different natural coordinates associated with them.
I think it depends on what you mean by an "observer". If you idealize an observer as a point moving through spacetime on a timelike path, then that doesn't pin down a unique coordinate system. You can constrain it a little by requiring that:
1. There is one timelike coordinate and three spacelike coordinates.
2. For each spacelike coordinate $x^j$, the corresponding component of the observer's proper velocity is zero.
Is that what you had in mind?

• Demystifier
If an observer measures the energy of a particle. What he measures is the inner product of his 4-velocity with that of the particle. If we choose a frame where the timelike vector is the 4-velocity of the observer and three spacelike orthogonal to it, that inner product will be the first component of the 4-velosity of the particle. If we choose a different one it will be a different component or combination of components, but that is just a different representation of the same invariant thing.
For a different observer the energy of the particle will be in general different. Not because of choice of coordinates but because it is a different inner product, a different invariant.
For the case of energy, I perfectly agree. But can you describe Lorentz contraction in a similar way?

It is very different when it comes to the connection coefficients.
Suppose that general relativity is just an approximation of a more fundamental theory. And suppose that action of the fundamental theory is the Einstein-Hilbert action plus some small diffeomorphism non-invariant term in which connection is treated as a fundamental quantity. If the additional term is sufficiently small, in practice all currently existing measurements would produce the same results as predicted by standard theory without the additional term. And yet, according to the fundamental theory, any such measurement would really be a measurement of connection. Indeed, Einstein said that it is theory that determines what is measurable. I would rephrase him by saying that it is our theoretical prejudices that determine what is measurable. If one has a theoretical prejudice that diffeomorphism symmetry is fundamental, then one cannot measure connection. If one has a theoretical prejudice that connection is fundamental, then one can measure connection. And yet, the two guys with different prejudices observe the same natural phenomena. They only differ in the theoretical interpretation of the observed phenomena.

Nontensor such as Christoffel symbol or energy-momentum pseudo tensor is NOT tensor. They do not have invariant values under change of coordinate system. However, in some coordinate they seem to exist e.g. centrifuge force, Coriolis force, downward gravity in accelerating rocket or gravity potential energy mgh.
To this list I would add gravitational waves, the energy-momentum of which is the energy-momentum pseudo-tensor.

Let me also add that it is possible to construct a true tensor of gravitational energy-momentum, as I explained in http://lanl.arxiv.org/abs/1407.8028 . The referees rejected my paper, but they did not object that it's wrong. They objected that such a true tensor energy-momentum is physically useless.

stevendaryl
Staff Emeritus
Suppose that general relativity is just an approximation of a more fundamental theory. And suppose that action of the fundamental theory is the Einstein-Hilbert action plus some small diffeomorphism non-invariant term in which connection is treated as a fundamental quantity.
I think it's a little tricky to say what it would mean for physics to NOT be diffeomorphism-invariant. There is the passive sense of diffeomorphism: a coordinate change. You can always rewrite your laws of physics so that they have the same form in every coordinate system. Then there is the active sense, which I think is a little harder to understand. Basically, the way I would put it is that if you distort the manifold (spacetime) itself, by moving points around to squash them closer together or pull them farther apart, then it's possible to adjust the various scalar, vector and tensor fields so that the result is observationally the same as it was. What I think is another way to put it is that a point in the manifold has no identity other than the values of scalar, vector and tensor fields at that point.

So what would be an example of an intrinsically non-diffeomorphism-invariant theory? I think that some people would characterize it by saying that there are fundamental scalar, vector or tensor fields that are non-dynamic. They affect the dynamics of particles and fields, but are not affected by it. In Newtonian physics, universal time is a fundamental nondynamic scalar field, and the spatial metric is a fundamental nondynamic tensor field. But is there really a sense in which something is intrinsically non-dynamic? I would think that you could always add some unobservable dynamics.

stevendaryl
Staff Emeritus
To this list I would add gravitational waves, the energy-momentum of which is the energy-momentum pseudo-tensor.
Gravitational waves can be characterized in a covariant way, can't they? They're an example of a nonvanishing, nonstatic Weyl curvature tensor, right?

I would think that you could always add some unobservable dynamics.
But in this way you can add any symmetry to the theory you like. You can add extra unobservable dimensions, extra unobservable gauge fields (to enlarge the SM gauge group U(1) times SU(2) times SU(3)), etc.

Gravitational waves can be characterized in a covariant way, can't they?
In principle yes, but that's quite complicated and you cannot define their energy-momentum.

martinbn