## Are Christoffel symbols measurable?

It's not really a problem of "right and wrong". What we are talking about here is just an analogy. E&M and GR are different theories. There are various analogies within them, but one should not expect a perfect 1 to 1 correspondence (or else they'd be the same theory!).

In the context of classical GR, one normally associates with A, the 4-vector potential from E&M, with g, the metric from GR. This is because in the linearized limit, both A and g exhibit some sort of "gauge symmetries" in that one can change the "gauge" (with appropriate definition of "gauge") and not change physical observables.

From a QFT point of view, however, the A is often used as a "principle connection" in a "covariant derivative" (where now the curvature appears in the vector bundle rather than the space-time manifold itself), and in that sense is analogous to the Christoffel symbols in GR.

The analogy is not perfect. It sort of depends on what one's purposes are.

Recognitions:
 Quote by waterfall So what is the counterpart of Ricci curvature (or R^c (ab)) in electromagnetism if it is not E and B which is already taken up by the Christoffel symbols (and how can Cromwell be wrong in the table in page 173 so maybe I can tell him).
The Christoffel symbols can be thought of as a GL(4) gauge connection (i.e., analogous to the gauge potential A in Yang-Mills theory). Then the Riemann tensor is the curvature of this GL(4) gauge connection (analogous to the field strength F in Yang-Mills theory). The metric itself only turns up because of the zero-torsion condition, which relates the metric to the Christoffel symbols.

As is usual in Yang-Mills theories, the gauge potential (i.e. Christoffel symbols) is not directly observable; only gauge-invariant quantities are observable. The Riemann tensor is gauge-covariant, but in order to give us a measurable quantity, we need something gauge-invariant; hence, we need to make a scalar somehow.

This coincides with Mentz and PAllen, that all observables are scalars.

So how do we make a scalar out of the Riemann tensor?

Locally, we have a natural orthonormal frame given by our own rest frame. We have a timelike vector that points to our future, and in a local spatial slice, we can define three mutually orthogonal axes; call them x, y, z. Once we define a system of units, we can define four orthogonal vectors of length 1 in whatever units we've chosen; call these vectors T, X, Y, Z. Now, the Riemann tensor has four "slots" which accept vectors, so now we can take our collection of four vectors, and fill the slots using various combinations, such as

R(T,X,Y,Z), R(X,Y,X,Y), R(T,X,Y,X), etc.

Each of these objects is a scalar, and hence measurable.

Notice that I've made no mention at all of coordinate systems. I've only talked about defining a local orthonormal frame, centered at our current position. One might imagine that there is a coordinate system, defined nearby, such that the four vectors T, X, Y, Z are given by displacements along some coordinates we'll call t, x, y, z. But the catch is that there are infinitely many coordinate systems that satisfy this property at our specific location. We don't have enough information, locally, to specify a single coordinate system; we are only able to specify a local orthonormal frame.

In particular, we are always free to choose a local coordinate system, compatible with our local orthonormal frame, in which the Christoffel symbols vanish at our specific location. So they're zero! Problem solved.

But if the Christoffel symbols vanish, then where did the curvature go? The point is that the curvature depends on derivatives of the Christoffel symbols, put together in just such a way that the invariants

R(T,X,Y,Z), R(X,Y,X,Y), R(T,X,Y,X), etc.

don't care what coordinate system we use.

However, it is not correct to say that the Christoffel symbols are a fictitious quantity; after all, they carry all the curvature information. But it is only gauge-invariant combinations of Christoffels that can be measured. In particular, this means we can measure any scalars made from the Riemann tensor.

The reason we can only measure scalars is this: Coordinate systems are just collections of labels. Real, physical processes are things that happen in the universe, and they do not care how we choose to label things. If there is a star sitting at point P on a manifold, it is sitting at point P whether I label that point (0,0,0,0) or (1,3,5,42). Therefore quantities that can be measured, which correspond to real, physical processes, must be scalars with respect to coordinate changes.

Measurements of quantities that have directions associated to them (such as vectors and tensors) are always made by holding up a collection of vectors in some known directions and comparing. This corresponds to contracting all the available free indices, making a scalar. For example, to measure the velocity in the z direction, you hold up a unit velocity vector in the z direction and take its dot product with the tangent to a particle's motion.

The Christoffel symbols, on the other hand, can be made to vanish at a given point by merely relabeling things. It is not enough to hold up a collection of vectors and contract, because the result is arbitrary.
 "So what is the counterpart of Ricci curvature (or R^c (ab)) in electromagnetism if it is not E and B " The Ju four vector (charge density, current density) is the source for an equation involving second derivatives of the electromagnetism Au four-vector, the stress energy tensor is the source for an equation involving second derivatives of the metric. The Ricci tensor minus (1/2 guv times the contracted Ricci tensor) is like the quantity del squared A minus the second time derivative of A. (Actually I should not really put it in the Lorenz gauge, but it is simpler that way.) You should study the structure of the linear field equations of General Relativity--it is clear that the metric corresponds to the Au four-vector in electromagnetism. It is not subtle.
 "This coincides with Mentz and PAllen, that all observables are scalars. So how do we make a scalar out of the Riemann tensor? " The Rieman tensor already is an observable--masny of its components correspond to the componen6ts of the tidal force.

Recognitions:
 Quote by ApplePion "This coincides with Mentz and PAllen, that all observables are scalars. So how do we make a scalar out of the Riemann tensor? " The Rieman tensor already is an observable--masny of its components correspond to the componen6ts of the tidal force.
Once you choose a frame, you can measure its components in that frame, which are scalars.

Did you read beyond the two lines you quoted?
 "Real, physical processes are things that happen in the universe, and they do not care how we choose to label things. If there is a star sitting at point P on a manifold, it is sitting at point P whether I label that point (0,0,0,0) or (1,3,5,42)." I can only imagine what your answer would be if someone asked to what the velocity of an object dropped from a height of 100 meters (in the Earth's coordinate system) is when it hits the ground.
 Me: "The Rieman tensor already is an observable--masny of its components correspond to the componen6ts of the tidal force." Ben: "Once you choose a frame, you can measure its components in that frame, which are scalars" The components of the Riemann tensor are not scalars. Perhaps you think scalars means "single numbers". That is not what it means. Scalars do not change values under coordinate transformations. The components of the Riemann tensor can change values under coordinate transformations.

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Gold Member
 Quote by ApplePion "Real, physical processes are things that happen in the universe, and they do not care how we choose to label things. If there is a star sitting at point P on a manifold, it is sitting at point P whether I label that point (0,0,0,0) or (1,3,5,42)." I can only imagine what your answer would be if someone asked to what the velocity of an object dropped from a height of 100 meters (in the Earth's coordinate system) is when it hits the ground. You have gotten carried away with and are misapplying the Principle of Covariance. What you are doing is no longer physics. Or science.
What, exactly, do you find objectionable about Ben's quote? I fail to see how anyone could disagree with it.

Recognitions:
 Quote by ApplePion I can only imagine what your answer would be if someone asked to what the velocity of an object dropped from a height of 100 meters (in the Earth's coordinate system) is when it hits the ground.
The quantities you've mentioned (vertical component of velocity on impact, location B of Earth's surface, location A at a height 100m above Earth's surface, initial condition of zero velocity, free-fall motion from point A to point B) are all perfectly unambiguous, coordinate-independent things. There is no issue answering such a question.

Recognitions:
 Quote by ApplePion Me: "The Rieman tensor already is an observable--masny of its components correspond to the componen6ts of the tidal force." Ben: "Once you choose a frame, you can measure its components in that frame, which are scalars" The components of the Riemann tensor are not scalars. Perhaps you think scalars means "single numbers". That is not what it means. Scalars do not change values under coordinate transformations. The components of the Riemann tensor can change values under coordinate transformations.
If I choose some vector fields, call them $a^\mu, b^\mu, c^\mu$, then the quantity

$$R_{\mu\nu\rho\sigma} a^\mu b^\nu c^\rho b^\sigma$$
is certainly a scalar, and it measures the components of the Riemann tensor along the given vector fields.

This is analogous to computing matrix elements in quantum mechanics, if you've done that. Matrix elements are numbers, not operators; but they tell you how to construct an operator in a given basis.
 Blog Entries: 4 Recognitions: Gold Member Does anybody have any simple examples? This appears to be one of the simplest possible examples on pages 4-6 here: http://brucel.spoonfedrelativity.com...erivatives.pdf It gives three nonzero Christoffel symbols for a 2D polar coordinate system. and six Christoffel symbols for a 3D spherical coordinate system.
 Simple examples of what? Christoffel symbols?
 So it is only in linearized GR where the vector potential corresponds to the metric guv... while in QFT it is the phase that corresponds to the metric guv as Cromwell right describes. Therefore Applepion how can you say Cromwell is wrong. He was talking in terms of QFT. So they are both right. In Aharonov-Bohm effect, it is only the magnetic flux that is measurable, not directly the magnetic vector potential. So in the QFT correspondence, the Christoffel symbols correponding to the vector potential means it is not directly measureable. So what is the equivalent of magnetic flux in the Christoffel symbols case? In the linearized gravity correspondence, the Christoffel symbols correponding to the E and B field means it is directly measurable. Do everyone agree with the above summary (including Matterwave and Applepion)?

Recognitions:
 Quote by waterfall So it is only in linearized GR where the vector potential corresponds to the metric guv... while in QFT it is the phase that corresponds to the metric guv as Cromwell right describes. Therefore Applepion how can you say Cromwell is wrong. He was talking in terms of QFT. So they are both right. In Aharonov-Bohm effect, it is only the magnetic flux that is measurable, not directly the magnetic vector potential. So in the QFT correspondence, the Christoffel symbols correponding to the vector potential means it is not directly measureable. So what is the equivalent of magnetic flux in the Christoffel symbols case? In the linearized gravity correspondence, the Christoffel symbols correponding to the E and B field means it is directly measurable. Do everyone agree with the above summary (including Matterwave and Applepion)?
No, take a look at Ben Niehoff's and Matterwave's statements that you can't compare exactly. There are two meanings of "gauge". The first sense is just that different mathematical expressions can describe the same physical situation. The metric is a gauge field in this sense. The second meaning is that it is like a Yang-Mills theory. Gravity is not a gauge theory in this sense.

http://arxiv.org/abs/1106.2037 (haven't read it, but looks good)

 Quote by atyy No, take a look at Ben Niehoff's and Matterwave's statements that you can't compare exactly. There are two meanings of "gauge". The first sense is just that different mathematical expressions can describe the same physical situation. The metric is a gauge field in this sense. The second meaning is that it is like a Yang-Mills theory. Gravity is not a gauge theory in this sense. http://arxiv.org/abs/1106.2037 (haven't read it, but looks good)
There are two kinds of gauge? I'm only familiar with that in gauge transformation. Maybe you are referring to the abelian (QED) and non-abelian (Electroweak) sense? In the first case of linearized gravity, it's abelian and in the case of GR.. it's non-abelian (where internal rotations won't take you to the same place). Is this what you are referring to?

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