Are Christoffel symbols measurable?

[ApplePion]

Perhaps this might be helpful to some people. The LAWS of physics are invariant under coordinate transformations. But the physical quantities whose behavior is specified by the laws are not necessarily invariant. For example, the physical law that a particle not acted upon by a force will not change its four-velocity is a law that is true in all coordinate systems. But the physical quantity whose behavior is specified is not the same in all coordinate systems. For example, the four-velocity has different numerical components in the Lorentz frame where the particle is at rest compared to a Lorentz frame where the particle is moving.

 

[atyy]

Please go to page 173 of the interesting gauge comparision table:

http://www.lightandmatter.com/genrel/genrel.pdf

Are you saying the table in page 173 is wrong??

I tend to agree with ApplePion. The usual comparison of gauge fields in electromagnetism and gravity is the electromagnetic potential A and the spacetime metric g.

If we do that, is the E field like the Christoffel symbol Г? The Г field can be considered a gravitational field in the sense of the equivalence principle in which acceleration = gravity, because both gravity and acceleration lead to non-zero Г (as Crowell himself has pointed out many times).

However, it's also bit different. If the E field is zero in an inertial frame, then we do say there is no electric force. OTOH, if Г is zero in a local inertial frame, there could still be tidal gravity, due to the derivatives of Г being non-zero.

The fact that local inertial frames in which Г is zero always exist is an implementation of the equivalence principle that says that gravity can always be canceled away. The fact that although Г ("gravitational field") is zero, its derivatives ("tidal gravity") can still be non-zero shows that the "local" qualification is very important in the equivalence principle - a nonlocal experiment that looks at the derivatives of Г can still detect non-zero tidal gravity which cannot be canceled away.

Also, usage differs. For example, it's also often said that the metric field is the "gravitational field". These are just terminology differences. Also, there are two meanings of "gauge field". Gravity is a "gauge field" only in one of the two sense of the word.

Also, the Aharonov-Bohm effect is a quantum effect, so the classical electromagnetic potential really only has effects when E and B are non-zero - there is no classical Aharonov-Bohm effect.

 

[PAllen]

We're getting into philosophy here. In my view (which is one common view, certainly not universal) the fields in classical field theories, the wave function in QM, the histories in sum over histories QFT, are not observables, and the question of their existence is philosophy, not physics. Observables are measurements we can make. Theoretical constructs of a theory (with usage rules) allow us to predict measurements. Measurements are invariant not only relative to internal representational features of the theory (coordinates, frames of reference) but even across theories (the theoretical constructs underpinning the behavior of a meter as we move it near an electric current, have radically changed 3 times in the last 150 years; the behavior of the meter has not changed at all).

 

[waterfall]

I tend to agree with ApplePion. The usual comparison of gauge fields in electromagnetism and gravity is the electromagnetic potential A and the spacetime metric g.

If we do that, is the E field like the Christoffel symbol Г? The Г field can be considered a gravitational field in the sense of the equivalence principle in which acceleration = gravity, because both gravity and acceleration lead to non-zero Г (as Crowell himself has pointed out many times).

However, it's also bit different. If the E field is zero in an inertial frame, then we do say there is no electric force. OTOH, if Г is zero in a local inertial frame, there could still be tidal gravity, due to the derivatives of Г being non-zero.

The fact that local inertial frames in which Г is zero always exist is an implementation of the equivalence principle that says that gravity can always be canceled away. The fact that although Г ("gravitational field") is zero, its derivatives ("tidal gravity") can still be non-zero shows that the "local" qualification is very important in the equivalence principle - a nonlocal experiment that looks at the derivatives of Г can still detect non-zero tidal gravity which cannot be canceled away.

Also, usage differs. For example, it's also often said that the metric field is the "gravitational field". These are just terminology differences. Also, there are two meanings of "gauge field". Gravity is a "gauge field" only in one of the two sense of the word.

Also, the Aharonov-Bohm effect is a quantum effect, so the classical electromagnetic potential really only has effects when E and B are non-zero - there is no classical Aharonov-Bohm effect.

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).

 

[Matterwave]

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.

 

[Ben Niehoff]

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.

 

[ApplePion]

"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.

 

[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.

 

[Ben Niehoff]

"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?

 

[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.

 

[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.

 

[elfmotat]

"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.

 

[Ben Niehoff]

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.

 

[Ben Niehoff]

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.

 

[JDoolin]

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/GR2c-Derivatives.pdf

It gives three nonzero Christoffel symbols for a 2D polar coordinate system.
and six Christoffel symbols for a 3D spherical coordinate system.

 

[Matterwave]

Simple examples of what? Christoffel symbols?

 

[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)?

 

[atyy]

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)

 

[waterfall]

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?

 

[atyy]

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?

No, I was saying there are two meanings of gauge, that's all. The first meaning is less specific, and just means different mathematical expressions describe the same physics. The second meaning is more specific and means a field that has the structure of a Yang-Mills field. Gravity is only a gauge theory in the first sense.

But on second thoughts, in the linear case, there is an analogy between gravity and electromagnetism. http://arxiv.org/abs/gr-qc/0311030v2

 

[waterfall]

No, I was saying there are two meanings of gauge, that's all. The first meaning is less specific, and just means different mathematical expressions describe the same physics. The second meaning is more specific and means a field that has the structure of a Yang-Mills field. Gravity is only a gauge theory in the first sense.

But on second thoughts, in the linear case, there is an analogy between gravity and electromagnetism. http://arxiv.org/abs/gr-qc/0311030v2

But gravity as a fundamental force is a gauge field. Because what defines a fundamental force is it must have internal gauge transformation like U(1) in electromagnetism, SU(2)xU(1) in Electroweak, SU(3) in strong force. Are you saying it is possible the gravity force has no SU(N) terms?

Also I saw in wiki that "Yang–Mills theory is a gauge theory based on the SU(N) group". For the electromagnetism U(1) that is not based on the SU(N) group. What theory do you call it then? Feynman Theory?

 

[atyy]

But gravity as a fundamental force is a gauge field. Because what defines a fundamental force is it must have internal gauge transformation like U(1) in electromagnetism, SU(2)xU(1) in Electroweak, SU(3) in strong force. Are you saying it is possible the gravity force has no SU(N) terms?

Also I saw in wiki that "Yang–Mills theory is a gauge theory based on the SU(N) group". For the electromagnetism U(1) that is not based on the SU(N) group. What theory do you call it then? Feynman Theory?

Gravity and Yang-Mills are gauge fields, but gravity is not like a Yang-Mills field in detail.

 

[Matterwave]

Electromagnetism is an "abelian" gauge field theory called Quantum Electro-dynamics. It is not a Yang-Mills theory. A Yang-Mills theory is a non-abelian gauge field theory like QCD or the Electro-weak theory.

There is no well established quantum field theory for gravity, so I'm not sure how you want us to answer the first part of your question.

 

[waterfall]

Electromagnetism is an "abelian" gauge field theory called Quantum Electro-dynamics. It is not a Yang-Mills theory. A Yang-Mills theory is a non-abelian gauge field theory like QCD or the Electro-weak theory.

There is no well established quantum field theory for gravity, so I'm not sure how you want us to answer the first part of your question.

For gravity to be a force. It has to have gauge transformation equivalent. So we still don't know what it is and it is not the metric guv nor the Christoffel symbols. This is what you guys are saying, correct? (say yes for record purposes)

In essence, we don't know what part or what is the gauge representation of the graviton. But what's weird is this. Electroweak has 3 gauge bozons, strong force has 8. If gravity is part of a larger gauge group. Why does it only have one boson?

Maybe gravity is not really a force at all. Maybe it is pure geometry. Remember in GR there is no force of any kind. Just geometry. So if the AsD/CFT has a correlate in our world. Then GR is just a classical limit that equates to pure information in the AsD/CFT world that isn't based on force and geometry. Do you agree?

 

[atyy]

For gravity to be a force. It has to have gauge transformation equivalent. So we still don't know what it is and it is not the metric guv nor the Christoffel symbols. This is what you guys are saying, correct? (say yes for record purposes)

In essence, we don't know what part or what is the gauge representation of the graviton. But what's weird is this. Electroweak has 3 gauge bozons, strong force has 8. If gravity is part of a larger gauge group. Why does it only have one boson?

Maybe gravity is not really a force at all. Maybe it is pure geometry. Remember in GR there is no force of any kind. Just geometry. So if the AsD/CFT has a correlate in our world. Then GR is just a classical limit that equates to pure information in the AsD/CFT world that isn't based on force and geometry. Do you agree?

All the gauge fields are geometrical. This is what Matterwave was saying about a principal bundle in post #35.
Witten, The Problem Of Gauge Theory

 

[waterfall]

All the gauge fields are geometrical. This is what Matterwave was saying about a principal bundle in post #35.
Witten The Problem Of Gauge Theory

Thanks for this crucial idea. This was why I kept encountering the idea of fiber bundles when I studied the Maxwell Equations before and didn't know the connection. Thought they were proposing the Faraday field lines as fiber bundles. So this is also how Weyl united GR and EM by proposing a new 5th dimension which the String Theory took advantage of right now...

 

[Matterwave]

Kaluza was the one who proposed a fifth dimension on which the curvature gives you the Maxwell's equations. Klein later proposed a mechanism by which this fifth dimension could exist without us realizing it (compactification). Thus, this 5-D GR+E&M theory is called "Kaluza Klein theory". String theory uses ideas from this (extra dimensions, and compactification), but is not the same as this.

I don't know what Weyl has to do with that...

 

[waterfall]

Kaluza was the one who proposed a fifth dimension on which the curvature gives you the Maxwell's equations. Klein later proposed a mechanism by which this fifth dimension could exist without us realizing it (compactification). Thus, this 5-D GR+E&M theory is called "Kaluza Klein theory". String theory uses ideas from this (extra dimensions, and compactification), but is not the same as this.

I don't know what Weyl has to do with that...

Yes, checking the Elegant Universe book, it was Klein, not Weyl.

But what Weyl did was this http://www.ams.org/notices/200607/fea-marateck.pdf

"In a 1918 article Hermann Weyl tried to combine electromagnetism and gravity by requiring the theory to be invariant under a local scale change of the metric, i.e., gμν → gμν e^α(x), where x is a 4-vector. This attempt was unsuccessful and was criticized by Einstein for being inconsistent with observed physical results. It predicted that a vector parallel transported from point p to q would have a length that was path dependent. Similarly, the time interval between ticks of a clock would also depend on the path on which the clock was transported.
The article did, however, introduce

• the term “gauge invariance”; his term was Eichinvarianz. It refers to invariance under his scale
change. The first use of “gauge invariance” in English3 was in Weyl’s translation4 of his famous
1929 paper.
• the geometric interpretation of electromagnetism.
• the beginnings of nonabelian gauge theory. The similarity of Weyl’s theory to nonabelian gauge theory is more striking in his 1929 paper."

Objections?

 

[Dale]

I am not so sure that all observables are scalars, but I am pretty sure that all observations are scalars.

 

[twofish-quant]

I am not so sure that all observables are scalars, but I am pretty sure that all observations are scalars.

I don't think that is true. Imagine a variable phi(A) where A is not a point in space but rather a region. Phi is not going to transform as a scalar field. You can also have thermodynamic quantities which are undefined at a specific point and require averaging to have meaning.

Volume is an observable, but it's certainly not a scalar. Wealth is a defined observable, but it's not a scalar.