Are Christoffel symbols measurable?

[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.

 

[twofish-quant]

Suppose I am a credit card company owner and I tell a customer that he owes me a thousand dollars. The customer says "If I give you 200 dollars I will only owe 800 dollars. Under those conditions it would no longer be 1,000 dollars. Me giving you 200 dollars and owing 800 is an equivalent situation as me owing 1000" So I say "OK that makes no difference in our situation, give me 200 dollars." Then he says, "But if I change it so that I give you 1,000 in cash and now had no debt we would be in an equivalent situation to me owing you 1,000 dollars. Therefore since I can make the debt vanish by an equivalence transformation, the debt does not have real meaning".

This doesn't work. In most common situations, wealth has scaling symmetry but not translational symmetry. I.e. if you have an economic situation, you can describe that situation equivalently by multiplying it by a scaling factor (i.e. do all your calculations in euros rather than dollars). Economics is not translationally symmetric. (I.e. if you add a constant amount to an economic situation, you are describing a different situation).

This has a number of implications

1) the important quantities are log-price rather than price
2) debt and credits are invariant quantities. If A is in debt to B, we can describe the amount of debt equivalently in dollars and euros, but we cannot by a change of coordinates eliminate the debt

A lot of the equations of finance can be derived from gauge theory.

 

[Dale]

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.

In both cases an observation of phi(A) or the average thermodynamic quantities is the interaction of some measuring device with the system of interest in order to produce a number. That number is the same, regardless of the coordinate system used to describe the experiment. Therefore the observation is a scalar.

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

I think you are correct here, so I will modify my above statement:

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

 

[twofish-quant]

That number is the same, regardless of the coordinate system used to describe the experiment. Therefore the observation is a scalar.

No it's not. I have a can of Coke that is 16 fluid ounces in one coordinate system and 473.18 mL in another. That's not a scalar.

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

I think you might have to define what is an observation. One thing about fluid measurement is that it doesn't correspond to a quantum mechanical operator, so if you argue that fluids can't be "observed" in a field theoretic sense, then I might be inclined to agree with you, but that defines observed in a what that's different enough from the normal meaning of the word that one has to be careful.

One thing here is that narrowing "observations" to things that can only be described in terms of fields is much too heavy a restriction. I take a coke can, fill it with water, and then dump out the water into a bucket of known volume. That doesn't fit well in field theory. For that matter prices are observations, but they don't fit into field theory and they certainly are not scalars (i.e. an observation of price gets you different numbers based on whether you are talking about dollars or euros).

 

[PAllen]

No it's not. I have a can of Coke that is 16 fluid ounces in one coordinate system and 473.18 mL in another. That's not a scalar.

.

That's units, not an issue of invariance. By that standard nothing is invariant. The norm of proper acceleration (for example) is a scalar if anything is, but it still has units that are purely conventional.

 

[twofish-quant]

Another observable that's not a scalar. Color. In order to specify color you need to include three components (R, G, B). If you have only one component, you've measured "redness", "greenness" or "blueness' but you haven't measured color. Also because of redshift, different observers in different coordinate systems will see different colors, and different people will see different colors in quantifiable and predictable ways (i.e. if you are color blind, the coordinate system changes).

Now you could argue that all observables can be decomposed into scalars, but that's something quite different.

 

[twofish-quant]

That's units, not an issue of invariance.

One way of thinking about differential geometry in terms of units of measure. Also I'm using the term "scalar" in a very mathematically narrow sense, and "observation" in a very broad sense. The reason I do that is that I can open a book on differential geometry and get a mathematically precise definition of "scalar" whereas there isn't an obvious mathematically precise definition for "observation."

By that standard nothing is invariant.

I think the question is "invariant with respect to what." In relativity, you typically keep thermodynamic quantities invariant, and then consider only coordinate transforms in space-time.

The norm of proper acceleration (for example) is a scalar if anything is, but it still has units that are purely conventional.

If you use the "tight" definition of scalar, then clearly it's not. Also, you **can** tell what's a scalar quantity by looking at the units. The fine-structure constant and pi are scalar with respect to everything. Rest mass is a scalar quantity. Charge is a scalar quantity.

This sounds like a massive nitpick. It is, but if you make these very fine distinctions then all sorts of useful things happen.

 

[twofish-quant]

My measurement of the can's volume is a scalar. If I measure it to be 16 fl oz then everyone in all coordinate systems will agree that I measured it to be 16 fl oz. If you measure it to be 473.18 mL then everyone in all coordinate systems will agree that you measured it to be 473.18 mL. That is a scalar.

I think we are have different definitions of what a "scalar" is. I'm defining it as a quantity that doesn't change when you change coordinate systems. I measure something in my coordinate system. You measure something in your coordinate system. We get the same number. There are some things that you can measure that have that characteristic (electric charge if you vary only space and time coordinates). There are some things that you can measure that *don't* have that characteristic (volume). Classifying things according to how they behave turns out to be useful.

 

[Dale]

I think you might have to define what is an observation.

Yes, it would be important to define what is an "observation" and also what is an "observable". I have a feeling that the disagreement in this thread is primarily due to poor definitions.

I think that Matterwave et al. are talking about "observations" being scalars and I think that ApplePion et al. are talking about "observables" not being scalars. And I think that the disagreement is that they are using the same words for two different concepts.

 

[twofish-quant]

Also I think that I've thought of an observable that clearly is not a scalar. Chriality. A particle is either left-handed or right-handed, and since this is a binary quantity. It's not a scalar.

For that matter, you flip a coin, the "headness" or "tailness" of the coin is a boolean quality which is not a scalar. For that matter any observation or observable that is binary isn't a scalar.

Finally, for the people that still insist that observations have to be a scalar, how do you know it's a scalar and not a pseudo-scalar? I have a feeling that "height" is a scalar, but "leftness" is a pseudoscalar.

 

[Dale]

Also I think that I've thought of an observable that clearly is not a scalar.

From my side there is not any disagreement that some observables are not scalars.

However, an observation of any of the quantities you have mentioned is a scalar (i.e. it is unchanged under diffeomorphisms) even if the corresponding observable is not.