# Are Christoffel symbols measurable?

by waterfall
Tags: christoffel, measurable, symbols
 P: 205 "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.
 P: 205 "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.
PF Gold
P: 1,594
 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?
 P: 205 "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.
 P: 205 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.
P: 261
 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.
PF Gold
P: 1,594
 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.
PF Gold
P: 1,594
 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.
 PF Gold P: 706 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.
 Sci Advisor P: 2,953 Simple examples of what? Christoffel symbols?
 P: 381 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)?
P: 8,802
 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)
P: 381
 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?
P: 8,802
 Quote by waterfall 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
P: 381
 Quote by atyy 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?
P: 8,802
 Quote by waterfall 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.
 Sci Advisor P: 2,953 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.
P: 381
 Quote by 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.
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?

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