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Are Christoffel symbols measurable? 
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#37
Feb1612, 02:07 PM

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 fourvector, 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 Relativityit is clear that the metric corresponds to the Au fourvector in electromagnetism. It is not subtle. 


#38
Feb1612, 02:15 PM

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 observablemasny of its components correspond to the componen6ts of the tidal force. 


#39
Feb1612, 02:17 PM

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PF Gold
P: 1,594

Did you read beyond the two lines you quoted? 


#40
Feb1612, 02:19 PM

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. 


#41
Feb1612, 02:23 PM

P: 205

Me: "The Rieman tensor already is an observablemasny 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. 


#42
Feb1612, 02:23 PM

P: 261




#43
Feb1612, 02:25 PM

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PF Gold
P: 1,594




#44
Feb1612, 02:30 PM

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PF Gold
P: 1,594

[tex]R_{\mu\nu\rho\sigma} a^\mu b^\nu c^\rho b^\sigma[/tex] 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. 


#45
Feb1612, 03:29 PM

PF Gold
P: 706

Does anybody have any simple examples?
This appears to be one of the simplest possible examples on pages 46 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. 


#47
Feb1612, 05:48 PM

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 AharonovBohm 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)? 


#48
Feb1612, 05:54 PM

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P: 8,802

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


#49
Feb1612, 06:22 PM

P: 381




#50
Feb1612, 06:51 PM

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P: 8,802

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


#51
Feb1612, 07:06 PM

P: 381

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? 


#52
Feb1612, 07:13 PM

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#53
Feb1612, 07:13 PM

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P: 2,953

Electromagnetism is an "abelian" gauge field theory called Quantum Electrodynamics. It is not a YangMills theory. A YangMills theory is a nonabelian gauge field theory like QCD or the Electroweak 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. 


#54
Feb1612, 07:23 PM

P: 381

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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