Are Christoffel symbols measurable?

In summary, the author says that in GR, all physical observable quantities are tensors. The Christoffel symbols are not physical like tensors and have a different property. They can be made to vanish by coordinate transformations, but that does not mean they cannot be measured. They are the gravitational field.
  • #1
waterfall
381
1
Is it true that in GR the gauge is described by Guv while the potential is the Christoffel symbols just like the gauge in EM is described by phase and the potential by the electric and magnetic scalar and vector potential and the observable the electromagnetic field and the Ricci curvature?

But GR is just geometry. Are the Christoffel symbols measurable or can it only occur in gauge transformation without observable effect? How do you vary the Christoffel symbols just like phase?
 
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  • #2
Christoffel symbols are not physical like tensors. In GR, all physical observable quantities are tensors. In some chosen frame (coordinates) the Christoffel symbols are 0. One can always choose this to be so locally at any point in space-time.
 
  • #3
In most fonts, Christoffel symbols are about 1/4 inch. :devil:
 
  • #4
Someone wrie "Christoffel symbols are not physical like tensors. In GR, all physical observable quantities are tensors"

This is just wrong. Tensors are objects that have a certain property under coordinate transformations. Christoffel symbols have a different property, and can even be made to vanish by coordinate transformations, but that does not mean they cannot be measured in a particular coordinate system. Essentially they are the gravitational field. So saying they have no observable properties is saying that in the theory of gravitation there are no gravitational fields. So does the writer think that jumping off a cliff will not have any physical effects?

I've seen this before--people going too extreme over the Principle of Covariance to the point where they do not understand that General Relativity is a PHYSICAL theory.
 
  • #5
I'm asking this because of the following analogy between electromagnetic and gravity.

The gauge of Electromagnetism is phase alpha while that of gravity is Guv.

And differentiation of this gives the gauge field Ab (magnetic vector potential) in electromagnetism and F^c (ab) (or Christoffel symbols (gravitational field)) in GR.

And a second differentiation gives the directly observable field(s) E and B in Electromagnetism and R^c (dab) (or Riemann tensor (curvature) in GR.

As gauge field, the Ab (magnetic vector potential) is not observable although it has an effect. So I think the counterpart Christoffel symbols (gravitational field)) is not observable too? How much are these gauge phase alpha and Guv identical and how do they differ?

In the gauge phase alpha, you can change the angle from 0 to 360 degrees. In the Christoffel symbols, what are the corresponding variable that you can change like the phase alpha in electromagnetic gauge?
 
  • #6
ApplePion said:
Someone wrie "Christoffel symbols are not physical like tensors. In GR, all physical observable quantities are tensors"

This is just wrong. Tensors are objects that have a certain property under coordinate transformations. Christoffel symbols have a different property, and can even be made to vanish by coordinate transformations, but that does not mean they cannot be measured in a particular coordinate system. Essentially they are the gravitational field. So saying they have no observable properties is saying that in the theory of gravitation there are no gravitational fields. So does the writer think that jumping off a cliff will not have any physical effects?

I've seen this before--people going too extreme over the Principle of Covariance to the point where they do not understand that General Relativity is a PHYSICAL theory.

This post sounds pretty condescending.
 
  • #7
PAllen said:
In most fonts, Christoffel symbols are about 1/4 inch. :devil:

This is a common mistake made by relativists. It is a font dependent quantity.:wink:

Matterwave said:
Christoffel symbols are not physical like tensors. In GR, all physical observable quantities are tensors. In some chosen frame (coordinates) the Christoffel symbols are 0. One can always choose this to be so locally at any point in space-time.

Observable quantities are scalars. Invariant scalars result from contractions of tensors, that's what makes them tensors. The CSs are not tensors but they are surely physical because they appear in the geodesic equations and the energy pseudo-tensor.
 
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  • #8
Someone wrote "Observable quantities are scalars."

This is quite incorrect. Scalars are quantities that have the same numerical value in different coordinate systems. But there are quantities that have different values in different coordinate systems that are quite observable. The electromagnetic field for example has different values in different coordinate systems. Are you saying it is not observable?

As for the person who claimed I was being condescending, this discussion struck a nerve with me because (outside of this board) I have been trying to discuss something serious with other physicists and they have been making some of the same mistakes made here, and are difficult to reason with. They think they are being "geometric" or that they have some deep understanding of the Principle of Covariance, but in reality they are failing to understand that physics describes the physical world.
 
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  • #9
Waterfall, it is not the metric that corresponds to the electromagnetism phase. The metric corresponds to the vector potential of electromagnetism. For example, A0 in electromagnetism is the Coulomb potential in electrostatics, while g00 is related to the Newtonian potential of gravitation. Likewise Fuv which is built from derivatives of Au is the electric and magnetic fields, while the Christoffel symbols which contain derivatives of the metric are the gravitational fields, including fields that are forces linear in the velocity of the particle acted upon (analogous to magnetism) and forces quadratic in the velocity.

If you work in the Linear Field Approximation of General Relativity there is a quantity closely related to the metric which can be transformed by a gauge transformation in a way analogous to the way the eletromagnetic four-vector can be transformed by the electromagnetism gauge transformation. You should look at the Linear Field Approximatio, and its gauge transformation, if you are not already familiar with it.

P.S. You should not refer to the metric as "Guv", because that looks like you are referring to the Einstein Tensor (Ruv - 1/2 guv R). You should refer to the metric as lower case "guv".
 
  • #10
The electric field and magnetic fields combine to form a tensor, the Faraday. By observable, I meant something we could measure with an instrument of some kind. It seems to me that observables should be tensors (more specifically the components of these tensors in any particular frame). How would you go about measuring the Christoffel symbols? It seems to me that one can only calculate them, given a coordinate system.

Surely, if I just took flat space-time with spherical polar coordinates describing the 3-D part, the Christoffel symbols "exist" in the sense that I can calculate them. However, there's nothing physical about that, they exist only because I chose some set of coordinates. The space-time is still flat. That's not gravity, that's just my coordinate system being non-Cartesian. What's physical is the Riemann tensor (or contractions thereof) which I can observe via the failure of a (e.g. angular momentum) vector to return to itself after parallel transport around a loop, as was seen by gravity probe B.

I guess you can call me "difficult to reason with", but you'll have a hard time convincing me that any quantity which is so obviously an artifact of a coordinate system choice is physical.
 
  • #11
Matterwave said:
The electric field and magnetic fields combine to form a tensor, the Faraday. By observable, I meant something we could measure with an instrument of some kind. It seems to me that observables should be tensors (more specifically the components of these tensors in any particular frame). How would you go about measuring the Christoffel symbols? It seems to me that one can only calculate them, given a coordinate system.

.

You can only calculate Christoffel symbols within a given coordinate system. But that is the case for tensors like the components of "Faraday"also.

ALL physical objects have values only within a given coordinate system. If you want to do a real experiment you have to set up a coordinate system. This is precisely the problen with the Principle of Covariance pseudo-scientists--they think that because objects have different values in different coordinate systems that these objects are not real.

The way you would measure Christoffel symbols would be as follows. To measure [1,00] you take an apple and drop it. The acceleration is to a good approximation [1,00]. This is not much different from how you would measure the E field in electromagnetism. To measure other components of the affine connections you would measure accelerations as a function of velocities just as you would measure magnetic fields. (The affine connection of course also has forces quadratic in the velocities)

Can you make the Christoffel symbols be different in a different coordinate system. Of course. But when you deal with a physical situation you choose a coordinate system and work within it. Yes, the same "things" happen if you choose a diffrent coordinate system, but that confuses the Principle of Covariance pseudo-scientists.

Let me give you another example. Suppose someone asks the speed of an object dropped from a height of 30 meters in the Earth's gravitational system ignoring air resistance. A physicist would calculate it as approximately square root of [(2) (9.8) (30)] within the coordinate system where the Earth is at rest. . A Principle of Covariance pseudo-scientist using his logic applied to this case would go on about velocity depending on the choice of coordinate system, and thus there is no answer. If I would say to him that I want to choose the coordinate system where the Earth is at rest, he would say to me that there is no preferred cordinate system. And he would have a smirk on his face, very impressed with himself thinking that he things "geometrically". And he would think he was thinking like Einstein. Einstein, of course would think the guy is not a scientist.

As I said before I actually am dealing with people like this. I don't know if according to the message board rules I can tell you their names here, but you can send me a message and I will.
 
  • #12
I guess you can call me "difficult to reason with" said:
If you really think that, you would be willing jump out of an airplane without a parachute, because the gravitational field is an artifact of a coordinate system--it can be made to vanish by a coordinate transformation.
 
  • #13
Suppose I was selling gum and I was asking for 10 pennies for a piece of gum. So I say "In pennies, the price is 10". Someone could say "But in nickels the price would be 2, and nickels are as valid a unit as pennies." So I would say "OK give me 2 nickels". He could say "But you can't say the price is 2, because in dimes the price is one. You see the gum does not have an actual price being that the price is different in different systems of coins. It would have to have the same price in any system of coins in order for the price to have meaning." I would then kick him out of my store.
 
  • #14
ApplePion said:
The way you would measure Christoffel symbols would be as follows. To measure [1,00] you take an apple and drop it. The acceleration is to a good approximation [1,00].

Huh? Neglecting air resistance if you measure acceleration of the apple with an accelerometer you don't get any acceleration. You are mixing the affine connections with something as trivial (well not so trivial for someone jumping out a cliff) as the relative acceleration between objects.
 
  • #15
In case you are thinking that it is the ability to make the Christoffel symbols go to *zero* by coordinate transformation that makes them not physically real, let's try this.

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".
 
  • #16
ApplePion said:
Waterfall, it is not the metric that corresponds to the electromagnetism phase. The metric corresponds to the vector potential of electromagnetism. For example, A0 in electromagnetism is the Coulomb potential in electrostatics, while g00 is related to the Newtonian potential of gravitation. Likewise Fuv which is built from derivatives of Au is the electric and magnetic fields, while the Christoffel symbols which contain derivatives of the metric are the gravitational fields, including fields that are forces linear in the velocity of the particle acted upon (analogous to magnetism) and forces quadratic in the velocity.

If you work in the Linear Field Approximation of General Relativity there is a quantity closely related to the metric which can be transformed by a gauge transformation in a way analogous to the way the eletromagnetic four-vector can be transformed by the electromagnetism gauge transformation. You should look at the Linear Field Approximatio, and its gauge transformation, if you are not already familiar with it.

P.S. You should not refer to the metric as "Guv", because that looks like you are referring to the Einstein Tensor (Ruv - 1/2 guv R). You should refer to the metric as lower case "guv".

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??
 
  • #17
TrickyDicky said:
Huh? Neglecting air resistance if you measure acceleration of the apple with an accelerometer you don't get any acceleration. .

Suppose I have a camera that takes pictures of the apple every nanosecond. From seeing the location of the apple every nanosecond I can to an excellent approximation measure the velocity as a function of time. From the velocity as a function of time I get the acceleration.

Do you realize you are actually arguing that the formula s = (1/2) at^2 (for the frame where the Earth is at rest) is not experimentally detectable or meaningful?

You are castrating physics. Physics, from your perspective, cannot make real scientific predictions.
 
  • #18
waterfall said:
Are you saying the table in page 173 is wrong??

Yes, it is clearly wrong. I explained why it is wrong in a way that should have been convincing. You no doubt will not examine my argument on its merits, because you saw something in a "book" saying otherwise.

I really do not like argument-by-authority, but as it turns out it is quite convenient to give you an argument in the sort of form that appeals to you.

Go to the 1994 edition of Ohanian and Ruffini, and look at Table 3.1 on page 144. They have it my way, in contradiction to what your book told you.

So now you have two books disagreeing. So what are you going to do?
 
  • #19
ApplePion said:
Suppose I was selling gum and I was asking for 10 pennies for a piece of gum. So I say "In pennies, the price is 10". Someone could say "But in nickels the price would be 2, and nickels are as valid a unit as pennies." So I would say "OK give me 2 nickels". He could say "But you can't say the price is 2, because in dimes the price is one. You see the gum does not have an actual price being that the price is different in different systems of coins. It would have to have the same price in any system of coins in order for the price to have meaning." I would then kick him out of my store.

What's funny about that example is that people do use it in finance. Google for "geometric arbitrage theory."

http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1113292

The basic idea is that you can convert gum -> 10 pennies -> 2 nickels -> gum, and this gives you zero curvature and zero profit. Now what you want to do is to look for situations where you can do gum -> 10 pennies -> 2.1 nickels -> 1.1 pieces of gum -> 11 pennies -> etc. etc.

So the idea is that you take a complicated network, use lots of math to characterize the network. Look very quickly for loops, and when you find them make very fast trades and profit.
 
  • #20
ApplePion said:
Someone wrote "Observable quantities are scalars."
This is quite incorrect. Scalars are quantities that have the same numerical value in different coordinate systems. .


No they are not. Lorentz scalars are invariant. I use the word scalar to indicate a single numerical value, which may be a component of a vector or tensor.

Not only are you condescending but you don't know what you are talking about.
I repeat - all measureable quantities are scalars.
 
  • #21
" I use the word scalar to indicate a single numerical value, which may be a component of a vector or tensor."

A component of a vector or a tensor cannot be a scalar (except in the most trivial case of a zeroth order tensor)

"Not only are you condescending but you don't know what you are talking about."

Interesting that you are claiming that.

"I repeat - all measureable quantities are scalars."

The electromagnetic field is measurable and it is not a scalar. The stress energy tensor is measurable and it is not a scalar. Indeed all "quantities" are measurable, or else they could not be quantities. This is physics, not "philosophy".
 
  • #22
ApplePion said:
Yes, it is clearly wrong. I explained why it is wrong in a way that should have been convincing. You no doubt will not examine my argument on its merits, because you saw something in a "book" saying otherwise.

I really do not like argument-by-authority, but as it turns out it is quite convenient to give you an argument in the sort of form that appeals to you.

Go to the 1994 edition of Ohanian and Ruffini, and look at Table 3.1 on page 144. They have it my way, in contradiction to what your book told you.

So now you have two books disagreeing. So what are you going to do?

I only have the 1976 copy of Ohanian Gravitation and Spacetime. Where is the Table 3.1 on the 1st edition? If you can shoot a cam of Table 3.1. Please send it to me. I can't find it anywhere. No preview in amazon or google book.
 
  • #23
Mentz114 said:
No they are not. Lorentz scalars are invariant. I use the word scalar to indicate a single numerical value, which may be a component of a vector or tensor.

I think the usual meaning of the word "scalar" is that it is a quantity or field phi such that under a certain transformation x --> x' one has
[tex]
\phi'(x') = \phi(x)
[/tex]
These transformations can be Galilei, Lorentz, general coordinate, SU(N) or whatever transformations. A component of a general tensor clearly doesn't fulfill this condition.
 
  • #24
waterfall said:
I only have the 1976 copy of Ohanian Gravitation and Spacetime. Where is the Table 3.1 on the 1st edition? If you can shoot a cam of Table 3.1. Please send it to me. I can't find it anywhere. No preview in amazon or google book.

Let's see if I can guide it to you--I think it was in the earlier versions also. In my version it is in the chapter called "Linear Approximation". The chapter is not super-long, so you can just start leafing thru it and you should see a chart with 2 columns. One column is labeled "Electromagnetism" and the other is labelled "Gravitation (linear approximation)". Then it shows various things, first the Electromagnetism version, and then the linear field gravitation analogue. The Au four-vector quantities for Electromagnetism match-up to the analogous huv 4-by-4 quantity huv, where huv is a simple quantity in terms of the metric (without derivatives). For example in electromagnetism they have a wave equation in Au, and then they have a similar wave equation in huv for linearized gravity.

Thanks.
 
  • #25
ApplePion said:
Let's see if I can guide it to you--I think it was in the earlier versions also. In my version it is in the chapter called "Linear Approximation". The chapter is not super-long, so you can just start leafing thru it and you should see a chart with 2 columns. One column is labeled "Electromagnetism" and the other is labelled "Gravitation (linear approximation)". Then it shows various things, first the Electromagnetism version, and then the linear field gravitation analogue. The Au four-vector quantities for Electromagnetism match-up to the analogous huv 4-by-4 quantity huv, where huv is a simple quantity in terms of the metric (without derivatives). For example in electromagnetism they have a wave equation in Au, and then they have a similar wave equation in huv for linearized gravity.

Thanks.

Ok. I found Table 3.1 in page 99. But it is a comparison between EM and linearized gravity. Not between EM and GR. Where is the Christoffel symbols in linearized gravity that must correspond to the E and B field. And how did the other book by Cromwell messed it up? His whole book is about GR and he got it wrong? Or maybe there are 2 camps of General Relativists with different ideas? Hope others can give an opinion concerning this too.. like Matterwave (maybe who belong to the other GR camp).
 
  • #26
haushofer said:
I think the usual meaning of the word "scalar" is that it is a quantity or field phi such that under a certain transformation x --> x' one has
[tex]
\phi'(x') = \phi(x)
[/tex]
These transformations can be Galilei, Lorentz, general coordinate, SU(N) or whatever transformations. A component of a general tensor clearly doesn't fulfill this condition.

The component of a general tensor is a single number and may be measureable. I did not imply it was invariant under transformation. I can't see the import of your remark.
When the term 'scalar field' is used, does that not mean there is a single value (scalar) at every point ? This is what I take as the usual meaning of the word.

I repeat - all measureable quantities are scalars ( single numbers).

Someone said,
The electromagnetic field is measurable and it is not a scalar. The stress energy tensor is measurable and it is not a scalar. Indeed all "quantities" are measurable, or else they could not be quantities. This is physics, not "philosophy".
Individual measurements are single numbers. For instance the x-component of velocity. Vectors and tensors (including the EMT) are collections of individual numbers. Vectors and tensors exist only in our minds and can never measure them except by collecting measurements of single numbers.

This is physics, not "philosophy".

Arrogant and condescending at the same time.
 
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  • #27
I would like to comment on some examples of alleged non-invariants proposed as measurements. First, a meta-comment. In physics, no matter what coordinates or observational point of view you use, everyone had better agree on what the result of a given instrument making a given measurement is. You cannot have observer dependence of whether a particular camera gets a picture of an apple or not.

You cannot measure electric field in the abstract. You measure electric field with a particular instrument which has a particular orientation and state of motion (world line). The measurements it makes are defined as contractions between the field and worldline and orientation of its tetrad. As such, in all coordinates, being contractions (invariant scalars), they come out the same.

A sequence of pictures of a falling apple is a record of the events of arrival of light signals from the apple following a geodesic and the world line (with proper acceleration) of someone sitting on earth. The proper time between a sequence of events along a world line is invariant.

Units have nothing to do with invariants.

I have seen numerous mistakes made in relativity by failure to phrase observable quantities as invariants.
 
  • #28
PAllen said:
I have seen numerous mistakes made in relativity by failure to phrase observable quantities as invariants.

Observables need not be invariants.

For example the electric field is a physically observable quantity but is not an invariant. Even the electomagnetic 4-by-4 tensor is not an invariant--tensors transform under coordinate transformations. The affine connection is not even a tensor, yet it is observable-- gravity exists.

This is quite an amazing experience.
 
  • #29
ApplePion said:
Observables need not be invariants.

For example the electric field is a physically observable quantity but is not an invariant. Even the electomagnetic 4-by-4 tensor is not an invariant--tensors transform under coordinate transformations. The affine connection is not even a tensor, yet it is observable-- gravity exists.

This is quite an amazing experience.

I disagree. None of these are actually observable. Observables are measurements. Measurements (in this class of theory) are predicted/computed as invariants constructed from the interaction of the instrument with the field. The measurements must be invariants. The fields are theoretical constructs whose purpose is to allow us to compute observables. It is only the invariant observables we can verify.
 
  • #30
ApplePion said:
This is quite an amazing experience.

But not as much as castrating physics, trust me on this.
 
  • #31
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.
 
  • #32
waterfall said:
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.
 
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  • #33
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).
 
  • #34
atyy said:
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).
 
  • #35
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.
 
<h2>1. What are Christoffel symbols and how are they related to measurement?</h2><p>Christoffel symbols are mathematical objects used in differential geometry to describe the curvature of a space. They are related to measurement in that they help us understand how distances and angles change as we move through a curved space.</p><h2>2. Can Christoffel symbols be measured directly?</h2><p>No, Christoffel symbols cannot be measured directly. They are abstract mathematical objects that represent the curvature of a space and are used in equations to calculate measurements.</p><h2>3. How are Christoffel symbols used in physics?</h2><p>Christoffel symbols are used in general relativity to describe the curvature of spacetime and how it is affected by the presence of matter and energy. They are also used in other areas of physics, such as in fluid dynamics and quantum mechanics.</p><h2>4. Are there any practical applications of Christoffel symbols?</h2><p>Yes, Christoffel symbols have many practical applications in fields such as engineering, computer graphics, and robotics. They are used to model and analyze the behavior of curved surfaces and objects.</p><h2>5. Are there any limitations to using Christoffel symbols in measurement?</h2><p>While Christoffel symbols are a useful tool in understanding the curvature of a space, they have limitations in certain situations. For example, they do not take into account quantum effects and cannot fully describe the behavior of black holes.</p>

1. What are Christoffel symbols and how are they related to measurement?

Christoffel symbols are mathematical objects used in differential geometry to describe the curvature of a space. They are related to measurement in that they help us understand how distances and angles change as we move through a curved space.

2. Can Christoffel symbols be measured directly?

No, Christoffel symbols cannot be measured directly. They are abstract mathematical objects that represent the curvature of a space and are used in equations to calculate measurements.

3. How are Christoffel symbols used in physics?

Christoffel symbols are used in general relativity to describe the curvature of spacetime and how it is affected by the presence of matter and energy. They are also used in other areas of physics, such as in fluid dynamics and quantum mechanics.

4. Are there any practical applications of Christoffel symbols?

Yes, Christoffel symbols have many practical applications in fields such as engineering, computer graphics, and robotics. They are used to model and analyze the behavior of curved surfaces and objects.

5. Are there any limitations to using Christoffel symbols in measurement?

While Christoffel symbols are a useful tool in understanding the curvature of a space, they have limitations in certain situations. For example, they do not take into account quantum effects and cannot fully describe the behavior of black holes.

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