Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Are Christoffel symbols measurable?

  1. Feb 15, 2012 #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?
     
  2. jcsd
  3. Feb 15, 2012 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  4. Feb 15, 2012 #3

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    In most fonts, Christoffel symbols are about 1/4 inch. :devil:
     
  5. Feb 15, 2012 #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.
     
  6. Feb 15, 2012 #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?
     
  7. Feb 16, 2012 #6

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    This post sounds pretty condescending.
     
  8. Feb 16, 2012 #7

    Mentz114

    User Avatar
    Gold Member

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

    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.
     
    Last edited: Feb 16, 2012
  9. Feb 16, 2012 #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.
     
    Last edited: Feb 16, 2012
  10. Feb 16, 2012 #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".
     
  11. Feb 16, 2012 #10

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  12. Feb 16, 2012 #11
    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.
     
  13. Feb 16, 2012 #12
     
  14. Feb 16, 2012 #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.
     
  15. Feb 16, 2012 #14
    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.
     
  16. Feb 16, 2012 #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".
     
  17. Feb 16, 2012 #16
    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??
     
  18. Feb 16, 2012 #17
    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.
     
  19. Feb 16, 2012 #18
    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?
     
  20. Feb 16, 2012 #19
    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.
     
  21. Feb 16, 2012 #20

    Mentz114

    User Avatar
    Gold Member


    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Are Christoffel symbols measurable?
  1. Christoffell symbols (Replies: 6)

  2. Christoffel symbols (Replies: 5)

  3. Christoffel symbols (Replies: 4)

Loading...