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Question about affine connection definition, Weinberg's Gravitation

  1. Feb 8, 2014 #1
    page 71 he appears to define the affine connection in terms of derivatives on the locally inertial coordinates with respect to the laboratory coordinates

    and then the very next page claims that all you need is the affine connection and metric tensor to determine the locally inertial coordinate system

    he says you get a differential equation for the locally inertial coordinate system in terms of laboratory coordinates, but this differential equation comes from the definition on the previous page

    I guess my question is what is the logic here? This seems circular, no wait... it IS circular. We need something else for this "differential equation" to be useful.
  2. jcsd
  3. Feb 8, 2014 #2


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    The best advice I can offer is to get a better book on GR. Weinberg's book is terrible. It's so terrible I almost felt like crying after having seen it.
  4. Feb 8, 2014 #3
    lol really? What book would you suggest? I've been reading Wald mainly but he started confusing me when he started talking about special and general covariance. He was saying "in our view the affine connection is a tensor" which makes no sense because I always thought it wasn't because of how it transforms. I thought Weinberg might be more explicit and he has so far but still some confusion in the logic. Again, any suggestions?
  5. Feb 8, 2014 #4


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    Yeah Wald isn't any better if your intent is to get a grasp of tensor calculus in the context of GR. I just hate Weinberg's book because he uses coordinates for everything like oh my god what's wrong with defining things geometrically like they're meant to be defined and performing coordinate-free calculations whenever possible. Mathematicians have come up with elegant geometric definitions of objects such as affine connections so that we don't have to deal with pointless and confusing coordinate-based definitions of everything. This is how I felt after perusing the book: http://stream1.gifsoup.com/view3/1156567/angry-panda-smash-o.gif

    Do you have Springer access through a university by any chance? If so I would really recommend the following book: https://www.amazon.com/General-Relativity-Graduate-Texts-Physics/dp/9400754094 it basically bleeds geometry and it's impossible not to love it.
  6. Feb 8, 2014 #5
    Oh cool I have never heard of that one but I will check it out.

    That panda picture is hilarious by the way. I definitely share that feeling from time to time while reading this book. lol
  7. Feb 9, 2014 #6


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    There's nothing circular. He just says that you can go either way. Given the inertial coordinates you can use Eq (3.2.4) to calculate Γ. Conversely, given the values of g and Γ you can use Eq (3.2.11) to determine the inertial coordinates.
  8. Feb 9, 2014 #7
    It's circular because all the equation is not really a differential equation, it's a statement that 1 = 1. The point is that the equation comes from the definition so it doesn't tell me anything unless I can determine the affine connection some other way.

    He defined the affine connection in terms of the differential and then says we can use the differential to find the affine connection. There needs to be something else to determine the affine connection so this statement is not trivial/circular.
    Last edited: Feb 9, 2014
  9. Feb 9, 2014 #8


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    I don't see anything circular. A<=>B means that if you know A, you can get B and viceversa.
  10. Feb 9, 2014 #9
    Yes A<=>B means that if you know A, you can get B and viceversa.

    However, relabeling A as B does not tell me new information. And saying I can get A given B or B given A is useless if I have arbitrarily DEFINED the relationship between A and B. The point is he explicitly DEFINES the affine connection and then says "oh look I can invert it so the label is now on the other side. "
  11. Feb 9, 2014 #10
    If I say A [itex]\equiv[/itex] B + 1

    and then say look, if I'm given A I can find B by...

    A - 1 = B

    but by definition A [itex]\equiv[/itex] B + 1

    So, B + 1 - 1 = B

    Oh yippee everything checks out B = B. True and useless. I need something besides B to determine A for this to be useful.


    From wald he gives the affine connection in terms of a covariant derivative that when acting on the metric gives zero, maybe this is the other condition required to make this useful.
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