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

Affine connection transformation

  1. Jul 4, 2014 #1
    Dear All,


    I am teaching myself tensors for the first time. I am using D'Inverno's book and got stuck at page 73. Basically, he says: demand that the first term on the left of the equation to be a type (1,1) tensor. Then he gets the affine connection transformation.

    I basically wrote the first term as a second rank mixed tensor transformation. Then I got stuck. I am not sure on how to isolate (?) the affine connection and show how it transforms. I tried many times but failed due to my lack of knowledge of tensors. Could someone help me understand this please?

    [itex]\nabla_{c}X^{a}[/itex]= [itex]\partial_{c}[/itex][itex]X^{a}[/itex]+[itex]\Gamma_{bc}^{a}[/itex][itex]X^{b}[/itex]

    Thanks in advance! (Sorry if my post isn't very informative as I have to go for 6 hours. When I come back I will be more than happy to upload pictures of my attempts)
     
    Last edited: Jul 4, 2014
  2. jcsd
  3. Jul 4, 2014 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    You can make a coordinate transformation on the term:

    $$\partial_c X^a +\Gamma^a_{bc}X^b$$

    You know how ##\partial_c X^a## transform (equation 6.1), and it does not transform as a tensor, so now you need that the transformation of ##\Gamma^a_{bc}X^b## to cancel out the "wrong terms".

    Are you having troubles with the details?
     
  4. Jul 5, 2014 #3

    haushofer

    User Avatar
    Science Advisor

    Carroll does this in detail in his gr notes, if i remember correctly.
     
  5. Jul 5, 2014 #4
    Oh I didn't think of it in that way. Yes my trouble was in the details. I keep messing up the dummy variables and didn't know that I can leave the connection alone while transforming the tensors I know. Like the way Carroll did.

    Edit: I managed to derive it finally :smile: but I have a question. In equation 6.1, why do we make [itex]\frac{\partial}{\partial x^{'c}}[/itex]= [itex]\frac{\partial x^{d}}{\partial x^{'c}}[/itex][itex]\frac{\partial}{\partial x^{d}}[/itex] ??? at first I didnt do this and it didnt work :frown:
     
    Last edited: Jul 5, 2014
  6. Jul 5, 2014 #5
    This indeed helped me a lot. Made me discover where I need improvements in. Thanks :smile:
     
  7. Jul 5, 2014 #6

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    This is just the chain rule from multi-variable calculus.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook