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

Index Notation - Quick one-line working

  1. Jan 31, 2015 #1
    I was reading my lecturer's notes on GR where I came across the geodesic equation for four-velocity. There is a line which read:

    Summing them up,
    [tex]\partial_i g_{aj} u^i u^j - \frac{1}{2} \partial_a g_{ij} u^i u^j = \frac{1}{2} u^i u^j \partial_a g_{ij} [/tex]

    I'm trying to understand how LHS = RHS, surely the indices ##a## and ##i## are different, how can you simply combine them?

    I tried writing them out:

    [tex]\partial_i g_{aj} u^i u^j - \frac{1}{2} \partial_a g_{ij} u^i u^j[/tex]
    [tex] = g_{aj} \partial_i u^i u^j + \left( u^i u^j \partial_i g_{aj} - \frac{1}{2} u^i u^j \partial_a g_{ij} \right) - \frac{1}{2} g_{ij} \partial_a u^i u^j [/tex]


    Source: http://physicspages.com/2013/04/02/geodesic-equation-and-four-velocity/
     
  2. jcsd
  3. Jan 31, 2015 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor


    This equation doesn't appear anywhere in the link you gave. The closest thing I can find at that link is:

    $$
    \partial_i g_{aj} u^i u^j + g_{aj} \frac{d u^j}{d \tau} - \frac{1}{2} \partial_a g_{ij} u^i u^j = 0
    $$

    which gives, expanding out the second term and moving it to the RHS,

    $$
    \partial_i g_{aj} u^i u^j - \frac{1}{2} \partial_a g_{ij} u^i u^j = - g_{aj} u^i \partial_i u^j
    $$

    So where does the equation you wrote come from?​
     
  4. Jan 31, 2015 #3


    Starting from the geodesic equation:

    [tex] \partial_i g_{aj} u^i u^j + g_{aj} \frac{d u^j}{d \tau} - \frac{1}{2} \partial_a g_{ij} u^i u^j = 0 [/tex]

    I'm confused by this step when 'summation' was mentioned. Without multiplying ##u^a## we get:

    [tex] LHS = g_{aj} \frac{d u^j}{d \tau} + \partial_i g_{aj} u^i u^j - \frac{1}{2} \partial_a g_{ij} u^i u^j [/tex]

    [tex] = g_{aj} \frac{d u^j}{d \tau} + \frac{1}{2} u^i u^j \partial_a g_{ij} [/tex]


    Comparing 2nd and 3rd terms, that's where I couldn't get my head around it.
     
  5. Jan 31, 2015 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    If you don't multiply by ##u^a## then the derivation doesn't work; that multiplication is crucial because it means there are no free indexes in any of the terms, so you can swap dummy indexes at will. That is what allows the ##a## and ##i## indexes to be swapped so that you can combine terms.

    That said, I'm not entirely sure the article you linked to has everything correct. It would be easier if it stuck to index notation everywhere. I don't have time now to dig into it in more detail, unfortunately.
     
  6. Jan 31, 2015 #5
    How does the multiplication of ##u^a## change anything at all? Also, what do you mean by 'it would be easier if it stuck to index notation everywhere. ' ? I thought this was all in index notation.
     
    Last edited: Jan 31, 2015
  7. Jan 31, 2015 #6

    ChrisVer

    User Avatar
    Gold Member

    I don't understand the problem.

    [itex] u^a g_{aj,i} u^i u^j + g_{aj} u^a d_\tau u^j - \frac{1}{2} u^a g_{ij,a} u^i u^j [/itex]

    The middle (2nd term) remains the same, but they rewrite the 1st and 3rd term as:

    [itex] u^a g_{aj,i} u^i u^j -\frac{1}{2} u^a g_{ij,a} u^i u^j = \frac{1}{2} g_{ij,a} u^i u^j u^a [/itex]

    Why is this done? because of symmetry reasons he rewrites the 1st term above by interchanging i with a [or if you like the other phrase better "by renaming them as null-indices"]...

    Also you sum when you write the same indices... that's what the writter means by summing up ....in this case summing the a-index,
     
  8. Jan 31, 2015 #7
    I'm still not convinced that you can simply swap ##i## with ##a##..
     
  9. Jan 31, 2015 #8

    ChrisVer

    User Avatar
    Gold Member

    do you know that you can write:
    [itex] u^i u^j x_{ij} = u^j u^i x_{ji} [/itex]
    simply by renaming indices?
    If not, let's say that i,j run from 1 to 2...

    [itex] u^i u^j x_{ij} = u^1 u^1 x_{11} + u^1 u^2 x_{12} + u^2 u^1 x_{21} + u^2 u^2 x_{22} [/itex]
    renaming the null indices i->j and j->i you get the same result:
    [itex] u^j u^i x_{ji} =u^1 u^1 x_{11} + u^1 u^2 x_{12} + u^2 u^1 x_{21} + u^2 u^2 x_{22} [/itex]

    As I said, you just have to rename null [summed up]-indices...

    Otherwise you can use the symmetry, to say that since [itex]u^i u^j[/itex] is symmetric under the interchange of [itex]i \leftrightarrow j[/itex] then [itex]x_{ij}[/itex] is also symmetric [or better put, only its symmetric part contributes].

    In that case you would have to symmetrize the first term in (i,a) and write:
    [itex]g_{aj,i} u^i u^a u^j - \frac{1}{2} g_{ij,a} u^a u^i u^j= \frac{1}{2} ( g_{aj,i} + g_{ij,a}) u^i u^a u^j - \frac{1}{2} g_{ij,a} u^a u^i u^j= \frac{1}{2} g_{aj,i} u^a u^i u^j[/itex]

    and rename the indices again [although you don't have to, because both expressions with renaming or not are equivalent*]...Of course that would be a very lame thing to do.... since you could have done the renaming from the begining.

    * the author could as well write as a result:
    [itex]g_{aj,i} u^i u^a u^j - \frac{1}{2} g_{ij,a} u^a u^i u^j= \frac{1}{2} g_{mn,r} u^m u^n u^r[/itex]
    or
    [itex]g_{aj,i} u^i u^a u^j - \frac{1}{2} g_{ij,a} u^a u^i u^j= g_{mn,r} u^r u^m u^n - \frac{1}{2} g_{mn,r} u^r u^m u^n =\frac{1}{2} g_{mn,r} u^m u^n u^r[/itex]



    I'm sorry, but it can't get more basic... The next thing one would have to do for illustrating what is going on, is to expand the summation ... but that would be really, really awful to read, so it's better to do that yourself on a scrap of paper.
     
    Last edited: Jan 31, 2015
  10. Feb 1, 2015 #9
    Got it, thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Index Notation - Quick one-line working
  1. Help with index notation (Replies: 28)

Loading...