1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Einstein Summation Convention / Lorentz Boost

  1. Dec 22, 2007 #1
    Einstein Summation Convention / Lorentz "Boost"

    1. The problem statement, all variables and given/known data

    I'm struggling to understand the Einstein Summation Convention - it's the first time I've used it. Would someone be able to explain it in the following context?

    Lorentz transformations and rotations can be expressed in matrix notation as

    [tex]x^{\mu'} = \Lambda^{\mu'}\!_{\mu}\:x^{\mu}[/tex]

    Coordinates are defined by [tex]x^{\mu}[/tex] with [tex]\mu = 0,1,2,3[/tex], such that [tex](x^{0}, x^{1}, x^{2}, x^{3}) = (ct, x, y, z)[/tex]

    I'm seeking clarification on the meanings of the various [tex]\mu, \mu'[/tex] indices in the matrix notation equation. Any help would be massively appreciated!
  2. jcsd
  3. Dec 22, 2007 #2
    Summation rule

    [tex]\mu, \acute{\mu}[/tex] means that these are all different indices which can have 0,1,2,3 and cannot contract with each other. I mean for example, while [tex]\mu = 0[/tex], the other one can be [tex]\acute{\mu} = 3[/tex].
    Einstein summation rule is that you have to sum the terms with same indices. So you should scan all the values of the indices.
    [tex]x^{\acute{\mu}} = \Lambda^{\acute{\mu}}\!_{\mu}\:x^{\mu}[/tex] means that

    [tex]x^0 = \Lambda^{0}\!_{0}\:x^{0}+\Lambda^{0}\!_{1}\:x^{1}+\Lambda^{0}\!_{2}\:x^{2}+\Lambda^{0}\!_{3}\:x^{3}[/tex]
    [tex]x^1 = \Lambda^{1}\!_{0}\:x^{0}+\Lambda^{1}\!_{1}\:x^{1}+\Lambda^{1}\!_{2}\:x^{2}+\Lambda^{1}\!_{3}\:x^{3}[/tex]
    [tex]x^2 = \Lambda^{2}\!_{0}\:x^{0}+\Lambda^{2}\!_{1}\:x^{1}+\Lambda^{2}\!_{2}\:x^{2}+\Lambda^{2}\!_{3}\:x^{3}[/tex]
    [tex]x^3 = \Lambda^{3}\!_{0}\:x^{0}+\Lambda^{3}\!_{1}\:x^{1}+\Lambda^{3}\!_{2}\:x^{2}+\Lambda^{3}\!_{3}\:x^{3}[/tex]

    Don't forget that each term is actually a matrix element. So write the values of the matrix elements above and obtain the Lorentz transformation equations.
  4. Dec 22, 2007 #3
    Ah, so am I correct in thinking that whenever an expression contains one index as a superscript and the same one as a subscript, a summation over those values is implied?

    Also, does the [tex]\Lambda^{\mu'}\!_{\mu}[/tex] just mean a matrix with [tex]\mu'[/tex] representing the row number and [tex]\mu[/tex] representing the column number?

    so, in essence,

    [tex]\Lambda^{0'}\!_{0} \Lambda^{0'}\!_{1} \Lambda^{0'}\!_{2} \Lambda^{0'}\!_{3}[/tex]
    [tex]\Lambda^{1'}\!_{0} \Lambda^{1'}\!_{1} \Lambda^{1'}\!_{2} \Lambda^{1'}\!_{3}[/tex]
    [tex]\Lambda^{2'}\!_{0} \Lambda^{2'}\!_{1} \Lambda^{2'}\!_{2} \Lambda^{2'}\!_{3}[/tex]
    [tex]\Lambda^{3'}\!_{0} \Lambda^{3'}\!_{1} \Lambda^{3'}\!_{2} \Lambda^{3'}\!_{3}[/tex]

    is the [tex]\Lambda^{\mu'}\!_{\mu}[/tex] matrix produced from your equations?
    Last edited: Dec 22, 2007
  5. Dec 22, 2007 #4
    Yes you'r right.
    [tex]\Lambda^{\musingle-quote}\!_{\mu}[/tex] matrix cannot produce from above equation. We'r talking about a new representation of Lorentz transforms.

    You can find the [tex]\Lambda^{\musingle-quote}\!_{\mu}[/tex] matrix here --> http://en.wikipedia.org/wiki/Lorentz_transformation
  6. Dec 22, 2007 #5


    User Avatar
    Homework Helper
    Gold Member

  7. Dec 22, 2007 #6


    User Avatar
    Science Advisor

    More correctly [itex]\Lambda^{\mu'}_\mu[/itex] means a tensor that can, in a given coordinates system, be represented by such a matrix.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook