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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


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  7. Dec 22, 2007 #6


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    More correctly [itex]\Lambda^{\mu'}_\mu[/itex] means a tensor that can, in a given coordinates system, be represented by such a matrix.
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