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Deciding if a transformation is a Lorentz transformation

  1. Dec 31, 2015 #1
    1. The problem statement, all variables and given/known data

    Is the transformation ##Y:(t,x,y,z)\rightarrow (t,x,-y,z)## a Lorentz transformation? If so, why is it not considered with P and T as a discrete Lorentz transformation? If not, why not?

    2. Relevant equations

    3. The attempt at a solution

    A Lorentz transformation ##\Lambda## satisfies the relation ##\Lambda^{T}g\Lambda = \Lambda##, where ##g## is the Minkowski metric.

    In our case, the transformation ##Y = \text{diag}(1,1,-1,1)##.

    Therefore, ##Y^{T}gY = \text{diag}(1,1,-1,1) \cdot{\text{diag}(1,-1,-1,-1)}\cdot{\text{diag}(1,1,-1,1)} = \text{diag}(1,-1,-1,-1) =## the parity operator.

    Therefore, the relation ##Y^{T}gY = Y## is not satisfied.

    Therefore, ##Y## is not a Lorentz transformation.

    Is my answer correct?
     
  2. jcsd
  3. Dec 31, 2015 #2

    strangerep

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    Science Advisor

    Are you sure about that? Are you sure it's not ##\Lambda^{T}g\Lambda = g## ? :wink:

    Alternatively, you could just check whether the Minkowski line element ##ds^2 = t^2 - x^2 - y^2 - z^2## is preserved.
     
  4. Jan 1, 2016 #3
    Oh, right! In that case, the transformation ##Y## satisfies ##Y^{T}gY=g##, so the transformation ##Y## is a Lorentz transformation.

    Second way:

    An inner product of Lorentz four-vectors is Lorentz invariant.

    (A Lorentz four-vector ##V^{\mu}## is a four-vector which transforms under the Lorentz transformation ##{\Lambda^{\mu}}_{\nu}## with the transformation law ##V'^{\mu} = {\Lambda^{\mu}}_{\nu}V^{\nu}##.)

    (A mathematical object (i.e. Lorentz scalar, Lorentz four-vector, Lorentz tensor and combinations thereof) is Lorentz invariant if it does not change under a Lorentz transformation. Not all mathematical objects are necessarily Lorentz invariant.)

    The only four-vectors mentioned in the problem are the position four-vectors ##x^{\mu}## before and ##x'^{\mu}## after the transformation ##Y##.

    We know that the position four-vector is a Lorentz four-vector, so the inner products ##x^{\mu}x_{\mu}## before and ##x'^{\mu}x'_{\mu}## after a Lorentz transformation must be equal to each other.

    So, let's take the inner products ##x^{\mu}x_{\mu}## before and ##x'^{\mu}x'_{\mu}## after the transformation ##Y## and check if the product is invariant:

    Before the transformation ##Y##, ##x^{\mu}x_{\mu}=g_{\mu\nu}x^{\mu}x^{\nu}=g_{00}x^{0}x^{0}+g_{11}x^{1}x^{1}+g_{22}x^{2}x^{2}+g_{33}x^{3}x^{3} = (x^{0})^{2}-(x^{1})^{2}-(x^{2})^{2}-(x^{3})^{2}=(t)^{2}-(x)^{2}-(y)^{2}-(z)^{2}##.

    After the transformation ##Y##, ##x'^{\mu}x'_{\mu}=g_{\mu\nu}x'^{\mu}x'^{\nu}=g_{00}x'^{0}x'^{0}+g_{11}x'^{1}x'^{1}+g_{22}x'^{2}x'^{2}+g_{33}x'^{3}x'^{3} = (x'^{0})^{2}-(x'^{1})^{2}-(x'^{2})^{2}-(x'^{3})^{2}=(t)^{2}-(x)^{2}-(-y)^{2}-(z)^{2}=(t)^{2}-(x)^{2}-(y)^{2}-(z)^{2}##.

    The inner products are the same under the transformation ##Y##, so the transformation ##Y## is a Lorentz transformation.


    Is my solution accurate?
     
    Last edited: Jan 1, 2016
  5. Jan 1, 2016 #4

    strangerep

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    Science Advisor

    Yes, but you don't need to do all that work if you realize that ##Y^T = Y^{-1}## for a Lorentz transformation (since this is true for all "SO(...)" transformations). Just write out the inner product in matrix notation...
     
  6. Jan 2, 2016 #5
    So, do you mean that,

    ##x'^{\mu}x'_{\mu} = x'^{T}x' = (Yx)^{T}(Yx) = x^{T}Y^{T}Yx = x^{T}x##,

    so that the inner product is invariant?
     
  7. Jan 2, 2016 #6

    strangerep

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    That's the general idea, but your details are not quite right. For arbitrary 4-vectors ##x,y## we have
    $$y'^T g' x' ~=~ y^T Y^T \; g' \; Y y ~,$$ so if ##Y^T g' Y = g##, then the rhs is ##y^T g x##, showing that the inner product between 2 arbitrary 4-vectors is preserved.
     
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