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I Momentum cut-off regularisation & Lorentz invariance

  1. Mar 23, 2017 #1
    Why is it that introducing a hard cut-off ##p^{2}=\Lambda^{2}## breaks Lorentz invariance? Is it simply that it introduces an energy scale and energy is not a Lorentz invariant quantity?

    Sorry if this is a trivial question, but I just want to make sure I understand the reasoning as I've heard/read it being stated, but never fully appreciated why.
     
    Last edited: Mar 23, 2017
  2. jcsd
  3. Mar 23, 2017 #2

    Demystifier

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    Well, if the cut-off is defined by
    $$p^{2}=\Lambda^{2}. . . . . (Eq. 1)$$
    then it does not break Lorentz invariance, because
    $$p^{2}=p_0^2-{\bf p}^2. . . . . (Eq. 2)$$
    is Lorentz invariant. However such a cutoff does not really make physical quantities finite because, with finite positive ##\Lambda^{2}##, (Eq. 1) allows ##p_0## and ##|{\bf p}|## to be arbitrarily big. To bound ##p_0## and ##|{\bf p}|## one can replace (Eq. 2) with a Euclidean scalar product
    $$p^{2}_E=p_0^2+{\bf p}^2. . . . . (Eq. 3)$$
    but then it is no longer Lorentz invariant.
     
  4. Mar 23, 2017 #3
    Is it instead "Euclidean" invariant? Is it possible to show that the cut-off can't also be Lorentz invariant? Is there any physical intuition as to why a cut-off breaks Lorentz invariance?
     
  5. Mar 23, 2017 #4

    Demystifier

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    Yes.

    I think I have just shown that.

    Sure. Suppose that there is some maximal possible energy-momentum ##p^{\mu}##, and suppose that it is a time-like vector. Then there is a Lorentz frame in which
    $$p^{\mu}=(\Lambda,0,0,0)$$
    But if ##p^{\mu}## has this form in one Lorentz frame, then it cannot have this form in other Lorentz frames. Therefore there is a special Lorentz frame in which ##p^{\mu}## takes this special form. But if there is a special Lorentz frame, then there is no Lorentz invariance.
     
  6. Mar 23, 2017 #5
    I guessing because if it is "Euclidean" invariant then it cannot be simultaneously Lorentz invariant?!

    So would it be correct to say that a hard cut-off introduces an energy scale, which is not Lorentz invariant (since energy is not a Lorentz invariant quantity)?
     
  7. Mar 24, 2017 #6

    Demystifier

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    Yes.

    It wouldn't be wrong.
     
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