# I Momentum cut-off regularisation & Lorentz invariance

Tags:
1. Mar 23, 2017

### Frank Castle

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. Mar 23, 2017

### Demystifier

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.

3. Mar 23, 2017

### Frank Castle

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?

4. Mar 23, 2017

### Demystifier

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.

5. Mar 23, 2017

### Frank Castle

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)?

6. Mar 24, 2017

### Demystifier

Yes.

It wouldn't be wrong.