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My professor told me that a 4-dimensional Minkowski - Space Integral I was working on can be written as the product of a metric tensor and a scalar:

$\int d^4 k \frac{k^\mu k^\nu}{(k^2-m_1^2)(k^2-m_2^2)} = g^{\mu \nu} B_2$

where $B_2$ is a scalar.

I am surprised that apparently for $\mu = \nu = 0$ , the sign of the integral is opposite to the sign for, say, $\mu = \nu = 1$. This seems wrong to me, and in an Eucledian Space it certainly would be. If I had been asked to guess, I would have written the above equation with $ g^\mu_\nu B_2$ on the right hand side instead..

Can someone bring clarity?

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1 Answer 1

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The key insight is that the integral you wrote down is a Lorentz-invariant tensor, so whatever it evaluates to must also be Lorentz-invariant.

To illustrate this, let's be a bit more general and consider any integral of the form \begin{align} I^{\mu\nu}[f] = \int d^4k\, f(k^2)k^\mu k^\nu \end{align} for any admissible function $f$. Then notice that for any Lorentz transformation $\Lambda = (\Lambda^\mu_{\phantom\mu\nu})$ we have \begin{align} \Lambda^\alpha_{\phantom\alpha\mu}\Lambda^\beta_{\phantom\beta\nu}\,I^{\mu\nu}[f] &= \int d^4k\, f(k^2)(\Lambda^\alpha_{\phantom\alpha\mu}k^\mu)( \Lambda^\beta_{\phantom\beta\nu}k^\nu)\\ &= \int d^4 u f(u^2) u^\alpha u^\beta \\ &= I^{\alpha\beta}[f] \end{align} where in the second equality we made the change of variables $u^\alpha = \Lambda^\alpha_{\phantom\alpha\mu}k^\mu$, and we noted that the term $f(k^2)$ becomes $f(u^2)$ because $k^2$ is Lorentz-invariant, and the measure $d^4 k$ is Lorentz-invariant.

Addendum. (Following comments on answer v1)

The above argument demonstrates that provided $f$ is such that $I^{\mu\nu}[f]$ is well-defined, the integral is a Lorentz-invariant two-tensor and therefore is of the form $g^{\mu\nu}B[f]$ for some scalar functional $B$ as you wrote.

However, as is it is written, the integral you wrote down is divergent both because the integrand is singular (it has poles at $k^2 = m_1^2$ and $k^2 = m_2^2$), and by power counting which shows that the integrand scales linearly with $k$ for large $k$ so it is UV divergent. This ill-definedness leads to all sorts of apparent "paradoxes." For example, as pointed out by user10001, if we were to set $m_1 = m_2$, then the integrand is manifestly positive unless $k^\mu = 0$, so how could $I^{00}$ and $I^{ii}$ have different signs?

The resolution is to note that this integral comes from QFT where one uses the so-called $i\epsilon$ prescription (for good physics reasons) which removes the poles from the integrand by shifting them away from the real axis. Moreover, one fixes the UV divergence by regularizing the integral (using e.g. dimensional regularization). The object one then needs to compute is \begin{align} I^{\mu\nu}(d) = \lim_{\epsilon\to 0}\int_{\mathbb R^4} d^dk \frac{k^\mu k^\nu}{(k^2 - m_1^2 + i\epsilon)(k^2-m_2^2+i\epsilon)}, \tag{$\star$} \end{align} for $d\neq 4$ and then analytically continue this to a function $I^{\mu\nu}(z)$ on the complex plain (minus some isolated singular points) which allows one to parametrize the divergence near $d=4$ by plugging in $z = 4-\epsilon$.

Note that $(\star)$ is not plagued by any of the apparent paradoxes that concerned us in the comments.

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    $\begingroup$ Your reasoning is perfectly fine but there is still a paradox here that I am not able to understand. From the OP's integral we can see that for $m_1=m_2$ the integral is positive for all $\mu=\nu$, simply because the integrand is positive and also the measure $d^4 k =dk^0 dk^1 dk^2 dk^3$ is positive; whereas if the value of integral were equal to $g^{\mu\nu} B_2$ then the sign of integral for $\mu=\nu=0$ will be opposite to that for $\mu=\nu=i,\: i=1,2,3$! $\endgroup$
    – user10001
    Commented May 30, 2014 at 4:06
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    $\begingroup$ I am right there with user10001. Your argument is very nicely elaborated and tells us what has to be the result by means of an "outer" argument. But how do we resolve what is seemingly a paradox, the fact that all integrals, calculated explicitly, seem to be positive? $\endgroup$ Commented May 30, 2014 at 6:09
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    $\begingroup$ @user10001 Yes, that's a great observation, but I think we have to be very careful here for the following reason. Notice that the integrand in the question has two poles at $k^2 = m_1^2$ and $k^2 = m_2^2$, so I'm guessing this is a loop integral in QFT in which case you're missing $i\epsilon$'s, otherwise the integral is ill-defined (divergent). Once one does this, the integrand is no longer manifestly non-negative, and the objection goes away. Moreover, even after you do this, I think you need to regularize the integral since the integrand scales like $k$ for $k$ large. $\endgroup$ Commented May 30, 2014 at 8:33
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    $\begingroup$ @Konstantin In any event, I'll try to incorporate my comments above addressed to user10001 into an addendum to the answer that more explicitly addresses these issues (particularly inclusion of the $i\epsilon$ prescription) as soon as I can. $\endgroup$ Commented May 30, 2014 at 8:54
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    $\begingroup$ @Josh Yeah that seems to resolve this paradox. More generally, it appears that the integrals of Lorentz covariant real integrands can never be finite on Minkowski space {in particular, integrals of the form $\int d^4 p f(p^2)^2 p^\mu p^\nu$ can not be finite}, and we necessarily have to go to Euclidean space to make sense of these integrals by assuming p0 to be a complex variable and making use of Wick rotation. The reason that they are not finite is may be because any function of the form f(p^2) doesn't go to zero at infinty on Minkowski space. $\endgroup$
    – user10001
    Commented May 30, 2014 at 8:58

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