# A How does Lorentz invariance help evaluate tensor integrals?

#### gjj

Summary
In evaluating a single loop diagram in the photon self-energy we get (among other things) a 2nd rank tensor integral (see Schwartz Quantum Field Theory and the Standard Model p. 830 Appendix B.3.4). We want to reduce this integral to a scalar integral and we're told that to do this "it must be proportional to the only tensor around, gµν." Why not some other tensor?
We're trying to reduce the tensor integral $\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}{\rm{ }}$ to a scalar integral (where ${{\Delta ^2}}$ is a scalar). We're told that the tensor integral is proportional to ${g^{\mu \nu }}$, so we have:
$C{g^{\mu \nu }} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$. Now we can turn the original integral into a scalar integral:
$C{g^{\mu \nu }}{g_{\mu \nu }} = {g_{\mu \nu }}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k_\nu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$
$C = \frac{1}{4}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$
$\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = C{g^{\mu \nu }} = {g^{\mu \nu }}\frac{1}{4}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$
It all works out very nicely, but if all we're after is Lorentz covariance what gives us the right to pick the one tensor (${g^{\mu\nu}}$) that doesn't change under a Lorentz transformation when any old tensor would give us Lorentz covariance. What does Schwartz mean when he says it's the only one around?

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#### haushofer

Well, how many rank-2 tensors do you know which are symmetric (!) in the indices mu and nu? Can you give some examples?

So, all Schwarz is saying, is

$$k^{\mu}k^{\nu} = C g^{\mu\nu}$$

for some constant C, which can be derived by contracting both sides with the metric:

$$k^{\mu}k^{\nu}g_{\mu\nu} = C g^{\mu\nu}g_{\mu\nu} \rightarrow k^2 = 4C \rightarrow C = \frac{1}{4}k^2$$

In the beginning, this seems like intuive guess work, but after a while, you kinda know which tensors are around and more comfortable to make these kinds of lucky guesses.

Maybe this topic has a better place in the high energy subforum, btw.

#### gjj

Thanks for reply. I think I understand what you're saying, but I'd like to be sure.
What Schwartz says is "Since the integral is a tensor under Lorentz transformations but only depends on the scalar Δ, it must be proportional to the only tensor around, gμν." I interpreted that statement to mean that the tensor integral is proportional to the metric tensor, $C{g^{\mu \nu }} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$. Looking it that way I wasn't clear on why he picked the metric tensor, but you're saying that's not the easiest way to look at it. What I should do is recognize the equality ${k^\mu }{k^\nu } = \frac{1}{4}{k^2}{g^{\mu \nu }}$ that you derived above and replace the ${{k^\mu }{k^\nu }}$ with $\frac{1}{4}{k^2}{g^{\mu \nu }}$ in the integrand of $\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$. In this case there's no question of why the ${g^{\mu \nu }}$ entered the calculation. Did I get it right?

#### vanhees71

Gold Member
The integrals appearing in the Feynman rules can be traced back to the scalar one
$$I(p)=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 p}{(2 \pi)^4} \frac{1}{(m^2-k^2-2p \cdot k -\mathrm{i} 0^+)^{\alpha}}.$$
You get all kinds of tensor integrals by taking partial derivatives with respect to $p^{\mu}$. Above I used the west-coast convention of the metric, $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$.

For details of the calculation, see p. 145 ff of

There everything is calculated in dimensionally regularized form for convenience.

#### haushofer

Thanks for reply. I think I understand what you're saying, but I'd like to be sure.
What Schwartz says is "Since the integral is a tensor under Lorentz transformations but only depends on the scalar Δ, it must be proportional to the only tensor around, gμν." I interpreted that statement to mean that the tensor integral is proportional to the metric tensor, $C{g^{\mu \nu }} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$. Looking it that way I wasn't clear on why he picked the metric tensor, but you're saying that's not the easiest way to look at it. What I should do is recognize the equality ${k^\mu }{k^\nu } = \frac{1}{4}{k^2}{g^{\mu \nu }}$ that you derived above and replace the ${{k^\mu }{k^\nu }}$ with $\frac{1}{4}{k^2}{g^{\mu \nu }}$ in the integrand of $\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}$. In this case there's no question of why the ${g^{\mu \nu }}$ entered the calculation. Did I get it right?
Yes, because there are "no other constant symmetric 2-tensors around" :)

"How does Lorentz invariance help evaluate tensor integrals?"

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