performing the integral over the polar angle $z_{13}$ is never going to yield a logarithm. It even turns out that, while the length of a vector contains a square root of the Feynman parameter $x$, the integral only depends on the length squared, and therefore we are left with an integral of the form

$$ \int_0^1 dx \frac{\sum_{i=0}^{k-1} a_i x^i}{(b_0 + b_1 \cdot x + b_2 \cdot x^2)^k} \qquad (B.35) $$

This is a well known class of textbook integrals, and can be easily computed, for example with the method of partial fraction decomposition, or Mathematica.

Therefore, the method of integrating structures of type $I_2$ is a combination of applying Feynman parameters and performing an appropriate $O(3)$ rotation.

### B.3.3 $I_3$

Consider the integral

$$ \int d\Omega I_3 \supset \int d\Omega \frac{s_{23}^m}{s_{53}^a} \qquad (B.36) $$

From the expressions before we have already seen that it is most convenient to eliminate any azimuthal angle $\varphi_{13}$ in the denominator. We achieve this as before by rotating $p_{3,com}$ in a suitable way.

However, now the numerator is not independent of $p_{3,com}$ and will be affected by the rotation. This is easily dealt with by acting on the vector $\vec{p}_3$ with the standard rotation matrix $R_3$. Since one can in general not choose the matrix $R_3$ in such a way that any dependence on $\varphi_{13}$ disappears, we still have $\varphi_{13}$ in the numerator. Therefore, we have

$$ \int d\Omega I_3 \supset \int_0^\pi d\theta_{13} \sin(\theta_{13}) \int_0^{2\pi} d\varphi_{13} \frac{[d_0 + d_2 \cdot \sin(\theta_{13}) \cos(\varphi_{13}) + d_1 \cdot \cos(\theta_{13})]^m}{[c_0 + c_1 \cdot \cos(\theta_{13})]^a} \qquad (B.37) $$

We perform the integration over the azimuthal angle $\varphi_{13}$ first. When we expand the numerator, we get a series with terms proportional to

$$ \int_0^{2\pi} d\varphi_{13} [\sin(\theta_{13}) \cos(\varphi_{13})]^k \qquad (B.38) $$

This integral vanishes for all odd powers of $k$. This is in so far convenient that keeping only the even powers eliminates all occurrences of $\sin(\theta_{13}) = \sqrt{1 - z_{13}^2}$, which is not straightforward to integrate in general. Therefore after performing the azimuthal integration we are left with the integration over the polar angle, which takes the form

$$ \int_{-1}^{+1} dz_{13} \frac{\sum_m a_m z_{13}^m}{[c_0 + c_1 \cdot z_{13}]^a} \qquad (B.39) $$

As in the previous subsection, this is a well known class of textbook integrals, and can be easily computed, for example with the method of partial fraction decomposition, or Mathematica.

Therefore, the method of integrating structures of type $I_3$ is performing an appropriate $O(3)$ rotation and then proceeding with the integration over the angles as described above.
C More on the computation of $\int d\Omega |\mathcal{M}|^2_{3\to 2}$

C.1 Detailed results of our computation

In this appendix we report the detailed results of our computation of $\int d\Omega |\mathcal{M}|^2$ for all possible $3 \to 2$ processes involving in the initial state at least one gauge boson $V$, scattering with other $V$'s and/or fermions and/or scalars charged under the gauge group, where for completeness in the scalar case we included the self-coupling $\mathcal{L} \supset \lambda |\phi|^4$ in addition to the scalar gauge coupling to $V$. We report $\int d\Omega |\mathcal{M}|^2$ in terms of scalar products of the 4-momenta in the center-of-mass frame $p_1 + p_2 + p_5 = p_3 + p_4$, including for definiteness only the terms that are leading order in $p_i \cdot p_j / \mu^2$. In case at least one initial particle belongs to the bath, we remind that one has the hierarchy $\mu^2 \ll p_1 \cdot p_2 \ll p_1 \cdot p_5 \simeq p_2 \cdot p_5$, leading to the expansions reported in tables 5 (case of one fermion charged under the gauge group), 6 (case of one scalar charged under the gauge group), 7 (case of both a fermion and a scalar charged under the gauge group). In case all initial particles belong to the shell, then all scalar products are of the same order, leading to the leading-order results in table 8. The reader finds in the main text, for each case of interest, the values of the scalar products and the parametric dependence of the leading terms in $\int d\Omega |\mathcal{M}|^2$ (table 4).

C.2 Simple upper estimates of the integrated amplitudes

In this appendix we give useful formulae to estimate an upper limit on the integrated spin-averaged squared matrix element, without actually having to compute the amplitude and performing the final phase space integration. In general, these estimates are very conservative in the sense that they overestimate the amplitudes. These estimates can then serve as a guidance on the computation of new amplitudes.

C.2.1 Equilibration between bath and shell

Assume that we have a sizeable spread of particles, i.e. $p_1 \cdot p_2 \gtrsim \mu^2$.

We take the following basis

$$ p_{1,com} = \begin{pmatrix} p_I \\ 0 \\ 0 \\ p_I \end{pmatrix}, \quad p_{2,com} = \begin{pmatrix} p_J \\ p_J \sin(\theta_{12}) \\ 0 \\ p_J \cos(\theta_{12}) \end{pmatrix}, \quad p_{3,com} = \begin{pmatrix} \sqrt{p_I^2 + 2p_I p_J \cos(\theta_{12}) + p_J^2} \\ -p_J \sin(\theta_{1,2}) \\ 0 \\ -p_I - p_J \cos(\theta_{12}) \end{pmatrix}, \quad (C.1) $$

$$ p_{4,com} = \begin{pmatrix} p_F \\ p_F \sin(\theta_{13}) \cos(\varphi_{13}) \\ p_F \sin(\theta_{13}) \sin(\varphi_{13}) \\ p_F \cos(\theta_{13}) \end{pmatrix}, \quad p_{5,com} = \begin{pmatrix} p_F \\ -p_F \sin(\theta_{13}) \cos(\varphi_{13}) \\ -p_F \sin(\theta_{13}) \sin(\varphi_{13}) \\ -p_F \cos(\theta_{13}), \end{pmatrix} \quad (C.2) $$

which is the one from appendix B.2, with all particles massless. It is helpful to define

$$ p_{IJ} = \sqrt{p_I^2 + 2p_I p_J \cos(\theta_{12}) + p_J^2} \ge 0, \quad (C.3) $$
<table><caption>**Table 5.** Scatterings involving vectors and fermions, where $g = g_D$ is the gauge coupling and $\mu$ is the IR cut-off from eq. (2.38).</caption><tr><td>a(p<sub>1</sub>) b(p<sub>2</sub>) c(p<sub>5</sub>)<br/>→ XY</td><td>∫ dΩ |M|^2: G<sub>V</sub> = U(1)<br/>∫ dΩ |M|^2: G<sub>V</sub> = SU(N)</td></tr><tr><td>VVV → f<span style="text-decoration: overline">f</span></td><td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">48π(2 ln(p<sub>1</sub>·p<sub>2</sub>) + 5 ln 2)</div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div><br/>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">π/N<sup>2</sup>(N<sup>2</sup>−1)<sup>2</sup></div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div> [9N<sup>2</sup>(N<sup>2</sup>−1) ln(p<sub>1</sub>·p<sub>5</sub>/µ<sup>2</sup>) + 2(N<sup>4</sup>−7N<sup>2</sup>+6) ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + N<sup>4</sup>(21 ln 2−15) + N<sup>2</sup>(9−51 ln 2) + 30 ln 2]</td></tr><tr><td>VVf → Vf<br/>VV<span style="text-decoration: overline">f</span> → V<span style="text-decoration: overline">f</span></td><td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">24π(2 ln(p<sub>1</sub>·p<sub>2</sub>) + 5 ln 2)</div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div><br/>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">N<sup>2</sup></div><div style="padding: 0 5px;">N<sup>2</sup>−1</div></div> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">36π p<sub>1</sub>·p<sub>5</sub>(5 ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + 6 ln 2)</div><div style="padding: 0 5px;">(p<sub>1</sub>·p<sub>2</sub>)<sup>2</sup></div></div></td></tr><tr><td>f<span style="text-decoration: overline">f</span>V → VV</td><td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">32π(ln(p<sub>1</sub>·p<sub>2</sub>) + 3 ln 2)</div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>5</sub></div></div><br/>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">8π p<sub>1</sub>·p<sub>5</sub>(2N<sup>2</sup>−3) ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + 3(N<sup>2</sup>−1) ln 2</div><div style="padding: 0 5px;">(p<sub>1</sub>·p<sub>2</sub>)<sup>2</sup></div></div></td></tr><tr><td>f<span style="text-decoration: overline">f</span>V → f<span style="text-decoration: overline">f</span></td><td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">4π(ln(p<sub>1</sub>·p<sub>5</sub>/µ<sup>2</sup>) + 18 ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) − 1 + 29 ln 2)</div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div><br/>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">π/6N<sup>2</sup></div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div> [3N(N<sup>2</sup>−1) ln(p<sub>1</sub>·p<sub>5</sub>/µ<sup>2</sup>) + 6(10N<sup>3</sup> + N<sup>2</sup>−10N−1) ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + N<sup>3</sup>(99 ln 2−5) + 12N<sup>2</sup> ln 2 + N(3−99 ln 2) − 12 ln 2]</td></tr><tr><td>ffV → ff<br/><span style="text-decoration: overline">f</span>fV → <span style="text-decoration: overline">f</span><span style="text-decoration: overline">f</span></td><td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">16π(14 ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + 15 ln 2)</div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div><br/>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">N<sup>2</sup>−1</div><div style="padding: 0 5px;">N<sup>2</sup></div></div> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">π(4(5N−2) ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + 30N ln 2)</div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div></td></tr><tr><td>VfV → Vf<br/>V<span style="text-decoration: overline">f</span>V → V<span style="text-decoration: overline">f</span></td><td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">π(10 ln(p<sub>1</sub>·p<sub>5</sub>/µ<sup>2</sup>) + 4 ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + 5 + 34 ln 2)</div><div style="padding: 0 5px;">p<sub>1</sub>·p<sub>2</sub></div></div><br/>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center;"><div style="border-bottom: 1px solid black; padding: 0 5px;">10π p<sub>1</sub>·p<sub>5</sub>(2(4N<sup>2</sup>−1) ln(p<sub>1</sub>·p<sub>2</sub>/µ<sup>2</sup>) + 3(3N<sup>2</sup>−1) ln 2)</div><div style="padding: 0 5px;">(p<sub>1</sub>·p<sub>2</sub>)<sup>2</sup></div></div></td></tr></table>
<table><caption>**Table 6.** Scatterings involving vectors and scalars, where $g = g_D$ is the gauge coupling and $\mu$ is the IR cut-off from eq. (2.38).</caption>
  <tbody>
    <tr>
      <td>a(p<sub>1</sub>) b(p<sub>2</sub>) c(p<sub>5</sub>)<br/>→ XY</td>
      <td>
        &int; d&Omega;|M|^2: G<sub>V</sub> = U(1)<br/>
        &int; d&Omega;|M|^2: G<sub>V</sub> = SU(N)
      </td>
    </tr>
    <tr>
      <td>VVV → &phi;&phi;&#773;</td>
      <td>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">8&pi; (9 ln(p<sub>1</sub>&middot;p<sub>5</sub>/&mu;<sup>2</sup>) + 11 ln 2)</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>5</sub></td></tr>
          </tbody>
        </table>
        <br/>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">2N<sup>2</sup> &minus; 3</td></tr>
            <tr><td style="padding: 0 0.2em;">2(N<sup>2</sup> &minus; 1)<sup>2</sup></td></tr>
          </tbody>
        </table> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">3&pi;</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>2</sub></td></tr>
          </tbody>
        </table>
      </td>
    </tr>
    <tr>
      <td>VV&phi; → V&phi;<br/>VV&phi;&#773; → V&phi;&#773;</td>
      <td>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">4&pi; (9 ln(p<sub>1</sub>&middot;p<sub>5</sub>/&mu;<sup>2</sup>) &minus; 9 + 11 ln 2)</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>5</sub></td></tr>
          </tbody>
        </table>
        <br/>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">8(4N<sup>2</sup> &minus; 1) ln(p<sub>1</sub>&middot;p<sub>2</sub>/&mu;<sup>2</sup>) + 12(3N<sup>2</sup> &minus; 1) ln 2</td></tr>
            <tr><td style="padding: 0 0.2em;">(p<sub>1</sub>&middot;p<sub>2</sub>)<sup>2</sup></td></tr>
          </tbody>
        </table>
      </td>
    </tr>
    <tr>
      <td>&phi;&phi;&#773;V → VV</td>
      <td>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">16&pi; (ln(p<sub>1</sub>&middot;p<sub>2</sub>/&mu;<sup>2</sup>) + 3 ln 2)</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>5</sub></td></tr>
          </tbody>
        </table>
        <br/>
        g<sup>6</sup> &pi; p<sub>1</sub>&middot;p<sub>5</sub> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">4(2N<sup>2</sup> &minus; 3) ln(p<sub>1</sub>&middot;p<sub>2</sub>/&mu;<sup>2</sup>) + 12(N<sup>2</sup> &minus; 1) ln 2</td></tr>
            <tr><td style="padding: 0 0.2em;">(p<sub>1</sub>&middot;p<sub>2</sub>)<sup>2</sup></td></tr>
          </tbody>
        </table>
      </td>
    </tr>
    <tr>
      <td>&phi;&phi;&#773;V → &phi;&phi;&#773;</td>
      <td>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">&pi;</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>2</sub></td></tr>
          </tbody>
        </table>
        <br/>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">2N<sup>2</sup> &minus; 3</td></tr>
            <tr><td style="padding: 0 0.2em;">24N</td></tr>
          </tbody>
        </table> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">&pi;</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>2</sub></td></tr>
          </tbody>
        </table>
      </td>
    </tr>
    <tr>
      <td>&phi;&phi;V → &phi;&phi;</td>
      <td>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">96&pi; (ln(p<sub>1</sub>&middot;p<sub>2</sub>/&mu;<sup>2</sup>) + ln 2)</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>5</sub></td></tr>
          </tbody>
        </table>
      </td>
    </tr>
    <tr>
      <td>&phi;&#773;&phi;&#773;V → &phi;&#773;&phi;&#773;</td>
      <td>
        <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">(N + 1)(4&lambda;g<sup>4</sup>N + g<sup>6</sup>(N &minus; 1)) ln(p<sub>1</sub>&middot;p<sub>5</sub>/&mu;<sup>2</sup>) &minus; 8(N<sup>2</sup> &minus; 1)(4&lambda;g<sup>4</sup>N &minus; g<sup>6</sup>(3N<sup>2</sup> &minus; 1)) ln(p<sub>1</sub>&middot;p<sub>2</sub>/&mu;<sup>2</sup>)</td></tr>
            <tr><td style="padding: 0 0.2em;">+ 16&lambda;<sup>2</sup>g<sub>D</sub><sup>2</sup> (N<sup>3</sup>(3 ln 2 &minus; 1) + N<sup>2</sup>(3 ln 2 &minus; 2)) &minus; 8&lambda;g<sup>4</sup> (N<sup>3</sup>(5 ln 2 + 1) + N<sup>2</sup> &minus; N(5 ln 2 + 2))</td></tr>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">+ g<sup>6</sup> (N<sup>3</sup>(35 ln 2 &minus; 1) + 13N<sup>2</sup> ln 2 &minus; N(35 ln 2 &minus; 3) &minus; 13 ln 2 &minus; 2)</td></tr>
          </tbody>
        </table>
      </td>
    </tr>
    <tr>
      <td>V&phi;V → V&phi;</td>
      <td>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">8&pi;</td></tr>
            <tr><td style="padding: 0 0.2em;">p<sub>1</sub>&middot;p<sub>2</sub></td></tr>
          </tbody>
        </table>
      </td>
    </tr>
    <tr>
      <td>V&phi;&#773;V → V&phi;&#773;</td>
      <td>
        g<sup>6</sup> <table style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2;">
          <tbody>
            <tr><td style="border-bottom: 1px solid black; padding: 0 0.2em;">8(4N<sup>2</sup> &minus; 1) ln(p<sub>1</sub>&middot;p<sub>2</sub>/&mu;<sup>2</sup>) + 12(3N<sup>2</sup> &minus; 1) ln 2</td></tr>
            <tr><td style="padding: 0 0.2em;">(p<sub>1</sub>&middot;p<sub>2</sub>)<sup>2</sup></td></tr>
          </tbody>
        </table>
      </td>
    </tr>
  </tbody>
</table>
<table><caption>**Table 7.** Scatterings involving vectors, fermions, and scalars, where $g = g_D$ is the gauge coupling and $\mu$ is the IR cut-off from eq. (2.38).</caption><tr><td>a(p<sub>1</sub>) b(p<sub>2</sub>) c(p<sub>5</sub>)<br/>→ XY</td><td>∫ dΩ |M|^2: G<sub>V</sub> = U(1)<br/>∫ dΩ |M|^2: G<sub>V</sub> = SU(N)</td></tr><tr><td>φ&#773;φV → f&#773;f</td><td>g<sup>6</sup> (2π (ln(p<sub>1</sub>·p<sub>5</sub>) / μ<sup>2</sup>) + 3 ln 2 - 1) / p<sub>1</sub>·p<sub>2</sub><br/>g<sup>6</sup> (3(N<sup>2</sup> - 1) ln(p<sub>1</sub>·p<sub>5</sub>) / μ<sup>2</sup>) + 9(N<sup>2</sup> - 1) ln 2 - 5N<sup>2</sup> + 3 / p<sub>1</sub>·p<sub>2</sub></td></tr><tr><td>f&#773;fV → φ&#773;φ</td><td>g<sup>6</sup> (2π) / p<sub>1</sub>·p<sub>2</sub><br/>g<sup>6</sup> (2N<sup>2</sup> - 3) / (12N) (π / p<sub>1</sub>·p<sub>2</sub>)</td></tr><tr><td>φfV → φf<br/>φ&#773;fV → φ&#773;f</td><td>g<sup>6</sup> (8π (2 ln(p<sub>1</sub>·p<sub>2</sub>) / μ<sup>2</sup>) + 3 ln 2) / p<sub>1</sub>·p<sub>2</sub><br/>g<sup>6</sup> (N<sup>2</sup> - 1) (π (2 ln(p<sub>1</sub>·p<sub>2</sub>) / μ<sup>2</sup>) + 3 ln 2) / p<sub>1</sub>·p<sub>2</sub></td></tr><tr><td>φ&#773;fV → φ&#773;f<br/>φ&#773;fV → φ&#773;f</td><td>g<sup>6</sup> (8π (2 ln(p<sub>1</sub>·p<sub>2</sub>) / μ<sup>2</sup>) + 3 ln 2) / p<sub>1</sub>·p<sub>2</sub><br/>g<sup>6</sup> (N<sup>2</sup> - 1) (π (2 ln(p<sub>1</sub>·p<sub>2</sub>) / μ<sup>2</sup>) + 3 ln 2) / p<sub>1</sub>·p<sub>2</sub></td></tr><tr><td>VVV → VV</td><td>0<br/>g<sup>6</sup> (N<sup>3</sup> / (N<sup>2</sup> - 1)<sup>2</sup>) (144π p<sub>1</sub>·p<sub>5</sub> (5 ln(p<sub>1</sub>·p<sub>2</sub>) / μ<sup>2</sup>) + 6 ln 2) / (p<sub>1</sub>·p<sub>2</sub>)<sup>2</sup></td></tr></table>

such that energy conservation implies

$$2p_F = p_I + p_J + p_{IJ}. \qquad (C.4)$$

We find these scalar products,

$$p_1 \cdot p_5 = \frac{1}{2} [(p_I + p_{IJ})^2 - p_J^2] \qquad (C.5)$$

$$p_2 \cdot p_5 = \frac{1}{2} [(p_J + p_{IJ})^2 - p_I^2] \qquad (C.6)$$

$$p_1 \cdot p_2 = \frac{1}{2} [(p_I + p_J)^2 - p_{IJ}^2] \qquad (C.7)$$
<table><caption>**Table 8.** Self-thermalization within the shell, where $g = g_D$ is the gauge coupling and $\mu$ is the IR cut-off from eq. (2.38).</caption>
  <thead>
    <tr>
      <th></th>
      <th>∫ dΩ |M|^2: G<sub>V</sub> = U(1)</th>
      <th>∫ dΩ |M|^2: G<sub>V</sub> = SU(N)</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>a(p<sub>i</sub>) b(p<sub>j</sub>) c(p<sub>k</sub>)<br/>→ XY</td>
      <td></td>
      <td></td>
    </tr>
    <tr>
      <td>VVV → f&#773;f</td>
      <td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">24π(5 ln(p<sub>i</sub>·p<sub>j</sub>) + 5 ln 3 - ln 2)</span><span style="display: block; padding: 0 0.2em;">p<sub>i</sub>·p<sub>j</sub></span></div></td>
      <td></td>
    </tr>
    <tr>
      <td>VVV → φ&#773;φ</td>
      <td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">π</span><span style="display: block; padding: 0 0.2em;">2N<sup>2</sup>(N<sup>2</sup>-1)<sup>2</sup></span></div> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">1</span><span style="display: block; padding: 0 0.2em;">p<sub>i</sub>·p<sub>j</sub></span></div> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; padding: 0 0.2em;">15(N<sup>4</sup> - 3N<sup>2</sup> + 2) ln(p<sub>i</sub>·p<sub>j</sub>)</span><span style="display: block; padding: 0 0.2em;">+ N<sup>4</sup>(29 ln 2 - 3) + N<sup>2</sup>(-9 + 105 ln 2 - 90 ln 3) + 24 ln 3 - 6 ln 2</span></div></td>
      <td></td>
    </tr>
    <tr>
      <td>VVV → VV</td>
      <td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">36π(ln(p<sub>i</sub>·p<sub>j</sub>) + 3 ln 2 - ln 3)</span><span style="display: block; padding: 0 0.2em;">p<sub>i</sub>·p<sub>j</sub></span></div></td>
      <td></td>
    </tr>
    <tr>
      <td>VVV → VV</td>
      <td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">π</span><span style="display: block; padding: 0 0.2em;">4N<sup>2</sup>(N<sup>2</sup>-1)<sup>2</sup></span></div> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">1</span><span style="display: block; padding: 0 0.2em;">p<sub>i</sub>·p<sub>j</sub></span></div> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; padding: 0 0.2em;">9(N<sup>4</sup> - 3N<sup>2</sup> + 2) ln(p<sub>i</sub>·p<sub>j</sub>)</span><span style="display: block; padding: 0 0.2em;">+ N<sup>4</sup>(3 + 11 ln 2) + N<sup>2</sup>(9 - 129 ln 2 + 54 ln 3) - 18 ln 3 + 54 ln 2</span></div></td>
      <td></td>
    </tr>
    <tr>
      <td>VVV → VV</td>
      <td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">0</span><span style="display: block; padding: 0 0.2em;">N<sup>3</sup>(N<sup>2</sup>-1)<sup>2</sup></span></div></td>
      <td></td>
    </tr>
    <tr>
      <td>VVV → VV</td>
      <td>g<sup>6</sup> <div style="display: inline-block; vertical-align: middle; text-align: center; line-height: 1.2em;"><span style="display: block; border-bottom: 1px solid black; padding: 0 0.2em;">2π(1323 ln(p<sub>i</sub>·p<sub>j</sub>) - 3 - 396 ln 3 + 1987 ln 2)</span><span style="display: block; padding: 0 0.2em;">p<sub>i</sub>·p<sub>j</sub></span></div></td>
      <td></td>
    </tr>
  </tbody>
</table>

Note, that two scalar products are the same if the two associated energies are the same, i.e. $p_1 \cdot p_5 = p_2 \cdot p_5$ if $p_I = p_J$.

We do not look at interference terms, since they cannot be more divergent than a single diagram squared. We have the principal form of the spin-averaged squared amplitude,

$$ |\mathcal{M}|^2 = \alpha^3 \sum \frac{\prod s_{mn}}{(s_{ij} - \mu)^a (s_{kl} - \mu)^b}. \qquad (C.8) $$

One case is given by

$$ |\mathcal{M}|^2 \supset \frac{1}{(s_{13} - \mu^2)^a (s_{45} - \mu^2)^b}, \qquad (C.9) $$

where if $p_1 \parallel p_2$, then also $p_1 \parallel p_5$, and both denominators become divergent at the same time. Since they are thermally distributed, this is not exactly the case, so they diverge at different integration regions. Therefore we find the suppression by the IR regulator (Debye mass) to be at maximum

$$ (\mu^2)^{-2+1}. \qquad (C.10) $$

We also have

$$ |\mathcal{M}|^2 \supset \frac{1}{(s_{12} - \mu^2)^a (s_{45} - \mu^2)^b}, \qquad (C.11) $$

which after integrating scales at worst like

$$ \frac{1}{(p_1 \cdot p_2)^2} (\mu^2)^{-2+1}. \qquad (C.12) $$
Therefore we find

$$ \int d\Omega |\mathcal{M}|^2 \approx \alpha^3 \frac{1}{p_1 \cdot p_5} \left(\frac{p_1 \cdot p_5}{p_1 \cdot p_2}\right)^k \left(\frac{p_1 \cdot p_5}{\mu^2}\right)^l, \quad k \le 2, l \le 1. \qquad (C.13) $$

Here we inserted the only leftover scale $p_1 \cdot p_5$ to fix the correct mass dimension, which is conservative from the point of view of finding an upper limit on the amplitude squared. For simplicity we have also neglected any appearances of logarithms, since they are not much larger than $O(1)$ factors.

Comparing eq. (C.13) with the calculated processes in tables 5, 6, 7, we find that we do not always have maximal divergence with $k=2, l=1$. This is due to cancellations at amplitude level depending on the specific physical process.

### C.2.2 Equilibration within the shell

For equilibration within the shell itself consider the following momentum basis,

$$ p_{1,com} \approx p_{2,com} \approx p_{5,com} \approx \begin{pmatrix} \mu \\ 0 \\ 0 \\ 0 \end{pmatrix}, \qquad (C.14) $$

$$ p_{3,com} = \begin{pmatrix} p_f \\ p_f \sin(\theta_{13}) \\ 0 \\ p_f \cos(\theta_{13}) \end{pmatrix}, \quad p_{4,com} = \begin{pmatrix} p_f \\ -p_f \sin(\theta_{13}) \\ 0 \\ -p_f \cos(\theta_{13}) \end{pmatrix}, \qquad (C.15) $$

such that $p_f = \frac{3}{2}\mu$. This holds under the assumption that the spread is small, which implies that in the COM-frame each initial particle is approximately at rest with sub-leading momentum component.

Then we find

$$ p_1 \cdot p_2 \approx p_1 \cdot p_5 \approx p_2 \cdot p_5 \approx m_{c,h}^2 \qquad (C.16) $$

$$ p_{1,2,5} \cdot p_3 \approx \frac{3}{2} \mu^2 (1 - z_{13}) \qquad (C.17) $$

$$ p_{1,2,5} \cdot p_4 \approx \frac{3}{2} \mu^2 (1 + z_{13}) \qquad (C.18) $$

$$ p_3 \cdot p_4 \approx 2p_f^2 = \frac{9}{2}\mu^2 \qquad (C.19) $$

An important observation is that $m_{c,h} \gg \mu$, i.e. having spread gives us a fictitious suppression, compared to the maximal possible upper limit. The maximal amplitude is therefore given diagrams scaling like

$$ |\mathcal{M}|^2 \propto \frac{1}{(p_1 \cdot p_3)^2} \frac{1}{(p_2 \cdot p_4)^2} \qquad (C.20) $$
which, since they only contain one energy scale, gives us immediately the scaling

$$ \int d\Omega |\mathcal{M}|^2 \sim \frac{1}{\mu^2}. \qquad (C.21) $$

Comparing eq. (C.21) to the calculated processes in table 8, we find that we not always have maximal divergence. As before, this is due to cancellations at amplitude level depending on the specific physical process.

While the estimates in eq. (C.13) and eq. (C.21) serve as a good guideline and a valuable sanity check for the full computation, we find that they are in most cases too conservative. Due to less divergent pole structures of specific amplitudes one expects to find a result which is smaller by orders of magnitudes by performing a full computation of the specific process. We refer again to the results for a few selected processes in tables 5, 6, 7, 8, for which we have performed the full calculation.

**Open Access.** This article is distributed under the terms of the Creative Commons Attribution License (CC-BY4.0), which permits any use, distribution and reproduction in any medium, provided the original author(s) and source are credited.

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