Boosting your spectra

Table of Contents

Starting from Rest Frame

Suppose we know the energy spectrum \(\dv*{N_{f}}{E_{1}}\) of some product \(f\) from the decay of a state \(I\) in the rest frame of \(I\). From this spectum, we would like to obtain the corresponding spectrum, \(\dv*{N_{f}}{E_{2}}\) in a boosted frame. First, assume that the four-momentum of \(f\) in the initial frame is:

$$ \begin{align} p^{\mu}_{1} & = (E_{1}; \mathbf{p}_{1}) \end{align} $$

where \(\mathbf{p}_{1}\) is the three-momentum of \(f\). without loss of generality, let’s assume that \(\mathbf{p}_{1}\) lies in the \(xz\)-plane:

$$ \mathbf{p}_{1} = (|{\mathbf{p}_{1}}| \tilde{z}_{1}, 0, |\mathbf{p}_{1}|z_{1}), $$

with \(z_{1}\) equal to the cosine of the angle \(\mathbf{p}_{1}\) makes with the \(z\) axis and \(\tilde{z}_{1} = \sqrt{1-z_{1}^2}\). Assume we boost along the \(z\)-axis, we can translate this four-momentum in the new frame using \(p^{\mu}_{2} = {\Lambda}{^\mu_\nu}p^{\nu}_{1}\) with

$$ {\Lambda}{^\mu_\nu} = \begin{pmatrix} \gamma & 0 & 0 & \gamma\beta \\ 0&1&0&0\\ 0&0&1&0\\ \beta\gamma&0&0&\gamma \end{pmatrix} $$

Explicity, \(p^{\mu}_{2}\) is given by:

$$ \begin{align} p^{\mu}_{2} & = (E_{2}; \mathbf{p}_{2}) \\ E_{2} & = \gamma E_{1} + \beta\gamma|{\mathbf{p}_{1}}|z_{1} \\ \mathbf{p}_{2} & = (|{\mathbf{p}_{1}}|\tilde{z}_{1}, 0, \beta\gamma E_{1} + \gamma|{\mathbf{p}_{1}}|z_{1}) \end{align} $$

Notice that \(E_{2}\) explicity depends on \(z_{1}\), which we have integrated over in obtaining \(\dv*{N_{f}}{E_{1}}\). To reintroduce \(z_{1}\), we use:

$$ \begin{align} {\dv{N_{f}}{E_{1}}}(E_{1}) & = \int_{-1}^{1}\dd{z} \pdv{N_{f}}{E_{1}}{z_{1}}, & \pdv{N_{f}}{E_{1}}{z_{1}} & = \frac{1}{2} \dv{N_{f}}{E_{1}} \end{align} $$

In order to translate this expression into \(\dv*{N_{f}}{E_{2}}\), we have two options. The first is to introduce a \(\delta\)-function enforcing the correct relation between \(E_{2}\) and \(E_{1}\) and integrate over \(E_{1}\) in addition to \(z_{1}))¹ . The trick is to first use:

$$ \begin{align} N_{f} = \int\dd{z_{1}}\dd{E_{1}}\pdv{N_{f}}{E_{1}}{z_{1}} \end{align} $$

and insert:

$$ \begin{align} 1 & = \int\dd{E_{2}}\delta(E_{2} - E_{2}(E_{1}, z_{1})) \end{align} $$

where \(E_{2}(E_{1},z_{1}) = \gamma E_{1} + \beta\gamma|{\mathbf{p}_{1}}|z_{1}\) Inserting this factor of unity, we obtain:

$$ \begin{align} N_{f} = \int\dd{E_{2}}\int\dd{z_{1}}\int\dd{E_{1}}\pdv{N_{f}}{E_{1}}{z_{1}} \delta(E_{2} - E_{2}(E_{1}, z_{1})) \end{align} $$

Differentiating with respect to \(E_{2}\), we find:

$$ \begin{align} \dv{N_{f}}{E_{2}} = \int\dd{z_{1}}\int\dd{E_{1}}\pdv{N_{f}}{E_{1}}{z_{1}} \delta(E_{2} - E_{2}(E_{1}, z_{1})) \end{align} $$

The second method is to use one of the following (which is equivalent; see ²)

$$ \begin{align} \dd[2]{N} = \pdv{N_{f}}{E_{1}}{z_{1}}\dd{E_{1}}\dd{z_{1}} = \pdv{N_{f}}{E_{1}}{z_{1}} \mathcal{J} \dd{E_{2}}\dd{z_{2}} = \pdv{N_{f}}{E_{1}}{z_{1}} \tilde{\mathcal{J}} \dd{E_{2}}\dd{z_{1}} \end{align} $$

with the Jacobians, \(\mathcal{J}\) or \(\tilde{\mathcal{J}}\), given by:

$$ \begin{align} \mathcal{J} & = \mqty| \pdv{E_{1}}{E_{2}} & \pdv{E_{1}}{z_{2}} \ \pdv{z_{1}}{E_{2}} & \pdv{z_{1}}{z_{2}} |, & \tilde{\mathcal{J}} & = \mqty| \pdv{E_{1}}{E_{2}} | \end{align} $$

depending on if one wishes to convert \(z_{1}\) into \(z_{2}\) (since the angular varibles will ultimately be integrated over, either choice is fine.) We will focus on the \(\delta\)-function approach since it is easiest to deal with.

In computing \(\dv*{N_{f}}{E_{2}}\), we need to perform one integration. But we have the choice of either integration over \(E_{1}\) or \(z_{1}\). The differences in the two will appear in the limits of integration. Suppose we integrate over \(z_{1}\). Then, the \(\delta\)-function can be casted to

$$ \begin{align} \delta(\gamma E\_{1} + \beta\gamma|{\mathbf{p}_{1}}|z_{1} - E_{2}) & = \frac{1}{\gamma\beta\abs{\mathbf{p}_{1}}}\delta(z_{1} - z^{0}_{1}), & z^{0}_{1} & = \frac{E_{2}-\gamma E_{1}}{\gamma\beta\abs{\mathbf{p}_{1}}} \end{align} $$

As a side note, if we used the Jacobian for $ (E_{1},z_{1}) \to (E_{1},E_{2})$, which is given by:

$$ \begin{align} \dd{E_{1}}\dd{z_{1}} = \mqty|\pdv{z_{1}}{E_{2}}|\dd{E_{1}}\dd{E_{2}} = \frac{1}{\gamma\beta\abs{\mathbf{p}}_{1}}\dd{E_{1}}\dd{E_{2}} \end{align} $$

then we get the same factor of $ 1/\gamma\beta\abs{\mathbf{p}_{1}}$.

Since $ -1 < z_{1} < 1$, the $ \delta$-function only has support if

$$ \begin{align} -1 < \frac{E_{2}-\gamma E_{1}}{\gamma\beta\abs{\mathbf{p}_{1}}} < 1 \end{align} $$

or

$$ \begin{align} \gamma\qty(E_{2}-\beta\sqrt{E_{2}^{2}-m_{f}^{2}}) < E_{1} < \gamma\qty(E_{2}+\beta\sqrt{E_{2}^{2}-m_{f}^{2}}) \end{align} $$

Defining $E^{\pm}_{1} = \gamma\qty(E_{2}\pm\beta\abs{\mathbf{p}_{2}})$ with $\abs{\mathbf{p}_{2}} = \sqrt{E_{2}^{2}-m_{f}^{2}}$, we find:

$$ \begin{align} \boxed{ \dv{N_{f}}{E_{2}} = \frac{1}{2\gamma\beta}\int^{E^{+}_{1}}_{E^{-}_{1}} \frac{\dd{E_{1}}}{\abs{\mathbf{p}_{1}}} \dv{N_{f}}{E_{1}} = \frac{1}{2\gamma\beta}\int^{E^{+}_{1}}_{E^{-}_{1}} \frac{\dd{E_{1}}}{\sqrt{E_{1}^{2}-m^2_{f}}} \dv{N_{f}}{E_{1}} } \end{align} $$

Note that the bounds may be different that $E^{\pm}_{1}$, depending on $E_{2}$. The function $\dv*{N_{f}}{E_{1}}$ itself has limits $E^{\mathrm{min}}_{1}$ and $E^{\mathrm{max}}_{1}$. Thus, the actual limits of integration are:

$$ \begin{align} \mathrm{max}(E^{\mathrm{min}}_{1}, E^{-}_{1}) \leq E_{1} \leq \mathrm{min}(E^{\mathrm{max}}_{1}, E^{+}_{1}) \end{align} $$

Boost Integral in terms of Scaleless Varibles

It is sometimes advantagous to deal with scaleless varibles when working with spectra. For example, it is common to define $x_{i} \equiv 2E_{i}/Q_i$, where $Q_i$ is the center-of-mass energy. If, in addition, we define a scaleless mass $\mu_i = 2m_{f}/Q_i$, then we can write:

$$ \begin{align} \dv{N_{f}}{x_{i}} & = \frac{Q_{i}}{2}\dv{N_{f}}{E_{i}} \end{align} $$

and hence:

$$ \begin{align} \dv{N_{f}}{x_{i}} & = \frac{Q_{2}}{2}\frac{1}{2\gamma\beta}\int^{x^{+}_{1}}_{x^{-}_{1}} \frac{(Q_{1}/2)\dd{x_{1}}}{\sqrt{(Q_{1}^2/4)x_{1}^{2}-(Q_{1}^2/4)\mu^2_{f}}} \frac{2}{Q_{1}}\dv{N_{f}}{x_{1}} \end{align} $$

Simplifying and using $ Q_{2} = \gamma Q_{1}$, we find:

$$ \begin{align} \boxed{ \dv{N_{f}}{x_{2}} = \frac{1}{2\beta}\int^{x^{+}_{1}}_{x^{-}_{1}} \frac{\dd{x_{1}}}{\sqrt{x_{1}^{2}-\mu^2_{1}}} \dv{N_{f}}{x_{1}} } \end{align} $$

In this case, the integration bounds are given by:

$$ \begin{align} x^{-}_{1} & \equiv \mathrm{max}\qty(x^{\mathrm{min}}_{1}, \gamma^{2}\qty(x_{2}-\beta\sqrt{x_{2}^2-\mu_{2}^2})) \\ x^{+}_{1} & \equiv \mathrm{min}\qty(x^{\mathrm{max}}_{1}, \gamma^{2}\qty(x_{2}+\beta\sqrt{x_{2}^2-\mu_{2}^2})) \end{align} $$

Examples

Dirac-Delta Function Spectrum

In the case where $\dv*{N_{f}}{E_{1}} = N \delta(E_{1}-\bar{E})$, we can exactly perform the integral in \EqnRef{eqn:boost_e}. In this case, we find

$$ \begin{align} \dv{N_{f}}{E_{2}} = \frac{1}{2\gamma\beta}\int^{E^{+}_{1}}_{E^{-}_{1}} \frac{\dd{E_{1}}}{\sqrt{E_{1}^{2}-m^2_{f}}} N \delta(E_{1} - \bar{E}) = \frac{1}{2\gamma\beta}\frac{N}{\sqrt{\bar{E}^{2}-m^2_{f}}} \begin{cases} 1 & \qq*{if} E^{+}_{1}\leq \bar{E} \leq E^{+}_{1}, \\ 0 & \mathrm{otherwise} \end{cases} \end{align} $$

As an example, take $ \pi^{0} \to \gamma + \gamma$. In this case, $ \dv*{N_{f}}{E_{1}} = 2\delta(E_{1} - m_{\pi^{0}}/2)$, and the spectrum is

$$ \begin{align} \dv{N_{f}}{E_{2}} = \frac{2}{\gamma\beta m_{\pi^{0}}} \begin{cases} 1 & \qq*{if} \gamma E_{2}(1-\beta)\leq m_{\pi^{0}}/2 \leq \gamma E_{2}(1+\beta), \\ 0 & \mathrm{otherwise} \end{cases} \end{align} $$

Muon Decay

The muon decays via $ \mu^{\pm} \to e^{\pm} + \nu_{e} + \nu_{\mu}$. This decay can also proceed radiatively. The result has been worked out in ³. The result for an unpolarized muon is:

$$ \begin{align} \dv{N}{x} & = \frac{\alpha}{3\pi}\sum_{n=0}^{4}\qty(a_{n} + b_{n}L(y))x^{n-1} \end{align} $$

where $ x\equiv 2E_{\gamma}/m_{\mu}$, $ L(x) \equiv \log((1-x)/r)$, $ r = (m_{e}/m_{\mu})^2$ and

$$ \begin{align} a_0 & = -\frac{17}{2}, & a_1 & = \frac{37}{3}, & a_2 & = -\frac{49}{4}, & a_3 & = 13, & a_4 & = -\frac{55}{12} \\ b_0 & = 3, & b_1 & = -5, & b_2 & = 6, & b_3 & = -6, & b_4 & = 2 \end{align} $$

Note the limits on $ x$ are $ 0 \leq x \leq 1 - r$. Since the photon is massless, \EqnRef{eqn:boost_x} takes the form:

$$ \begin{align} \dv{N_{f}}{x_{2}} & = \frac{1}{2\beta}\int^{x^{+}}_{x^{-}} \frac{\dd{x_{1}}}{x_{1}} \dv{N_{f}}{x_{1}} \end{align} $$

with

$$ \begin{align} x^{-} & \equiv \gamma^{2}x_{2}\qty(1 - \beta), & x^{+} & \equiv \mathrm{min}\qty(1-r, \gamma^{2}x_{2}\qty(1 + \beta)) \end{align} $$

First, the integral of the polynomial part of $ \dv*{N}{x}$ is given by:

$$ \begin{align} \frac{1}{2\beta}\sum_{n=0}^{4}a_{n}\int^{x^{+}}_{x^{-}} \dd{x_{1}}x_{1}^{n-2} = \frac{1}{2\beta}\qty[ a_0\frac{x^{+} - x^{-}}{x^{+}x^{-}} + a_1\log(\frac{x^{+}}{x^{-}}) + \sum_{n=2}^{4}a_n \frac{ \qty(x^{+})^{n} - \qty(x^{-})^{n} }{n}] \end{align} $$

The log terms have the form:

$$ \begin{align} \int^{x^{+}_{1}}_{x^{-}_{1}} \dd{x_{1}}x_{1}^{n-2} L(x_1) \end{align} $$

To integrate, we use:

$$ \begin{align} \int\dd{x}x^{-2}L(x) & = \log(1-x) - log(x) - \frac{1}{x}L(x) \\ \int\dd{x}x^{-1}L(x) & = L(x)\log(x) + \mathrm{Li}_{2}(1-x) \\ \int\dd{x}L(x) & = -x - (1-x) L(x) \\ \int\dd{x}x L(x) & = \frac{1}{4}x (2 + x) -\frac{1}{2}\log(1-x) + \frac{1}{2}x^2 L(x) \\ \int\dd{x}x^2L(x) & = -\frac{1}{18}x (6 + 3 x + 2 x^2) - \frac{1}{3}\log(1-x) + \frac{1}{3}x^3 L(x) \end{align} $$

Using the definitions of $ b_{n}$, we find:

$$ \begin{align} & \frac{1}{2\beta}\sum_{n=0}^{4}b_{n}\int^{x^{+}_{1}}_{x^{-}_{1}} \dd{x_{1}}x_{1}^{n-2} L(x_1) \\ & \quad = \frac{1}{2\beta}\bigg{[} -3\log(x) + \frac{16}{3}\log(1-x) - \frac{1}{18} x (66 - 21 x + 4 x^2) - 5 \mathrm{Li}_{2}(1-x) \\ & \hspace{2cm} \log(\frac{1-x}{r})\qty(-6 - \frac{3}{x} + 6 x - 3 x^2 + \frac{2}{3}x^3 - 5\log(x))\bigg{]}\notag\bigg{|}^{x^{+}}_{x^{-}} \end{align} $$