I'm considering an ultra-relativistic cosmic proton colliding with a CMB photon, creating a neutral pion, as depicted by this equation: $$\tag{1} p + \gamma \rightarrow p + \pi. $$ This process is translated into four-vectors (I'm using units such that $c \equiv 1$ and $\hbar \equiv 1$, and metric signature $\eta = (1, -1, -1, -1)$): $$\tag{2} \mathbf{p}_1 + \mathbf{k} = \mathbf{p}_2 + \mathbf{q}, $$ where $\mathbf{q}$ is the pion's four-momentum. Taking the invariant square of both sides, assuming an ultra-relativistic proton (so that $p_1 \approx E_1$), and considering the threshold reaction, I get the proton's initial energy (this is the GZK limit on cosmic protons): $$\tag{3} E_1 \approx \frac{m_{\pi} (2 m_p + m_{\pi})}{4 \omega} \approx 1 \times 10^{20} \, \mathrm{eV}. $$ Of course, $\omega \approx 6.625 \times 10^{-4} \, \mathrm{eV}$ is the energy of the CMB photon, in the CMB isotropy frame. Now, my trouble is to find the energy variation of the proton: $\Delta E = E_2 - E_1$. I'm assuming an head on collision between the proton and the photon, and I suppose that the proton stays ultra-relativistic all the way (this may be false), without any deflection. Equation (2) could be written like this: $$\tag{4} \mathbf{p}_1 - \mathbf{p}_2 + \mathbf{k} \equiv -\, \Delta\mathbf{p} + \mathbf{k} = \mathbf{q}. $$ Now, I take the invariant square of both sides, using $p_1 \approx E_1$, $p_2 \approx E_2$ and the following (this is the source of my problem): \begin{align} \Delta\mathbf{p}^2 &= p_1^a \, p_{1 a} + p_2^a \, p_{2 a} - 2 p_1^a \, p_{2 a} \\[2ex] &= 2 m_p^2 - 2 \bigl(E_1 \, E_2 - p_1 \, p_2 \cos \alpha \bigr) \\[2ex] &\approx 2 m_p^2 - 2 \bigl(E_1 \, E_2 - E_1 \, E_2 \cos (0^{\circ}) \bigr) = 2 m_p^2. \tag{5} \end{align} $\alpha = 0^{\circ}$ is the angle between the proton's initial and final momentum (no deflection). Then, from the square of (4), I get something wrong ($\Delta E$ should be negative, but $m_p > m_{\pi}$): $$\tag{6} \Delta E = \frac{2 m_p^2 - m_{\pi}^2}{4 \omega} > 0. $$ If I use $\Delta\mathbf{p}^2 \approx 0$ instead, then $$\tag{7} \Delta E \approx -\, \frac{m_{\pi}^2}{4 \omega} < 0, $$ but $\Delta\mathbf{p}^2 \equiv p_1^a \, p_{1 a} + p_2^a \, p_{2 a} - 2 p_1^a \, p_{2 a}$ seems inconsistent with $\Delta\mathbf{p}^2 \approx 0$, since $p_1^a \, p_{1 a} + p_2^a \, p_{2 a} = 2 m_p^2$. I don't think that using a light-like four-vector is right, for a proton.

So what am I doing wrong, and what should be the proper way to find $\Delta E$ ?

  • $\begingroup$ At first glance, I would suspect the assumptions, e.g. $p_1 \approx E_1$ and $p_2 \approx E_2$. Consider doing the exact problem, then make approximations only when you need to. It could be that your approximations are too much, and you may need to retain another order in the approximation. $\endgroup$
    – robphy
    Jun 8, 2022 at 2:23
  • 1
    $\begingroup$ For instance, since $m^2=E^2-p^2$, then $p=\sqrt{E^2+m^2}\approx E(1+(1/2)(m/E)^2)$. $\endgroup$
    – robphy
    Jun 8, 2022 at 4:11
  • $\begingroup$ Are you focussed on the threshold reaction where, in the cm frame, the pπ are at rest? $\endgroup$ Jun 9, 2022 at 15:22
  • $\begingroup$ @CosmasZachos, yes. I would like to check the value given on the Wikipedia page that the proton loses 20% of its energy while producing a neutral pion. I'm finding 12.6%, not 20%. For the production of an electron-positron pair, I confirm their 0.1%. See here, in the "computation" paragraph: en.wikipedia.org/wiki/… $\endgroup$
    – Cham
    Jun 9, 2022 at 18:20
  • 1
    $\begingroup$ Your formula is right. I just dropped the subdominant term in the denominator. $\endgroup$ Jun 9, 2022 at 20:07

3 Answers 3


Since the desired result was provided in another answer, I thought it might be helpful to visualize the conservation-of-total-4-momenta as a problem in hyperbolic trigonometry, which reveals some physically interesting quantities that are not normally featured.

The goal is to get setup the problem exactly,
then systematically apply simplifying approximations to obtain an approximate formula (which could be useful for other problems).
---It's about the process, not just this approximate formula.

My approach is based on my answer to a similar problem
Transformation of the Lorentz factor when a relativistic particle partially absorbs energy from a photon?


The conservation of total-4-momenta can be drawn as a polygon on an energy-momentum diagram (with energy running upward). \begin{align} \tilde {\bf p_1} + \tilde{\bf \gamma} &= \tilde {\bf p_2} + \tilde{\bf q}\\ \tilde {\bf OM} + \tilde{\bf MN} &= \tilde {\bf OM'} + \tilde{\bf M'N}\\ \end{align}

The OP is looking for the lab-frame's energy-component of the change in the 4-momentum of the proton, $\tilde{\bf MM'}$, which is a spacelike chord on the mass-shell of the proton.

The "threshold condition" means that the outgoing particles are at rest with respect to each other. Vectorially, this means $\tilde {\bf p_2} =f \tilde{\bf q}$ (their 4-momenta are parallel), where $f=m_p/m_{\pi}$, the ratio of their rest-masses. Geometrically, this means "$O$, $M'$, and $N$ are collinear", and thus the polygon that would have been a quadrilateral reduces to a triangle in this case. Furthermore, this means that the "invariant mass" of the system is equal to the sum of the rest-masses of the outgoing particles: $N=m_p+m_{\pi}$.

In the spirit of the Bondi k-calculus, since $\tilde{\bf\gamma} =\tilde{\bf MN}$ is future-lightlike and $\tilde{\bf OM}$ and $\tilde{\bf ON}$ are future-timelike, we can immediately write \begin{equation}N=\exp(\theta_i -\theta_f) m_p,\end{equation} where $\exp(\theta_i -\theta_f)$ is the reciprocal of $k_{rel}=\exp(\theta_f -\theta_i)$, the relative-Doppler factor written in terms of the relative-rapidity. In my opinion, the beauty of this is that $$k_{rel}=\exp\theta_{rel}=\frac{1}{\exp(\theta_i-\theta_f)}=\frac{m_p}{N}=\frac{m_p}{m_p+m_{\pi}}$$ Since $k=\sqrt{\frac{1+v}{1-v}}$, we have $v=\frac{k^2-1}{k^2+1}$. So, $k_{rel}<1$ implies $v_{rel}<0$--that is, the final particle is slower.
In hindsight, it is interesting to note that $$k_{rel}-1= \left(-\frac{m_{\pi}}{m_p+m_{\pi}}\right).$$

Sidebar: proof of the Bondi-like relation

  • Conservation of 4-momentum componentwise: \begin{align} m_p\cosh\theta_f + m_{\pi}\cosh\theta_f &= m_p\cosh\theta_i + (\phantom{-}\epsilon) \\ m_p\sinh\theta_f + m_{\pi}\sinh\theta_f &= m_p\sinh\theta_i + (- \epsilon) \end{align} By addition, $$(m_p+m_{\pi}) \exp\theta_f = m_p \exp\theta_i,$$ which is equivalent to $N=\exp(\theta_i -\theta_f) m_p$... essentially the Doppler effect.
    By subtraction, $$(m_p+m_{\pi}) \exp(-\theta_f) = m_p \exp(-\theta_i)+2\epsilon.$$ When this is multiplied with the "equation gotten from addition", one obtains essentially the square magnitude of the total 4-momentum.
    These are essentially the conservation laws written in light-cone coordinates (an eigenbasis of the Lorentz boost).

Back to the OP's problem:

\begin{align} \frac{\Delta E}{E}=\frac{(MM')_t}{(OM)_t} &=\frac{\epsilon-m_{\pi}\cosh\theta_f}{m_p\cosh\theta_i} &=\frac{\epsilon-m_{\pi}\cosh(\theta_i+\theta_{rel})}{m_p\cosh\theta_i} \end{align} is the exact result, but is likely rather messy to proceed. And we might need information about $\theta_i$ (about the initial proton velocity in the lab frame).

  • (UPDATE) Here is a slightly different way to proceed before making an approximation, as was done in the ORIGINAL section.
    \begin{align} \frac{\Delta E}{E} &=\frac{\epsilon-m_{\pi}\cosh(\theta_i+\theta_{rel})}{m_p\cosh\theta_i}\\ &=\frac{\epsilon}{m_p\cosh\theta_i} -\frac{m_{\pi}}{m_p} \left(\frac{\cosh\theta_i \cosh\theta_{rel}+\sinh\theta_i \sinh\theta_{rel}}{\cosh\theta_i}\right)\\ &=\frac{\epsilon}{m_p\cosh\theta_i} -\frac{m_{\pi}}{m_p} \left(\cosh\theta_{rel}+\tanh\theta_i \sinh\theta_{rel} \right) \end{align} Now invoke the ultra-relativistic condition so that $\tanh\theta_i \approx 1$ and use the identity $\exp\theta=\cosh\theta+\sinh\theta$ to get \begin{align} \frac{\Delta E}{E} &\approx\frac{\epsilon}{m_p\cosh\theta_i} -\frac{m_{\pi}}{m_p} \exp\theta_{rel} \end{align}
  • (ORIGINAL) However, if we now invoke the ultra-relativistic condition, then $\theta_i, \theta_f \gg 1$,
    which implies that $\exp\theta\gg 1$ and thus $\exp\theta \approx 2\cosh\theta$. \begin{align} \frac{\Delta E}{E}\approx \frac{\epsilon-m_{\pi}2\exp(\theta_i+\theta_{rel})}{m_p 2\exp\theta_i} &= \frac{\frac{1}{2}\epsilon\exp(-\theta_i)-m_{\pi}\exp(\theta_{rel})}{m_p} \end{align}

Using the condition that $\epsilon\exp(-\theta_i) \ll m_{\pi}$ (either small $\epsilon/m_{\pi}$, initially ultra-relativistic proton, or both), we further approximate as \begin{align} \frac{\Delta E}{E}\approx \frac{-m_{\pi}\exp(\theta_{rel})}{m_p} =\frac{-m_{\pi}\left( \frac{m_p}{m_p+m_{\pi}} \right)}{m_p} \stackrel{\checkmark}{=}\frac{-m_{\pi}}{m_p+m_{\pi}}. \end{align} (Maybe $k_{rel}-1$ can be shown to naturally here, together with a physical interpretation.)

Hopefully, this is useful (it was to me) to see the approximate formula arise from an exact formulation of the spacetime geometric interpretation of the conservation laws applied in this particle process.


Since you asked, it is a two liner, and you appear justified in contesting the 20% energy loss number. On the line of the proton-photon collision, you get the GKZ threshold contrasting the cosmic frame to the cm frame. The problem is essentially one-dimensional, and safely light-like before and after the collision, since $E_1\approx 10^{11}$GeV.

In the cm frame, you have a pion and a proton at rest. Transforming to the cosmic frame, you have both of these moving in the same direction with a common velocity, so $$ v=\frac{p_2}{\sqrt{m_p^2+p_2^2}} = \frac{q}{\sqrt{m_\pi^2+ q^2}} ~~~\implies \\ q= p_2 {m_\pi \over m_p}, $$ so, from conservation of momentum, $p_1\approx p_2+ q$, and hence the proton energy loss is $$\Delta E\approx \frac{p_2-p_1}{p_1}={-q\over q+p_2}={-1\over m_p/m_\pi +1}, $$ as you observe, which is not close to 20%. I am not sure what the WP statement is trying to say.

  • $\begingroup$ This answer is very elegant. Thanks! $\endgroup$
    – Cham
    Jun 9, 2022 at 23:36

This is a comment addressing the probability of a relativistic proton to interact with a low energy photon and produce as a two body reaction a pi0.

If you look at "photoproduction of pi0" a number of references come up, for example this


The model fitting the data has a number of feynman diagrams but note that it is a nuclear physics field theory. This means that for the quarks composing the pi0 in order to bind to it the energies must be low, nuclear physics energies so that there is a meaning to exchanges of rho and pi.

At the energies you are contemplating there is very very low probability for any quarks generated to bind into a pi0

  • $\begingroup$ Are you sure that the probability of (1) is very low at high energy? The Wikipedia article en.wikipedia.org/wiki/… is implying that the probability is high. And what about the process of an electron-positron pair production? $$p + \gamma \rightarrow p + e^{-} + e^{+}$$ $\endgroup$
    – Cham
    Jun 8, 2022 at 14:09
  • $\begingroup$ I am commenting on the reaction you have written down in your question, the two body to two body. Generally pi0 are produced in the interactions where heavier hadrons are produced and pi0 complete the conservation laws for quark interactions. pair production is a normal process for gamma scattering at relativistic energies, and other pair productions. $\endgroup$
    – anna v
    Jun 8, 2022 at 15:01
  • $\begingroup$ I do recall $\pi^0$ being more common than $\pi^\pm$, but all three interactions happen (though usually with either intermediate states or with extra particles). See also physics.stackexchange.com/q/406680/25301, physics.stackexchange.com/q/406680/25301, etc $\endgroup$
    – Kyle Kanos
    Jun 8, 2022 at 16:18
  • $\begingroup$ @KyleKanos sure, but they are inclusive channels, not two body ones, there should be a pronton +gamma---> proton+pi0 +X in the notation. that is why I say it is comment, $\endgroup$
    – anna v
    Jun 8, 2022 at 17:16
  • $\begingroup$ Yes, my comment was intended as supporting your comment/answer posted. It definitely should be $p+\gamma\to p+\pi^{0,\pm}+X$ $\endgroup$
    – Kyle Kanos
    Jun 8, 2022 at 18:22

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