Is $A_{pp}(s,t)=A_{p\bar p}(t,s)$ true based on crossing symmetry?

Consider $pp$ and $p\bar p$ elastic colissions ($p + p \rightarrow p + p$ and $p + \bar p \rightarrow p + \bar p$). The scattering amplitudes are related by crossing in the following way:

1) $A_{pp}(s,t)=A_{p\bar p}(u,t)=A_{p\bar p}(4m^2-s-t,t) \simeq A_{p\bar p}(-s-t,t)$ (energy large compared to $4m^2$)


$A$, scattering amplitude.

$s,t,u$ Mandelstam variables.

$p$ proton $\bar p$ antiproton.

I don´t have any problem with this.

However, unless I made a huge mistake (the more I look at my graphs the more convinced I am that I haven't), crossing should also impose:

2) $A_{pp}(s,t)=A_{p\bar p}(t,s)$ which is very, very difficult for me to accept because of the limit t=0 ($pp$ scattering amplitude equal to $p\bar p$ pure annihilation plus later $p\bar p$ pair creation??

How on earth can they interact if they're not approaching?? Is this a $p\bar p$ resonance?? Moreover, there's no energy for them to scatter off.

$t=0$ makes no sense to me on the right hand side of the equation but it does on the left hand side. Unless $p$ and $\bar p$ are moving in the same direction, instead of head-on. Could this be it? Would this explain a 0 C.O.M frame $s$, but a huge $t$? I think this could be the explanation.

It would be interesting to check if $A_{pp}(4m^2,\epsilon)\simeq A_{p\bar p}(\epsilon,4m^2)$ , $\epsilon$ very small compared to $4m^2$.

This data must be avialable somewhere.

$m$ proton mass.

Can anybody tell me if this latter relationship is wrong?

By the way:

1) has a very interesting implication in the $t=0$ limit, that could, perhaps, be easily checked with the existing models:

$tg^{-1}\frac{1}{\rho^{pp}}(s,t=0)-tg^{-1}\frac{1}{\rho^{p\bar p}}(4m^2-s,t=0)=|2n\pi|$, where:

$n$ is a non-specified natural number.


$A$, scattering amplitude.

I will elaborate a little further, assuming that both 1) and 2) are right (I don't know if 2) is):

From 1) and 2) it is immediate to deduce that:

3) $A_{p\bar p}(u,t)=A_{p\bar p}(t,s) \Rightarrow A_{p\bar p}(4m^2-s-t,t)=A_{p\bar p}(t,s)$.

and making $t=0$,

4) $A_{p\bar p}(4m^2-s,0)=A_{p\bar p}(0,s)$ and something qualitatively new must happen at $s^\frac{1}{2}=2m\simeq 2 GeV$ and,

It really does!!

a) $p + p$ total cross-section has a minimum at that energy!! (about $20 mb$).

b) There is a resonance that suddenly rises the total cossection up to $50 mb$!!

c) Non elestic $pp$ collission do have a $2 GeV$ threshold!!

I guess that if you already know nuclear physics this is hardly surprising, but this is a purely mathematical result.

From the point of view of $p + \bar p$ things look very messy, however. $A_{p\bar p}(0,0)=A_{p\bar p}(0, 2 GeV)$.

Well, not really, a $p + \bar p$ pair that transfers $q\leq 2 GeV$, being both at rest, is just impossible (partial annhihilation is just not an option) and a $p + \bar p$ whose $C.O.M$ frame energy is less than $2 GeV$ is also impossible.

Well, it looks like 1) and 2) are telling us, at the very least, things that are known to be true on physical grounds using pure mathematical considerations. Maybe 2) is right after all.

I will elaborate a little further because I think there might be experimental evidence that supports/discards (2):

$A_{pp}(s,t)=A_{p\bar p}(4m^2-s-t,t) \Rightarrow$ (t=0)

$\Rightarrow A_{pp}(s,0)=A_{p\bar p}(4m^2-s,0)$.

If the (global) minimum of the total $pp$ cross-section happens, more or less, at $s^\frac{1}{2}$=$2m$ + protonium binding energy$\simeq$ $2 Da + 0.102 Da$ (theoretical estimate)=$1.97 GeV$ then (2) should be OK.

$A_{p\bar p}(4m^2-s,0)$. For $s=1.97$ there is going to be a resonance and a subsequent $p\bar p$ anihilation. The chances of this process producing another $p\bar p$ afterwards cannot be lower (global minimum) and, therefore, the same thing happens with $A_{pp}(s,0)$.


Page 11, up left. I'm right!

Oh, I can hardly believe it, but that pronounced dip at 2$GeV$ does indeed tell a lot!

  • $\begingroup$ It's been hell. $\endgroup$ – Carlos L. Janer Jul 27 '18 at 16:27
  • $\begingroup$ But thanks, I understand what you mean. $\endgroup$ – Carlos L. Janer Jul 27 '18 at 16:28
  • $\begingroup$ One more headache: The s-channel CM momentum has to be timelike, but the t-channel momentum exchange has to be spacelike. When you turn the diagram 90 degrees to explore crossing symmetry, you find yourself in an experimentally inaccessible regime. Good luck with analytic continuation. $\endgroup$ – Bert Barrois Feb 28 '19 at 14:20

I've already found the answer to my question:


Slide 5. Equation 2) is indeed true and is a direct consequence of crossing symmetry (although crossing symmetry assumes PT=1 which is not exactly true in EW interactions). Sorry for the bother.

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