# Mandelstam variables 1 positive 2 negative

The three Mandelstam-variables are defined as: $$s=(p_A+p_B)^2=(p_C+p_D)^2,$$$$t=(p_A-p_C)^2=(p_B-p_D)^2$$$$u=(p_A-p_D)^2=(p_B-p_C)^2.$$ Where A and B are the incoming particles and C and D are the outgoing particles, and the second equation follows from the conservation of 4-momentum.

The Mandelstam-variable $s$ gives the center-of-mass energy and is thus always positive, now for the Mandelstam-variables $t$ and $u$ the demands can be found by looking at them in the laboratory-frame, after an easy calculation it can be shown that:$$s\geq \max[m_A^2+m_B^2,m_C^2+m_D^2],$$ $$t\leq \min[m_A^2+m_C^2,m_B^2+m_D^2],$$ $$u\leq \min[m_A^2+m_D^2,m_B^2+m_C^2].$$

This means that there should exist a situation in which the three Mandelstam-variables are positive. I was wondering is such a situation exists, or that I am simply overlooking some facts and that indeed only ONE Mandelstam-variable can be positive?

To elaborate more on the calculation which lead me to this conclusion, I've calculated these variables in the frame in which particle A is at rest (since $s$, $t$ and $u$ are invariant I can do this). This yields for the momenta: $$p_A=(m_A,0),$$$$p_B=(E_B,\vec{p}_B),$$$$p_C=(E_C,\vec{p}_C),$$$$p_D=(E_D,\vec{p}_D).$$ So for my Mandelstam variables I would get that: $$s=(p_A+p_B)^2=p_A^2+p_B^2+2p_A\cdot p_B = m_A^2+m_B^2+2m_AE_B,$$ $$t=(p_A-p_C)^2=p_A^2+p_C^2-2p_A\cdot p_C = m_A^2+m_C^2-2m_AE_C,$$ $$t=(p_A-p_D)^2=p_A^2+p_D^2-2p_A\cdot p_D = m_A^2+m_D^2-2m_AE_D,$$ where I used the first equality in the definition of the Mandelstam variables. I could do the same calculation with the second equality, if I would then discard the terms of the form $2m_AE$ I would get the inequalities above.

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Edit: I know that the second equality doesn't give simple terms $2mE$, but terms of the form $$E_CE_D-\vec{p_C}\cdot\vec{p_D}=E_CE_D-|\vec{p_C}||\vec{p_D}|\cos(\theta),$$ and since $E^2=m^2+|\vec{p}|^2$ we have that $E>|\vec{p}|$, so these terms are positive.

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Edit 2: I've calculated these for the non-elastic proces $e^-e^+\rightarrow\mu^-\mu^+$ (since inelastic problems are the source of the problem) and I found that: $$s=2m(m+E_B),$$ $$t=m^2+M^2-2mE_C,$$ $$u=m^2+M^2-2mE_D,$$ where $m$ is the mass of the electron and $M$ is the mass of the muon, particle $A$ (either an electron or positron) is standing still and for particle $B$ (electron or positron) and $C$ and $D$ (the 2 muon particles) the energies are $E_B$, $E_C$ and $E_D$. The variable $s$ is obviously positive, to check if $t$ of $u$ are positive I summed them and used the conservation of energy: $m+E_B=E_C+E_D$, which gives: $$t+u=2(m^2+M^2)-2m(m+E_B)=2(m^2+M^2)-s,$$ now in order to be able to create 2 muons the energy should be high enough, hence $s=(2M)^2$, filling this in yields: $$t+u=2(m^2-M^2).$$ Which brings me back to the question: ''Should only one of the three variables be positive (just as most books claim) or are there special cases''? In my derivation above I used the conservation laws, but as you can see in this example the possibility of $t$ and $u$ being positive (for small enough $E_B$) vanishes by demanding that de center of mass energy $\sqrt{s}$ is large enough, will this always be the case ?

• From your equations in the first line: $s=(p_A+p_B)^2,\,t=(p_A-p_C)^2 \text{ and }u=(p_A-p_D)^2.$ This states that s, t and u are all positive since they are squares of real numbers.
– LDC3
Commented Apr 13, 2014 at 3:33
• @LDC3 The $p$s that appear are 4-momenta. So no, you cannot guarantee that the squares are positive. Commented Apr 13, 2014 at 3:49
• @LDC3, you can see this if you would calculate $s$, $t$ and $u$ in the center of mass- (or momentum)-frame for an elastic collision (so the masses stay conserved). In that case you would get the specific demand that $s\geq (m_1+m_2)^2$, $t\leq 0$ and $u\leq 0$ for the proces to be physical.
– Nick
Commented Apr 13, 2014 at 13:35

I) Yes, e.g. all three Mandelstam variables

$$s~:=~(p_1+p_2)^2~=~m_1^2+m_2^2+2 p_1\cdot p_2 ~\approx~ (m_1+m_2)^2 + m_1m_2 ({\bf v}_1-{\bf v}_2)^2 ~>~0,$$

$$t~:=~(p_1-p_3)^2~=~m_1^2+m_3^2-2 p_1\cdot p_3~\approx~ (m_1-m_3)^2 - m_1m_3 ({\bf v}_1-{\bf v}_3)^2 ~>~0,$$

$$u~:=~(p_1-p_4)^2~=~m_1^2+m_4^2-2 p_1\cdot p_4~\approx~ (m_1-m_4)^2 - m_1m_4 ({\bf v}_1-{\bf v}_4)^2 ~>~0,$$

are strictly positive in the non-relativistic limit

$$|{\bf v}_i|~\ll~ c, \qquad i~\in~\{1,2,3,4\},$$

of massive particles

$$m_i~>~ 0, \qquad i~\in~\{1,2,3,4\},$$

with unequal (rest) masses

$$i~\neq~j~~\Rightarrow~~ m_i~\neq~ m_j, \qquad i,j~\in~\{1,2,3,4\}.$$

Here we have used units where $c=1$, and the non-relativistic formulas

$${\bf p}_i~\approx~m_i{\bf v}_i, \qquad E_i~=~\sqrt{m_i^2+{\bf p}_i^2}~\approx~m_i\left(1+ \frac{{\bf v}_i^2 }{2}\right), \qquad i~\in~\{1,2,3,4\},$$

and

$$p_i\cdot p_j~=~E_i E_j - {\bf p}_i\cdot {\bf p}_j~\approx~~m_i m_j \left(1+ \frac{1}{2}\left({\bf v}_i-{\bf v}_j\right)^2 \right), \qquad i,j~\in~\{1,2,3,4\}.$$

II) By the way $s+t+u=\sum_{i=1}^4 m_i^2 \geq 0$ implies that it is impossible to have all Mandelstam variables $s,t,u<0$ negative. So at least one of the three sectors are physical.

• thanks for the great and clear answer. I assume for the variables $t$ and $u$ the second term (with the velocities) can be made sufficiently small ? Also doesn't the fact that the masses differ from eachother impose extra conditions on the system above ?
– Nick
Commented Apr 17, 2014 at 14:20
• The pairwise different masses is an easy realizable and sufficient condition. I did not analyze the question of necessity carefully. Commented Apr 21, 2014 at 20:09