# Mandelstam variables 1 positive 2 negative

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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 ?

This post imported from StackExchange Physics at 2014-04-14 16:19 (UCT), posted by SE-user Nick
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 The $p$s that appear are 4-momenta. So no, you cannot guarantee that the squares are positive.
@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.
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