Mandelstam variables sign

I am self-studying the book "Quantum Field Theory and the Standard Model" by Schwartz, on page 99 (paragraph "Mandelstam variables"), the context is the $$2\rightarrow 2$$ scattering in the CM frame.

He states that $$s>0$$ and $$t<0$$, $$u<0$$.

How can he state that $$t<0$$?

As far as I can see that is not always the case, though $$t<0$$ when $$m_1=m_2=m_3=m_4$$.

Below I report my calculation:

\begin{align} t&=(p_1-p_3)^2 \\ &= (E_1-E_3)^2-|\vec{p_1}-\vec{p_3}|^2 \\ &= (E_1-E_3)^2 -\left(|\vec{p_1}|^2+|\vec{p_3}|^2-2\vec{p_1}\cdot\vec{p_3}\right) \\ &= (E_1-E_3)^2-\left(E_1^2-m_1^2+E_3^2-m_3^2-2\vec{p_1}\cdot\vec{p_3}\right) \\ &= m_1^2+m_3^2-2E_1E_3+2\vec{p_1}\cdot\vec{p_3}\end{align}

using $$2\vec{p_1}\cdot\vec{p_3}\leq |\vec{p_1}|^2 + |\vec{p_3}|^2 = E_1^2-m_1^2+E_3^2-m_3^2$$

we have:

$$t\leq (E_1-E_3)^2$$

In the CM frame we have $$\vec{p_1}=-\vec{p_2}$$ and $$\vec{p_3}=-\vec{p_4}$$ and since $$E_1+E_2=E_3+E_4$$, we have

$$\sqrt{m_1^2+|\vec{p_1}|^2} + \sqrt{m_2^2+|\vec{p_1}|^2} = \sqrt{m_3^2+|\vec{p_3}|^2} + \sqrt{m_4^2+|\vec{p_3}|^2}$$

Now, if $$m_1=m_2=m_3=m_4$$, the equation above gives $$|\vec{p_1}|^2=|\vec{p_3}|^2$$, then $$E_1=E_3$$ (cause also $$m_1=m_3$$), so we have $$t\leq 0$$.

However, in general if the masses are not all equal, we have $$E_1\neq E_3$$ then $$(E_1-E_3)^2 > 0$$ and so $$t\leq (E_1-E_3)^2$$ doesn't imply that $$t \leq 0$$.

Have I made an error somewhere? Or is there another reason why the $$t<0$$ condition is satisfied?

• I'd advise doing this in the center of momentum frame to make your life easier and to make the answer more apparent. Your final result being the square of the energy difference isn't correct. Commented Jul 11 at 18:08
• It's done in CM, as yuo can see $\vec{p_1}+\vec{p_2}=0$. What am I missing? Commented Jul 11 at 19:11
• Your text may be imposing the same mass for that point, in his example. For different masses, take $m_1=m$, $m_3=0$, and normalize t by $m^2$ and both momenta by m. You then see t may be easily made positive. Commented Jul 11 at 20:31