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How does one prove the following equation (67.5) from the BLP Quantum Electrodynamics book? The q's are the 4 momenta, and h is the sum of all four masses. Two q's written after one another in the determinant means a Lorentz inner product with the minus two metric convention. The s channel pertains to suffixes (1,2) and (3,4), the t channel to (1,3) and (2,4), and the u to (1,4) and (2,3). The variables s, t and u are defined as squares of sums of 4 momenta q with these suffixes. For example, s = (q1 +q2)^2 = (q3+q4)^2 and t = (q1 +q3)^2. There are no minus signs in the definition of the s, t and u variables in terms of the q 4-vectors.

Inserting the s, t and u variables in the determinant in (67.4) did not work out for me, neither did working backwards from the result (67.5). Is there anything I am missing?

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Urgh I hate that chapter of Landau, so dull. Anyway, this isn't the grossest calculation in that book at least. After a trivial expansion we obtain \begin{align} \begin{vmatrix} q_1^2& q_1\cdot q_2&q_1\cdot q_3 \\ q_1\cdot q_2& q_2^2& q_2\cdot q_3\\ q_1\cdot q_3& q_2\cdot q_3& q_3^2 \end{vmatrix} =q_1^2q_2^2q_3^2+2(q_1\cdot q_2)(q_1\cdot q_3)(q_2\cdot q_3)-q_2^2(q_1\cdot q_3)^2-q_3^2(q_1\cdot q_2)^2-q_1^2(q_2\cdot q_3)^2 \end{align} But we know that $q_i^2=m_i^2$, and $s=m_1^2+m_2^2+2q_1q_2$, $t=m_1^2+m_3^2-2q_1q_3$, and $u=m_2^2+m_3^2-2q_2q_3$, so our determinant equals \begin{align} m_1^2m_2^2m_3^2+\frac{1}{4}(s-m_1^2-m_2^2)(m_1^2+m_3^2-t)(m_2^2+m_3^2-u)-\frac{1}{4}m_2^2(m_1^2+m_3^2-t)^2-\frac{1}{4}m_3^2(s-m_2^2-m_1^2)^2-\frac{1}{4}m_1^2(m_2^2+m_3^2-u)^2\geq 0 \end{align} Rejigging this a bit, since we want $stu$ on its own, we get \begin{align} stu \geq -4m_1^2m_2^2m_3^2 &-(s-m_1^2-m_2^2)(t-m_1^2-m_3^2)(u-m_2^2-m_3^2)+stu+m_2^2(t-m_1^2-m_3^2)^2\\ &+m_3^2(s-m_2^2-m_1^2)^2+m_1^2(u-m_2^2-m_3^2)^2 \end{align} Now we eliminate the quadratic terms (since our desired answer is linear in $s,t,u$), upon which we'll discover the RHS is an homogeneous linear polynomial. Finding the coefficients is then trivial.

The way we do this is use the fact that the sum of the variables is the sum of the square masses $s+t+u=m_1^2+m_2^2+m_3^2+m_4^2=h$, from which it follows \begin{align} s^2=hs-st-su \end{align} and similarly for $t^2$ and $u^2$. (A hint that we needed to do this also comes from the fact there was $m_4^2$ in the final result, yet $q_4$ is nowhere to be seen in the determinant - so it better come from some identity involving the variables, and $s+t+u$ is the only obvious candidate.)

The quadratic terms on the RHS are \begin{align} m_1^2s^2+m_2^2t^2+m_3^2u^2+(m_1^2+m_2^2)tu+(m_1^2+m_2^3)su+(m_2^2+m_3^2)st \end{align} But when we sub in the above expression for $s^2,t^2,u^2$ it becomes linear! \begin{align} m_1^2\left(hs-st-su\right)+m_2^2\left(ht-st-tu\right)+m_3^2\left(hu-su-tu\right)+(m_1^2+m_2^2)tu+(m_1^2+m_2^3)su+(m_2^2+m_3^2)st \end{align} for instance in the above the coefficient of $tu$ is $\left(-m_1^2-m_2^2+m_1^2+m_2^2\right)=0$. The quadratic terms therefore reduce to \begin{align} h\left(m_1^2s+m_2^2t+m_3^2u\right) \end{align} So much for the quadratic terms. Now you might think that what we're left with is something that looks like $a s+bt+c u +d$ but this cannot be so because if $s=t=u=0$ then all the masses separately vanish and the constant term must then vanish - so what we have is secretly of the form $ a s+b t+c u$ because of the constraint relating the $m_i$s and the Mandelstams. The result essentially follows; to find for instance $a$, we can just set $t=u=0$, $s=h$, and find using our third equation \begin{align} ah&=-(m_3^2+m_4^2)(m_1^2+m_3^2)(m_2^2+m_3^2)+m_2^2(m_1^2+m_3^2)^2+m_1^2(m_2^2+m_3^2)^2+m_3^2(m_3^2+m_4^2)^2-4m_1^2m_2^2m_3^2 \\ &=(m_1^2m_2^2-m_3^2m_4^2)(m_1^2+m_2^2-m_3^2-m_4^2). \end{align} The values of $bh$ and $ch$ are trivially obvious by symmetry, but it's no additional effort to arrive at them a la the above.

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  • $\begingroup$ I tried the same thing you did, and I failed where you succeeded. Thank you for the answer. $\endgroup$ – AMV Nov 27 '18 at 2:28
  • $\begingroup$ No worries, hope all that was clear $\endgroup$ – user213887 Nov 27 '18 at 3:46
  • $\begingroup$ Actually I have found a little problem, about the s=t=u=0 bit. I set these values in the third equation of your answer, and found that a bunch of positive valued terms involving squares of masses add up to zero! I did combine the first term on the rhs with a minus sign with the rest. Do you not think that is somewhat contradictory? I hesitated to proceed in my calculation before asking the question, and this was one of the sources of my discouragement. Could you say something about that ? Does that mean that all the masses identically vanish due to that substitution? $\endgroup$ – AMV Nov 27 '18 at 4:41
  • $\begingroup$ No, because the masses aren't independent variables - they satisfy a constraint $s+t+u=m_1^2+m_2^2+m_3^2+m_4^2$. So if you set $s=t=u=0$ in my third equation, yes you get a bunch of terms involving the masses. But there's a simultaneous equation $m_1^2+m_2^2+m_3^2+m_4^2=0$ the only solution to which is $m_1=m_2=m_3=m_4=0$ and therefore the constant you get is zero. $\endgroup$ – user213887 Nov 27 '18 at 4:49
  • $\begingroup$ That's the one slightly tricky aspect of the problem - you have some algebraic mess involving $m$s and $s$/$t$/$u$, but those variables are actually all related, and you can use it like I did in my fourth equation to replace $m$s with Mandelstams in all sorts of complicated ways. That is why the easiest thing to do is work out that there exists a way of using that constraint to make the expression linear, then solve for the coefficients. $\endgroup$ – user213887 Nov 27 '18 at 4:52

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